Arithmetic Functions.

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MATHEMATICS RESEARCH DEVELOPMENTS

ARITHMETIC FUNCTIONS

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MATHEMATICS RESEARCH DEVELOPMENTS

ARITHMETIC FUNCTIONS

JÓZSEF SÁNDOR AND

KRASSIMIR TODOROV ATANASSOV

Copyright © 2021 by Nova Science Publishers, Inc. All rights reserved. No part of this book may be reproduced, stored in a retrieval system or transmitted in any form or by any means: electronic, electrostatic, magnetic, tape, mechanical photocopying, recording or otherwise without the written permission of the Publisher. For permission to use material from this book please contact us: Telephone 631-231-7269; Fax 631-231-8175 Web Site: http://www.novapublishers.com

NOTICE TO THE READER The Publisher has taken reasonable care in the preparation of this book, but makes no expressed or implied warranty of any kind and assumes no responsibility for any errors or omissions. No liability is assumed for incidental or consequential damages in connection with or arising out of information contained in this book. The Publisher shall not be liable for any special, consequential, or exemplary damages resulting, in whole or in part, from the readers’ use of, or reliance upon, this material. Any parts of this book based on government reports are so indicated and copyright is claimed for those parts to the extent applicable to compilations of such works. Independent verification should be sought for any data, advice or recommendations contained in this book. In addition, no responsibility is assumed by the publisher for any injury and/or damage to persons or property arising from any methods, products, instructions, ideas or otherwise contained in this publication. This publication is designed to provide accurate and authoritative information with regard to the subject matter covered herein. It is sold with the clear understanding that the Publisher is not engaged in rendering legal or any other professional services. If legal or any other expert assistance is required, the services of a competent person should be sought. FROM A DECLARATION OF PARTICIPANTS JOINTLY ADOPTED BY A COMMITTEE OF THE AMERICAN BAR ASSOCIATION AND A COMMITTEE OF PUBLISHERS. Additional color graphics may be available in the e-book version of this book.

Library of Congress Cataloging-in-Publication Data ISBN: H%RRN

Published by Nova Science Publishers, Inc. † New York

Contents Glossary of Symbols

vii

Preface

xi

1 On Standard Arithmetic Functions ϕ, ψ and σ 1 1.1. A. Mullin’s Inequality and Its Modification . . . . . . . . . . . 1 1.2. Inequalities Related to ϕ, ψ and σ-Functions . . . . . . . . . . . 6 1.3. A Modification of Sivaramakrishnan-Venkataraman’s Inequality 28 1.4. On the Composition of Some Arithmetic Functions . . . . . . . 29 1.5. On the Equation ϕ(n) + d(n) = n and Related Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 2 Perfect and Related Numbers 2.1. On (m, n)-Super-Perfect Numbers . . . . . . . . . . . . . . 2.2. On a Modification of Perfect Numbers . . . . . . . . . . . . 2.3. On Multiplicatively Perfect Numbers . . . . . . . . . . . . . 2.4. Other Modifications of the Concept of Perfect Number . . . 2.5. A New Point of View on Perfect and Other Similar Numbers 2.6. On Bi-Unitary Harmonic Numbers . . . . . . . . . . . . . . 2.7. On Modified Hyperperfect Numbers . . . . . . . . . . . . . 2.8. On Balanced Numbers . . . . . . . . . . . . . . . . . . . .

. . . . . . . .

. . . . . . . .

55 55 63 71 78 83 100 111 115

3 On Modifications and Extensions of the Arithmetic Functions ϕ, ψ and σ 125 3.1. On an Arithmetic Function, Related to Operation “Logarithm” . 125 3.2. Irrational Factor: Definition, Properties and Problems . . . . . . 128

vi

Contents 3.3. Converse Factor: Definition, Properties and Problems . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4. Restrictive Factor: Definition, Properties and Problems . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5. On an Arithmetic Function, Related to Operation “Derivative” 3.6. On an Arithmetic Function Related to Function σ . . . . . . . 3.7. Extension Factor: Definition, Properties and Problems . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8. Extensions of Restrictive and Extension Factors . . . . . . . . 3.8.1. First Round of Generalizations . . . . . . . . . . . . . 3.8.2. Second Round of Generalizations . . . . . . . . . . . 3.8.3. Additive Analogues . . . . . . . . . . . . . . . . . . .

. 131 . 133 . 135 . 142 . . . . .

145 160 160 166 171

4 Arithmetic Functions of Other Types 175 4.1. A Digital Arithmetic Function . . . . . . . . . . . . . . . . . . 175 4.2. On an Inequality of Klamkin . . . . . . . . . . . . . . . . . . . 193 4.3. Some Representations Related to Arithmetic Function “Factorial”204 4.4. Some Representations Concerning the Product of Divisors of n . 209 4.5. A Note on Certain Euler–Mascheroni Type Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 4.6. On Two Conjectures by K. Kashihara on Prime Numbers . . . . 214 4.7. A Generalization of J. S´andor and F. Luca’s Theorem . . . . . . 217 About the Authors

237

Index

239

Glossary of Symbols Some of the used in the book notations are given below. • N is the set of all positive integers. If n > 1 and n ∈ N , then n has the k form n = ∏ pαi i i=1

that is called a canonical factorization of n, where k, α1 , . . ., αk ∈ N , n > 1, p1 , . . ., pk are different primes; in some cases, it is suitable the order p1 < · · · < pk to be valid; • set(n) = {p1 , p2 , . . ., pk }, k

• γ(n) = ∏ pi , i=1

k

• βs (n) = ∑ psi , for a fixed natural number s ≥ 1, i=1

• d/n means that d divides n; • d 6 |n means that d does not divide n; • ω(n) is the number of the distinct prime divisors of n ∈ N , n > 1 and ω(1) = 0; k

• Ω(n) = ∑ αi ; i=1

k pαi +1 −1 i pi −1 i=1

• σ(n) = ∏

is the sum of all different divisors of n and σ(1) = 1,

viii

J´ozsef S´andor and Krassimir Atanassov • σl (n) is the sum of l-th powers of all different divisors of n ∈ N , n > 1, where l is a real number, and σl (1) = 1, i.e., σl (n) = ∑ d l , d/n

where here and further ∑ means that the sum is over divisors d of n. d/n

Function σl is multiplicative and σ0 is denoted by d, where d(n) is the number of all different divisors of n ∈ N and k

d(n) = ∏(αi + 1) i=1

for n > 1 and d(1) = 1. As usually, function σ1 is denoted by σ. k

• ϕ(n) = ∏ pαi i −1 .(pi − 1), ϕ(1) = 1 is Euler’s totient function; i=1

• ϕs is Jordan’s totient function, where s 6= 0 is a real number. Function ϕs is multiplicative and  k  1 s ϕs (n) = n . ∏ 1 − s pi i=1 for n > 1, n ∈ N and ϕs (1) = 1. Function ϕ1 coincides with ϕ; • ψ is Dedekind’s function, which is multiplicative and  k  1 ψ(n) = n. ∏ 1 + pi i=1 for n > 1, n ∈ N and ψ(1) = 1; • function ψs is a generalization of function ψ, where s 6= 0 is a real number. It is a multiplicative and  k  1 s ψs (n) = n . ∏ 1 + s pi i=1 for n > 1, n ∈ N and ψs (1) = 1. Function ψ1 coincides with ψ, while ψ0 (n) = 2ω(n) for n ∈ N ;

ix

Glossary of Symbols

• µ is the M¨obius function. It is multiplicative, µ(1) = 1 and for n > 1, n ∈

N

µ(n) =

 ω(n)   (−1) ,  

0,

if max αi = 1 1≤i≤k

;

otherwise

• λ is the Liouville’s function. It is a completely multiplicative function, λ(1) = 1 and for n > 1: k

λ(n) = (−1)∑i=1 αi . It is well-known that

∑ λ(d) = d/n

  1, 

0,

if n is a square ; otherwise

• (m, n) is g.c.d. (greatest common divisor) of m and n; • [m, n] is l.c.m. (least common multiple) of m and n; • n! is the factorial of n; • pn is the n-th prime; • π(n) is the number of primes smaller or equal to n.

Preface In the mid 1980s, the research paths of the two authors were running in parallel: Krassimir Atanassov, born in Bulgaria in 1954, published his first papers on number theory in the “Bulletin of Number Theory and Related Topics” in 1984, while J´ozsef S´andor, born in Romania in 1956, published his ones a year later. In 1987, S´andor contacted Atanassov to inquire about his current areas of research, because several times by then S´andor had submitted results to the “Bulletin” only to receive a response that similar findings by Atanassov were already in press. In the years to come, the situation turned around, and Atanassov started seeing in the “Bulletin” results he was going to submit already published by S´andor. Thus, both authors came to the idea to join forces and write a common paper [1]. It was followed by other papers, and thus in 2008 the idea about the present book was born. It took us many years of active email correspondence to unify, at least partially, our styles of exposition and the notations used in our individual papers. Although we, together with Prof. Anthony G. Shannon (Australia) are the Editors of the international journal “Notes on Number Theory and Discrete Mathematics” already 26 years, and we live just 600-700 kilometers apart, we have not met yet. But we hope this will happen one day, as well. This, in short, is the story of the present book, and we hope that you will find reading it as interesting as we found writing it. The aim of this book is to present some new results of the authors in the theory of arithmetic functions. Arithmetic functions are very important in many parts of theoretical and applied sciences, and many mathematicians have devoted great interest in this field. For example, the classical Euler totient function, and some other related functions appears almost in all major domains of mathematics or its applications. The famous Riemann hypothesis, which is one of the most difficult unsolved problems of our time, can be stated as an inequal-

xii

J´ozsef S´andor and Krassimir Atanassov

ity in terms of the sum-of-divisors function. One of the main aims of the authors is to study some interesting new properties of these classical arithmetic functions. Many relations are proved, especially inequalities connected with these functions. Besides the above two arithmetic functions, some functions appear naturally as the number of divisors of a number, the distinct or total number of prime divisors of a number, Dedekind’s arithmetic function, the prime counting function, the core function, and many other classical functions. Also, famous numbers related to such functions are studied, such as perfect numbers and their variations, which actually contain many unsolved and difficult problems. One of the interesting features of the book is the introduction and study of certain new arithmetic functions. These new functions have been considered by the authors separately or together, and their importance is shown in many connections with the classical arithmetic functions or in their applications to other problems. This book may be of interest to students, researchers, teachers and professors working in number theory and related fields as well as to anyone who discovers connections with algebra, mathematical analysis, or other domains. It can be used in seminars or in formulation of new problems on arithmetic functions and their various applications. Also, it could become the starting point for new research on the themes introduced in these chapters. J´ozsef S´andor Krassimir Atanassov

Chapter 1

On Standard Arithmetic Functions ϕ, ψ and σ In this Chapter, some basic properties of the well-known arithmetic functions ϕ, ψ and σ will be discussed. Equalities and inequalities involving these functions will be given.

1.1. A. Mullin’s Inequality and Its Modification In [2] A. Mullin proposed several hypotheses, one of which (Problem 1) was proved by K. Atanassov in [3]. He essentially improved (and shortened) the proof in [4] and modified Mullin’s hypothesis in [5]. Problem 1 from [2] is the following: Prove or disprove the statement that for every fixed natural number d > 1: (a(a + d) − 1)2 − ϕ(a(a + d)).σ(a(a + d)) = d 2 iff 1 a and a + d are simultaneously prime numbers. k

Following [4] for the natural number n = ∏ pαi i , where k, α1 , α2 , . . . αk ≥ 1 i=1

are natural numbers and p1 , p2 , . . ., pk are different prime numbers, we define the following function A(n) = n2 − β2 (n) + 1 − ϕ(n)σ(n). 1 Everywhere

in the book, we shall use “iff” instead of “if and only if”.

2

J´ozsef S´andor and Krassimir Atanassov Obviously, for every natural number n, Ω(n) ≥ ω(n) . In these terms, Mullin’s hypothesis has the form: A(n) = 0 iff Ω(n) = ω(n) = 2.

Theorem 1.1.1. For every natural number n, for which Ω(n) ≥ 3 and ω(n) ≥ 2, it is valid that: A(n) > 0. Proof. Let Ω(n) = 3, i.e., n = pqr (p, q, r ≥ 2 are prime numbers), where p 6= q 6= r 6= p, or p = q 6= r. For the first case: A(n) = p2 q2 r2 − p2 − q2 − r2 + 1 − (p2 − 1)(q2 − 1)(r2 − 1) = p2 q2 + p2 r2 + q2 r2 − 2(p2 + q2 + r2 − 1) > 0. For the second case: A(n) = p4 r2 − p2 − r2 + 1 − p(p3 − 1)(r2 − 1) = p4 + pr2 − p2 − r2 − p + 1 > 0. Let us suppose that for every natural number n, if 3 ≤ Ω(n) ≤ k and ω(n) ≥ 2, then the above assertion is valid. Let p be a prime number and n0 = np. Therefore, Ω(n0) = k + 1 and ω(n0) ≥ 2. Two cases are possible for p: First case: p 6∈ set(n). Then A(n0 ) = n2 p2 − β2 (n) − p2 + 1 − ϕ(n)σ(n)(p2 − 1) and β2 (n) ≥ 22 + 32 = 13. By induction assumption A(n) > 0 and hence: A(n0 ) = n2 p2 − β2 (n) − p2 + 1 − ϕ(n)σ(n)(p2 − 1) > (ϕ(n)σ(n) + β2 (n) − 1)p2 − β2 (n) − p2 + 1 − ϕ(n)σ(n)(p2 − 1) = ϕ(n)σ(n)p2 + p2 β2 (n) − 2p2 + 1 − β2 (n) − ϕ(n)σ(n)(p2 − 1) = ϕ(n)σ(n) + β2 (n)(p2 − 1) − 2p2 + 1 > 13(p2 − 1) − 2p2 + 1 > 0.

On Standard Arithmetic Functions ϕ, ψ and σ

3

Second case: p ∈ set(n). Then ω(n0) = ω(n) ≥ 2 and n0 = m.pa , where Ω(m) ≤ k − 1, a ≥ 1 and A(n0) = m2 p2a+2 − β2 (m) − p2 + 1 − ϕ(m)σ(m)pa (pa+2 − 1). From set(n0 ) = set(m) ∪ {p}

and from ω(n0) ≥ 2 it follows that ω(m) ≥ 1. If ω(m) = 1, i.e., if m = qk−a , for k − a ≥ 1, then A(n0) = q2k−2a p2a+2 − p2 − q2 + 1 − qk−a−1 (qk−a+1 − 1)pa (pa+2 − 1) = p2a+2 qk−a−1 + pa q2k−2a − p2 − q2 − pa (qk−a−1 + 1) > 0. If ω(n) ≥ 2, then, as above, β2 (n) ≥ 13 and by the induction assumption A(n) > 0 it follows that: A(n0) > ϕ(m)σ(m)p2a+2 + p2a+2 β2 (m) − p2a+2 − β2 (m) − p2 + 1 −ϕ(m)σ(m)pa (pa+2 − 1)

= ϕ(m)σ(m)pa + β2 (m)(p2a+2 − 1) − p2a+2 − p2 + 1 > 13(p2a+2 − 1) − p2a+2 − p2 + 1 > 0

which proves Theorem 1.1.1. It is obvious that if ω(n) = Ω(n) = 2, i.e., if n = pq for two different prime numbers p and q, then: A(n) = 0; and if Ω(n) ≥ ω(n) = 1, i.e., if n = pk for some prime number p and for the natural number k ≥ 1, then: A(n) = p2k − p2 + 1 − pk−1 (pk+1 − 1)   > 0, for k ≥ 3 = pk−1 − p2 + 1 .  < 0, for k ≤ 2

In such a way, it is easy to see that equality is reached only in the case d(n) = ω(n) = 2, which proves the validity of A. Mullin’s hypothesis.

4

J´ozsef S´andor and Krassimir Atanassov Now we formulate a modification of Theorem 1.1.1, following [6].

Theorem 1.1.2. For every natural number n, such that Ω(n) ≥ 2, ϕ(n).ψ(n) ≤ n2 − (ω(n) − 2).γ(n) − (Ω(n) − 1).β2(n) + 1. Proof. Let Ω(n) = 2. Then (a) if n = p2 for some prime number p, then ω(n) = 1 and n2 + γ(n) − β2 (n) − ϕ(n).ψ(n) + 1 = p4 + p − p2 − p.(p − 1).p.(p + 1) + 1 = p + 1 > 0; (b) if n = pq for some prime numbers p and q, then ω(n) = 2 and n2 − β2 (n) − ϕ(n).ψ(n) + 1 = p2 .q2 − p2 − q2 − (p − 1).(q − 1).(p + 1).(q + 1) + 1 = 0, i.e., the assertion is valid for d(n) = 2. Now, let us assume that the assertion is valid for all natural numbers n such that Ω(n) ≤ m for some natural number m ≥ 2. We prove that the assertion is valid for the numbers n0 with Ω(n0) = m + 1, too. Let p be a prime number. We construct the number n0 = n.p, for which Ω(n0) = d(n) + 1 = m + 1 and we prove that the assertion is valid for n0 . Let A ≡ n02 − (ω(n0 ) − 2).γ(n0) − (d(n0 ) − 1).β2 (n0 ) + 1 − ϕ(n0 ).ψ(n0). There are two possibilities for p and n: (a) if p 6∈ set(n), then: (a.1) if ω(n) = 1, i.e., n = qm for some prime number q, some natural number m ≥ 1 and ω(n0 ) = 2, then, having in mind the obvious inequality q2m−2 ≥ 22m−2 ≥ 2m: A = p2 .q2m − m.(p2 + q2 ) + 1 − (p2 − 1).q2m−2.(q2 − 1) ≥ p2 .q2m−2 − m.p2 − m.q2 + 1 + q2m − q2m−2 ≥ p2 .q2m−2 − m.p2 − q2m−2 ≥ 2m(p2 − 1) − m.p2

On Standard Arithmetic Functions ϕ, ψ and σ

5

= m.p2 − 2m > 0.

(a.2) if ω(n) = 2, the proof is analogous. (a.3) if ω(n) ≥ 3, then β2 (n) ≥ 22 + 32 + 52 = 38 and by induction

A = n2 .p2 − (ω(n) − 1).γ(n).p − d(n).(β2(n) + p2 ) + 1 −ϕ(n).ψ(n).(p2 − 1)

≥ p2 .(ϕ(n).ψ(n) + (ω(n) − 2).γ(n) + (d(n) − 1).β2(n) − 1)

−(ω(n) − 1).γ(n).p − d(n).(β2(n) + p2 ) + 1 − ϕ(n).ψ(n).(p2 − 1) 1 > p.γ(n).((ω(n) − 2).p − ω(n) + 1) + .(p2 .(d(n) − 1).β2(n) 2 1 2 −2.d(n).β2(n)) + .(p .(d(n) − 1).β2(n) − 2.p2 .(d(n) + 1)) + 1 2 1 ≥ p.γ(n).(2.(ω(n) − 2) − ω(n) + 1) + .(4.(d(n) − 1) − 2.d(n)).β2(n) 2 1 2 + .p .(4.(d(n) − 1) − 2.(d(n) + 1)) + 1 2 ≥ p.γ(n).(ω(n) − 3) + (d(n) − 1).β2(n) + p2 .(d(n) − 3) > 0,

because, it can be easily seen that each of the three terms is a non-negative number. (b) if p ∈ set(n), then ω(n0 ) = ω(n), β2(n0 ) = β2 (n) and (b.1) if ω(n) = 1, i.e., n = pm then: A = p2m+2 + p − m.p2 + 1 − p2m .(p2 − 1) = p2m + p − m.p2 + 1 > 0. (b.2) if ω(n) = 2, the proof is analogous. (b.3) if ω(n) ≥ 3, then β2 (n) ≥ 22 + 32 + 52 = 38, ω(n) − 2 ≥ 1, γ(n) ≥ 2.3.5 = 30 and A = n2 .p2 − (ω(n) − 2).γ(n) − d(n).β2(n) + 1 − p2 .ϕ(n).ψ(n) ≥ p2 .(ϕ(n).ψ(n) + (ω(n) − 2).γ(n) + (d(n) − 1).β2(n) − 1) −(ω(n) − 2).γ(n) − d(n).β2(n) + 1 − p2 .ϕ(n).ψ(n)

= (p2 − 1).(ω(n) − 2).γ(n) + p2.(d(n) − 1).β2(n) − p2 − d(n).β2 (n) + 1 > 30(p2 − 1) + (p2 − 1).d(n).β2(n) − p2 .β2 (n) − p2 + 1 > β2 (n).(3(p2 − 1) − p2 ) > 0,

i.e., the assertion is valid for d(n0 ) = m + 1, which completes the proof of Theorem 1.1.2.

6

J´ozsef S´andor and Krassimir Atanassov

1.2. Inequalities Related to ϕ, ψ and σ-Functions Theorem 1.2.1. For every natural number n ≥ 19: ϕ(n)ψ(n) > σ(n)n .

(1.2.1)

Proof. First, let us put A(n) = ϕ(n)ψ(n) − σ(n)n . We see directly that A(n) < 0 for n ∈ S, where S = {1, 2, . . ., 10, 12, 14, 16,18} and A(11), A(13),A(15),A(17) > 0. Second, for every x > 1, we define the function f (x) = (x + 1) ln(x − 1) − x ln(x + 1). Then

x+1 x − ln(x + 1) − x−1 x+1 x − 1 3x + 1 = ln + x + 1 x2 − 1

f 0 (x) = ln(x − 1) +

(see, e.g., [7])

  3x + 1 2 4 8 = 2 − + + +... x −1 x + 1 2(x + 1)2 3(x + 1)2   2 4 8 3x + 1 − + + + . . . > 2 x −1 x + 1 (x + 1)2 (x + 1)2 =

3x + 1 2 1 − . 2 2 x − 1 x + 1 1 − x+1

3x + 1 2 − 2 x −1 x−1 x−1 1 = 2 = > 0. x −1 x+1 Therefore, f is an increasing function. We can check that f (9) > 0, while f (1), f (2), . . ., f (8) < 0. Therefore, for x ≥ 9: =

(x + 1) ln(x − 1) − x ln(x + 1) > 0

On Standard Arithmetic Functions ϕ, ψ and σ

7

and hence e(x+1) ln(x−1) − ex ln(x+1) > 0, i.e., (x − 1)x+1 > (x + 1)x .

(1.2.2)

Now, let n ≥ 19 and let n be a prime number. Then, from (1.2.2) ϕ(n)ψ(n) − σ(n)n = (n − 1)n+1 − (n + 1)n > 0. Let n = 2a . Then a

a

ϕ(2a )ψ(2 ) − σ(2a )2 = 2(a−1).3.2 > 2(a−1).3.2

a−1

a

b

a−1

a

− (2a+1 − 1)2 = a

− 2(a+1)2 ≥ 0

for a ≥ 5. Let n = 2a .3b . Then a

ϕ(2a .3b )ψ(2 .3 ) − σ(2a .3b )2 .3 = (2

a−1

.2.3

b−1 3.2a−1 .22 .3b−1

)

> (2a .3b−1)2

a+1 .3b



− (2

a+1

b

3b+1 − 1 − 1). 2 a

2a .3b

b

− (2a .3b+1 )2 .3 > 0,

because a.2a+1.3b > a.2a.3b and from (b − 1).2 ≥ (b + 1), for b ≥ 3, it follows that (b − 1).2a+1.3b ≥ (b + 1).2a.3b . Let n = 2a .3b .5c. Then

a

ϕ(2a .3b .5c)ψ(2 .3

b .5c )

− σ(2a .3b .5c)2 a−1

2

b−1

a .3b .5c

c−1

= (2a−1 .2.3b−1.22 .5c−1 )3.2 .2 .3 .2.3.5 2a .3b .5c  3b+1 − 1 5c+1 − 1 a+1 . − (2 − 1). 2 4 > (2a+2 .3b−1.5c−1)2

a+2 .3b+1 .5c−1

− (2a−2 .3b+1 .5c+1)2

a .3b .5c

because 12(a + 2).2a.3b .5c−1 > 5(a − 2).2a.3b .5c−1 > 0

> 0,

8

J´ozsef S´andor and Krassimir Atanassov

for each natural number a, 12(b − 1).2a.3b .5c−1 > 5(b + 1).2a.3b .5c−1 > 0 for each natural number b ≥ 3, 12(c − 1).2a .3b .5c−1 > 5(c + 1).2a .3b .5c−1 > 0 for each natural number c ≥ 3. Finally, let n = 2a .3b .5c.7d , where a, b, c, d ≥ 0 are natural numbers. Now, (1.2.1) has the form a

ϕ(2a .3b .5c.7d )ψ(2 .3

b

.5c .7d )

a

− σ(2a .3b .5c.7d )2 .3 2

a−1

b−1

b

.5c .7d

c−1

3

d−1

= (2a−1.2.3b−1.22 .5c−1 .2.3.7d−1)3.2 .2 .3 .2.3.5 .2 .7  2a .3b .5c .7d 3b+1 − 1 5c+1 − 1 7d+1 − 1 a+1 − (2 − 1). . . 2 4 6 > (2a+3.3b .5c−1 .7d−1)2 because 2(a+3).2

a+5 .3b+1 .5c−1 .7d−1

a+5.3b+1 .5c−1 .d d−1

− (2a−3 .3b .5c+1 .7d+1)2 a

− 2(a−3).2 .3

b .5c .7d

a .3b .5c .7d

> 0,

>0

for each natural number a, 96b > 35b for each natural number b, 96(c − 1) > 35(c + 1) for each natural number c ≥ 3 and 96(d − 1) > 35(d + 1) for each natural number d ≥ 3. These checks exclude all elements of the set S. Let us assume that (1.2.1) is valid for some natural number n 6∈ S so that Ω(n) = k ≥ 1. Let p be an arbitrary prime number. For p there are two cases. Case 1: p 6∈ set(n). Then, using (1.2.2) we obtain: ϕ(np)ψ(np) − σ(np)np = (ϕ(n).(p − 1))ψ(n).(p+1) − (σ(n).(p + 1))np = ϕ(n)ψ(n).(p+1).(p − 1)ψ(n).(p+1) − σ(n)np .(p + 1)np

> ϕ(n)ψ(n).p .(ϕ(n)ψ(n).(p − 1)ψ(n).(p+1) − (p + 1)np ) > ϕ(n)ψ(n).p.((p − 1)ψ(n).(p+1) − (p + 1)np )

> ϕ(n)ψ(n).p .((p − 1)(n+1).(p+1) − (p + 1)np ) > 0.

On Standard Arithmetic Functions ϕ, ψ and σ

9

Case 2: p ∈ set(n). Then, n = m.pa for some a ≥ 1 and m ≥ 1. Now, σ(np) = σ(mpa+1 ) = σ(m).

pa+2 − 1 pa+2 − 1 = σ(n). a+1 . p−1 p −1

Therefore, 

pa+2 − 1 ϕ(np) − σ(np) = (ϕ(n).p) − σ(n). a+1 p −1   a+2 np  p −1 ≥ σ(n)np . pψ(n).p − a+1 p −1 np    1 np ψ(n).p , ≥ σ(n) . p − p+ p+1 because for each a ≥ 1: ψ(np)

ψ(n).p

np

np

pa+2 − 1 p3 − 1 1 ≤ = p+ . a+1 2 p −1 p −1 p+1

Now, we see that p

ψ(n).p

= pψ(n).p − pnp



np 1 − p+ p+1 n  p(p+1)! p+1  1 1+ p(p + 1) n

> pψ(n).p − pnp .3 p+1

(because, for every two natural numbers x ≥ 2 and y ≥ 1: x(x+1) > 3 and hence y y x > 3 x+1 .) > pψ(n).p − pnp .pn = pψ(n).p − pn(p+1)

= pψ(mp

a ).p

− pmp

a (p+1)

> 0,

because ψ(mpa ).p−mpa (p+1) = ψ(m).pa(p+1)−mpa(p+1) = pa(p+1)(ψ(m)−m) > 0. Therefore, Theorem 1.2.1 is proved. By analogy, we can prove Theorem 1.2.2. [8, 9] For every natural number n ≥ 5 :

10

J´ozsef S´andor and Krassimir Atanassov

(a) σ(n)ϕ(n) < nn < ϕ(n)σ(n) (b) ψ(n)ϕ(n) < nn < ϕ(n)ψ(n). Now, we can see that the following inequalities ψ(n)ϕ(n) ≤ σ(n)ϕ(n) < nn < ϕ(n)ψ(n) ≤ ϕ(n)σ(n) are valid, too. The following two inequalities follow directly: ψ(n)ϕ(n) < ϕ(n)σ(n), σ(n)ϕ(n) < ϕ(n)ψ(n). Therefore, the inequalities ϕ(n). log(ψ(n)) < σ(n). log(ϕ(n)), ϕ(n). log(σ(n)) < ψ(n). log(ϕ(n)). hold, as well. Hence, we obtain (ψ(n).σ(n))ϕ(n) < ϕ(n)ψ(n)+σ(n). We must prove that for any natural number n that is not prime, √ ψ(n) > n + 2 n.

(1.2.3)

Let n = p.q for some prime numbers p and q. Then √ √ ψ(n) = (p + 1).(q + 1) = p.q + p + q + 1 > p.q + 2 p.q = n + 2 n. Let us assume that the assertion is valid for some n, and let p be a prime number. If p 6∈ set(n), then √ √ ψ(np) = (p + 1).ψ(n) > (p + 1).n + 2 n > n.p + 2(p + 1) np. If p ∈ set(n), then

√ √ ψ(np) = p.ψ(n) > p.n + 2.p. n > n.p + 2 np.

On Standard Arithmetic Functions ϕ, ψ and σ

11

Therefore, (1.2.3) is valid. Theorem 1.2.3. [10] For every natural number n ≥ 2 ϕ(n)ψ(n)σ(n) ≥ n3 + n2 − n − 1.

(1.2.4)

Proof. Let the natural number n be prime. Then ϕ(n)ψ(n)σ(n) = (n − 1)(n + 1)2 = n3 + n2 − n − 1 and (1.2.4) holds. Let Ω(n) = 2. Then, for n there are two cases. In the first case, n = pq for two distinct primes p and q. Let p > q. Then ϕ(n)ψ(n)σ(n) = ϕ(pq)ψ(pq)σ(pq) = (p3 + p2 − p − 1)(q3 + q2 − q − 1) = p3 q3 + p2 q3 − pq3 − q3 + p3 q2 + p2 q2 − pq2 − q2 − p3 q − p2 q + pq + q − p3 − p2 + p + 1 = p3 q3 + p2 q2−pq−1 + p3 (q2−q−1) + p2 (q3−q−1)−p(q3 + q2−2q−1)−q3 −q2 + q + 2

(from p ≥ q + 2) ≥ p3 q3 + p2 q2 − pq − 1 + p((q + 2)2 (q2 − q − 1) − q3 − q2 + 2q + 1) +(q + 2)2 (q3 − q − 1) − q3 − q2 + q + 2

= p3 q3 + p2 q2 − pq − 1 + p(q4 + 2q3 − 2q2 − 6q − 3) − q3 − q2 + 2q + 1) +q5 + 4q4 + 2q3 − 6q2 − 7q − 2

(from q ≥ 2)

> (pq)3 + (pq)2 − pq − 1 = n3 + n2 − n − 1

i.e., (1.2.4) holds, too. In the second case, n = p2 for a prime number p. Then ϕ(n)ψ(n)σ(n) = ϕ(p2 )ψ(p2 )σ(p2 ) = p(p − 1)p(p + 1)

p3 − 1 p−1

= p2 (p + 1)(p3 − 1) = p6 + p5 − p3 − p2 (from p ≥ 2) ≥ p6 + 2p4 − p3 − p2 > p6 + p4 − p2 − 1 = n3 + n2 − n − 1, i.e., (1.2.4) is true.

12

J´ozsef S´andor and Krassimir Atanassov

Let us assume that (1.2.4) is valid for every natural number n with Ω(n) = m for some natural number m ≥ 2. Let p be a prime number. Then, Ω(np) = Ω(n) + 1. For p there are two cases. In the first case, p 6∈ set(n). Then, ϕ(np)ψ(np)σ(np) − (np)3 − (np)2 + np + 1 = ϕ(n)ψ(n)σ(n)(p3 + p2 − p − 1) − (np)3 − (np)2 + np + 1

≥ (n3 + n2 − n − 1)(p3 + p2 − p − 1) − (np)3 − (np)2 + np + 1

= n3 (p2 − p − 1) + n2 (p3 − p − 1) − n(p3 + p2 − 2p − 1) − (p3 + p2 − p − 2) (by assumption, n ≥ 4, but it is enough that n ≥ 2) ≥ 8(p2 − p − 1) + 4(p3 − p − 1) − 2(p3 + p2 − 2p − 1) − (p3 + p2 − p − 2) = p3 + 5p2 − 7p − 8 > 0 for p ≥ 2. In the second case, p ∈ set(n). Then, from σ(np) > pσ(n) we obtain ϕ(np)ψ(np)σ(np) − (np)3 − (np)2 + np + 1 > p3 ϕ(n)ψ(n)σ(n) − (np)3 − (np)2 + np + 1

> p3 (n3 + n2 − n − 1) − (np)3 − (np)2 + np + 1 = p3 n2 − p3 n − p3 − (np)2 + np + 1

(the smallest value of n is 4) ≥ 11p3 − 16p2 + 4p + 1 > 0 for p ≥ 2. Therefore, we proved the validity of (1.2.4) for the natural number np. The text to the end of this Section follows [11]. It is well-known that for n = pa (p prime, a ≥ 1 integer) functions ϕ, ψ and σ satisfy the equalities:   ϕ(pa ) = pa 1 − 1p ,   (1.2.5) ψ(pa ) = pa 1 + p1 , σ(pa ) =

pa+1 −1 p−1 .

On Standard Arithmetic Functions ϕ, ψ and σ

13

In what follows, we will need also the unitary analogues of the functions ϕ and σ; namely the arithmetical functions ϕ∗ (n) and σ∗ (n) (connected with the “unitary divisors” of n; see e.g., [1] for many properties and references). These functions are also multiplicative, and for prime powers they take the values ϕ∗ (pa ) = pa − 1, (1.2.6) σ∗ (pa ) = pa + 1. In [11] J. S´andor formulated and proved the following refinement of Theorem 1.2.3: Theorem 1.2.4. For all n ≥ 1, one has the inequalities ϕ(n)ψ(n)σ(n) ≥ ϕ∗ (n)(σ∗(n))2 ≥ (n − 1)(n + 1)2 .

(1.2.7)

There is an equality in the first relation of (1.2.7) only when n is squarefree, or n = 1, while in the second one only when n is a prime power. Proof.

For the first term of inequality (1.2.7), note that both members r

are multiplicative functions. So, if n =

∏ pai

i

is the prime factorization of

i=1

n > 1, it will be sufficient to prove the inequality for a prime power pai i . Then, the general result follows by a term-by-term multiplication of these inequalities. Let us for simplicity denote pa ≡ pai i . Then, we have to prove the relation (by using (1.2.5) and (1.2.6)): p2a−2 (p + 1)(pa+1 − 1) ≥ (pa − 1)(pa + 1)2 .

(1.2.8)

After elementary transformations, (1.2.8) may be written also as: p3a−1 + pa + 1 ≥ p2a + p2a−1 + p2a−2 .

(1.2.9)

We shall prove this inequality by induction upon a ≥ 1. For a = 1, the relation is true (in fact, there is an equality in (1.2.9)). Assuming (1.2.9) for a, let us try to prove it for a + 1. By multiplying both sides of (1.2.9) by p2 , we get p3a+1 + pa+2 + p2 ≥ p2a+2 + p2a+1 + p2a = A, and note that A is in fact the right side of (1.2.9) for a := a + 1.

14

J´ozsef S´andor and Krassimir Atanassov

Therefore, it will be sufficient to prove that the left-hand side of (1.2.9) for a := a + 1 satisfies the inequality: p3a+2 + pa+1 + 1 ≥ p3a+1 + pa+2 + p2 .

(1.2.10)

This may be written also as p3a+1(p − 1) ≥ pa+1 (p − 1) + p2 − 1, i.e., p3a+1 ≥ pa+1 + p + 1.

(1.2.11)

Now, inequality (1.2.11) is trivial, since it equivalently states that pa+1 (p2a − 1) ≥ p + 1, and the left-hand side contains also p2a − 1 = (pa + 1)(p − 1) ≥ (p + 1)(p − 1) ≥ p + 1 (the inequality is in fact strict). We mention that the above proof shows in fact that inequality (1.2.9) is strict for a > 1. Thus, one has an equality in (1.2.8) only for a = 1, and this implies that there is an equality for n > 1 in the left-hand side of (1.2.7) only when n is a product of distinct primes, i.e., n is squarefree. Now, the second inequality of (1.2.7), when r

r

i=1

i=1

n = ∏ pai i = ∏ xi > 1, can be rewritten as: (x1 − 1) . . .(xr − 1)(x1 + 1)2 . . .(xr + 1)2 ≥ (x1 . . .xr − 1)(x1 . . .xr + 1)2 , (1.2.12) where r ≥ 1 and xi = pai i . Clearly, there is an equality in (1.2.12) for r = 1 (i.e., when n is a prime power); we shall prove that for r > 1 there is a strict inequality. First, we prove the inequality for r = 2. The general case – via mathematical induction – will be reduced essentially to this case. Put for simplicity x1 = x, x2 = y when the inequality becomes (x − 1)(x + 1)2 (y − 1)(y + 1)2 > (xy − 1)(xy + 1)2 .

(1.2.13)

On Standard Arithmetic Functions ϕ, ψ and σ

15

Here x ≥ 2 and y ≥ 3 (as p1 ≥ 2, p2 ≥ 3 are distinct primes). As (x − 1)(x + 1)2 = x3 + x2 − x − 1, etc.; (1.2.13) may be written also as (x3 + x2 − x − 1)(y3 + y2 − y − 1) > x3 y3 + x2 y2 − xy − 1, or x3 (y2 − y − 1) + x2 (y3 − y − 1) > x(y3 + y2 − 2y − 1) + y3 + y2 − y − 2. (1.2.14) Write this as x[x(y3 − y − 1) − (y3 + y2 − 2y − 1)] + x3 (y2 − y − 1) > y3 + y2 − y − 2. (1.2.15) Here, x(y3 − y − 1) − (y3 + y2 − 2y − 1) ≥ 2(y3 − y − 1) − (y3 + y2 − 2y − 1) = y3 − y2 − 1 > 0 by x ≥ 2. Thus, the left-hand side of (1.2.15) is ≥ 2(y3 − y2 − 1) + 8(y2 − y − 1) > y3 + y2 − y − 2, as this is y3 + 5y2 − 7y − 8 > 0. Now, y(y2 + 5y − 7) ≥ 3(9 + 15 − 7) = 51 > 8, and this proves (1.2.15), i.e., (1.2.13). Now, assuming (1.2.12) for r > 1, let us try to prove it for r + 1, i.e. (x1 − 1) . . .(xr − 1)(xr+1 − 1)(x1 + 1)2 . . .(xr + 1)2 (xr+1 + 1)2 > (x1 . . .xr xr+1 − 1)(x1 . . .xr xr+1 + 1)2 .

(1.2.16)

By multiplying both sides of (1.2.12) with (xr+1 − 1)(xr+1 + 1)2 , it is sufficient to prove that (x1 . . .xr − 1)(x1 . . .xr + 1)2 (xr+1 − 1)(xr+1 + 1)2 > (x1 . . .xr xr+1 − 1)(x1 . . .xr xr+1 + 1)2

(1.2.17)

16

J´ozsef S´andor and Krassimir Atanassov

Let x1 . . .xr = x, xr+1 = y. Then it is immediate that inequality (1.2.17) becomes exactly (1.2.13). This completes the proof of Theorem 1.2.4. Other inequalities, connecting ϕ∗ (n) and σ∗ (n) were proved in [1] (in more general forms); for example 6 2 · n < ϕ∗ (n) · σ∗ (n) < n2 for n > 1, π2

(1.2.18)

ϕ∗ (n) + σ∗ (n) ≤ nd ∗ (n) for n ≥ 1,

(1.2.19)

∗

ϕ∗ (n) + d ∗ (n) ≤ σ∗ (n) for n ≥ 1, ∗

∗

2

2

d (n) · n ≤ ϕ (n)(d (n)) ≤ n for n ≥ 1,

(1.2.20) (1.2.21)

where d ∗ (n) = 2ω(n) is the number of unitary divisors of n. Many inequalities on the arithmetical functions ϕ, σ, d, ϕ∗ , σ∗ , d ∗ are proved in J. S´andor’s paper [12]. For example, we quote the relations: σ∗ (n) ≤ d ∗ (n)ϕ(n) for any n ≥ 3 odd,

σ∗ (n) ≤ 23 d ∗ (n)ϕ(n) for n ≥ 2 even.

(1.2.22)

It is easy to see that ϕ(n) ≤ ϕ∗ (n)

σ(n) ≥ σ∗ (n)

d(n) ≥ d ∗ (n)

for n ≥ 1.

(1.2.23)

On the other hand, one has:

σ∗ (n) ≤ ϕ∗ (n)(d ∗ (n))α, for n ≥ 1,

(1.2.24)

where α = log2 3 (thus 1 < α < 2). Clearly, inequalities (1.2.18)–(1.2.21) or (1.2.22)–(1.2.24) can be connected to relation (1.2.7). The right-hand side of (1.2.18) implies n2 σ∗ (n) > ϕ∗ (n)(σ∗(n))2 ≥ (n − 1)(n + 1)2 (n > 1).

(1.2.25)

Another example is ϕ∗ (n)(d ∗ (n))2(ϕ(n))2 ≥ ϕ∗ (n)(σ∗ (n))2 ≥ (n − 1)(n + 1)2 for n ≥ 3 odd, (1.2.26)

On Standard Arithmetic Functions ϕ, ψ and σ which is a consequence of (1.2.22) and (1.2.7), etc. In paper [13], it is proved that √ σ(n) > n + (ω(n) − 1) n for n ≥ 2.

17

(1.2.27)

As an application of (1.2.27), it is shown that √ σ(n) > n + n iff n 6= prime, (1.2.28) √ √ σ(n) > n + n + 3 n iff n is not a prime and n is not a square of prime. (1.2.29) It is immediate that σ(n) ≥ ψ(n) for any n ≥ 1. In paper [14], it is shown that π2 · ψ(n) for n ≥ 1. (1.2.30) σ(n) < 6 2

As π6 < 1.7 < 2, particularly we get σ(n) < 2ψ(n). A stronger inequality than the last one – which is however not comparable with (1.2.30) – is due to Ch. Wall: σ(n) + σ∗ (n) ψ(n) ≥ . (1.2.31) 2 By (1.2.23) we get σ(n) + σ∗ (n) ≥ σ∗ (n), (1.2.32) 2 which particularly shows that ψ(n) lies between σ∗ (n) and σ(n). By using relation (1.2.5) and (1.2.6), one can deduce the following formulas: ! r 1 2 ϕ(n)σ(n) = n · ∏ 1 − a +1 , (1.2.33) pi i i=1 ! r 1 (1.2.34) ϕ∗ (n)σ∗ (n) = n2 · ∏ 1 − 2ai , pi i=1  r  1 2 ϕ(n)ψ(n) = n · ∏ 1 − 2 , (1.2.35) pi i=1 σ(n) ≥ ψ(n) ≥

r

where n = ∏ pai i is the prime factorization of n > 1. i=1

Theorem 1.2.5. For all n > 1 one has ϕ(n)ψ(n) ≤ ϕ(n)σ(n) ≤ ϕ∗ (n)σ∗ (n) ≤ n2 − 1,

(1.2.36)

18

J´ozsef S´andor and Krassimir Atanassov n ≤ n2 − 1, γ(n)   n 2 2 ≤ n2 − 1. ϕ(n)ψ(n) ≤ n − γ(n) ϕ(n)σ(n) ≤ n2 −

(1.2.37) (1.2.38)

Proof. As 2 ≤ ai + 1 ≤ 2ai , the first two inequalities of (1.2.36) are consequences of relations (1.2.33)–(1.2.35). For the last inequality of (1.2.36) use the classical (Weierstrass-type) inequality: (x1 − 1)(x2 − 1) . . .(xr − 1) ≤ x1 x2 . . .xr − 1,

(1.2.39)

where r ≥ 1 is integer, and xi > 1 (i = 1, 2, . . ., r) are arbitrary real numbers. Api ply now (1.2.39) for xi = p2a i in order to deduce the last inequality of (1.2.36). Applying (1.2.39) for xi = pai i +1 , we get the first inequality of (1.2.37). By n ≥ γ(n), clearly the last relation of (1.2.37) follows, too. Finally, apply (1.2.39) for xi = p2i for the proof of (1.2.38). Theorem 1.2.6. holds true:

For any n > 1, the following refinement of (1.2.29) 6 ψ(n) ϕ(n)ψ(n) > > 2. σ(n) n2 π

(1.2.40)

Proof. The first inequality of (1.2.40) follows by ϕ(n)σ(n) < n2 , which is contained particularly in (1.2.36). For the second inequality of (1.2.40) remark that by (1.2.35),    n  1 1 2 2 ϕ(n)ψ(n) = n · ∏ 1 − 2 > n · ∏ 1 − 2 , p pi p prime i=1 where p runs through the set of all prime numbers. It is well-known, from the Euler product representation of the zeta function that  −1 ∞ 1 1 ζ(s) = ∑ k = ∏ 1− s . p k=1 n p prime Particularly, ζ(2) =

  1 −1 1 − , ∏ 2 p p prime

On Standard Arithmetic Functions ϕ, ψ and σ i.e.,



1 ∏ 1 − p2 p prime



=

19

1 6 = 2 ζ(2) π

∞

1 π2 = ∑ 2 6. k=1 n This proves the second inequality of (1.2.40). We must mention that by (1.2.36), the right-hand side of (1.2.40) offers also a strong refinement of the left-hand side of (1.2.18). The lower bound from the second inequality of (1.2.40) is the best possible in a sense, that there exists a sequence (nk ) such that by the Euler series

lim

k→∞

ϕ(nk )ψ(nk ) 6 = 2, 2 π nk

namely nk = p1 p2 . . . pk , where pk now is the k-th prime number. For certain particular values of n, however, better lower bounds will be provided by the following theorem. Theorem 1.2.7. Let p(n), respectively P(n) denote the smallest, respectively the largest prime factors of n. Then    ϕ(n)ψ(n) 1 1 ≥ 1 − 1 + . (1.2.41) n2 p(n) P(n) Proof. We will use the inequality pi + 1 pi+1 ≥ , for i = 1, 2, . . ., k − 1, pi pi+1 − 1

(1.2.42)

where 2 ≤ p1 < p2 < . . . < pk are the distinct prime factors of n. Indeed, (1.2.42) is in fact pi+1 − pi ≥ 1. Now, by ψ(n) p1 + 1 p2 + 1 pk−1 + 1 pk + 1 = · ... · n p1 p2 pk−1 pk and (1.2.42), we can write ψ(n) p2 pk pk + 1 p1 − 1 pk + 1 ≥ ... · = · n p2 − 1 pk − 1 pk p1 pk



 p1 pk ... . p1 − 1 pk − 1

20

J´ozsef S´andor and Krassimir Atanassov

Since p1 = p(n), pk = P(n), inequality (1.2.41) follows. Corollary 1.2.1. If n ≥ 3 is odd, then   ϕ(n)ψ(n) 2 1 2 6 ≥ 1+ > > 2. 2 n 3 P(n) 3 π

(1.2.43)

1 Proof. Since 1 − p(n) ≥ 1 − 13 (by p(n) ≥ 3), from (1.2.43) we get the first inequality. The last inequality holds, as π2 > 9. We will mention that if n ≥ 2 is even, we get   ϕ(n)ψ(n) 1 1 ≥ 1+ . (1.2.44) n2 2 P(n) 2

π The right-hand side of (1.2.44) is > π62 only if P(n) < 12−π 2 = 4.6, . . ., so a b P(n) ≤ 3, i.e., when n is of the form n = 2 · 3 (a ≥ 1, b ≥ 0 integers). If n ≥ 3 is odd, not divisible by 3, then (1.2.43) may be refined as   ϕ(n)ψ(n) 4 1 4 2 ≥ 1+ > > . n2 5 P(n) 5 3

Theorem 1.2.8 [15]. For each natural number n ≥ 2: ψ(n)n > σ(n)ϕ(n).

(1.2.45)

Proof. Let n be a prime number. Then from (4) we obtain: ψ(n)n − σ(n)ϕ(n) = (n + 1)n − (n + 1)n−1 > 0. Let for the natural number n ≥ 2, the inequality (1.2.45) be valid. Let p be a prime number. For it there are two possibilities. Case 1. Let p 6∈ set(n). Then, ψ(np)np − σ(np)ϕ(np) = (ψ(n)(p + 1))np − (σ(n)(p + 1))ϕ(n)(p−1) = ψ(n)np .(p + 1)np − σ(n)ϕ(n)(p−1).(p + 1)ϕ(n)(p−1)

= (ψ(n)n) p .(p + 1)np − (σ(n)ϕ(n)) p−1 .(p + 1)ϕ(n)(p−1) (by the induction assumption) = (ψ(n)n ) p .(p + 1)np − (ψ(n)n ) p−1 .(p + 1)ϕ(n)(p−1) > 0.

On Standard Arithmetic Functions ϕ, ψ and σ

21

Case 2. Let p ∈ set(n). Then n = pa m for the natural numbers a, m ≥ 1, where (m, p) = 1. First, obviously, for q ≥ 3 1 1 q > (1 + )q−1 > (1 + 2 )q−1 . q q

(1.2.46)

Therefore, for each prime number q ≥ 3: 1 1 q2q−1 = q.q2q−2 > q2q−2(1 + )q−1 = (q2 (1 + ))q−1 q q Second, we see that for q ≥ 3 and for a ≥ 1: q+ =

1 q(qa+1 − 1) =

Third,

1 qa+2 − 1 q2 + 1 qa+2 − 1 − = − a+1 q qa+1 − 1 q q −1 ((qa+3 + qa+1 − q2 − 1) − (qa+3 + q)) 1

q(qa+1 − 1)

(qa+1 − q2 + q − 1) > 0.

σ(np) = σ(mpa+1 ) = σ(m)

pa+2 − 1 pa+2 − 1 = σ(n) a+1 . p−1 p −1

Now, we obtain sequentially: ϕ(n)p  pa+2 − 1 ψ(np)np − σ(np)ϕ(np) = (ψ(n)p)np − σ(n) a+1 p −1 np np

= ψ(n) p − σ(n)

ϕ(n)p



(by the induction assumption) np np

> ψ(n) p − ψ(n) = ψ(n)np pnp −

np





pa+2 − 1 pa+1 − 1

pa+2 − 1 pa+1 − 1

pa+2 − 1 pa+1 − 1

ϕ(n)p

ϕ(n)p

ϕ(n)p !

(1.2.47)

22

J´ozsef S´andor and Krassimir Atanassov  a+2 (n−1)p ! p − 1 ≥ ψ(n)np pnp − pa+1 − 1

(from (1.2.47)) ≥ ψ(n)np



1 pnp − p + p

(n−1)p !



1 pp − 1 + 2 p

= ψ(n)np p(n−1)p

(n−1)p !

(from (1.2.46)) > 0. This completes the proof. Theorem 1.2.9 [15]. For each natural number n ≥ 2: σ(n)n < ψ(n)σ(n).

(1.2.48) 1

Proof. It is well-known that the function f (x) = x x is strictly decreasing for x ≥ e – Euler’s number. As 3 > e, particularly we get that 1

1

σ(n) σ(n) ≤ ψ(n) ψ(n) , as σ(n) ≥ ψ(n) ≥ 3 for n ≥ 2. Now, as ψ(n) > n for n ≥ 2, we get that 1

1

σ(n) σ(n) < ψ(n) n , which implies (1.2.48). This completes the proof. Suggested by a paper [16], V. Kannan and R. Srikanth [17] have discovered the following inequality ϕ(n)ϕ(n).ψ(n)ψ(n) > nϕ(n)+ψ(n)

(1.2.49)

for n > 1, where ϕ and ψ denote the Euler totient function, resp. Dedekind arithmetic function. Following [18], below, we give inequalities of type (1.2.49), with strong refinements, as well as their converses and analogues relations.

On Standard Arithmetic Functions ϕ, ψ and σ

23

First, we formulate some auxilary assertions. Lemma 1.2.1. Let xi > 0, λi > 0 (i = 1, 2, ..., n) be real numbers such that λ1 + λ2 + ... + λn = 1. Then one has xλ1 1 xλ2 2 . . .xλn n ≤ λ1 x1 + λ2 x2 + · · · + λn xn .

(1.2.50)

There is inequality only for x1 = ... = xn = 1. Proof. This is the well-known weighted arithmetic mean-geometric mean inequality, see, e.g., [19]. Lemma 1.2.2. Let x, y > 0 and λ, µ > 0 with λ + µ = 1. Then xλ yµ ≤ λx + µy

(1.2.51)

1 . + µy

(1.2.52)

and xλ yµ ≥

λ x

Proof. Let n = 2, x1 = x, x2 = y in (1.2.50). Then (1.2.51) follows. Now, letting n = 2, x1 = 1x , x2 =

1 y

in (1.2.50), we get (1.2.52).

Remark 1.2.1. The inequality (1.2.51) is called also as the weighted arithmetic-geometric inequality for two numbers; while (2.1.52) as the weighted geometric-harmonic inequality for two numbers. Lemma 1.2.3. For any a, b > 0 one has 

a+b 2

a+b

a

b

≤ a .b ≤



a2 + b2 a+b

a+b

(1.2.53)

with equality only for a = b. a b Proof. Put x = a, y = b, λ = a+b , µ = a+b in (1.2.51). Then the right side of (1.2.53) follows. Apply now inequality (1.2.52) for the same numbers.

The left side of (1.2.53) follows.

24

J´ozsef S´andor and Krassimir Atanassov

Lemma 1.2.4. For any a, b > 0 one has 

ab(a + b) a2 + b2

a+b

b

a

≤ a .b ≤



2ab a+b

a+b

,

(1.2.54)

with equality only in a = b. Proof. Let now x = b, y = a, λ = a

b

a a+b .b a+b ≤

a b a+b , µ = a+b

in (9). Then we get

a b 2ab .b + .a = , a+b a+b a+b

and the right side of (1.2.54) is proved. By applying inequality (1.2.52) with the same selections, we get 1

a

b

a a+b .b a+b ≥

a b b(a+b) + a(a+b)

=

ab(a + b) , a2 + b2

and we are done with the left side of (1.2.54). Remark 1.2.2. As

√ 2ab ≤ ab, a+b

clearly one has 

2ab a+b

a+b

≤ (ab)

a+b 2

.

(1.2.55)

Theorem 1.2.10. For any integer n > 1 one has

n

ϕ(n)+ψ(n)


1: ϕ(n)ϕ(n)σ(n)σ(n) > nϕ(n)+σ(n)

(1.2.59)

26

J´ozsef S´andor and Krassimir Atanassov

and ϕ(n)σ(n)σ(n)ϕ(n) < nϕ(n)+σ(n).

(1.2.60)

Theorem 1.2.12. Let n no , A = n | n ∈ N & f (n) ≤ 2

where f is an arithmetic function and for every natural number n f (n) > 0. Then, for any arithmetic function g so that g(n) > 0 one has f (n)g(n).g(n) f (n) ≤ n f (n)+g(n)

(1.2.61)

for n ∈ A. Proof. We apply the right side of (1.2.54) for a = f (n), b = g(n) and using the fact that the inequality 2 f (n)g(n) ≤n f (n) + g(n) may be rewritten as g(n)(2 f (n) − n) ≤ n f (n). Now, since f (n) > 0, g(n) > 0, this is true, if 2 f (n) − n ≤ 0. Thus, if n ∈ A, then (1.2.61) holds true. Remark 1.2.6. In particular, we get ϕ(n)g(n).g(n)ϕ(n) ≤ nϕ(n)+g(n)

(1.2.62)

for every even number n and for any arithmetic function g so that g(n) > 0. If g(n) 6= ϕ(n), then the inequality (1.2.62) is strict. Indeed, it is well-known that ϕ(n) ≤ n2 for any even number n. So, for A - the set of positive even integers, (1.2.62) follows. Theorem 1.2.13. Let the arithmetical functions f , g and h satisfy the following conditions: (i) f (n).g(n) < n2 for n > 1; (ii) n + 1 ≤ h(n) ≤ g(n) for n ≥ 2.

On Standard Arithmetic Functions ϕ, ψ and σ Then one has

n2

(g(n)) f (n) < (g(n)) g(n) < (h(n))n

27

(1.2.63)

Proof. The first inequality of (1.2.63) follows by condition (i) and the remark that g(n) > 1 by (ii). Now, for the proof of the second inequality , we will use 1 the known fact that for x > 0 the real function F(x) = x x is strictly decreasing for x ≥ e (Eulers constant). Therefore, one has by (ii) that 1

1

(g(n)) g(n) ≤ (h(n)) h(n) , by remarking that (ii) can be applied, as n + 1 ≥ 3 > e for n ≥ 2. Now, as 1

1

h(n) > n, we get (h(n)) h(n) < (h(n)) n , and the result follows.

Remark 1.2.7. 1) By selecting f (n) = ϕ(n), g(n) = σ(n) and h(n) = ψ(n), we get from (21) the following two inequalities: (σ(n))ϕ(n) < (ψ(n))n and (σ(n))n < (ψ(n))σ(n), which are Theorems 1.2.8 and 1.2.9. 2) Select f (n) = ϕ∗ (n), g(n) = σ∗ (n), and h(n) = n + 1, where ϕ∗ is the unitary analogue of the Euler totient function, and σ∗ is the unitary analogue of the sigma function. Now, by ([20], relation (1.2.57)), condition (i) is satisfied. The condition (ii) reduces to σ∗ (n) ≥ n + 1, which is well-known. Then we get the inequalities: ∗ (n)

(σ∗ (n))n < (n + 1)σ and

∗

(σ∗ (n))ϕ (n) < (n + 1)n .

(1.2.64) (1.2.65)

Finally, we would like to mention that while preparing [18], we found that in paper [17] there are some mistakes. For example, the inequality in the last line of page 20 is false. It is asserted that

28

J´ozsef S´andor and Krassimir Atanassov     1 1 ∑ ln 1 + p > ∑ ln 1 − p , p|n p|n

where p runs through the prime divisors of n. This is not correct. Let for example, n = 6. Then the prime divisors of n are p = 2 and p = 3. We should have         1 1 1 1 . + ln 1 + > ln 1 − + ln 1 − ln 1 + 2 3 2 3 After simple computations, this reduces to ln2 > ln3, which is false.

1.3. A Modification of SivaramakrishnanVenkataraman’s Inequality In [21] it is given the following inequality of R. Sivaramakrishnan and C. Venkataraman: √ σ(n) ≥ d(n). n, k

where n = ∏ pαi i , k, α1 , α2 , . . . , αk ≥ 1 are natural numbers and p1 , p2 , . . ., pk i=1

are different prime numbers. Here, we prove Theorem 1.3.1. For each natural number n: √ ψ(n) ≥ d(n). n.

(1.3.1)

Obviously, for every natural number n: σ(n) ≥ ψ(n). For Ω(n) = 1, (1.3.1) is valid, because in this case n is a prime number and √ √ ψ(n) − d(n). n = n + 1 − 2. n ≥ 0. First, we prove that (1.3.1) is valid for every odd number. Let us assume that (1.3.1) is valid for every n such that s ≥ Ω(n) ≥ 1. Let n be a fixed natural number, such that Ω(n) = s and let p be a fixed prime number. For p there are two cases. Case 1. p 6∈ set(n). Then Ω(n.p) = Ω(n) + 1 and √ √ ψ(n.p) − d(n.p). n.p = ψ(n).(p + 1) − 2.d(n). n.p

On Standard Arithmetic Functions ϕ, ψ and σ √ √ √ √ ≥ (p + 1).d(n). n − 2.d(n). n.p = d(n). n.(p + 1 − 2. n.p) ≥ 0.

29

Case 2. p ∈ set(n). Then Ω(n.p) = Ω(n)+1, n+ pa .m for some natural numbers a, m ≥ 1, m satisfies (1.3.1), because Ω(m) ≤ s, and √ √ ψ(n.p) − d(n.p). n.p = ψ(n).p − d(m.pa+1). n.p √ √ √ = ψ(n).p − d(m).(a + 2). n.p ≥ p.d(n). n − d(m).(a + 2). n.p √ √ √ √ = d(m). n.p((a + 1). p − (a + 2)) ≥ d(m). n.p((a + 1). 3 − (a + 2)) > 0.

Second, we shall prove that (1.3.1) is valid for every even number, too. Let n = 2.m, where m is an odd number and let for m (1.3.1) be vaild. Then √ √ ψ(2.m) − d(2.m). 2.m = 3.ψ(m) − 2.d(m). 2.m √ √ √ √ ≥ 3.d(m). m − 2.d(m). 2.m = d(m). m(3 − 2. 2) > 0.

Let n = 2a .m, where a ≥ 2 is a natural number, m is an odd number, and let for m (1.3.1) be valid. Then √ √ ψ(2a .m) − d(2a .m). 2a .m = 3.2a−1 .ψ(m) − (a + 1).d(m). 2a .m

√ √ √ a−2 ≥ 3.2a−1 .d(m). m − (a + 1).d(m). 2a .m = d(m). 2a .m(3.2 2 − (a + 1)) ≥ 0.

The check of the last inequality for a ≥ 2 is trivial. This completes the proof.

1.4. On the Composition of Some Arithmetic Functions In this section we will follow the paper [22]. It is well-known that n is called perfect if σ(n) = 2n. Euclid and Euler [23, 24] have determined all even perfect numbers, by showing that they are of the form n = 2k (2k+1 − 1), where 2k+1 − 1 is a prime and k ≥ 1 is a natural number. The primes of the form 2k+1 − 1 are the so-called Mersenne primes, and at this moment there are known exactly 51 such primes. The 51th Mersenne prime is 282589933 − 1, see [25]. It is possible that there are infinitely many Mersenne primes, but the proof of this result seems unattackable at present. On the other hand, no odd perfect number is known, and the existence of such numbers is one of the most difficult open problems of mathematics.

30

J´ozsef S´andor and Krassimir Atanassov

D. Suryanarayana [26] defined the notion of a superperfect number, i.e., a number n with the property σ(σ(n)) = 2n, and he and H.J. Kanold [26, 27] have obtained the general form of the even superperfect numbers. These are n = 2k , where 2k+1 − 1 is a prime. Numbers n with the property σ(n) = 2n − 1 have been called almost perfect, while that of σ(n) = 2n + 1, quasi-perfect. For many results and conjectures on this topic, see [28], and J. S´andor’s book [24]. For an arithmetic function f , the number n is called f -perfect, if f (n) = 2n. Thus, the superperfect numbers will be in fact the σ ◦ σ-perfect numbers where “◦” denotes composition. Here, we use the following (equivalent) forms of functions ϕ and ψ:   ϕ(n) = n ∏ 1 − p1 , p|n   (1.4.1) ψ(n) = n ∏ 1 + 1p . p|n

It is well known that functions ϕ, σ and ψ are multiplicative, i.e., they satisfy the functional equation f (mn) = f (m) f (n) for (m, n) = 1. For results on ψ ◦ ψperfect, ψ ◦ σ-perfect, σ ◦ ψ-perfect, and ψ ◦ ϕ-perfect numbers, see the first part of [29]. Let σ∗ (n) be the sum of unitary divisors of n, given by σ∗ (n) =

∏ (pα + 1),

(1.4.2)

pα ||n

where pα ||n means that for the prime power pα one has pα |n, but pα+1 6 |n. By convention, let σ∗ (2) = 1. In [29] almost and quasi σ∗ ◦ σ∗ -perfect numbers (i.e., satisfying σ∗ (σ∗ ((n)) = 2n∓1) are studied, where it is shown that for n > 3 there are no such numbers. This result has been rediscovered by V. Sitaramaiah and M.V. Subbarao [30]. In 1964, A. Makowski and A. Schinzel [31] conjectured that σ(ϕ(n)) ≥

n 2

(1.4.3)

for all n ≥ 1. The first results after the Makowski-Schinzel’s paper were proved by J. S´andor [32, 33]. He proved that (1.4.3) holds if and only if σ(ϕ(m)) ≥ m,

(1.4.4)

for all odd m ≥ 1 and obtained a class of numbers satisfying (1.4.3) and (1.4.4). But (1.4.4) holds iff it is true for a squarefree n, see [29, 33]. This has been

On Standard Arithmetic Functions ϕ, ψ and σ

31

rediscovered by G.L. Cohen and R. Gupta ([34]). Many other partial results have been discovered by C. Pomerance [35], G.L. Cohen [34], A. Grytczuk, F. Luca and M. Wojtowicz [36, 37], F. Luca and C. Pomerance [38], K. Ford [14]. See also [1, 39, 40]. K. Ford proved that n σ(ϕ(n)) ≥ 39.4 for all n. In 1988 J. S´andor [32, 41] conjectured that ψ(ϕ(m)) ≥ m,

(1.4.5)

for all odd m. He showed that (1.4.5) is equivalent to ψ(ϕ(n)) ≥

n 2

(1.4.6)

for all n, and obtained a class of numbers satisfying these inequalities. In 1988, J. S´andor [41] conjectured also that ϕ(ψ(n)) ≤ n

(1.4.7)

for any n ≥ 2 and V. Vitek [42] verified this conjecture for n ≤ 104 . In 1990, P. Erd¨os [13] expressed his opinion that this new conjecture could be as difficult as the Makowski–Schinzel conjecture (1.4.3). Nonetheless, as we will see, conjectures (1.4.5), (1.4.6) and (1.4.7) are not generally true, and it will be interesting to study the classes of numbers for which this is valid. Following J. S´andor’s paper [22], we study this conjecture and certain new properties of the above and related composite functions. Now, we formulate some basic asssertions. Lemma 1.4.1. For any a, b ≥ 2 ϕ(ab) ≤ aϕ(b)

(1.4.8)

with equality only if set(a) ⊂ set(b). Proof. We have ab =

∏ p|a,p|b

pα .

∏ q|a,q|b

qβ .

∏ r|b,r|a

rγ,

32

J´ozsef S´andor and Krassimir Atanassov

so

      ϕ(ab) 1 1 1 = ∏ 1− . ∏ 1− . ∏ 1− ab p q r p q r     1 1 ϕ(b) . ∏ 1− = , ≤ ∏ 1− q r b q r

so ϕ(ab) ≤ aϕ(b), with equality if “p does not exist”, i.e., p with the property p|a, p 6 |b. Thus for all p|a one also has p|b.

Lemma 1.4.2. If set(a) 6⊂ set(b), then for any a, b ≥ 2 one has ϕ(ab) ≤ (a − 1)ϕ(b),

(1.4.9)

ψ(ab) ≥ (a + 1)ψ(b),

(1.4.10)

and Proof. We give only the proof of (1.4.9). Let a = ∏ pα .

∏ qβ ,

b = ∏ rγ .

0

∏ qβ ,

where the q-s are the common prime factors, and the p-s (p ∈ set(a)) are such that p 6∈ set(b), i.e., suppose that α ≥ 1. Clearly β, β0, γ ≥ 0. Then   ϕ(ab) 1 = a. ∏ 1 − ≤ a−1 ϕ(b) p iff

∏



1 1− p

Now, 1−

∏ pα

by α ≥ 1. The inequality



≤ 1−

1 1 = 1− . α a ∏ p . ∏ qβ

1 1 1 ≥ 1− ≥ 1− α β . ∏q ∏p ∏p

  1 1 1− ≥ ∏p 1− p ∏p

is trivial, since by putting, e.g., p − 1 = u, one gets

∏(u + 1) ≥ ∏ u,

On Standard Arithmetic Functions ϕ, ψ and σ

33

and this is clear, since u > 0. There is an equality only when there is a single u, i.e., if the set of p such that set(a) 6⊂ set(b) has a single element, at the first power, and all β = 0, i.e., when a = p 6 |b. Indeed, ϕ(pb) = ϕ(p)ϕ(b) = (p − 1)ϕ(b). It is well-known (see, e.g., [32, 29]) that σ(ab) ≥ aσ(b). Lemma 1.4.3. For all a, b ≥ 1 ψ(ab) ≥ aψ(b).

(1.4.11)

Proof. Let u|v. Then       ψ(u) 1 1 1 ψ(v) = ∏ 1+ ≤ ∏ 1+ . ∏ 1+ = , u p p q|u,q|v q v p|u p|u,p|v with equality if q does not exist with q|u, q 6 |v. Put v = ab and u = b. Then ψ(v) ψ(u) u ≤ v becomes exactly (1.4.11). There is an equality if for each p|a one also has p|b, i.e., set(a) ⊂ set(b). Therefore, there is a similarity between the inequalities (1.4.8) and (1.4.11). The validity of the following assertion is given in [32]. Lemma 1.4.4. If set(a) 6⊂ set(b), then for any a, b ≥ 2 one has σ(ab)ψ(a).σ(b). Theorem 1.4.1. There are infinitely many n such that ψ(ϕ(n)) < ϕ(ψ(n)) < n.

(1.4.12)

For infinitely many m one has ϕ(ψ(m)) < ψ(ϕ(m)) < m.

(1.4.13)

There are infinitely many k such that 1 ϕ(ψ(k)) = ψ(ϕ(k)) < k. 2

(1.4.14)

34

J´ozsef S´andor and Krassimir Atanassov

Proof. We prove that (1.4.12) is valid for n = 3.2n for any a ≥ 1. This follows from ϕ(3.2a ) = 2a , ψ(2a ) = 3.2a−1, ψ(3.2a) = 3.2a+1, ϕ(3.2a+1) = 2a+1 , so 3.2a > ϕ(ψ(3.2a )) > ψ(ϕ(3.2a)). For the proof of (1.4.13), put m = 2a .5b (b ≥ 2). Then an easy computation shows that ψ(ϕ0 (m)) = 2a+1.32 .5b−2 , and ϕ(ψ(m)) = 2a+2 .3.5b−2 and the inequalities (1.4.13) will follow. For h = 3s remark that ϕ(ψ(h)) = 49 .h and ψ(ϕ(h)) = 43 .h, so ϕ(ψ(h)) < h < ψ(ϕ(h)),

(1.4.15)

which in a certain sense completes (1.4.12) and (1.4.13). Finally, for k = 2a .7b (b ≥ 2) one can deduce ψ(ϕ(k)) =

48 24 .k, ϕ(ψ(k)) = .k, 49 49

so (1.4.14) follows. We note that since ψ(ϕ(k)) < k,

(1.4.16)

by (1.4.14) and (1.4.16) one can say that k ϕ(ψ(k)) < , 2

(1.4.17)

for the above values of k. Note also that for h in (1.4.15) one has 1 ϕ(ψ(h)) = ψ(ϕ(k)). 3

(1.4.18)

For the values m given by (1.4.13) one has 2 ϕ(ψ(m)) = ψ(ϕ(m)). 3 For n = 2a .3b (b ≥ 2) one can see that ϕ(ψ(n)) = ψ(ϕ(n)). More generally, one can prove:

(1.4.19)

On Standard Arithmetic Functions ϕ, ψ and σ

35

Theorem 1.4.2. Let 1 < n = pα1 1 pα2 2 . . . pαr r be the prime factorization of n and suppose that the odd part of n is squarefull, i.e., li ≥ 2 for all i with pi ≥ 3. Then, ϕ(ψ(n)) = ψ(ϕ(n)) is true iff set((p1 − 1) . . .(pr − 1)) ⊂ {p1 , . . ., pr } and set((p1 + 1) . . .(pr + 1)) ⊂ {p1 , . . ., pr }

(1.4.20)

Proof. Since ϕ(n) = pα1 1 −1 pα2 2 −1 . . . pαr r −1 (p1 − 1)(p2 − 1) . . .(pr − 1) and ψ(n) = pα1 1 −1 pα2 2 −1 . . . pαr r −1 (p1 + 1)(p2 + 1) . . .(pr + 1) one can write ψ(ϕ(m)) =

pα1 1 −1 . . . pαr r −1 (p1 − 1) . . .(pr − 1).

α −1

s|(p1 1

∏ ...pαr r −1 (p1 −1)...(pr −1))

  1 1+ s

and ϕ(ψ(m)) =

pα1 1 −1 . . . pαr r −1 (p1 + 1) . . .(pr + 1). α −1

q|(p1 1

∏ ...pαr r −1 (p1 +1)...(pr +1))

  1 . 1− q

Since αi ≥ 1 when pi ≥ 3, the equality ψ(ϕ(n)) = ϕ(ψ(n)), by     1 1 (p1 − 1) . . .(pr − 1) 1 + ... 1+ p1 pr     1 1 ... 1− = (p1 + 1) . . .(pr + 1) 1 − p1 pr can also be written as

∏

t|(p1 −1)...(pr −1)



1 1+ t



= q|(p1

  1 1− . ∏ q +1)...(p +1) r

36

J´ozsef S´andor and Krassimir Atanassov

Since 1 + 1t > 1 and 1 − 1q < 1, this is impossible, in general. It is possible only if all prime factors of (p1 − 1) . . .(pr − 1) are among p1 , . . ., pr , and also the same for the prime factors of (p1 + 1) . . .(pr + 1). For example, n = 2a .3b .5c with a ≥ 1, b ≥ 2, c ≥ 2 satisfies (1.4.20). Indeed set((2 − 1)(3 − 1)(5 − 1)) = {2} ⊂ {2, 3} = set((2 + 1)(3 + 1)(5 + 1)). Similar examples are n = 2a .3b .5c .7d , n = 2a .3b .5c .11d , n = 2a .3b.7c .13d , n = 2a .3b .5c.7d .11e.13 f ,

etc. Theorem 1.4.3. Let n be squarefull. Then inequality (1.4.7) is true. Proof. Let n = pα1 1 pα2 2 . . . pαr r with αi ≥ 2 for all i = 1, 2, . . ., r. Then ϕ(ψ(n)) = ϕ(pα1 1 −1 pα2 2 −1 . . . pαr r −1 (p1 + 1)(p2 + 1) . . .(pr + 1)) ≤ (p1 + 1)(p2 + 1) . . .(pr + 1)ϕ(pα1 1 −1 pα2 2 −1 . . . pαr r −1 ),

by Lemma 1.4.2. But

ϕ(pα1 1 −1 pα2 2 −1 . . . pαr r −1 ) = pα1 1 −2 pα2 2 −2 . . . pαr r −2 (p1 − 1)(p2 − 1) . . .(pr − 1),

since αi ≥ 2. Then

so

ϕ(ψ(n)) ≤ (p21 − 1)(p22 − 1) . . .(p2r − 1)pα1 1 −2 pα2 2 −2 . . . pαr r −2     1 1 = pα1 1 pα2 2 . . . pαr r 1 − 2 . . . 1 − 2 , pr p1     1 1 ϕ(ψ(n)) ≤ n. 1 − 2 . . . 1 − 2 . pr p1 There is an equality in (1.4.21) if set((p1 + 1)(p2 + 1) . . .(pr + 1)) ⊂ {p1 , p2 , . . ., pr }. Clearly, the inequality (1.4.21) is the best possible one, and by     1 1 1 − 2 . . . 1 − 2 < 1. pr p1

(1.4.21)

On Standard Arithmetic Functions ϕ, ψ and σ

37

it implies inequality (1.4.7). Theorem 1.4.4. For any n ≥ 2 one has    ψ(n) ϕ n < n. n

(1.4.22)

Proof. It is immediate that   1 ϕ(n)ψ(n) = , 1, 1 − ∏ n2 p2 p|n so ϕ(n)ψ(n) ≤ n2 for any n ≥ 2. Now, by (1.4.8) one can write      ψ(n) ψ(n) ψ(n) ϕ n ≤ ϕ(n) ≤ ϕ(n) < n, n n n by the relation proved above. h i If n|ψ(n), i.e., when ψ(n) = ψ(n) n n , relation (1.4.22) gives inequality (1.4.7), i.e. ϕ(n)ψ(n) < n. For the study of the equation ψ(n) = k.n,

(1.4.23)

we use a notion and a method proposed by C. Wall [43]. We say that n is ωmultiple of m if m|n and set(m) = set(n). We need a simple result, stated as the following Lemma 1.4.5. If m and n are squarefree, and ψ(n) ψ(m) = , n m then n = m. Proof. Without loss of generality we may suppose (m, n) = 1; m, n > 1; m = q1 . . .q j (q1 < · · · < q j ); n = p1 . . . pk (p1 < · · · < pk ).

38

J´ozsef S´andor and Krassimir Atanassov Then the assumed equality has the form n(1 + q1 ) . . .(1 + q j ) = m(1 + p1 ) . . .(1 + pk ).

Since pk |n, the relation pk |(1+ p1 ) . . .(1+PK−1 )(1+ pk ) implies pk |(1+ pk) for some i ∈ {1, 2, . . ., k}. Here 1 + p1 < · · · < 1 + pk−1 < 1 + pk , so we must have pk |(1 + pk−1 ). This may happen only when k = 2, p1 = 2, p2 = 3; j = 2, q1 = 2, q3 = 3 (since for k ≥ 3, pk − pk−1 ≥ 2, so pk 6 |(1 + pk−1)). In this case (n, m) = 6 > 1, which is a contradiction. Therefore, k = j and pk = q j . Theorem 1.4.5. Assume that the least solution nk of (1.4.23) is a squarefree number. Then, all solutions of (1.4.23) are given by the ω-multiples of nk . Proof. If n is an ω-multiple of nk , then clearly ψ(n) ψ(nk) = = k, n nk by (1.4.1). Conversely, if n is a solution, set m = greatest squarefree divisor of n. Then ψ(n) ψ(m) ψ(nk ) = =k= . n m nk By Lemma 1.4.5, m = nk , i.e., n is an ω-multiple of nk . Theorem 1.4.6. Let n ≥ 3, and suppose that n is ω-deficient, i.e., ψ(n) < 2n. Then, inequality (1.4.7) holds. Proof. First remark that for any n ≥ 3, ψ(n) is an even number. Indeed, if n = 2a , then ψ(n) = 2a−1 .3, which is odd only for a = 1, i.e., n = 2. If n has at least one odd prime factor p, then by (1.4.1), ψ(n) will be even. Now, applying Lemma 1.4.2 for b = 2, one obtains ϕ(2a) ≤ a, i.e., ϕ(u) ≤ 2u for u = 2a (even). Here, an equality occurs only when u = 2k (k ≥ 1). Now, ψ(n) ϕ(ψ(n)) ≤ 2 , ψ(n) being even, and since n is ω-deficient, the Theorem follows.

On Standard Arithmetic Functions ϕ, ψ and σ

39

We must mention that the inequality ϕ(ψ(n)) ≤

ψ(n) 2

(1.4.24)

is the best possible, since we have equality for ψ(n) = 2k . Let n = pα1 1 pα2 2 . . . pαr r . Then pα1 1 −1 pα2 2 −1 . . . pαr r −1 (p1 + 1)(p2 + 1) . . .(pr + 1) = 2k is possible only if α1 = · · · = αr = 1, and p1 + 1 = 2a1 , . . ., pr + 1 = 2ar , i.e., when p1 = 2a1 − 1, . . ., pr = 2ar − 1, are distinct Mersenne primes, and n = p1 . . . pr . So, there is an equality in (1.4.24) iff n is a product of distinct Mersenne primes. Since by Theorem 1.4.5 one has ψ(n) = 2n iff n = 2a .3b (a, b ≥ 1), if one assumes ψ(n) ≤ 2n, then by (1.4.24), inequality (1.4.7) follows again. Therefore, in Theorem 1.4.6 one may assume ψ(n) ≤ 2n. Theorem 1.4.6 and the above remark impliy that when n is even and ω(n) ≤ 2, (1.4.7) is true. Indeed, 1 + 12 = 32 < 2, and (1 + 12 )(1 + 13 ) = 2. So, e.g., when n = pα1 1 pα2 2 , then       ψ(n) 1 1 1 1 = 1+ 1+ ≤ 1+ 1+ = 2. n p1 p2 2 3 On the other hand, if n is odd, and ω(n) ≤ 4, then (7) is valid. Indeed,      1 1 1 1 4 6 8 12 2304 1+ 1+ 1+ 1+ = . . . = < 2. 3 5 7 11 3 5 7 11 1155 Another remark is the following. If 2 and 3 do not divide n, and n has at most six prime factors, then ϕ(ψ(n)) < n. If 2, 3 and 5 do not divide n, and n has at most 12 prime factors, then the same result holds true. If 2, 3, 5 and 7 do not divide n, and n has at most 21 prime factors, then the inequality is true. If 2 and 3 do not divide n, we prove that ψ(n) < 2n, and by the presented method the results will follow. For example, when n is not divisible by 2 and 3, then the least prime factor of n could be 5, so ψ(n) 6 8 12 14 18 20 24 30 32 < . . . . . . . . 21 . Clearly p1 ≥ 2, p2 ≥ 3, . . ., ps ≥ s + 1, so it is sufficient to prove that     1 1 1 (1.4.26) 1 − 2 . . . 1 − s+1 > . 2 2 2 In the proof of (1.4.26) we will use the classical Weierstrass inequality s

s

∏ (1 − ak ) > 1 − ∑ ak ,

k=1

(1.4.27)

k+1

where ak ∈ (0, 1) (see e.g., [19]). 1 Put ak = 2k+1 in (1.4.27). Since s

∑

k=1 2

1 k+1

  1 1 1 1 1 1 − 2k 2k − 1 = . 1 + + · · · + k−1 = . = . 4 2 2 4 1 − 21 2k + 1 k

k

−1 Therefore, (1.4.26) becomes equivalent to 1 − 22k +1 > 12 . Hence, 12 > 22k −1 , +1 k k i.e., 2 > 2 − 1, that is true. Therefore, (1.4.26) follows, and the Theorem is proved. Related to the above theorems is the following result:

On Standard Arithmetic Functions ϕ, ψ and σ

41

Theorem 1.4.8. Let n be even, and suppose that the greatest odd part m of n is ψ-deficient, and that 3 6 |ψ(m). Then (1.4.7) is true. Proof. Let n = 2k m, when ϕ(ψ(n)) = ϕ(2k−1.3ψ(m)) = 2.ϕ(2k−1.ψ(m)), since (3, 2k−1.ψ(m)) = 1. But ϕ(2k−1.ψ(m)) ≤ 2k−2 .ψ(m) ≤ 2k−1 .m, so ϕ(ψ(n)) < 2k .m = n. In [29], it is proved that for all n ≥ 2 even, one has ϕ(σ(n)) > 2n,

(1.4.28)

with equality only if n = 2k , where 2k+1 − 1 is prime. Since σ(m) ≥ ψ(m), clearly this implies σ(σ(n)) > 2n (1.4.29) with the above equalities. So, the Suryanarayana-Kanold theorem (see [26, 27]) is reobtained, in an improved form. In [29], it is also proved that for all n ≥ 2 even, one has σ(ψ(n)) > 2n,

(1.4.30)

with equality only for n = 2. What are the odd solutions of σ(ψ(n)) = 2n? We now prove the following Theorem 1.4.9. Let n = 2k .m be even (k ≥ 1, m > 1 odd), and suppose that m is not a product of distinct Fermat primes, and that m satisfies (1.4.5). Then n σ(ϕ(n)) > n − m ≥ . (1.4.31) 2 Proof. First remark that if m is not a product of distinct Fermat primes, then ϕ(m) is not a power of 2. Indeed, if m = pa11 pa22 . . . par r then pa11 −1 pa22 −1 . . . par r −1 (p1 − 1)(p2 − 1) . . .(pr − 1) = 2s iff (since pi ≥ 3),

a1 − 1 = · · · = ar − 1 = 0

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J´ozsef S´andor and Krassimir Atanassov

and p1 − 1 = 2s1 , . . ., pr − 1 = 2sr , i.e., p1 = 2s1 + 1, . . ., pr = 2sr + 1 are distinct Fermat primes. Thus, there exists at least one odd prime divisor of ϕ(m). Now, σ(ϕ(2k .m)) = σ(2k−1.ϕ(m)) ≥ ψ(ϕ(m)).σ(2k−1) ≥ m.(2k − 1) = n − m, by relation (1.4.5). The last inequality of (1.4.31) is trivial, since n k−1 m ≤ 2 = 2 .m, where k − 1 ≥ 0. We shall note that the relation (1.4.28) gives an improvement of (1.4.3) for certain values of n. Theorem 1.4.10. Let p be an odd prime. Then ϕ(ψ(p)) ≤

p+1 p

(1.4.32)

with equality only if p is a Mersenne prime, and ψ(ϕ(p)) ≥ 32 .(p − 1), with equality only if p is a Fermat prime. Proof. ψ(p) = p + 1 and p + 1 being even, ϕ(p + 1) ≤ p+1 2 , with equality only if p + 1 = 2k , i.e., when p = 2k − 1 is a Mersenne prime. Since 3 2 .(p − 1) ≥ p, this inequality is better than (1.4.5) for n = p. Similarly, ϕ(p) = p − 1 is even, so ψ(p − 1) ≥ 23 .(p − 1), on basis of the following Lemma 1.4.6. If n ≥ 2 is even, then 3 ψ(n) ≥ .n, 2

(1.4.33)

with equality only if n = 2a . Proof. If n = 2a .N, with N odd, 3 ψ(n) = ψ(2a .N) = ψ(2a ).ψ(N) = 2a−1 .3.ψ(N) ≥ 2a−1.3.N = .n. 2

On Standard Arithmetic Functions ϕ, ψ and σ

43

Equality occurs only when N = 1, i.e., when n = 2a . Since p − 1 = 2a implies p = 2a + 1 – a Fermat prime, (1.4.32) is completely proved. Since 23 (p − 1) ≥ p, this inequality is better than (1.4.5) for n = p. 3 It is easily seen that for p ≥ 5 one has p+1 2 < p < 2 (p − 1), so (1.4.32) implies, as a corollary that ϕ(ψ(p)) < p < ψ(ϕ(p))

(1.4.34)

for p ≥ 5 prime. This is connected to relation (1.4.15). If n is even, and n 6= 2a , then since ψ(N) ≥ N + 1, with equality only when N is a prime, (1.4.33) can be improved to 3  n ψ(n) ≥ . n + , 2 N with equality only for n = 2a .N, where N is a prime. Theorem 1.4.11. Let a, b ≥ 1 and suppose that a|b. Then ϕ(ψ(a))|ϕ(ψ(b)) and ψ(ϕ(a))|ψ(ϕ(b)). In particular, if a|b, then ϕ(ψ(a)) ≤ ϕ(ψ(b)) and ψ(ϕ(a)) ≤ ψ(ϕ(b)).

(1.4.35)

Proof. The proof follows immediately from the following Lemma 1.4.7. If a|b, then ϕ(a)|ϕ(b)

(1.4.36)

ψ(a)|ψ(b)

(1.4.37)

and Proof. This follows from (1.4.1), see e.g., [29, 32]. Now, if a|b, then ψ(a)|ψ(b) by (1.4.37), so by (1.4.36), ϕ(ψ(a))|ϕ(ψ(b)). Similarly, a|b implies ϕ(a)|ϕ(b) by (1.4.36), so by (1.4.37), ψ(ϕ(a))|ψ(ϕ(b)). The inequalities in (1.4.33) are trivial consequences. Let a = p be a prime such that p 6 |k, and put b = k p−1 − 1. By Fermat’s Little Theorem, one has a|b, so all results of (1.4.35) are correct in this case. For example, ψ(ϕ(a)) ≤ ψ(ϕ(b)) gives, in the case of (1.4.35), and Theorem 1.4.9: 3 ψ(ϕ(k p−1 − 1)) ≥ ψ(ϕ(p)) ≥ .(p − 1), 2

44

J´ozsef S´andor and Krassimir Atanassov

for any prime p 6 |k, and any positive integer k > 1. Let (n, k) = 1. Then, by Euler’s divisibility theorem, one similarly has: ψ(ϕ(kϕ(n) − 1)) ≥ ψ(ϕ(n)) for any positive integers n, k > 1 such that (n, k) = 1. Let n > 1 be a positive integer, having as distinct prime factors p1 , . . ., pr . Then, it is immediate that ϕ(n)|ψ(n) (1.4.38) iff (p1 − 1) . . .(pr − 1)|(p1 + 1) . . .(pr + 1). For example, (1.4.38) is true for n = 2m , n = 2m.5s (m, s ≥ 1), etc. Now assuming (1.4.38), by (1.4.36) one can write the following inequalities: ϕ(ψ(ϕ(n))) ≤ ϕ(ψ(ψ(n))) and ψ(ϕ(ψ(n))) ≤ ψ(ϕ(ψ(n))). By studying the first 100 values of n with the property (1.4.38), the following interesting example may be remarked: ϕ(13) = ϕ(14) = 8, ψ(13) = ψ(14) = 24 and ϕ(13)|ψ(13). Similarly ϕ(70) = ϕ(72) = 24, ψ(70) = ψ(72) = 144, with ϕ(70)|ψ(70). Are there infinitely many such examples? Are there infinitely many n such that ϕ(n) = ϕ(n + 1) and ψ(n) = ψ(n + 1)? Or ϕ(n) = ϕ(n + 2) and ψ(n) = ψ(n + 2)? Let a = 8, b = σ(8k − 1). Then a|b (see e.g., [29] for such relations), and since ψ(ϕ(8)) = 6, ϕ(ψ(8)) = 12, by (1.4.35) we obtain the divisibility relations 6|ψ(ϕ(σ(8k − 1))) and 12|ϕ(ψ(σ(8k − 1))) for k ≥ 1. The second relation implies e.g., that if ϕ(ψ(σ(n))) = 2n, then n 6≡ −1(mod 8) and if ϕ(ψ(σ(n))) = 4n, then n 6≡ −1(mod 24). Theorem 1.4.12. Inequality (1.4.7) is true for n ≥ 2 if it is true for the squarefree part of n. Inequality (1.4.5) is true for an odd m ≥ 3 if it is true for the squarefree part of m.

On Standard Arithmetic Functions ϕ, ψ and σ

45

Proof. We give here the proof for the sake of completeness. Let n0 be the squarefree part of n, i.e., if n = pα1 1 pα2 2 . . . pαr r , then n0 = p1 p2 . . . pr and ϕ(ψ(n)) = ϕ(pα1 1 −1 pα2 2 −1 . . . pαr r −1 (p1 + 1)(p2 + 1) . . .(pr + 1)) ≤ pα1 1 −1 pα2 2 −1 . . . pαr r −1 ϕ((p1 + 1)(p2 + 1) . . .(pr + 1)) n = 0 .ϕ(ψ(n0)) n by inequality (1.4.8). Thus ϕ(ψ(n)) ϕ(ψ(n0 )) ≤ . n n0 Therefore, if

ϕ(ψ(n0 )) n0

< 12, then

ϕ(ψ(n)) n

< 1. Similarly, one can prove that

ψ(ϕ(n)) ψ(ϕ(m0 )) ≥ , m n0

(1.4.39)

so if (1.4.5) is true for the squarefree part m0 of m, then (1.4.5) is also true for m. As a consequence, (1.4.7) is true for all n iff it is true for all squarefree n. As we have stated previously, (1.4.5) is not generally true for all m. Let, e.g., m = 3.F, where F > 3 is a Fermat prime. Indeed, put F = 2k + 1, then ϕ(m) = 2k+1 , so ψ(ϕ0 (m)) = 2k .3 < 3.(2k + 1) = 3.F = m, contradicting (1.4.5). However, if m has the form m = 5.F, where F > 5 is again a Fermat prime, then (1.4.5) is valid, since in this case ψ(ϕ0 (m)) = 6.2k > 5.(2k + 1) = m. More generally, we prove now: Theorem 1.4.13. Let 5 ≤ F1 < · · · < Fs be Fermat primes. Then inequality (1.4.5) is valid (with strict inequality) for m = F1a1 . . .Fsas with arbitrary ai ≥ 1 (i = 1, 2, . . ., s). b

Proof. Let Fi = 1 + 22 i (i ≥ 1) be Fermat primes, where b1 ≥ 1. Since

46

J´ozsef S´andor and Krassimir Atanassov

b1 < b2 < · · · < bs, clearly bi ≥ i for any i = 1, 2, . . ., s. By (1.4.39) it is sufficient to prove the result for m0 = F1 . . .Fs , when (1.4.5) becomes, after some elementary computations:     1 3 1 1 + b1 . . . 1 + 2bs ≤ . (1.4.40) 2 2 2 2 We prove that (1.4.40) holds with a strict inequality. By the classical Weierstrass inequalities one has s

∏ (1 + ak )
0.

48

J´ozsef S´andor and Krassimir Atanassov

Let now p be a prime such that   1 δ ∏ 1 − q < c(ε) q≤p   (this is possible, since ∏ 1 − 1q → 0 as p → ∞). q≤p

Then, if x is large, then for all n < x, except perhaps a number of η.x + ε.x integers one has ψ(n) < c(ε).n and ψ(n) ≡ 0(mod q) for any q ≤ p, (η > 0). But for these exceptions one has ϕ(ψ(n)) < δ.n, and this completes the proof; η, ε > 0 being arbitrary. It can be proved similarly that ψ(ϕ(n)) > δ.n, except perhaps a set of density zero. Theorem 1.4.14 implies that lim inf

n→∞

ϕ(ψ(n)) n

(1.4.45)

= 0, and so, one has

ψ(ϕ(n)) = +∞. n n→∞

lim sup

For another proof of these results, see [32]. We cannot determine the following ψ(ϕ(n)) ϕ(ψ(n)) values of lim inf n and lim sup n . n→∞

n→∞

However, we can prove Theorem 1.4.15.   ψ(ϕ(k)) 1 ψ(ϕ(n)) ≤ inf : k is a multiple of 4 < . n→∞ n k 2

lim inf

Proof. Let k be a multiple of 4, and p > k2 . Then       1 k k p−1 ϕ kp = ϕ ϕ(p) = 2ϕ . , 2 2 2 2 since 2| 2k . Now, by ψ(ab) ≤ ψ(a)ψ((b)) one can write      1 p−1 ψ ϕ kp ≤ ψ(ϕ(k))ψ . 2 2

(1.4.46)

On Standard Arithmetic Functions ϕ, ψ and σ 49    Since ψ p−1 ≤ σ p−1 , and by the known result of Makowski-Schinzel: 2 2

lim inf



σ( p−1 2 ) p−1 2

= 1, from the above one can write:

ψ(ϕ( 12 kp)) ψ(k) ψ( p−1 ψ(ϕ(k)) 2 ) . lim inf ≤ , ≤ 1 p−1 p→∞ p→∞ k k 2 kp 2

lim inf

and now relation (1.4.46) follows, by taking inf after k. Since 232 − 1 = F0 .F1 .F2 .F3 .F4 , k

where Fk = 22 + 1, and all Fi (0 ≤ i ≤ 4) are primes, it follows, that ϕ(232 − 1) = 21 .22 .24 .28 .216 = 231 . Thus, ϕ(4(232 − 1)) = 232 , by ϕ(5) = 2. Since ψ(232) = 231 .3, by letting 231 .3 1 in (1.4.46) k = 4(232 − 1), we get the inf ≤ 4(2 , where θ > 3.2130 . 32 −1 < 2( 4 −θ) 3

In any case, we get in (1.4.46) that lim inf < 12 , and the fact that a value slightly greater than 2.14 = 38 . 3

In [32], it is asked the value of liminf ψ(σ(n)) ≤ 1. We now prove n Theorem 1.4.16. lim inf ψ(σ(n))

ψ(σ(n)) = 1. n

(1.4.47)

σ(n)

Proof. Since n ≥ n ≥ 1, clearly this limit is higher or equal to 1. By the above inequality, the result follows. However, we give here a new proof of this fact. We remark that, since ϕ(N) ≤ ψ(N)σ(N), and by the known result ϕ(N(a, p)) σ(N(a, p)) = lim = 1, p→∞ N(a, p) p→∞ N(a, p) lim

where N(a, p) =

a p −1 p−1 ,

(a > 1, p prime) we easily get ϕ(N(a, p)) = 1. p→∞ N(a, p) lim

(1.4.48)

Now, let a = q be an arbitrary prime in (1.4.48). We remark that N(q, p) = = σ(q p−1 ). Now,

q p −1 q−1

σ(q p−1 ) qp − 1 q = → , p−1 p−1 q (q − 1).q q−1

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J´ozsef S´andor and Krassimir Atanassov

as p → ∞, from (1.4.48) we can write: ψ(σ(q p−1 )) q = < 1 + ε, p→∞ q p−1 q−1 lim

(1.4.49)

for q ≥ q(ε), ε > 0. Now by (1.4.49), (1.4.47) follows. In [32], it is proved, by assuming the infinitude of Mersenne primes, that lim inf

n→∞

ψ(ψ(n)) 3 = . n 2

(1.4.50)

Can we prove (1.4.50) without any assumptions? We have conjectured in [32] that the following limit is true, but in the proof we have used the fact that there are infinitely many Mersenne primes. Now we prove this result without any assumptions. Theorem 1.4.17. We have lim inf

ψ(ψ(n)) 3 = . n 2

Proof. Since ψ(n) ≥ 32 n for all even n, and ψ(n) ≥ n for all n, clearly ψ(ψ(n)) ≥ 3 3 2 .n for all n. Therefore, it will be sufficient to find a sequence with limit 2 . By using theorems on primes in arithmetical progressions, it can be proved, as in Makowski-Schinzel [31], that lim sup

ϕ(a) σ(a) = lim inf =1 a a

as p tends to infinity, where a = p+1 2 and p ≡ 1(mod 4). Since p+1 is odd, we get 2     p+1 p+1 σ(p + 1) = σ 2. = 3.σ , 2 2 implying that lim inf σ(p+1) = 23 . Since ψ(n) ≤ σ(n), we can write that p

lim inf ψ(p+1) ≤ 32 . By ψ(p+1) > 32 , this yields lim inf ψ(p+1) = 32 , completing p p p the proof of the theorem.

On Standard Arithmetic Functions ϕ, ψ and σ

51

1.5. On the Equation ϕ(n) + d(n) = n and Related Inequalities The present Section is based on [45]. In paper [46], we have proved the following inequality ϕ(n) + d(n) ≤ n + 1,

(1.5.1)

for the positive integer n ≥ 2, with equality only for n = 4 or n - prime. In fact, (1.5.1) was a consequence of the stronger relation (see [46]): ϕ(n) + d(n) ≤ n,

(1.5.2)

for and n 6= 4, different than prime. However, the cases of equality for (1.5.2) are not studied in [46]. Here, we consider also the case of equality. Certain related new inequalities will be pointed out, too. Theorem 1.5.1. n = 8 and n = 9.

The equality ϕ(n) + d(n) = n had the only solutions

Proof. Case 1) Let n be an even number. Then it is well-known that √ ϕ(n) ≤ n2 . Using the relation d(n) < 2 n (see, e.g., [47]), we get ϕ(n) + d(n)
1 is composite, then √ ϕ(n) ≤ n − n. Lemma 1.5.2. For any n ≥ 1262 one has d(n)
1. Let d ∗ (n) denote the number of unitary divisors of n (see [20]). Then, as d ∗ (n) ≥ ω(n) + 1, we get, by relation (1.5.2) that: n + 1 ≥ ϕ(n) + d(n) ≥ ϕ(n) + d ∗ (n) ≥ f (n) + ω(n) + 1 Thus, particularly, the equation ϕ(n) + ω(n) = n has the only solutions as n=primes. Many other equations are studied in our paper [49].

Chapter 2

Perfect and Related Numbers 2.1. On (m, n)-Super-Perfect Numbers It is well known that a number n is called perfect, if σ(n) = 2n. The Euclid-Euler theorem (see [50]) gives the general form of even perfect numbers: n = 2k .p, where p = 2k+1 − 1 is a prime (“Mersenne prime”). No odd perfect numbers are known. The number n is said to be super-perfect if σ(σ(n)) = 2n. The Suryanarayana-Kanold theorem (see [26, 27]) gives the general form of even super-perfect numbers n = 2k , where p = 2k+1 − 1 is a prime. No odd superperfect numbers are known. For new proofs of these results see [22, 51, 52]. Two numbers a and b are called amicable if σ(a) = σ(b) = a + b. For many results on perfect, super-perfect, amicable and related numbers, see Chapter 1 of the recent book [52]. In an unpublished paper, I. Z. Ruzsa (see [52]) called the numbers a and b a lovely-pair, if σ(a) = 2b and σ(b) = 2a. All such pairs with a and b both even are given by a = 2k .(2q+1 − 1), b = 2q .(2k+1 − 1), where 2q+1 − 1 and 2k+1 − 1 are both Mersenne primes. For new proofs and extensions, see [22, 24]. Note that when a = b, we reobtain the even perfect numbers, so Ruzsa’s theorem is again a generalization (as the case of amicable numbers is, too) of perfect numbers. In [22] it is proved also that a super lovelypair (a, b), satisfying σ(σ(a)) = 2b and σ(σ(b)) = 2a with a and b even, implies a = b, so one reobtains the super-perfect numbers of Suryanarayana-Kanold.

56

J´ozsef S´andor and Krassimir Atanassov

Here, we will study an extension of the super-perfect numbers, so the effect that for fixed m and n, the numbers a satisfy the relation σ(mσ(a)) = na

(2.1.1)

Definition 2.1.1. Given two positive integers m and n, a number a is said to be (m, n)-super-perfect, if it satisfies equation (2.1.1). We note that the (1, 2)-super-perfect numbers are in fact the classical superperfect numbers. There are also many other problems which lead to an equation of type (2.1.1). For example, searching for lovely-pairs (a, b), with b – odd, by σ(a) = 2b, and σ(b) = 2a one gets σ(σ(a)) = σ(2b) = 3σ(b) = 6a, so a is an (1, 6)-super-perfect number. If a and b are both odd members of a lovely-pair, then a and b are both (1, 6)-super-perfect numbers. Though the equation σ(σ(a)) = 6a (2.1.2) seems quite difficult, we note that by a computer search one can find in the range a ≤ 109 only 8 even numbers, the least being 42, and the largest being 218 .3.7. There is a single odd solution, namely a = 3.5.7.132.31. We now give a general result on even (m, n)-super-perfect numbers. Theorem 2.1.1. Let a be an even (m, n)-super-perfect number, where m ≥ 2. Let 2k ||a, and assume that (m, 2k+1 − 1) = 1. Then n ≥ 2(m + 1).

(2.1.3)

If n = 2(m + 1), then a = 2k , where 2k+1 − 1 is a Mersenne prime. If m = 1, then in place of (2.1.3) one has n ≥ 2m = 2.

(2.1.4)

We needs the following Lemma 2.1.1. For all positive integers a and b one has σ(ab) ≥ aσ(b)

(2.1.5)

Perfect and Related Numbers

57

with equality only for a = 1. For all positive integers a ≥ 2 one has σ(a) ≥ a + 1

(2.1.6)

with equality only for a being prime. Similarly, σ(a) ≥ a

(2.1.7)

with equalty only for a = 1. This Lemma is well-known (see e.g., [51]). Proof of Theorem 2.1.1. Let a = 2k .A, (A ≥ 1, odd) be an even (m, n)-superperfect number. Then, σ(a) = (2k+1σ(A)), so σ(mσ(a)) = σ((2k+1 − 1)mσ(A)) ≥ σ(A).σ((2k+1 − 1)m) ≥ A.σ(2k+1 − 1)σ(m) by (2.1.5), (2.1.7) and the multiplicative property of σ-function (since (m, 2k+1 − 1) = 1). On the other hand, by relation (2.1.6) one has σ(m) ≥ m + 1 for m ≥ 2 and σ(2k+1 − 1) ≥ 2k + 1. Thus, if a satisfies equation (2.1.1), then one can write na = n.2k .A ≥ A.2k+1.(m + 1), (2.1.8) implying n ≥ 2(m + 1), i.e., relation (2.1.3). If n = 2(m + 1), then in the above chain of inequalities, all inequalities become equalities. This means that A = 1, 2K+1 − 1 is a prime and m is a prime. When m = 1, the above proof shows only that n ≥ 2, so for n = 2 the general solutions are given again (i.e., the Suryanarayana-Kanold theorem is reobtained). Remark 2.1.1. The case n = 2(m + 1) considered above forces that m is a prime. However, we can state a more general result (with the same proof), in which this is not necessary. Theorem 2.1.2. Assuming the conditions from Theorem 2.1.1, one has n ≥ 2σ(m).

(2.1.9)

58

J´ozsef S´andor and Krassimir Atanassov If n = 2σ(m), then a = 2k , where 2k+1 − 1 is a prime.

Remark 2.1.2. For m = 1, n = 2, thus the Suryanarayana-Kanold theorem is reobtained. Remark 2.1.3. For m = 2, relation (2.1.9) implies n ≥ 6, thus there are no even (2, k)-super-perfect numbers for 1 ≤ k ≤ 5. The general form of the even (2, 6)-super-perfect numbers is a = 2k , where 2k+1 − 1 is a prime. For m = 3 one obtains n = 8, so the even (3, 8)-super-perfect numbers are determined again, etc. We now give an example, when all solutions can be determined. Theorem 2.1.3. The single (2, 3)-super-perfect number is a = 1. Proof. As we have seen in Remark 2, there are no even solutions for σ(2σ(a)) = 3a.

(2.1.10)

The number a = 1 is an odd solution, as σ(2) = 3. But we can prove generally that, this is the single solution of (2.1.10). Namely, by inequality (2.1.5) we can write that σ(2σ(a)) ≥ σ(a).σ(2) = 3σ(a) ≥ 3a, with equality only σ(a) = 1 and a = 1. Thus, Theorem 2.1.3 follows. More generally (with the same proof) it is valid the following Theorem 2.1.4. The single (m, σ(m))-super-perfect number is a = 1. Remark 2.1.4. Therefore, the single (3,4), (4,7), (5,6), (6,12), (7,8),. . . super-perfect number is a = 1. Related to Theorem 2.1.2, for the odd solutions of (m, 2σ(m))-super-perfect numbers, one can prove Theorem 2.1.5. Assume that a is an odd (m, σ(m))-super-perfect number satisfying σ(σ(a)) ≤ 2a. Then a must be a perfect square.

Perfect and Related Numbers

59

Proof. We shall use the following well-known lemma. Lemma 2.1.2. For all a, b ≥ 1, one has σ(ab) ≤ σ(a)σ(b),

(2.1.11)

with equality only for (a, b) = 1. Let now a satisfy equation (1) with n = 2σ(m). Then, by relation (2.1.11) one has 2σ(m)a = σ(mσ(a)) ≤ σ(m)σ(σ(a)). This implies σ(σ(a)) ≥ 2a, which combined with the assumption σ(σ(a)) ≤ 2a yields σ(σ(a)) = 2a. That is, a is an odd super-perfect number. It is known (see [52]) that then a must be a perfect square. A lovely-pair (a, b) satisfies the system σ(a) = 2b, σ(b) = 2a. One can obtain many variations of these numbers. For example, by considering the system σ(a) = b2 (2.1.12) σ(b) = 2a one can ask for general solutions of (2.1.12). Since b = 2σ(a), one obtains in fact σ(2σ(a)) = 2a, i.e., a is a (2, 2)-super-perfect number. Such numbers do not exist, as by (2.1.5) and (2.1.7) one has σ(2σ(a)) ≥ σ(a)σ(2) = 3σ(a) = 2a, i.e., 2a ≥ 3a, which is a contradiction. Replacing the second equation of (2.1.12) as σ(b) = 6a, one obtains, however, another system σ(a) = 2b (2.1.13) σ(b) = 6a leading to the equation σ(2σ(a)) = 6a, i.e., a (2-6)-super-perfect number. As we have seen, the even solutions are a = 2k , with 2k+1 − 1 being a prime. This gives

60

J´ozsef S´andor and Krassimir Atanassov

b = 2σ(a) = 2(2k+1 − 1), so all even pairs of system (2.1.13) are determined. The determination of pairs (a, b) with a being odd and b being even numbers, however, seems difficult. Indeed, suppose that b has the form b = 2B, with B being an odd number. Then σ(a) = B, so σ(2B) = 3σ(B) = 3σ(σ(a)) = 6a gives σ(σ(a)) = 2a, i.e., a must be an odd super-perfect number. As we have mentioned in the introduction, although it is known that a must be a perfect square, it is not known the existence of such numbers. In any case, system (2.1.13) is closely connected to the super-perfect numbers of SuryanarayanaKanold. Another possibility of extending the lovely-pairs is to replace the function σ with other arithmetic functions. The unitary and other cases were studied in [52]. we now introduce another analogue, with the arithmetical function σ+ , counting the sum of even divisors of its argument. Definition 2.1.2. For an even number n, let us denote by σ+ (n) the sum of even divisors of n. We shall say that (a, b) is a +-lovely-pair, if it satisfies the system σ+ (a) = 2b (2.1.14) σ+ (b) = 2a The case a = b has been recountly studied in [53]. Then, we obtain the so-called +-perfect numbers a satisfying σ+ (a) = 2a.

(2.1.15)

It was shown in [53] that a must have one of the following forms: 1. a = 2N, where N is an odd ordinary perfect number, i.e., σ(N) = 2N; 2. a = 2k (2k − 1), where 2k − 1 is a prime. Before studying system (2.1.14), we need the following auxiliary lemma Lemma 2.1.3. If n = 2k .N with k ≥ 1, N odd, positive integers, then σ+ (n) = 2.(2k − 1).σ(N).

(2.1.16)

61

Perfect and Related Numbers

Proof. Each even divisor of n may be written as d = 2a .A, where 1 ≤ a ≤ k, and A|N. Thus, σ+ (n) =

∑ a≤k,A|N

2a .A = ( ∑ 2a ).( ∑ A) = (2 + 22 + · · · + 2k ).σ(N) a≤k

A/N

= 2(2k − 1)σ(N), which completes the proof of (2.1.16). We now prove the following Theorem 2.1.6. Let (a, b) be an even pair of +-lovely-pairs. Then a, b must have one of the following forms: 1. a = 2k .(2q − 1), b = 2q .(2k − 1), where 2q − 1 and 2k − 1 are both primes; 2. a = 2A, b = 2B, where (A, B) is an odd lovely-pair (in Ruzsa’s sense); 3. a = 2A, b = 2q B, where q ≥ 2, A and B are satisfying A = (2q − 1)A1 and σ(B) = 2A1 , σ((2q − 1)A1 = 2q B. Proof. By (2.1.16), the system (2.1.14) becomes (2k − 1)σ(A) = 2q .B

(2q − 1)σ(B) = 2k .A

,

(2.1.17)

where a = 2k .A, b = 2q .B, with A, B being odd numbers. Then the first equaion of (2.1.17) implies B = (2k − 1).B0, while the second implies that A = (2q − 1).A0. Replacing, one gets σ((2q − 1)A0 ) = 2q .B0 σ((2q − 1)B0 ) = 2k .A0

.

(2.1.18)

Let k, q ≥ 2. Then, using Lemma 2.1.1, we can write subsequently that 2q .B0 ≥ A0 .σ(2q − 1) ≥ A0 .2q , with an equality only when 2q − 1 is a prime. Similarly, 2k .A0 ≥ B0 .σ(2k − 1) ≥ B0 .2k,

62

J´ozsef S´andor and Krassimir Atanassov

with an equality only when 2k − 1 is a prime. Thus B0 ≥ A0 ≥ B0 , which implies B0 = A0 , so we must have equalities in all the above inequalities. This yields A0 = B0 = 1, so A = 2q − 1 and B = 2k − 1, proving the first part of the Theorem. When k = 1, q = 1, then σ(A) = 2B, σ(B) = 2A, i.e., (A, B) is an odd lovely-pair in the Ruzsa’s sense. Finally, if k = 1, q ≥ 2 then A = (2q − 1)A1 , so σ(B) = 2A1 , which gives σ((2q − 1).A1 ) = 2q .B. Remark 2.1.4. Case 3 of Theorem 2.1.6 remains momentarily in doubt. Indeed, let, e.g., be q = 2. Then σ(B) = 2A1 and σ(3A1 ) = 4B. Assume that 3 6 | A1 . Then σ(3)σ(A1 ) = 4B, so σ(σ(A1 )) = 2A1 . This means that A1 must be a super-perfect number (in the Suryanarayana-Kanold sense), not divisible by 3. More generally, if 2q − 1 is a prime, and (2q − 1, A1 ) = 1, then σ(σ(A1 )) = 2A1 . Now, we discuss a special case of (m, n)-super-perfect numbers, when m is changed to the value of an arithmetic function of a. The first example is related to Euler’s totient function ϕ, so the equation with a super-perfect number obtains the form σ(ϕ(a).σ(a)) = n.a.

(2.1.19)

Let for a given natural number m > 1 P(m) be the largest prime divisor of m. Theorem 2.1.7. If n is an even perfect number, then a =

n is a soluP(n)

tion of (2.1.19). Proof. Let m be an even perfect number. First, we see that if m = 2k .(2k+1 − 1) for 2k+1 − 1 is a prime number, then P(m) = 2k+1 − 1. Hence      m m σ ϕ .σ = σ(ϕ(2k ).σ(2k )) P(m) P(m) = σ(2k−1 .(2k+1 − 1)) = (2k − 1).2k+1 m m = 2.(2k − 1). = n. , P(m) P(m) where n = 2.(2k − 1). The second example is related to Dedekind’s function ψ, so the equation with a super-perfect number obtains the form σ(ψ(a).σ(a)) = n.a.

(2.1.20)

Perfect and Related Numbers

63

Analogously to the above, we can prove Theorem 2.1.8. If n is an even perfect number, then a =

n is a soluP(n)

tion of (2.1.20). In a future authors’ research, next modifications of the concept of superperfect number will be discussed on the basis of [54, 55], where, for the first time, the following concepts, generalizing the notion “perfect number” are introduced and their basic properties are studied. The natural number n is a multiplicative perfect number iff Pd (n) ≡ ∏ d = n2 . d|n

Let g be a fixed arithmetic function. The natural number n is a g-additive perfect number iff ∑ g(d) = 2g(n). d|n

Obviously, for g(n) = n multiplicative perfect numbers are obtained. Let g be a fixed arithmetic function. The natural number n is a gmultiplicative perfect number iff

∏ g(d) = (g(n))2. d|n

Obviously, for g(n) = n multiplicative perfect numbers are obtained, as well again.

2.2. On a Modification of Perfect Numbers In [56] G.L. Cohen and H.J.J. te Riele called the natural number n (m, k)-perfect if σ(m) (n) = kn, where σ(0) (n) = n, σ(m)(n) = σ(σ(m−1)(n)) for m ≥ 1 or σ(m) (n) = σ(σ(. . .(σ(n) . . .)). | {z } m−times

64

J´ozsef S´andor and Krassimir Atanassov Table 2.2.1. n = 2 m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

σ(m) 3 4 7 8 15 24 60 168 480 1512 4800 15748 28748 46872 153600 507780 1956864 6534024 20736000 77297220

k×n 2×2 4×2 12×2 30×2 84×2 240×2 756×2 2400×2 7874×2 14311×2 23436×2 76800×2 253890×2 978432×2 3267012×2 10368000×2 38648610×2

In [56], only the case m ≤ 4 is discussed. Below we give a table with examples for m ≤ 20. Theorem 2.2.1. If n is an odd perfect number, then the following equalities are simultaneously valid: σ(2) (n) = 3!n, σ(3) (n) = 4!n, if n /≡ 0(mod 3),

σ(4) (n) = 5!n, if n /≡ 0(mod 3),

σ(5) (n) = 6!n, if n /≡ 0(mod 15).

65

Perfect and Related Numbers Table 2.2.2. n = 3 m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

σ(m) 4 7 8 15 24 60 168 480 1512 4800 15748 28748 46872 153600 507780 1956864 6534024 20736000 77297220

k×n 5×3 8×3 20×3 56×3 160×3 504×3 1600×3 15624×3 51200×3 169260×3 652288×3 2178008×3 6912000×3 25765740×3

Proof. Let n be an odd perfect number, i.e., σ(n) = 2n. Then, σ(2) (n) = σ(σ(n)) = σ(2n) = σ(2).σ(n) = 3.2.n = 3!n, σ(3)(n) = σ(σ(2) (n)) = σ(6n) = σ(6).σ(n) = 12.2.n = 4!n, σ(4) (n) = σ(σ(3) (n)) = σ(24n) = σ(24).σ(n) = 60.2.n = 5!n, σ(5) (n) = σ(σ(4) (n)) = σ(120n) = σ(120).σ(n) = 360.2.n = 6!n. Unfortunately, σ(6) (n) 6= 7!n for all natural numbers n. Theorem 2.2.2. For every n ≥ 3 σ(σ(n)) < n2 .

(2.2.1)

66

J´ozsef S´andor and Krassimir Atanassov Table 2.2.3. n = 4 m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

σ(m) 7 8 15 24 60 168 480 1512 4800 15748 28748 46872 153600 507780 1956864 6534024 20736000 77297220

k×n 2×4 6×4 15×4 42×4 120×4 378×4 1200×4 3937×4 7187×4 11718×4 38400×4 126945×4 489216×4 1633506×4 5184000×4 19324305×4

Proof. We shall use the following well-known inequality: σ(a.b) ≤ σ(a).σ(b)

(2.2.2)

√ σ(n) < n n

(2.2.3)

and will prove the inequality

which has different proofs, e.g., the following. When n is a prime number, then √ σ(n) = n + 1 < n n for n ≥ 3. If we assume that (2.2.3) is valid for some natural number n ≥ 3 and if p is a given prime number, then we discuss two cases: Case 1: p 6∈ set(n). Therefore, √ √ σ(np) = σ(n)(p + 1) < (n n).(p p).

Perfect and Related Numbers

67

Table 2.2.4. n = 5 m 1 2 3 4 5 6 7 8 9

σ(m) 6 12 28 56 120 360 1170 3276 10192

k×n 24 × 5 72 × 5 234 × 5 -

Table 2.2.5. n = 6 m 1 2 3 4 5 6 7 8 9

σ(m) 12 28 56 120 360 1170 3276 10192 24738

k×n 2×6 10 × 6 60 × 6 195 × 6 644 × 6 4123 × 6

Case 2: p ∈ set(n), i.e., n = mpa for some natural numbers a and m. σ(np) = σ(mpa+1 ) = σ(m)

√ pa+2 − 1 pa+2 − 1 √ = σ(n) a+1 < (n n).(p + 1) = np np. p−1 p −1

First, we shall discuss the case, when n is an odd number. Let n ≥ 3 be a prime number. Then, from (2.2.3) we obtain: √ σ(σ(n)) = σ(n + 1) < (n + 1) n + 1 < n2

68

J´ozsef S´andor and Krassimir Atanassov Table 2.2.6. n = 7 m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

σ(m) 8 15 24 60 168 480 1512 4800 15748 28748 46872 153600 507780 1956864 6534024 20736000 77297220

k×n 24 × 7 216 × 7 6696 × 7 72540 × 7 279552 × 7 933432 × 7 11042460 × 7

for each natural number n ≥ 3. Let us assume that (2.2.1) is valid for some natural number n ≥ 3 and let p 6∈ set(n) be a prime number. Then, from the assumption and from (2.2.2) and (2.2.3) it follows that: σ(σ(np)) = σ(σ(n)(p + 1)) < σ(σ(n)).σ(p + 1) < n2 .(p + 1)

i.e.,

p

p + 1 < n2 .p2 = (np)2,

σ(σ(np)) ≤ (np)2 .

(2.2.4)

Let p ∈ set(n). Therefore, n = m.pa for some natural numbers a and m for which (m, p) = 1. Then, from the assumption and from (2.2.2) and (2.2.3): σ(σ(np)) = σ(σ(m)σ(pa+1 )) < σ(σ(m)).σ(σ(pa+1)) < m2 .p2(a+1) = (np)2 . Second, we shall discuss the case when n is an even number.

Perfect and Related Numbers

69

Table 2.2.7. n = 8 m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

σ(m) 15 24 60 168 480 1512 4800 15748 28748 46872 153600 507780 1956864 6534024 20736000 77297220

k×n 3×8 21 × 8 60 × 8 189 × 8 600 × 8 6109 × 8 19200 × 8 244608 × 8 816753 × 8 2592000 × 8 -

Let n = 2a m, where a, m ≥ 1 are natural numbers and m is odd. Then from (2.2.3), (2.2.4) and (2.2.2) we obtain for a ≥ 3 that σ(σ(n)) = σ(σ(2a m)) = σ((2a+1 − 1)σ(m)) p < σ(2a+1 − 1)σ(σ(m)) < m2 (2a+1 − 1) 2a+1 − 1 √ < m2 2a+1 2a+1 < m2 22a = n2 . Finally, we directly see that σ(σ(4)) = σ(7) = 8 < 16 = 42 and σ(σ(2)) = σ(3) = 4 = 22 and only in the latter case we obtain an equality. With this the Theorem is proved.

70

J´ozsef S´andor and Krassimir Atanassov Table 2.2.8. n = 9 m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

σ(m) 13 14 24 60 168 480 1512 4800 15748 28748 46872 153600 507780 1956864 6534024 20736000 77297220

k ×n 168 × 9 5208 × 9 56420 × 9 2304000 × 9 8588580 × 9

Now we see that √ σ(σ(σ(n))) ≤ (σ(n))2 < (n n)2 = n3 σ(σ(σ(σ(n)))) ≤ (σ(σ(n))2 < (n2 )2 = n4 . Therefore, for instance by induction we can prove the validity of the following: Theorem 2.2.3. For every m ≥ 2 and n ≥ 3 σ(m) (n) ≤ nm.

Perfect and Related Numbers

71

Table 2.2.9. n = 10 m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

σ(m) 18 39 56 120 360 1170 3276 10192 24738 40932 106400 312480 1757984 3494988 11058432 39146688 139278360

k×n 12 × 10 36 × 10 117 × 10 10640 × 10 31248 × 10 13927836 × 10

2.3. On Multiplicatively Perfect Numbers As we mentioned in Section 2.1, the Suryanarayana-Kanold Theorem [26, 27] gives the general form of even super-perfect numbers: n = 2k , where 2k+1 − 1 = p is a prime. No odd super-perfect numbers are known. For new proofs of these results, see [29, 57]. Many open problems are stated e.g., in [28, 29]. The following research is based on [58]. Let T (n) denote the product of all divisors of n. There are many numbers n with the property T (n) = n2 , but none satisfying T (T (n)) = n2 . Let us call the number n > 1 multiplicatively perfect (or, for short, m-perfect; see Chapter 2 of [59]) if T (n) = n2 . (2.3.1) and multiplicatively super-perfect (m-super-perfect), if T (T (n)) = n2 .

(2.3.2)

72

J´ozsef S´andor and Krassimir Atanassov To begin with, we prove the following little result (cf. Chapter 2 of [59]).

Theorem 2.3.1. All m-perfect numbers n have one of the following forms: n = p1 p2 or n = p31 , where p1 , p2 are arbitrary distinct primes. There are no m-super-perfect numbers. Proof. Firstly, we note that if d1 , d2 , . . ., ds are all divisors of n, then   n n n {d1 , d2 , . . ., ds} = , , . . ., , d1 d2 ds implying that d1 d2 . . .ds = i.e.,

n n n ... , d1 d2 ds s

T (n) = n 2 ,

(2.3.3)

where s = d(n) denotes the number of (distinct) divisors of n. Let n = pα1 1 pα2 2 . . . pαr r be the prime factorization of n > 1. It is well-known that d(n) = (α1 + 1) . . .(αr + 1), so equation (2.3.1) combined with (2.3.3) gives (α1 + 1) . . .(αr + 1) = 4. Since αi +1 > 1, for r ≥ 2 we can only have α1 +1 = 2, α2 +1 = 2, implying that α1 = α2 = 1, i.e., n = p1 p2 . For r = 1, we have α1 + 1 = 4, i.e., α1 = 3, giving n = p3 . There are no other solutions n > 1 (n = 1 is a trivial solution) of equation (2.3.1)). On the other hand, let us remark that for n ≥ 2 one has d(n) ≥ 2, so T (n) ≥ n

(2.3.4)

with the equality occurring only for n being a prime. If n is not a prime, then it is immediate that d(n) ≥ 3, giving 3

T (n) ≥ n 2 .

(2.3.5)

Now, relations (2.3.4) and (2.3.5) together give 9

T (T (n)) ≥ n 4 .

Perfect and Related Numbers

73

Thus, by 49 > 2, there are no non-trivial (i.e., n 6= 1) m-super-perfect numbers. In fact, we have found that the equation   9 a T (T (n)) = n , a ∈ 1, 4 has no non-trivial solutions. Corollary 2.3.1. n = 6 is the only perfect number, which is also m-perfect. Indeed, n cannot be odd, since by a result of Sylvester, an odd perfect number must have at least five prime divisors. If n is even, then n = 2k p = p1 p2 , i.e., k = 1, when 2 = p1 and 22 − 1 = 3, when 3 = p2 . Thus, n = 2.3 = 6. In a similar manner, one can define k-m-perfect numbers by T (T (n)) = nk ,

(2.3.6)

where k ≥ 2 is given. Since the equation (α1 + 1) . . .(αr + 1) = 2k has a finite number of solutions, the general form of k-multiply perfect numbers can be determined. We collect certain particular cases in the following: Theorem 2.3.2. 1) All 3-m-perfect numbers have the forms n = p1 p22 or n = p51 ; 2) All 4-m-perfect numbers have the forms n = p1 p32 or n = p1 p2 p3 or n = p71 ; 3) All 5-m-perfect numbers have the forms n = p1 p42 or n = p91 ; 4) All 6-m-perfect numbers have the forms n = p1 p2 p23 , n = p1 p52 or n = p11 1 ; 5) All 7-m-perfect numbers have the forms n = p1 p62 or n = p13 1 ; 6) All 8-m-perfect numbers have the forms n = p1 p2 p3 p4 or n = p1 p2 p33 , n = p31 p32 or n = p15 1 ; 7) All 9-m-perfect numbers have the forms n = p1 p22 p23 , n = p1 p82 , n = p17 1 ; 8) All 10-m-perfect numbers have the forms n = p1 p2 p43 , n = p1 p92 , n = p19 1 , etc. (Here pi denotes certain distinct prime.)

74

J´ozsef S´andor and Krassimir Atanassov

Proof. We prove only the Case 6). By relation (2.3.3) we must solve the equation (α1 + 1) . . .(αr + 1) = 16, in αr and r. It is easy to see that the following four subcases are possible: i) α1 + 1 = 2, α2 + 1 = 2, α3 + 1 = 2, α4 + 1 = 2; ii) α1 + 1 = 2, α2 + 1 = 2, α3 + 1 = 4, α4 + 1 = 4; iii) α1 + 1 = 4, α2 + 1 = 4; iv) α1 + 1 = 16. This gives the general forms of all 8-m-perfect numbers, namely: i) α1 = α2 = α3 = α4 = 1 and n = p1 p2 p3 p4 ; ii) α1 = α2 = 1, α3 = 3 and n = p1 p2 p33 ; iii) α1 = α2 = 3 and n = p31 p32 ; iv) α1 = 15 and n = p15 1 . Corollary 2.3.2. 1) n = 28 is the single perfect and tri-perfect number. 2) There are no perfect and 4-perfect numbers; 3) n = 496 is the only perfect number which is 5-m-perfect; 4) There are no perfect numbers which are 6-m-perfect; 5) n = 8128 is the only perfect number which is 7-m-perfect. In fact, we have Theorem 2.3.3. Let p be a prime, with 2 p − 1 prime too (i.e., 2 p − 1 is a Mersenne prime). Then 2 p−1 (2 p − 1) is the only perfect number, which is p − m-perfect. Proof. By writing (α1 + 1) . . .(αr + 1) = 2p, where p is a prime, the following cases are only possible:

75

Perfect and Related Numbers i) α1 + 1 = 2, α2 + 1 = p; ii) α1 + 1 = 2p.

Then, n = p1 p2p−1 or n = p2p−1 are the general forms of p-m-perfect 1 p−1 numbers. By the Euclid-Euler theorem p1 p2 = 2 p−1 (2 p − 1) iff p2 = 2 and p1 = 2 p − 1 is prime. Remark 2.3.1. For p < 10000 the following Mersenne primes are known; namely for p = 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941. It is an open problem to show the existence of infinitely many Mersenne primes ([28]). As we have seen, the equation (2.3.2), i.e., T (T (n)) = n2 , has no non-trivial solutions. A similar problem arises for the equation T (T (n)) = nk ,

(2.3.8)

(n > 1, k ≥ 2, fixed). By (2.3.3), we can see that this is equivalent to d(n)d(T (n)) = k. 4

(2.3.9)

Let n = pα1 1 pα2 2 . . . pαr r > 1 be the canonical representation of n. By d(n) = (α1 + 1) . . .(αr + 1) and (2.3.3) we have α1 (α1 +1)...(αr +1) 2

T (n) = p1

αr (α1 +1)...(αr +1) 2

. . . pr

,

so (2.3.9) becomes equivalent to (α1 + 1) . . .(αr + 1)



   α1 (α1 + 1) . . .(αr + 1) αr (α1 + 1) . . .(αr + 1) ... = 4k, 2 2

and this, clearly has at most a finite number of solutions. Theorem 2.3.1. 1) Equation (2.3.8) is not solvable for k = 4, 5, 6; 2) For k = 3 the general solutions are n = p21 ; 3) For k = 7 the solutions are n = p31 ;

76

J´ozsef S´andor and Krassimir Atanassov 4) For k = 9 the solutions are n = p1 p2 (p1 6= p2 primes).

Proof. For k = 4, 5, 6, from (2.3.3) we must solve the corresponding equations for 16, 20, 24. It is a simple exercise to verify these impossibilities. For k = 3 we have the single equality 12 = 3.4, when α1 = 2, α1 (α21 +1) + 1 = 4. For k = 7, α1 = 3 by 3.4 2 + 1 = 7 and 4.7 = 28. For k = 9 we have 2.2.3.3 = 36 and α1 = α2 = 1. Corollary 2.3.2. n = 6 is the single perfect number which is also 9super-m-perfect. Indeed, p1 p2 = 2.(22 − 1) = 2.3 = 6 by Theorem 2.3.1 and the Euclid-Euler theorem. Remark 2.3.2. duce

By relation (2.3.5), by consecutive iteration we can de3k

T (T (...(T(n)...)) ≥ n 2k | {z } k

for n being non-prime. Since

3k

> 2k .k for all k ≥ 1 (by induction:

3k+1 = 3.3k > 3.2k.k > 2.2k (k + 1) = 2k+1 (k + 1)) we can obtain the following generalization of equation (2.3.2): T (T (...(T (n)...)) = nk | {z } k

has no non-trivial solutions. By relation (2.3.3) we have

logT (n) d(n) = . log n 2 Clearly, this implies lim inf

n→∞

lim sup

n→∞

logT (n) = 1, logn

logT (n) = +∞ logn

(2.3.10)

Perfect and Related Numbers

77

√ (take, e.g., n = p (prime); n = 2k (k ∈ N)). Since 2 ≤ d(n) ≤ 2 n (see, e.g., [60]) for n ≥ 2 we get logT (n) √ ≤ n. 1≤ logn By 2ω(n) ≤ d(n) ≤ 2Ω(n) (see, e.g., [12]) we can deduce: 2ω(n)−1 ≤

log T (n) ≤ 2Ω(n)−1 (n ≥ 2). logn

Since it is known by a theorem of Hardy and Ramanujan [61] that the normal order of magnitude of ω(n) and Ω(n) is loglog n, the above double inequality implies the normal order of magnitude of loglog T (n) is (1 + log2) loglogn. By a theorem of S. Wiegert [62] we have lim sup

n→∞

logd(n) loglogn = log2, log n

so by (2.3.10) we get: lim sup

n→∞

(loglogT (n))(loglogn) = log2. log n

In fact, by a result of Nicolas and Robin [63], for n ≥ 3 one has logd(n) logn ≤c (c ∼ 1.5379...), log2 loglogn we can obtain the following inequality: log logT (n) ≤ loglog n +

k logn − log 2, loglogn

where k = c log2 and n ≥ 3. This gives log logT (n) =0 n→∞ f (n) lim

n for any positive function with f (n)log → 0 for n → ∞. loglog n 0 By ϕ (n)d(n) ≥ n (see [64]) and ϕ(n)d 2 (n) ≤ n2 for n 6= 4 (see [47]) we get

n n ≤ d(n) ≤ p ϕ(n) ϕ(n)

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J´ozsef S´andor and Krassimir Atanassov

for n > 4 and from this, (2.3.10) yields n logT (n) n ≤ ≤ p . 2ϕ(n) logn 2 ϕ(n)

Hence, the arithmetic function T is connected to the other classical arithmetic functions. √ n+1 By n ≤ σ(n) d(n) ≤ 2 (see [21, 64, 65]), we get σ(n) logT (n) σ(n) ≤ ≤ √ . n+1 logn 2 n For infinitely many primes p we have   log p d(p − 1) > exp c loglog p (c > 0, constant, see [66]), so we have loglogT (p − 1) > loglog(p − 1) +

c log p − log2 loglog p

for infinitely many primes p, implying, e.g., lim sup

p→∞

and lim inf

n→∞

loglog T (p − 1) = +∞ loglog p

(loglogT (n))(loglogn) > 0. logn

2.4. Other Modifications of the Concept of Perfect Number Here we shall formulate some modifications of the perfect numbers and shall discuss some of their properties, giving a series of open problems, related to them. It is well known that for the natural number k

n = ∏ pαi i , i=1

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79

where k, α1 , α2 , ... αk ≥ 1 are natural numbers and p1 < p2 < ... < pk are different prime numbers, the Jordan’s function (ϕs ) (s being a natural number) is defined by: k

s(αi −1)

ϕs (n) = ∏ pi i=1

.(psi − 1).

Let Pd (n) and d(n) be, respectively, the product and the number of all divisors of n. It is well-known, that for every natural number n: p Pd (n) = nd(n). (2.4.1)

In Jelenski’s book [67] it is noted that some of the ancient Greek mathematicians called perfect numbers these numbers that are products of their proper divisors. For example, 6 = 1.2.3, 8 = 1.2.4, 10 = 1.2.5, 14 = 1.2.7, etc. These numbers can be called multiplicative perfect numbers, for instance of the additive perfect numbers, that are sum of their proper divisors. The following two definitions are more explicit.

Definition 2.4.1. number iff

The natural number n is an (ordinary) additive perfect σ(n) ≡ ∑ d = 2n. d|n

Definition 2.4.2. The natural number n is a multiplicative perfect number if Pd (n) ≡ ∏ d = n2 . d|n

Lemma 2.4.1. The necessary condition for a natural number n > 1 to be a multiplicative perfect number is d(4) = 4. Proof. The proof follows directly from (2.4.1) and Definition 2.4.2. Corollary 2.4.1. If n > 1 is a multiplicative perfect number, then n has exactly two different proper divisors larger than 1. Corollary 2.4.2. Number n > 1 is a multiplicative perfect number, iff n is a cube of a prime number or n is a product of two different prime numbers. Corollary 2.4.3. Multiplicative perfect numbers are infinitely many.

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J´ozsef S´andor and Krassimir Atanassov

We must note that for the additive perfect numbers no similar to the last result has been obtained up to now. Definition 2.4.3. Let g be a fixed arithmetic function. The natural number n is a g-additive perfect number iff

∑ g(d) = 2g(n).

(2.4.2)

d|n

Definition 2.4.4. Let g be a fixed arithmetic function. The natural number n is a g-multiplicative perfect number iff

∏ g(d) = (g(n))2. d|n

Obviously, Definition 2.4.1 is a partial case of Definition 2.4.3, and Definition 2.4.2 is a partial case of Definition 2.4.4 for g(n) = n. We must note that for each natural number n that is a g-multiplicative perfect number, if g(n) > 0, then for every positive real number c 6= 1 n is a logc g(n)additive perfect number. The following examples hold. Example 2.4.1. Let g(n) = cn for some positive real number c 6= 1. Then the g-multiplicative perfect numbers coincide with the classical additive perfect numbers (from Definition 2.4.1). Theorem 2.4.1. There are no ϕ2 -, ϕ3 - and ϕ4 -additive perfect numbers. Proof. Let us assume that the natural number n is a ϕ2 -additive perfect number. Since ∑ ϕs (d) = ns d|n

for every natural number s, then for s = 2 the Diophantine equation k

∏ p2α i

k

i

i=1

= 2 ∏ p2(αi −1)(p2i − 1) i=1

i.e., k

k

∏ p2i = 2 ∏(p2i − 1) i=1

i=1

(2.4.3)

81

Perfect and Related Numbers

ought to have a solution. Therefore, p1 = 2, because p1 is the smallest prime divisor of n and n is divided by 2. Hence, (2.4.3) has the form k

k

i=2

i=2

2. ∏ p2i = 3 ∏(p2i − 1).

(2.4.4)

All (k − 1 in number) multipliers (p2i − 1) (2 ≤ i ≤ k) are even numbers, while in the left-hand side we divide by 2, and not by 4. Therefore, k = 2. Hence, (2.4.4) has the form: 2p22 = 3(p22 − 1) or p22 = 3 that is impossible. Therefore, Theorem 2.4.1 is valid for the case of a ϕ2 additive perfect number. For the two other cases the proof is similar. Of interest is the following Open problem 2.4.1. Is there a natural number s > 1 for which there exists a ϕs -additive perfect number? Theorem 2.4.2. The natural number n is a ϕ-additive perfect number iff n = 2k , for each natural number k. Proof. Let n = 2k . Then: 2ϕ(n) = 2.ϕ(2k ) = 2.2k−1 = 2k = n = ∑ ϕ(d). d|n

Therefore, from (2.4.2) for g(n) = ϕ(n), we obtain that n is a ϕ-additive perfect number. Let us assume that the natural number n is a ϕ-additive perfect number. Therefore, ∑ ϕ(d) = 2ϕ(n). d|n

But

∑ ϕ(d) = n. d|n

Therefore, n = 2.ϕ(n).

(2.4.5)

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J´ozsef S´andor and Krassimir Atanassov

Hence, n is an even number. Let n = 2k .m, where m ≥ 1 is an odd number. From (2.4.5) and from the fact that ϕ is a multiplicative function, it follows that n = 2k .m = 2.ϕ(n) = 2.ϕ(2k.m) = 2.ϕ(2k ).ϕ(m) = 2k .ϕ(m), i.e., m = ϕ(m), which is possible only if m = 1. Therefore, n = 2k and Theorem 2.4.2 is proved. From Theorem 2.4.2 it follows the following Corollary 2.4.4. There are infinitely many ϕ-additive perfect numbers. Corollary 2.4.5. There are no odd ϕ-additive perfect numbers. Of course, we can note (for more details see, e.g., [68]) that currently it is not clear whether the classical additive perfect numbers are infinitely many. Also, we do not know whether there is at least one odd number n > 1 that is an additive perfect number. The following open problems are interesting. Open problem 2.4.2. Find all arithmetic (in particular, multiplicative) functions g, for which there exists no g-additive perfect number. As we see above, ϕ2 , ϕ3 and ϕ4 are such functions. Open problem 2.4.3. For each arithmetic (in particular, multiplicative) function g, to find the g-additive perfect numbers, if such exist. Is there an infinite set of numbers that are not g-additive perfect ones? Open problem 2.4.4. Find a procedure for obtaining all g-multiplicative perfect numbers for separate classical arithmetic functions. Finally, we can introduce the following Definition 2.4.5. Let g be a fixed arithmetic function and p be a fixed prime number. The natural number n is a (g, p)-power perfect number iff

∏ g(d) = pg(n). d|n

Perfect and Related Numbers

83

Theorem 2.4.3. n = 4 is the unique (2ϕ(n), 2)-power perfect number. The proof is similar to the above one. Open problem 2.4.5. Find a procedure for obtaining all (g, p)-power perfect numbers for separate classical arithmetic functions and for different prime numbers p.

2.5. A New Point of View on Perfect and Other Similar Numbers The text of the present section is based on the paper of M. Vassilev-Missana and K. Atanassov [69]. Initially, we introduce the following notations: εt is the completely multiplicative function that is given by εt (n) = nt for n ∈ N and t being a fixed real number; ζ is the Riemann’s zeta-function; f (n) f ± g, f .g, gf , f g are short expressions for f (n) ± g(n), f (n).g(n), g(n) ,

f (n)g(n) for n ∈ N , where f and g are arithmetic functions.

In [54], the following definitions are proposed. Definition 2.5.1. Let g be a fixed arithmetic function. A natural number n is said to be g-additive perfect number iff

∑ g(d) = 2g(n).

(2.5.1)

d/n

Definition 2.5.2. Let g be a fixed arithmetic function. A natural number n is said to be g-multiplicative perfect number iff

∏ g(d) = (g(n))2. d/n

(2.5.2)

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J´ozsef S´andor and Krassimir Atanassov

In 2003, in an unpublished paper [70], J. L. Pe has proposed the same definition (like Definition 2.5.1) to generalize the perfect numbers. We assume that J. Pe had not know about the paper [54]. We give the full answer to the Open problem 2.4.1 in Theorem 2.5.1, but our main goal is to propose some ways for generalization of g-perfect numbers, perfect numbers and their modifications. We will show that the perfect numbers, as well as their modifications and generalizations, admit another unexpected point of view. They are the fixed points of an appropriate arithmetic functions. Thus, we become to the numbers that we call “Omega-perfect numbers” (ΩF -perfect numbers). Relevant to ΩF -perfect numbers are the so-called [G, η]-perfect numbers. The latter look like eigen-vectors of G corresponding to η (considered as an eigenvalue) if we treat G as an operator, although G is an arithmetic function. A particular case of such points of view on “perfect numbers” and their modifications and generalizations are the so-called g f -perfect numbers, that generalize (directly) the g-perfect numbers. They are introduced with the help of Dirichlet’s convolution. Below, we give some results corresponding to g f -perfect numbers. Like a further generalization of g f -perfect numbers, the so-called g f r-multiperfect numbers are described below, that generalize harmonic divisor numbers, too. The assertion below gives an answer to the Open problem 2.4.1 that was formulated in [54]. Theorem 2.5.1. Let s ≥ 2 be a fixed real number. numbers do not exist.

Then, ϕs -perfect

Proof. Let us assume that n > 1, n ∈ N is a ϕs -perfect number. from the definition of ϕs -perfect number:

Then

∑ ϕs (d) = 2ϕs (n). d/n

Hence, ns = 2ϕs (n), since it is well-known that

∑ ϕs (d) = ns d/n

(2.5.3)

85

Perfect and Related Numbers (see [71]). The representation   1 ϕs (n) = n . ∏ 1 − s pi i=1 k

s

and equality (2.5.3) yield k

1 1 0 for each n ∈ N . Then, the only gg -perfect numbers are all primes. The above assertion follows immediately from (2.5.6) and (2.5.7). Let us consider the linear convolutional operators D and L that are given by: Df = µ∗ f, L f = ε0 ∗ f . They are both act on each arithmetic function f . It is not difficult to prove the following M¨obius formulas D(L f ) = f , (2.5.8) L(D f ) = f .

(2.5.9)

Using the commutative and associative laws of Dirichlet’s convolution, it is easy to check that for every two arithmetic functions f and g the following relations are fulfilled: D( f ∗ g) = (D f ) ∗ g = f ∗ (Dg),

(2.5.10)

L( f ∗ g) = (L f ) ∗ g = f ∗ (Lg).

(2.5.11)

With the help of the above relations we will prove the next Lemma 2.5.1. Let α 6= 0 and β be real numbers. Then, the double identity ϕα ∗ σβ = εα .σβ−α = εβ .σα−β (2.5.12) holds. Proof. Using that σβ = Lεβ and ϕα = Dεα

Perfect and Related Numbers

87

we obtain, due to (2.5.8) – (2.5.11), that ϕα ∗ σβ = (Dεα) ∗ (Lεβ ) = D(εα ∗ (Lεβ ))

= D(L(εα ∗ εβ )) = εα ∗ εβ

(2.5.13)

But for n ∈ N we have (εα ∗ εβ )(n) = ∑d/n ( dn )α.d β

= nα. ∑ d/nd β−α

= nα.σβ−α (n). Hence, εα ∗ εβ = εα .σβ−α . In the same manner, we obtain for n ∈ N that (εα ∗ εβ )(n) = ∑d/n( dn )β .d α

= nβ . ∑d/n d α−β

= nβ .σα−β (n). Hence, εα ∗ εβ = εβ .σα−β and (2.5.12) is proved, because of (2.5.13). Corollary 2.5.1. The double identity ϕ ∗ σ = σ ∗ ϕ = ε1 .d

(2.5.14)

(ϕ ∗ σ)(n) = (σ ∗ ϕ)(n) = n.d(n)

(2.5.15)

holds, i.e., holds for every n ∈ N . We obtain as a first corollary from (2.5.14)-(2.5.15) the following Proposition 2.5.2. ϕσ -perfect numbers do not exist.

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J´ozsef S´andor and Krassimir Atanassov

Proof. Let us assume that n ∈ N is a ϕσ -perfect number. Then, (σ ∗ ϕ)(n) = 2ϕ(n). Hence, n.d(n) = 2ϕ(n), since (2.5.15) is true. But the above equality is impossible, because n > ϕ(n) and d(n) ≥ 2. The assertion is proved. As a second corollary from (2.5.15) we obtain the following Theorem 2.5.2. The only σϕ -perfect number is n = 6. Proof. First, let us observe that n ∈ N is a σϕ -perfect number iff σ(n) n = d(n) 2

(2.5.16)

holds. Indeed, (2.5.16) directly follows from the defining equality of a σϕ perfect number (ϕ ∗ σ)(n) = 2σ(n) and from (2.5.15). Now, in order to prove the Theorem, we must consider the only possible cases: (i)

ω(n) ≥ 3,

(ii)

ω(n) = 1,

(iii) ω(n) = 2. Let (i) hold. Then, the inequality σ(n) ≤ d(n)

 ω(n) 3 .n 4

holds (see 3.32 on page 44 in [74]). From the inequalities  ω(n)  3 3 3 1 ≤ < 4 4 2

(2.5.17)

Perfect and Related Numbers

89

and from (2.5.17) we conclude that σ(n) n < . d(n) 2 Therefore, (2.5.16) is impossible. Let (ii) hold. Then n = pα , where α ∈ N and p is a prime. In this case, we rewrite (2.5.16) in the form σ(pα ) pα = . (2.5.18) d(pα ) 2 Using that σ(pα ) = and that

pα+1 − 1 p−1

d(pα ) = α + 1 we conclude that (2.5.18) is equivalent to (α − 1).pα+1 = (α + 1).pα − 2. But (2.5.19) is impossible for α = 1. Let α ≥ 2. Then, α+1 2 = 1+ ≤ 3. α−1 α−1 Therefore, if p ≥ 3, then α+1 . p> α−1 Hence (α − 1).pα+1 > (α + 1).pα and (2.5.19) is impossible. Let p = 2. Then, (2.5.19) yields (α − 1).3α+1 = (α + 1).3α − 2. Hence, (2α − 4).3α + 2 = 0. But for α ≥ 2 the last equality is impossible.

(2.5.19)

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J´ozsef S´andor and Krassimir Atanassov

Let (iii) hold. Then, n = pα .qβ , where p and q are different primes, such that 2 ≤ p < q and α, β ∈ N . In this case, we have: d(n) = (α + 1).(β + 1), σ(pα .qβ ) = σ(pα ).σ(qβ) = Therefore, σ(n) d(n)

pα+1 − 1 qβ+1 − 1 . . p−1 q−1

=

pα+1 − 1 qβ+1 − 1 . pα+1 − pα qβ+1 − qβ

=

pα+1 − pα + (pα − 1) qβ+1 − qβ + (qβ − 1) . pα+1 − pα qβ+1 − qβ

=

! ! qβ .(1 − q1β ) pα .(1 − p1α ) 1+ α . 1+ β . p .(p − 1) q .(q − 1)

Using the above equalities, we rewrite (2.5.16) in the form (α + 1).(β + 1) = Eα,β (p.q), 2 where Eα,β (p, q) = Let

Obviously,

1+

1 − p1α p−1

!

. 1+

1 − q1β q−1

(2.5.20) !

.



   1 1 D(p, q) = 1 + . 1+ . p−1 q−1 Eα,β (p, q) < D(p, q).

Let p ≥ 3. Since p < q we have     1 1 15 D(p, q) ≤ 1 + . 1+ = < 2. 3−1 5−1 8 Hence, Eα,β (p, q) < 2,

(2.5.21)

Perfect and Related Numbers

91

because of (2.5.21). Therefore, (2.5.20) is impossible, since the left-hand side of (2.5.20) is greater than or equal to 2. Let p = 2. Then, q ≥ 3 and !   1 − q1β 1 . Eα,β (p, q) = Eα,β (2, q) = 2 − α . 1 + 2 q−1 Let min(α, β) ≥ 2. Then, (α + 1).(β + 1) 9 ≥ . 2 2

(2.5.22)

On the other hand,       1 1 1 3 9 Eα,β (2, q) ≤ 2 − α . 1 + ≤ 2− α . < . 2 q−1 2 2 2 Hence, (α + 1).(β + 1) > Eα,β (2, q), 2 since (2.5.22) is true. Therefore, (2.5.20) is impossible when p = 2 and min(α, β) ≥ 2. Let min(α, β) = 1. We have two subcases: • (a1 ) min(α, β) = α, • (a2 ) min(α, β) = β. Let (a1 ) hold. Then, α = 1, β ≥ α, q ≥ 3. Hence, for p = 2, (2.5.20) takes the form ! 1 − q1β 3 β+1 = . 1+ . (2.5.23) 2 q−1 Hence, β
1 with bi |ai for i = 1, . . ., r. In [76] the following different modifications (and generalizations) of the perfect numbers are proposed: • (b1 ) A number n is r-multiperfect for r ∈ N , r > 2, if σ(n) = r.n; • (b2 ) A number n is superperfect, if σ(2) (n) ≡ σ(σ(n)) = 2n;

Perfect and Related Numbers

95

• (b3 ) A number n is an e-perfect, if σe (n) = 2n, where σe (n) is the sum of the e-divisors of n. There are no odd e-perfect numbers. The first few even are: 36, 180, 252,396, 468,... • (b4 ) A number n is e-superperfect (see [77]), if (2)

σe (n) ≡ σe (σe (n)) = 2n; • (b5 ) A number n is unitary perfect, if σ∗ (n) = 2n, where σ∗ (n) is the unitary divisor function. There are no odd unitary perfect numbers and it has been conjuctured that there are only a finite number of even ones. The first few unitary perfect numbers are: 6, 60, 90, 87360,...; • (b6 ) A number n is superunitary perfect, if (σ∗ )(2)(n) ≡ σ∗ (σ∗ (n)) = 2n. The first few superunitary perfect numbers are: 2,9,165,238,1640,...; • (b7 ) A number n is almost perfect (also it is known as a least deficient or slightly defective number), if σ(n) = 2n − 1. The only known almost perfect numbers are the powers of 2; • (b8 ) A number n is (α, β)-superperfect (see [78]), where α, β ∈ N , if σ(α.σ(n)) = β.n. Below, we shall show that all of the above cited kinds of perfect numbers are only particular cases of the introduced in the present paper ΩF -perfect numbers for an appropriate arithmetic function F. For each n ∈ N we put a specific form of function F, as follows.

96

J´ozsef S´andor and Krassimir Atanassov For (b1): 1 F(n) = σ(n). r For (b2): 1 F(n) = σ(2) (n). 2 For (b3): 1 F(n) = σe (n). 2 For (b4): 1 (2) F(n) = σe (n). 2 For (b5): 1 F(n) = σ∗ (n). 2 For (b6): 1 F(n) = (σ∗ )(2) (n). 2 For (b7): 1 F(n) = (1 + σ(n)). 2 For (b8): 1 F(n) = σ(α.σ(n)). β Thus, it is seen that the numbers from (b1 ) to (b8 ) are ΩF -perfect numbers. The case (b8 ) could be generalized in the following way: • (b9 ) A number n is it ( f , g)-superperfect, where f and g are given arithmetic functions, such that g(m) 6= 0 for each m ∈ N , if σ( f (n).σ(n)) = g(n).n.

For f (n) = α, g(n) = β for each n ∈ N (b9 ) becomes (b8 ). Now, it is seen that ( f , g)-superperfect numbers are ΩF -perfect numbers, too, because one may put for each n ∈ N : F(n) =

1 .σ( f (n).σ(n)). g(n)

Perfect and Related Numbers

97

It is seen that if g(n) > 0, for each n ∈ N , then g-perfect numbers are ΩF perfect numbers, too, because one may put for each n ∈ N : n. ∑ g(d) F(n) =

d/n

2g(n)

.

Also, it is seen that, if g(n) > 0 for each n ∈ N , then g f -perfect numbers are ΩF -perfect numbers, too, for F, which is given for each n ∈ N by: F(n) =

n.( f ∗ g)(n) . 2g(n)

Now, we introduce numbers that are relevant to ΩF -perfect numbers. Let G be a given arithmetic function and η 6= 0 be a real number. Definition 2.5.5. A number n ∈ N is said to be [G, η]-perfect iff n satisfies the equation G(n) = η.n (2.5.24) We note that if n satisfies (2.5.24) then, n looks like an eigenvector corresponding to the eigenvalue η, if we treat G like an operator, while G is an arithmetic function only. It is seen that, if n is a [G, η]-perfect number, then n is an ΩF -perfect number, too, for F that is given for each n ∈ N by: F(n) =

1 .G(n). η

It is seen that if n is an ΩF -perfect number, then n is a [G, η]-perfect number, too, since one may put for each n ∈ N G(n) = η.F(n). Therefore, [G, η]-perfect numbers are relevant to ΩF -perfect numbers. Below we give some examples. • (c1 ) [σ, 2]-perfect numbers coincide with perfect numbers; • (c2 ) [σ, r]-perfect numbers coincide with r-multiperfect numbers, when r ≥ 2.;

98

J´ozsef S´andor and Krassimir Atanassov • (c3 ) [σ(2) , 2]-perfect numbers coincide with superperfect numbers;

and etc. We can write the corresponding [G, η]-perfect numbers to the cases (b1 ) − (b6 ). It is seen that if g(n) > 0 for each n ∈ N , then [G, 2]-perfect numbers coincide with g-perfect numbers, when G is the function that is given for each n ∈ N by n. ∑ g(d) d|n (2.5.25) G(n) = g(n) If we choose for each n ∈ N

n. ∑ g(d)

G(n) =

d|n

(g(n))2

,

then [G, 2]-perfect numbers coincide with g-multiplicative perfect numbers (see (2.5.2) from Definition 2.5.2). It is seen that for s 6= 0, s is a real number, [ϕs .ε1−s, 21 ]-perfect numbers coincide with ϕs -perfect numbers, because of the relation

∑ ϕs (n) = ns. d|n

Thus, we proved the following result (cf. Theorem 2.5.1): Proposition 2.5.6. For s ≥ 2 [ϕs .ε1−s, 12 ]-perfect numbers do not exist. The only [ϕ, 12 ]-perfect numbers are the powers of 2. As a fact, we observe that if in (2.5.25) is put ϕs instead of g, then [G, s]perfect numbers coincide with ϕs -perfect numbers. But according to Theorem 2.5.1, we see that [G, 2]-perfect numbers do not exist for s ≥ 2, while for s = 1 they are the powers of 2. Below we introduce a generalization of g f -perfect numbers, giving the following Definition 2.5.6. Let r ≥ 2 and r, n ∈ N . g f − r-multiperfect number iff the equality ( f ∗ g)(n) = rg(n) holds, where “*” is given by (2.5.6).

A number n is said to be

Perfect and Related Numbers

99

When r = 2, we obtain g f − 2-multiperfect numbers that coincide with g f perfect numbers. Especially for f = ϕ and g = σ, these numbers coincide with the so-called balanced numbers proposed by M. V. Subbarao [79] with a single solution n = 6 (see also Theorem 2.5.2). When r ≥ 2 runs all possible integers, σϕ -r-multiperfect numbers describe all so-called harmonic divisor numbers or Ore numbers, that were introduced by O. Ore in [80] like the numbers n ∈ N for which n.d(n) is divisible by σ(n). It is known (see [80]) that the set of Ore numbers contains the set of all perfect numbers. Thus, we generalize Ore numbers, too. Finally, we observe that the g f -r-multiperfect numbers are a special case of Omega-perfect numbers, since, one may put for each n ∈ N : F(n) =

n.( f ∗ g)(n) . r.g(n)

The name “Omega-perfect” is adopted from the fact that the letter Ω is the last in the Greek alphabet just like the ΩF -perfect numbers, generalizing all known modifications and generalizations of the perfect numbers up to now, resemble a final generalization of the mentioned objects. Of course, it is possible to conduct a subsequent step of generalization, replacing the equation F(n) = n with a new equation of the kind G(n) = 0,

(2.5.26)

where G is an arithmetic function, since one may put G(n) = F(n) − n, and consider the perfect numbers and their generalizations and modifications to be zeros of equation (2.5.26). But this approach is not good, because in this case the arithmetic nature of the perfect or similar numbers is completely lost. For this reason, we suppose that the Omega-perfect numbers are the last possible generalization of the idea of the perfect and similar numbers.

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2.6. On Bi-Unitary Harmonic Numbers In 1948, O. Ore [80] considered numbers n whose divisors have an integral harmonic mean r H(n) = 1 , 1 d1 + dr where 1 = d1 < d2 < ... < dr = n are all the divisors of n. Since 1

1

n

1

∑ di = n ∑ di = n σ(n), i

i

and r = d(n) (with σ(n) and d(n) denoting the sum, respectively the number of divisors of n), clearly nd(n) H(n) = , σ(n) so H(n) is an integer if σ(n)|nd(n).

(2.6.1)

C. Pomerance [81] called a number n with property (2.6.1), a harmonic number. Ore proved that if n is perfect (i.e., σ(n) = 2n), then it is harmonic. Indeed, if n is perfect, then 2|d(n) is always true, i.e., n is not a perfect square. Ore also proved that if n is harmonic, then ω(n) ≥ 2, and Pomerance showed that the only harmonic numbers with two distinct prime factors are the even perfect numbers. In 1963, M. V. Subbarao [79] called the number n a balanced σ(n) number if d(n) = n2 , and proved that n = 6 is the single balanced number. Now, remark that a balanced number satisfies H(n) = 2, so it is a particular harmonic number. M. Garcia [82] extended the list of harmonic numbers to include all 45 which are < 107 , and found more than 200 larger ones. The least one, apart from 1 and the perfect numbers, is 140. All 130 harmonic numbers up to 2.109 are listed by G. L. Cohen [83]; and R. M. Sorli (see Cohen and Sorli [84]) has continued the list to 1010 . Ore conjectured that every harmonic number is even, but this probably is very difficult. Indeed, this result, if true, would imply that there are no odd perfect numbers. See also W. H. Mills [85], who proved that if there exists an odd harmonic number n, then n has a prime-power factor greater than 107 . In 1998, G. L. Cohen and M. Deng [86] introduced a generalization of harmonic numbers. Let k ≥ 1 be an integer and let σk (n) be the sum of the k-th

Perfect and Related Numbers

101

powers of the divisors of n. Then, n is called k-harmonic, if σk (n)|nkd(n).

(2.6.2)

They proved that for k > 1 there is no k-harmonic number in the range 1 < n < 1010 . A divisor d of n is called a unitary divisor, if (d, dn ) = 1. Let σ∗ (n) and ∗ d (n) denote, respectively, the sum and the number of unitary divisors of n. M. V. Subbarao and L. J. Warren [87] introduced the unitary perfect numbers n satisfying σ∗ (n) = 2n. They found the first four unitary perfect numbers, while the fifth one was discovered by C. Wall [88]. A number n is called unitary harmonic if σ∗ (n)|nd ∗(n). (2.6.3) This concept is introduced by K. Nageswara Rao [89], who showed that if n is unitary perfect, then it is also unitary harmonic. P. Hagis and G. Lord [90] proved that if H ∗ (x) is the counting function of these numbers, then for ε > 0 and large x one has log x 1 H ∗ (x) < 2.2x 2 .2(1+ε) log log x . The same result was obtained in 1957 by H.-J. Kanold [91] for the counting function of harmonic numbers. C. Wall [92] showed that there are 23 unitary harmonic numbers n with ω(n) ≤ 4, and claimed that there are 43 unitary harmonic numbers n ≤ 106 . However, Hagis and Lord [93] have shown this with 45 instead of 43. Recently, T. Goto and S. Shibata [94] have determined all harmonic numbers satisfying H(n) ≤ 300. Goto and K. Okeya have extended the study to H(n) ≤ 1200. For infinitary harmonic numbers, related to the concept of an “infinitary divisor”, see Hagis and Cohen [95]. For many results involving these topics, see also Chapter I of [24]. In this Section, we follow the paper [96]. A divisor d of n is called bi-unitary divisor if the greatest common unitary divisor of d and dn is 1. Let σ∗∗ (n) be the sum of all bi-unitary divisors of n. Wall [97] called a number n bi-unitary perfect, if σ∗∗ (n) = 2n, and proved that there are only three bi-unitary perfect numbers, namely 6, 60 and 90. It is not

102

J´ozsef S´andor and Krassimir Atanassov

difficult to verify that σ∗∗ is multiplicative, and for prime powers pa , σ∗∗ (pa ) = σ(pa ) =

σ∗∗ (pa ) =

pa+1 − 1 , if α is odd, p−1

a pa+1 − 1 − p 2 , if α is even. p−1

Let d ∗∗ (n) be the number of all bi-unitary divisors of n. Then it is also known (see D. Suryanarana [90]) that if n = pα1 1 ...pαr r is the prime factorization of n > 1, then ! ! d ∗∗ (n) =

∏

ai – even

ai

∏ (ai + 1)

(2.6.5)

ai – odd

We now introduce the main notion and results of this Section. Definition 2.6.1. The number n is called bi-unitary harmonic, if σ∗∗ (n)|nd ∗∗(n).

(2.6.6)

Theorem 2.6.1. Let k ≥ 1 be an integer and suppose that n is bi-unitary kperfect, i.e., σ∗∗ (n) = kn. Then n is bi-unitary harmonic iff k|d ∗∗ (n).

(2.6.7)

Particularly, 6, 60, 90 are bi-unitary harmonic numbers. Proof. (2.6.7) is a consequence of (2.6.6) and the Definition of bi-unitary k-perfect numbers. Remark that for k = 2, by relation (2.6.5), (2.6.7) is always true. By Wall’s result on bi-unitary perfect numbers it follows that 6, 60, 90 are also bi-unitary harmonic numbers. Corollary 2.6.1. If ω(n) ≥ 2 and n is bi-unitary 4-perfect, then n is biunitary harmonic number. Proof. Remark that for ω(n) ≥ 2, by (2.6.5), 4|d ∗∗(n) so by (2.6.7) the result follows.

Perfect and Related Numbers

103

Theorem 2.6.2. Let n = pα1 1 ...pαr r > 1 be the prime factorization of n, and suppose that all αi (i = 1, 2, ..., r) are odd. Then n is bi-unitary harmonic iff it is harmonic. Proof. Since all αi (i = 1, 2, ..., r) are odd, by (2.6.5), d ∗∗ (n) =

∏ (ai + 1) = d(n),

ai – odd

and by (2.6.4), σ∗∗ (n) =

∏ σ∗∗ (pα ) = ∏ σ(pα ) = σ(n).

α– odd

α– odd

Thus (2.6.6) is true if and only if (2.6.1) is true. Corollary 2.6.2. 1) Besides 1, the only squarefree bi-unitary harmonic number is 6. This follows by a result of Ore [80] on harmonic numbers. 2) If n is odd bi-unitary harmonic, with all odd, then n has a component exceeding 107 . If n is even, then ω(n) ≥ 3. This follows by the result of Mills stated in the Introduction, as well as by the fact that if ω(n) = 2, then n being harmonic, it must be a perfect number. All even perfect numbers have the form 22k p, where p is an odd prime. Since 2k is even, this leads to a contradiction. Remark 2.6.1. By Theorem 2.6.2, new bi-unitary harmonic numbers can be found. For example, n = 21 .33 .5 = 270, n = 25 .3.7 = 672, n = 25 .33 .5.7 = 30240, n = 23 .33 .53 .7.13 = 2457000, see [84]. A computer program, on the other hand, may be applied for the search of biunitary harmonic numbers. For example, there are 50 such numbers n ≤ 106 , but the search could be extended to 109 (see the Table at the end of the Section), etc. Theorem 2.6.3. Let n = pα1 1 ...pαr r > 1 be the prime factorization of n. If all αi (i = 1, 2, ..., r) are even, and n is bi-unitary harmonic, then ω(n) ≥ 2.

(2.6.8)

Proof. If n = p2a is a prime power, with even exponent 2a, then by (2.6.5), (2.6.6) is equivalent to (1 + p + p2 + · · · + pa−1 + pa+1 + · · · + p2a )|p2a(2a)..

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Clearly p2a is relatively prime to 1 + p + p2 + · · · + pa−1 + pa+1 + · · · + p2a , so we must have (1 + p + p2 + · · · + pa−1 + pa+1 + · · · + p2a )|(2a). But this is impossible, since the first term contains a number of 2a terms, each (excepting 1) greater than 1, so 1 + p + p2 + · · ·+ pa−1 + pa+1 + · · ·+ p2a > 2a. Therefore (2.6.8) follows. Corollary 2.6.3. If n is bi-unitary harmonic number, then ω(n) ≥ 2. Indeed, by Theorem 2.6.3, n cannot be of the form p2a . On the other hand, if n = p2a+1 (p prime), then it is harmonic, contradicting Ore’s result that ω(n) ≥ 2. If there are odd, as well as even exponents, the following particular result holds true. Theorem 2.6.4. There are no bi-unitary harmonic numbers of the form p3 .q2 (p, q distinct primes). Proof. If n = p2a+1 q2b , then d ∗∗ (n) = 2b(2a + 2) = 4b(a + 1), σ∗∗ (n) = (1 + p + · · · + p2a+1 )(1 + q + · · · + qb−1 + qb+1 + · · · + q2b ), so n is bi-unitary harmonic iff (1 + p + · · · + p2a+1 )(1 + q + · · · + qb−1 + qb+1 + · · · + q2b )|4b(a + 1)p2a+1q2b (2.6.9) For a = 1, b = 1 this becomes (1 + p + p2 + p3 )(1 + q2 )|8p3 q2 .

(2.6.10)

Since (1 + p + p2 + p3 , p3 ) = 1 and (1 + q2 , q2 ) = 1, it follows that (1 + p + p2 + p3 )|8q2 and (1 + q2 )|8p3 .

(2.6.11)

If p = 2, it follows that (1 + q2)|64, which is impossible for all q. Similarly, if q = 2, then 5|8p3 , so p = 5, and then the first relation of (2.6.11) is impossible. Thus, p, q ≥ 3. Remark that 1 + p + p2 + p3 = (1 + p)(1 + p2 )

Perfect and Related Numbers

105

and that (2.6.11) implies that (1 + p) and (1 + p2 ) can have only two distinct prime factors, namely: 2 and q. Let 1 + p = kq, i.e., p = kq − 1. Then p2 + 1 = k2 q2 − 2kq + 2 is divisible by q only if q = 2. But this is impossible. Clearly, k is even, k = 2s, so p2 + 1 = 2(2s2q2 − 2sq + 1), which cannot be a power of 2, since 2s2 q2 − 2sq + 1 is odd. Thus, (1 + p)(1 + p2 ) can have also other prime factors than 2 and q, contradicting (2.6.11). Remark 2.6.2. It can be proved similarly that there are no bi-unitary harmonic numbers of the form pq4 or p3 .q4 , and that the only one of the form pq2 is 5.32. Finally, we state the following result. Theorem 2.6.5. Let n be of the form r

s

i=1

j=1

r

s

i=1

j=1

2b

i +1 n = ∏ p2a ∏ q j j, i

and let

b −1

i +1 n1 = ∏ p2a ∏qjj i

and

s

,

b

n2 = ∏ q j j , j=1

where pi and q j are distinct primes; ai , b j are positive integers. Suppose that n1 is a harmonic number, while n2 is a unitary harmonic number. Then n = n1 n2 is a bi-unitary harmonic number. Proof. This follows from the fact that for the numbers n given above one has the identity H ∗∗ (n) = H(n1 )H ∗ (n2 ), (2.6.12) where H, H ∗ , H ∗∗ are the corresponding harmonic means (e.g., nd ∗∗ (n) H ∗∗ (n) = σ∗∗ (n) ). Identity (2.6.12) can be proved by using the definitions and the results (e.g., relation (2.6.5)) for the above functions. Remark 2.6.3. By examining the Table 2.6.1 with all bi-unitary harmonic numbers up to 109 , we can remark that there are in total 211 such numbers

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J´ozsef S´andor and Krassimir Atanassov

in this range. The first 12 bi-unitary harmonic numbers are all harmonic, or unitary harmonic; the first number without this property is n = 9072. There are only 5 bi-unitary harmonic numbers up to 109 which are powerful, namely n = 3307500, 9922500,23152500, 138915000, 555660000. From these, only n = 9922500 = 22 .34 .54 .72 is a perfect square. It is interesting to note that, the existence of perfect squares in the set of harmonic or unitary harmonic numbers, is an open problem up to now. As we have seen, the harmonic means of divisors, unitary divisors, and biunitary divisors are given explicitly by H(n) =

nd ∗ (n) nd ∗∗ (n) nd(n) ∗ ∗∗ , H (n) = , H (n) = . σ∗ (n) σ∗∗ (n) σ( n)

In what follows, the harmonic, unitary harmonic, respectively bi-unitary harmonic numbers will be called simply as H, H ∗ and H ∗∗ -numbers. This will be motivated also by the introduction of the following six new fractions and related numbers: H1 (n) = H4 (n) =

nd(n) nd ∗ (n) nd(n) , H (n) = , H3 (n) = ∗∗ , 2 ∗ σ (n) σ(n) σ (n)

nd ∗∗ (n) nd ∗ (n) nd ∗∗ (n) , H5 (n) = ∗∗ , H6 (n) = ∗ . σ(n) σ (n) σ (n)

When H1 (n) is an integer, we will say that n is a H1 -number, etc. By remarking that d ∗∗ (n) is always divis∗ ible by d (n), and that if n has the form n = pε11 . . . pεr r where ψi ∈ {1, 2} (i = 1, 2, . . ., r), pi are distinct primes, then d ∗∗ (n) = d ∗ (n), σ∗∗ (n) = σ∗ (n), we can state the following result. Theorem 2.6.6. In all cases, H ∗∗ (n) = H5 (n)kl (n), H4 (n) = H2 (n)k2 (n), H6 (n) = H ∗ (n)k3(n), where k1 (n), k2 (n), k3(n) are integers. If n has the form n = pε11 . . . pεr r , then H3 (n) = H1 (n), H5 (n) = H ∗ (n), H6 (n) = H ∗∗ (n) = H ∗ (n), H4 (n) = H2 (n).

Perfect and Related Numbers

107

If n has the form pa11 . . . par r with all ai odd (i = 1, 2, . . ., r), then H5 (n) = H2 (n), H6 (n) = H1 (n), H3 (n) = H4 (n) = H(n). Corollary 2.6.4. In all cases, a H2 -number is also a H4 -number; a H ∗ -number is also a H6 -number; a H5 -number is also a H ∗∗ -number. If n = pε11 . . . pεr r , then n is a H ∗∗ -number iff it is a H ∗ -, and a H6 -number; n is H5 -number, iff it is a H ∗ -number, and n is a H2 -number iff n is a H4 -number. If all ai are odd, then, respectively, the notions of H2 - and H5 -numbers; H, H3 and H4 -numbers; H1 and H6 -numbers, coincide. Remark 2.6.4. Since in Wall [92] there are stated all H ∗ -numbers with ω(n) ≤ 4, we can say from the Table 2.6.1 (see below), that the only H ∗∗ (i.e., bi-unitary harmonic)-number of the form p2 q is 32 .5 = 45; of the form p2 qr are 22 .3.5 = 60 and 32 .2.5 = 90; of the form p2 q2 r is 52 72 .13 = 15925; of the form p2 qrs are 22 .3.5.7 = 420, 32 .2.5.7 = 630; of the form p2 q2 rs is 22 .52 .7.13 = 9100; of the form p2 q2 r2 s is 32 .52 .132 .17 = 646425. (Here p, q, r, s are distinct primes). These complement some results of Theorem 2.6.6. Clearly, the deeper study of all of the above numbers cannot be done in this paper (but there are some results under preparation). We state only the following result Theorem 2.6.7. If n is a perfect number, then n is a H2 and H4 -number, too. If n > 1 is a H2 -number, then it cannot be a geometric number (i.e., a perfect square). Proof. Let σ(n) = 2n. Then, as d ∗ (n) = 2ω(n), and 2|d ∗∗ (n), clearly H2 (n) and H4 (n) will be integers. It is well-known that σ(n) is odd if n is a perfect square (i.e., n = m2 ). Then, if H2 (n) is an integer, then clearly σ(n) divides n, and this is possible only for n = 1. Remark 2.6.5. The similar problem in the case of H-numbers, i.e., if they are or not geometric, is a difficult open problem (see e.g., [84]). Remark 2.6.6. A number n > 1 is called friendly number (or Duffinian number), see, e.g., [24], if (n, σ(n)) = 1. Clearly, if n is friendly, then n cannot be a H2 - or a H4 -number. Indeed, in this case one must have σ(n)|d ∗ (n), or σ(n)|d ∗∗(n), but this is impossible for n > 1, since σ(n) > d ∗∗ (n) ≥ d ∗ (n) for

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J´ozsef S´andor and Krassimir Atanassov

n > 1. A similar result holds true for the H-numbers. Table 2.6.1.

n 1 6 = 2.3 45 = 32 .5 60 = 22 .3.5 90 = 2.32 .5 270 = 2.33 .5 420 = 22 .3.5.7 630 = 2.32 .5.7 672 = 25 .3.7 2970 = 2.33.5.11 5460 = 22 .3.5.7.13 8190 = 2.32.5.7.13 9072 = 24 .34 .7 9100 = 22 .52 .7 10080 = 25 .32 .5.7 15925 = 52 .72 .13 22680 = 23 .34 .5.7 22848 = 26 .3.7.17 27300 = 22 .3.52 .7.13 30240 = 25 .33 .5.7 40950 = 2.32 .52 .7.13 45360 = 24 .34 .5.7 54600 = 23 .3.52 .7.13 81900 = 22 .32 .52 .7.13 95550 = 2.3.52 .72 .13 99792 = 24 .34 .7.11 136500 = 22 .3.53 .7.13 163800 = 23 .32 .52 .7.13 172900 = 22 .52 .7.13.19 208656 = 24 .34 .7.23 245700 = 22 .33 .52 .7.13 249480 = 23 .34 .5.7.11 312480 = 25 .32 .5.7.31 332640 = 25 .33 .5.7.11 342720 = 26 .32 .5.7.17 385560 = 23 .34 .5.7.17

H ∗∗ (n) 1 2 3 4 4 6 7 7 8 11 13 13 12 10 16 7 18 16 15 24 15 20 20 18 14 22 25 24 25 23 27 33 31 44 32 34

n 409500 = 22 .32 .53 .7.13 472500 = 22 .33 .54 .7 491400 = 23 .33 .5 2.7.13 646425 = 32 .32 .132 .17 695520 = 25 .33 .5.7.23 708288 = 26 .3.7.17.31 716625 = 32 .53 .72 .13 791700 = 22 .3.52 .7.13.29 819000 = 23 .32 .53 .7.13 861840 = 24 .34 .5.7.19 900900 = 22 .32 .52 .7.11.13 955500 = 22 .3.5v3.72 .13 982800 = 24 .3 3.52 .7.13 1028160 = 26 .33 .5.7.17 1037400 = 23 .3.52 .7.13. 1187550 = 2.32 .52 .7.13.29 1228500 = 22 .33 .53 .7.13 1392300 = 22 .32 .52 .7.13.17 1421280 = 25 .33 .5.7.47 1433250 = 2.32 .53 .72 .13 1528800 = 25 .3.52 .72 .13 1571328 = 29 .32 .11 .31 1801800 = 23 .32 .5 2.11.13 2457000 = 23 .33 .53 .7.13 2579850 = 2.34 .52 .72 .13 2888704 = 210 .7.13.31 3307500 = 22 .33 .54 .72 3767400 = 23 .32 .52 .7.13.23 3878550 = 2.33 .52 .132 .17 4176900 = 22 .33 .52 .7.13.17 4291056 = 24 .34 .7.11.43 4299750 = 2.33 .53 .72 .13 4504500 = 22 .32 .53 .7.11.13 4713984 = 29 .33 .11.314 4961250 = 2.34 .54 .72 5405400 = 23 .33 .52 .7.11.13

H ∗∗ (n) 30 25 36 13 46 31 21 29 40 38 33 28 40 48 38 29 45 34 47 28 32 32 44 60 27 32 28 46 26 51 43 42 55 8 25 66

109

Perfect and Related Numbers Table 2.6.1. Continued

n 6168960 = 27 .34 .5.7.17 6397300 = 22 .52 .7.13.19.37 6688500 = 22 .3.53 .73 .13 7698600 = 23 .32 .52 .7.13.47 7780500 = 22 .32 .53 .7.13.19 7983360 = 28 .34 .5.7.11 8353800 = 23 .33 .52 .7.13.17 8666112 = 210 .3.7.13.31 9922500 = 22 .34 .54 .72 10032750 = 2.32 .53 .73 .13 10624320 = 26 .32 .5.7.17.31 10701600 = 25 .3.52 .73 .13 10999296 = 29 .32 .7.11.31 11302200 = 23 .33 .52 .7.13.23 11309760 = 26 .33 .5.7.11.17 11875500 = 22 .32 .53 .13.29 12899250 = 2.34 .53 .72 .13 13022100 = 22 .33 .52 .7.13.53 14303520 = 25 .33 .5.7.11.43 15561000 = 23 .32 .53 .7.13.19 18763200 = 24 .33 .52 .7.13.19 19061280 = 25 .32 .5.7.31.61 19845000 = 23 .34 .54 .72 20638800 = 24 .34 .52 .72.13 20884500 = 22 .33 .53 .7.13.17 22932000 = 25 .32 .53 .72 .13 23152500 = 22 .33 .54 73 23569920 = 29 .33 .5.11.31 23647680 = 26 .33 .5.7.17.23 24160500 = 22 .32 .53 .7.13.59 25798500 = 22 .34 .53 .72 .13 25832520 = 23 .34 .5.7.17.67 27027000 = 23 .33 .53 .7.11.13 29381625 = 32 .53 .72 .13.41 31872960 = 26 .32 .5.7.17.31 31888080 = 24 .34 .5.7.19.37 32997888 = 29 .33 .7.11.31

H ∗∗ (n) 64 37 49 47 57 64 68 48 30 49 62 56 56 69 88 58 45 53 86 76 76 61 40 48 85 64 49 80 92 59 54 67 110 41 93 74 84

n 34889400 = 23 .33 .52 .7.13.71 35626500 = 22 .33 .53 .7.13.29 38383800 = 23 .3.52 .7.13.19 38785500 = 22 .33 .53 .132.17 42997500 = 22 .33 .54 .72 .13 43205568 = 26 .3.7.17.31.61 43330560 = 210 .3.5.7.13.31 43857450 = 2.34 .52 .72 .13.17 46683000 = 23 .33 .53 .7.13.19 47297250 = 2.33 .53 .72 .11.13 47392800 = 25 .3.52 .72 .13.31 48323520 = 26 .33 .5.7.17.47 50213520 = 24 .37 .5.7.41 51597000 = 23 .34 .53 .72 .13 51979200 = 26 .3.52 .72 .13.17 56511000 = 23 .33 .53 .7.13.23 64701000 = 23 .32 .53 .7.13.79 68796000 = 25 .33 .53 .72 .13 71253000 = 23 .33 .53 .7.13.29 77477400 = 23 .32 .52 .7.11.13.43 77641200 = 24 .33 .52 .7.13.79 93284100 = 22 .32 .52 .7.13.17.67 95327232 = 210 .3.7.11.13.31 98993664 = 29 .34 .7.11.31 103194000 = 24 .34 .53 .72 .13 108421632 = 29 .33 .11.23.31 109147500 = 22 .34 .54 .72 .11 109336500 = 22 .33 .53 .7.13.89 129991680 = 210 .32 .5.7.13.31 133660800 = 27 .33 .52 .7.13.17 136732050 = 2.34 .52 .72 .13.53 138915000 = 23 .34 .54 .73 1429908429 = .32.7.11.13.31 144471600 = 24 .34 .52 .73 .13 144963000 = 23 .33 .53 .7.13.59 160254000 = 25 .32 .53 .73 .13 164989440 = 29 .33 .5.7.11.31

H ∗∗ (n) 71 87 74 52 52 61 80 51 114 77 62 94 72 72 64 115 79 96 116 86 79 67 88 90 80 92 55 89 96 128 53 70 104 84 118 112 140

110

J´ozsef S´andor and Krassimir Atanassov Table 2.6.1. Continued

n 172972800 = 28 .33 .52 .7.11.13 176289750 = 2.33 .53 .72. 13.41 188527500 = 22 .34 .54 .72 .19 191237760 = 27 .34 .5.7.17.31 199320576 = 210 .3.7.13.23.31 219287250 = 2.34 .53 .72 .13.17 221557248 = 29 .33 .11.31.47 227026800 = 24 .34 .52 .72 .11.13 232432200 = 23 .33 .52 .7.11.13.43 247484160 = 28 .34 .5.7.11.31 271498500 = 22 .33 .53 .7.132.17 283783500 = 22 .34 .53 .72 .11.13 287752500 = 22 .34 .54 .72 .29 287878500 = 22 .32 .53 .13.19.37 288943200 = 25 .34 .52 .73 .13 300982500 = 22 .33 .54 .7v3.13 325798200 = 23 .34 .52 .7.132.17 341775000 = 23 .32 .55 .72 .31 356879250 = 2.33 .53 .72 .13.83 361179000 = 23 .34 .53 .73 .13 363854400 = 26 .3.52 .73 .13.17 374078250 = 2.34 .53 .72 .13.29 377055000 = 23 .34 .54 .72 .19 389975040 = 210 .33 .5.7.13.31 390957840 = 24 .35 .5.7.132 .17 407307264 = 210 .3.7.13.31.47 421866900 = 22 .33.52.7.13.17.101 428972544 = 29 .33.7.11.13.31 434397600 = 25 .33.52.7.132.17 438574500 = 22 .34.53.72.13.17 447828480 = 33.5.11.19.31 467002900 = 22 .52 .7.13.19.37.73

H ∗∗ (n) 128 82 57 124 92 85 94 88 129 124 91 99 58 111 108 91 78 70 83 126 112 87 76 144 104 94 101 156 104 102 152 73

n 474692400 = 24 .34 .52 .72 .13.23 481572000 = 25 .33 .53 .73 .13 486319680 = 33.5.7.11.17.43 488697300 = 22 .35 .52 .7.132 .17 490990500 = 22 .32 .53 .7.11.13.109 494968320 = 29 .34 .5.7.11.31 513513000 = 23 .33 .53 .7.11.13.19 552348720 = 24 .5.7.11.41 555660000 = 25 .34 .54 .73 559704600 = 23 .33 .52 .7.13.17.67 567567000 = 23 .34 .53 .72 .11.13 575757000 = 23 .32 .53 .7.13.19.37 585427500 = 22 .34 .54 .72 .59 639802800 = 24 .34 .52 .72 .13.31 648083520 = 26 .32 .5.7.17.31.61 648784500 = 22 .3.53 .73 .13.97 690908400 = 24 .33 .52 .7.13.19.37 708107400 = 23 .33 .52 .7.11.13.131 710892000 = 25 .32 .53 .72 .13.31 722358000 = 24 .34 .53 .73 .13 756756000 = 25 .33 .53 .72 .11.13 758951424 = 29 .33 .7.11.23.31 779688000 = 26 32 .53 .72.13.17 793457920 = 27 34 .5.7.17.127 823280640 = 210 .3.5.7.13.19.31 953629840 = 24 37 .5.7.17.41 877149000 = 23 34 .53 .72 .13.17 879196500 = 22 32 .53 .7.13.19.113 970023600 = 24 34 .52 .72 .13.47 973176750 = 2.32 .53 .73 .13.97 977394600 = 23 35 .52 .7.132.17 992548080 = 24 38 .5.31.61

H ∗∗ (n) 92 168 172 81 109 150 209 132 100 134 132 148 59 93 122 97 148 131 124 140 176 161 128 127 152 136 136 113 94 97 108 81

Perfect and Related Numbers

111

2.7. On Modified Hyperperfect Numbers The following research is based on [98]. Let d be a positive divisor of the integer n > 1. If the greatest common unitary divisor of d and dn is 1, then d is called a bi-unitary divisor of n. If n = pa11 . . . par r > 1 is the prime factorization of n, a divisor d of n is called an exponential divisor (or e-divisor, for short), if d = pb11 . . . pbr r > 1 with bi |ai for i = 1, . . ., r. For the history of these notions, as well as the connected arithmetical functions, see e.g., [24, 28]. In what follows, σ(n), σ∗ (n), σ∗∗(n), σe(n) will denote the sum of divisors, -unitary divisors, -bi-unitary divisors, and e-divisors, respectively. It is well-known that a positive integer m is called n-hyperperfect (HP for short), if m = 1 + n(σ(m) − m − 1). (2.7.1) For n = 1 one has σ(m) = 2m, i.e., the 1-HP numbers coincide with the classical perfect numbers. For results on HP-numbers, see [24]. Let f : N ∗ → N ∗ = {1, 2, ...} be an arithmetical function. Then m will be called f-n-hyperperfect number, if m = 1 + n( f (m) − m − 1)

(2.7.2)

for some integer n ≥ 1. For f (m) = σ(m) one obtains the HP-numbers, while for f (m) = σ∗ (m), we get the unitary hyperperfect numbers (UHP) introduced by P. Hagis [24]. When f (m) = σ∗∗ (m), we get the Bi-unitary HyperPerfect numbers (BHP), introduced by J. S´andor [99]. For f (m) = σe (m), we get the ehyperperfect numbers (e-HP), introduced also by J. S´andor [100]. In what follows, m will be called a modified f-n-hyperperfect number if m = n( f (m) − m).

(2.7.3)

For f (m) = σ(m), we get the Modified HyperPerfect (MHP) numbers. Since for n = 1 one has in (2.7.3) f (m) = 2m, one obtains again a generalization of f -perfect numbers. First we prove Theorem 2.7.1. All MHP numbers are the classical perfect numbers, as well as the prime numbers.

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Proof. so

Since (2.7.3) implies n|m, put m = kn, giving k = f (kn) − kn,

For f ≡ σ, this gives

f (kn) = k(n + 1).

(2.7.4)

σ(kn) = k(n + 1).

(2.7.5)

For n = 1, (2.7.5) gives σ(k) = 2k, so m = k is the classical perfect number. For k = 1, relation (2.7.5) implies σ(n) = n + 1, which is possible only for n = p (prime), since σ(n) ≥ n + 1, with equality if n has only two distinct divisors – namely 1 and n -, so n is a prime. Thus, m = p is a modified phyperperfect number. Assume now that, k > 1, n > 1 in (2.7.5). Then it is well-known that σ(kn) > kσ(n) (see, e.g., [24]). Since σ(n) ≥ n + 1 for all n > 1, we can infer that σ(kn) > k(n + 1), in contradiction with (2.7.5). For the case of unitary divisors, one can state the following theorem. Theorem 2.7.2. All unitary MHP-numbers are unitary perfect numbers, as well as, the prime powers. Proof. Equality (2.7.4) now becomes σ∗ (kn) = k(n + 1).

(2.7.6)

For n = 1, we get σ∗ (k) = 2k, i.e., k is a unitary perfect number. For k = 1, we get σ∗ (n) = n+1, which is true only for n = pa (prime power), by σ∗ (n) = ∏ (Pa + 1). pa ||n

Let us now assume that n, k > 1. Since k(n + 1) = kn + k, and k is not only a divisor, but a unitary one of kn, one can write (k, nk k ) = 1, i.e., (k, n) = 1. But then, σ∗ being multiplicative, σ∗ (kn) = σ∗ (k)σ∗ (n) ≥ (k + 1)(n + 1) > k(n + 1) for k > 1, n > 1. This contradicts (2.7.6), so Theorem 2.7.2 is proved. For bi-unitary divisors we can state Theorem 2.7.3. All BMHP-numbers are 6, 60, 90, as well as all primes or squares of primes.

Perfect and Related Numbers

113

Proof. Equation (2.7.5) now becomes σ∗∗ (kn) = k(n + 1),

(2.7.7)

where σ∗∗ (n) denotes the sum of bi-unitary divisors of n. For n = 1, we get σ∗∗ (k) = 2k, so by a result of C. Wall (see [24]) one can write k ∈ {6, 60, 90}. For k = 1, we get σ∗∗ (n) = n + 1, which is possible only for n = p or n = p2 (p – prime). This is well-known, but we note that it follows also from σ∗∗ (pa ) = a σ(pa ), if a is odd (p is prime), σ∗∗ (pa ) = σ∗∗ (pa ) − p 2 if a is even; and the multiplicativity of σ. Let now k > 1, n > 1. Then, kn 6= k, kn 6= n, and   nk k, = (k, n)∗ = 1, k ∗ where (k, n)∗ denotes the greatest common unitary divisor of k and n. Since k 6= n, by (k, n)∗ = 1, and n is also a divisor of n, but not a bi-unitary one, by (2.7.6) (which means that the only bi-unitary divisors of kn are kn and k). But then (n, nk n )∗ 6= 1, i.e., (n, k)∗ 6= 1, in contradiction with (k, n)∗ = 1. Finally, the case of e-divisors is contained in Theorem 2.7.4. All modified exponentially n-hyperperfect numbers m q−1 are given by m = kn, where k = p1 p2 ...pr , n = p1 p2 ...pr , with p1 , p2 , ..., pr distinct primes, and q an arbitrary primes; as well as the e-perfect numbers. Proof. We have to study the equation: σe (kn) = k(n + 1).

(2.7.8)

For n = 1, we have σe (k) = 2k, i.e., the e-perfect numbers. For k = 1, we get σe (n) = n + 1. For n squarefree, one has σe (n) = n, while for n not squarefree, by the below lemma, σe (n) > n + 1, giving a contradiction.

n Lemma 2.7.1. If n > 1 is not squarefree, then σe (n) ≥ n + qa−1 , where a p q ||n and a ≥ 2. There is equality only for p1 ...pr q with p1 , ..., pr, q distinct primes, and p an arbitrary prime.

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Proof. Let n = pa11 ...par r . Therefore, there exists a ∈ {a1 , ..., ar} so that a ≥ 2. Thus σe (n) ≥ pa11 ...par r (qa + q) = n + pa11 ...par r q since σe (qa ) ≥ q1 + qa with equality only if a is a prime, while σe (pb ) ≥ pb , with equality only for b = 1. Corollary 2.7.1. σe (n) ≥ n + γ(n) for n not squarefree. Now, suppose that (n, k) = 1. Since σe is multiplicative, equation(2.7.8) becomes σe (n)σe(k) = k(n + 1). If k > 1 is squarefree, then σe (k) = k, so this is σe (n) = n + 1, which is impossible. If k is not squarefree, but n is squarefree, then σe (n) = n, so (2.7.7) becomes nσe (k) = k(n + 1). Since (n, k) = 1 and (n, n + 1) = 1, this is again impossible. Thus, if (n, k) = 1 for n > 1, k > 1, the equation is unsolvable. a0 a0 Let (n, k) > 1. Writing n = pa11 ...par r qb11 ...qbr r , k = p11 ...pr r γc11 ...γcrt , where pi , q j , γk are distinct primes, and ai , b j , ck are nonnegative integers (1 ≤ i ≤ r, 1 ≤ j ≤ s, 1 ≤ k ≤ t). Since by (2.7.8) written in the form σe (kn) = kn + k − k is an e-divisor of 0 a +a0 nk = p11 1 ...par r+ar qb11 ...qbr r γc11 ...γcrt , we must have b1 = ... = bs = 0. Also a01 |(a1 + a01 ), ..., a0r|(ar + a0r ), i.e., a1 = (m1 − 1)a01 , ..., ar = (mr − 1)a0r , with mi (1 ≤ i ≤ r) positive integers. We note that, since a1 ≥ 1, ..., ar ≥ 1, we have m1 > 1, ..., mr > 1. Since γc11 6= k is also an e-divisor of nk, we must have c1 = 0. Similarly, c2 = ... = cr = (m −1)a0

(m −1)a0

a0

a0

m a0

m a0

r 1 ...pr r , k = p11 ...pr r , nk = p1 1 1 ...pr r r . 0. Thus, n = p1 1 Remark that by m1 > 1, if one assumes a01 > 1, then 1|m1 a01 implies that p1 ...pr is also an e-divisor of nk , with p1 ...pk 6= k. Thus, we must necessarily mr 1 have a01 = ... = a0r = 1, so k = p1 ...pr and nk = pm 1 ...pr . Since n > 1, at least one of m1 , ..., mr is greater than 1. Put m1 > 1. Then, if at least one of m2 , ..., mr mr 2 is greater than 1, then p1 pm 2 ...pr 6= k is another e-divisor of nk. If m1 > 1 is not a prime, then m1 can also have a divisor 1 < a < m1 so pa1 p2 ...pr will be another e-divisor, which is a contradiction. Thus a = q is a prime, which finishes the proof of Theorem 2.7.4.

Perfect and Related Numbers

115

For results and/or open problems on perfect, unitary perfect, e-perfect numbers; as well as on hyperperfect or unitary hyperperfect numbers, see the monographs [24, 28].

2.8. On Balanced Numbers In 1963, M. V. Subbarao [79] considered the solvability of equation σ(n) n = , d(n) 2

(2.8.1)

where n ≥ 1 is an integer, while σ(n) and d(n) denote, respectively, the sum and the number of the divisors of n. A number n satisfying equation (1) has been called a balanced number. It is shown that n = 6 is the single balanced number. Around 2007, the two authors discovered independently (see [28, 100]) new proofs of this result. Let σk (n) denote the sum of the k-th powers of the divisors of n (k ≥ 1, integer). Let us call a number n k-balanced number if it satisfies the equality σk (n) nk = . d(n) 2

(2.8.2)

In [100], it was shown that for k > 1 there are no k-balanced numbers. The proofs, given in [100], are based on the inequality σk (n) nk ≤ , d(n) 2

(2.8.3)

for any k, n ≥ 1 and ω(n) ≥ 2, where ω(n) denotes the number of the distinct prime divisors of n. This inequality was published in 1990 in [24], and proved by the use of the Jensen–Hadamard (or Hadamard) integral inequalities of real analysis. Though not explicitly stated, in [100] the following results were pointed out. Theorem 2.8.1. The inequality (2.8.3) holds true in case k = 1 for any n > 1 if and only if n 6= 4 and n is a prime number. There is equality only for n = 6.

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Theorem 2.8.2. The inequality (2.8.3) holds true for any n > 1 if and only if n is a prime number. The aim of the present paper is to offer new proofs of these two theorems. Certain new inequalities, as well as connections with related problems will be considered, too. Theorem 2.8.3. For any integers n > 1 and k ≥ 1, one has   nk .d(n) 1 k σk (n) ≤ n 1 − k−1 + + 1. 2 2k When n ≥ 3 is an odd number, one has   1 nk .d(n) k σk (n) ≤ n 1 − k + + 1. 3 3k

(2.8.4)

(2.8.5)

There is an equality in (2.8.4) only for prime n, or n = 4; and there is an equality in (2.8.5) only for n prime, or n = 9. By letting k = 1, we get the following theorem. Theorem 2.8.4. For any integer n > 1 one has σ(n) ≤

n.d(n) + 1, 2

(2.8.6)

with equality only for n = prime, or n = 4. If n ≥ 3 is odd, then σ(n) ≤

n.(d(n) + 1) + 1, 3

(2.8.7)

with equality only for n prime, or n = 9. When k = 1, (2.8.6) is a simple corollary of (2.8.4), and, respectively, (2.8.7) is a simple corollary of (2.8.5). Thus, we shall give Proof of Theorem 2.8.3. d(n) = 2 and since

First, we remark that when n is prime, then σk (n) = nk + 1,

so there is equality (2.8.4), as well as (2.8.5). Let us suppose now that n > 1 is composite. If s is a divisor of n, then n = s.k, where k ≥ 1. Now, let us suppose

117

Perfect and Related Numbers

that s 6= 1 and s 6= n. Then, it is clear that k 6= 1, so k ≥ 2. This gives that d(n) > 2 and n n s= ≤ . k 2 This implies at once σk (n) = 1 + nk +

∑ 1 n can be written as d(n) < √ . The Now, the inequality 2 + d(n) n−2 √ √ n−3 inequality > n can be written as 2 n > 3, which is valid for n ≥ 3. d(n)

Therefore, if

d(n) < holds true, one has

√

n

σ(n) n−3 √ ≥ 2+ > n. d(n) d(n)

(2.8.20) (2.8.21)

In [48], it is proved that (2.8.20) is true for any n ≥ 1262. Thus, (2.8.21) holds for such value of n. Clearly, the second inequality of (2.8.21) holds true for any prime n. The inequality σ(n) √ > n d(n)

(2.8.22)

holds true for any n > 1 (see [99]), and (2.8.21) gives an improvement of this relation. 4. In 2009, in [102], J. S´andor proved the inequality (n > 1): √ σ(n) ≥ n + 1 + n(d(n) − 2), with equality only if n = p or n = p2 for p being prime. This is stronger than (2.8.18), as the inequality √ n + 1 + n(d(n) − 2) ≥ 2d(n) + n − 3

(2.8.23)

Perfect and Related Numbers is equivalent to d(n) ≥ 2, which is valid for any n ≥ 2. From (2.8.23), we get √ ( n − 1)2 √ σ(n) √ ≥ n+ > n d(n) d(n)

123

(2.8.24)

for n ≥ 1, which is another refinement of (2.8.22). 5. In paper [103], it has been shown that √ d(n) < 4 3 n

(2.8.25)

√ √ for any n > 1, which improves inequality (2.8.20) if 4 3 n < n, i.e., if n > 46 = 4096. We also note that (2.8.25) improves the classical Sierpinski’s inequality √ √ √ (see, [104]) d(n) < 2 n, if 4 3 n < n, i.e., n > 26 = 64. 6. In 2014, in paper [105], by using other methods, J. S´andor proved the following refinement of Theorem 2: For k > 1 and n > 1 not prime, one has σk (n) 2n d(n) < < k n n + ϕ(n) 2 where ϕ(n) is the Euler’s totient function.

Chapter 3

On Modifications and Extensions of the Arithmetic Functions ϕ, ψ and σ In a series of papers, the authors studied some properties of the well-known arithmetic functions ϕ, ψ and σ. With respect to this research, new functions were defined and some of their properties were described.

3.1. On an Arithmetic Function, Related to Operation “Logarithm” Firstly, following [8, 107], for the natural number n ≥ 2, k

n = ∏ pαi i , i=1

where k, α1 , α2 , ... αk ≥ 1 are natural numbers and p1 , p2 , ..., pk are different prime numbers, we define the following function: k

B(n) = ∑ αi .pi , B(1) = 1. i=1

In [8], this function is denoted by ζ, but back in the 1970s it was defined as B (see, e.g., [24]).

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Theorem 3.1.1. For every natural number n, B(n)ω(n) ≤ n + 4.

(3.1.1)

Proof. Let n be a prime number. Then, from (3.1.1) we obtain: n + 4 − B(n)ω(n) = n + 4 − n.1 = 4 ≥ 0. Let us assume that (3.1.1) is valid for an arbitrary natural number n with d(n) = k and let p be an arbitrary prime number. For p there are two cases. Case 1: p ∈ set(n). Therefore, ω(np) = ω(n) and np + 4 − B(np)ω(np) = n.p + 4 − (B(n) + p).ω(n) = n.(p − 1) + n + 4 − B(n).ω(n) − p.ω(n) (from (3.1.1)) ≥ n.(p − 1) − p.ω(n) n = .2.(p − 1) − p.ω(n) ≥ 0, 2 because, obviously, for every natural number n: n ≥ 2ω(n) and for every prime number p: 2(p − 1) ≥ p. Case 2: p 6∈ set(n). Therefore, ω(np) = ω(n) + 1. If n is a prime number, then np + 4 − B(np)ω(np) = n.p + 4 − 2.(n + p) ≥ 0 and we obtain equality only for n = 2. Let d(n) ≥ 2. Then, np + 4 − B(np)ω(np) = n.p + 4 − (B(n) + p).(ω(n) + 1). If p = 2, then by the condition that p 6∈ set(n), it follows that n is an odd number. Therefore, np + 4 − B(np)ω(np) = 2n + 4 − (B(n) + 2).(ω(n) + 1) = 2n + 4 − B(n)ω(n) − B(n) − 2.ω(n) − 2 (from (3.1.1)) ≥ n − B(n) − 2.ω(n) − 2

On Modifications and Extensions of the Arithmetic Functions ...

127

(for each natural number n ≥ 31 we can check that: B(n) ≥ 2.ω(n) + 6) ≥ n − 2.B(n) + 4 ≥ 0 (by assumption, n = q.m, where q ≥ 3 is a prime number and by inductive assumption m + 4 − B(m)ω(m) ≥ 0) = q.m − 2.B(q.m) + 4 = q.m − 2.(q + m) + 4 = q.(m − 2) − 2.m + 4 ≥ 3.(m − 2) − 2.m + 4 = m − 2 ≥ 0.

Finally, let p > 2. Then,

np + 4 − B(np)ω(np) = n(p − 1) + n + 4 − (B(n) + p).(ω(n) + 1) = n(p − 1) + n + 4 − B(n)ω(n) − B(n) − p.ω(n) − p

(from (3.1.1))

≥ n(p − 1) − B(n) − p.ω(n) − p

(we can directly check that B(n) ≥ ω(n) + 1)

≥ n(p − 1) − (p + 1).B(n) ≥ 0, because, obviously, for every natural number n: n ≥ 2.B(n) and for every prime number p ≥ 3: 2(p − 1) ≥ p + 1. Therefore, for n ≥ 31 the assertion is valid. By a direct check we see that n n + 4 − B(n)ω(n) 1 4 2 4 3 4 4 4 5 4 6 0 7 4 8 6 9 7 10 0 11 4 12 2 13 4 14 0 15 3

n n + 4 − B(n)ω(n) 16 4 17 4 18 6 19 4 20 6 21 5 22 0 23 4 24 10 25 19 26 0 27 22 28 10 29 4 30 4

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Hence, Theorem 3.1.1 is valid for every natural number n. If we like to construct an inequality with the same components, but in the opposite direction, it could have the form B(n)ω(n) ≥ n, but there are counterexamples, e.g., B(1000)ω(1000) = (3.2 + 3.5)2 = 212 < 1000. One of the possible inequalities is discussed in the following Theorem 3.1.2. For every natural number n B(n)d(n) ≥ n.

(3.1.2)

Proof. Let n be a prime number. Then, from (3.1.2) we obtain: B(n)d(n) − n = n1 − n = 0. Let us assume that (3.1.2) is valid for an arbitrary natural number n with d(n) = k, and let p be an arbitrary prime number. Then, B(n.p)d(n.p) − n.p = (B(n) + p)d(n)+1 − n.p (from (3.1.2)) ≥ (B(n) + p)d(n)+1 − B(n)d(n).p p d(n) = B(n)d(n)((B(n) + p).(1 + ) − p) B(n) > B(n)d(n)(B(n) + p − p) > 0. Therefore, (3.1.2) is valid.

3.2. Irrational Factor: Definition, Properties and Problems k

Following [108], let us juxtapose to the natural number n = ∏ pαi i the new i=1

number

n

1/αi

n0 = ∏ pi i=1

.

On Modifications and Extensions of the Arithmetic Functions ...

129

It can be easily seen that if for every i (1 ≤ i ≤ k) αi = 1, then n0 is a natural number and n0 = n. On the other hand, if there is at least one αi > 1, then n0 6= n and now n0 is not a natural and more general – a rational number. Therefore, in this case n0 is an irrational number. Let us denote n0 by IF(n) and let us name it “Irrational Factor” of n. On the other hand, IF(n) is a product of the solutions of the algebraic system of equations xα1 1 = p1 xα2 2 = p2 ... xαk k = pk Therefore, IF(n) is a product of algebraic numbers and hence it is an algebraic number. Theorem 3.2.1. IF is a multiplicative function. Proof. Let m and n be two natural numbers for which (m, n) = 1. Thereβ fore, if n has the above form and m = ∏li=1 qi i , where q1 , q2 , ..., qk are different prime numbers, for every i (1 ≤ i ≤ k) and for every j (1 ≤ j ≤ l) : pi 6= q j , and β1 , β2 , ..., βk ≥ 1 are natural numbers, then k

1 α

l

1 β

IF(n.m) = ∏ pi i . ∏ q j j = IF(n).IF(m). i=1

j=1

IF is not a monotonic function. Its first 40 values are given in Table 3.2.1. If k = l, p1 = q1 , ..., pk = qk and α1 ≥ β1 , α2 ≥ β2 , ...αk ≥ βl , then n ≥ m and IF(n) ≤ IF(m). The inequality will be strong if at least one of the inequalities between αi and βi (1 ≤ i ≤ k) is strong. On the other hand, if k < l, p1 = q1 , ..., pk = qk and α1 = β1 , α2 = β2 , ...αk = βl , then n < m and IF(n) < IF(m). Very interesting, but open, is the question of the relation between n and m when k < l, p1 = q1 , ..., ps = qs for 0 ≤ s ≤ k and there are no restrictions for the relations between the α− and β− powers.

130

J´ozsef S´andor and Krassimir Atanassov Table 3.2.1. n 1 2 3 4 5 6 7 8 9 10

IF (n) 1 2 3√ 2 5 6 7 √ 3 √2 3 10

n 11 12 13 14 15 16 17 18 19 20

IF(n) 11√ 3 2 13 14 15 √ 4 2 17√ 2 3 19√ 5 2

n 21 22 23 24 25 26 27 28 29 30

IF(n) 21 22 23√ 3√ 3 2 5 26 √ 3 3 √ 7 2 29 30

n 31 32 33 34 35 36 37 38 39 40

IF(n) 31 √ 5 2 33 34 35 √ 6 37 38 39√ 5 32

Obviously, IF(n) > 1 for every natural number n > 1. Theorem 3.2.2. For every two natural numbers n and m: p IF (nm) = m IF(n).

Moreover, we must note that for every natural number n there exists a natural number m such that IF(n)m is a natural number. The minimal such number is m = GCD(α1 , ..., αk). For the M¨obius function µ for every natural number n it is valid the equality |µ(n)| = sg(n − IF(n)), where for every integer number n ≥ 0:  0, sg(x) = 1, Let

if x > 0 if x = 0



 n J(n) = . IF(n)

Therefore J(n) ≤

n ≤ n. [IF(n)]

On Modifications and Extensions of the Arithmetic Functions ...

131

From the inequality, for every natural number n and every prime number p p.

pn+1 pn < p1/n p1/(n+1)

it follows that p.J(pn ) ≤ J(pn+1 ). The question of the relations between the multiplicative functions ϕ, ψ, σ and IF is important, but open in general. For example, we can calculate that J(37 ) = 1869 > 1458 = ϕ(37 ) and J(73 ) = 179 < 294 = ϕ(7943). When the following inequalities are valid for the natural number n: (a) J(n) ≤ ϕ(n), (b) n − J(n) ≥ ϕ(n), (c) J(n) + n ≤ ψ(n), (d) J(n).n ≥ σ(n), (e) ϕ(n).σ(n) ≤ n2 − J(n)?

3.3. Converse Factor: Definition, Properties and Problems Now, following [109], we juxtapose to the natural number n the natural number k

p

n0 = ∏ αi i . i=1

It can be easily seen that if for every i (1 ≤ i ≤ k) αi = 1, then n0 = 1. On the other hand, if there is at least one αi > 1, then n0 ≥ 1.

132

J´ozsef S´andor and Krassimir Atanassov Let us name n0 “Converse Factor” and let us denote it by CF(n). It can be seen that CF is a multiplicative function. Indeed, let m and n be two natural numbers, for which (m, n) = 1. Therefore, l

β

if n has the above form and m = ∏ q j j , where q1 , q2 , ..., q j are different prime j=1

numbers, for every i (1 ≤ i ≤ k) and for every j (1 ≤ j ≤ l) : pi 6= q j , and β1 , β2 , ..., βl ≥ 1 are natural numbers, then k

l

i=1

j=1

q

CF(n.m) = ∏ αipi . ∏ β j j = CF(n).CF(m). On the other hand, if for the prime numbers a, b, c: m = a.b and n = b.c, then CF(m.n) = CF(a.b2.c) = 2b > 1 = CF(ab).CF(bc) = CF(m).CF(n). CF is not a monotonic function. If k = l, p1 = q1 , ..., pk = ql and α1 ≥ β1 , α2 ≥ β2 , ...αk ≥ βl , then n ≥ m and CF(n) ≥ CF(m). The inequality will be strong if at least one of the inequalities between αi and βi is strong (1 ≤ i ≤ k) . If k < l, p1 = q1 , ..., pk = qk and α1 = β1 , α2 = β2 , ...αk = βk , then n < m and CF(n) ≤ CF(m). There is no relation between n and m when k 6= l. Obviously, CF(n) ≥ 1 for every natural number n > 1; and for every two natural numbers n and m: ! k  k m k  ∑ αi m pi αi m CF(n ) = CF ∏ pi =∏ = mi=1 .CF(n). i=1 i=1 αi In Section 3.1 we introduced function B for which: k

k

B(n) = ∑ αi pi = ∑ pi αi = B(CF(n)) i=1

i=1

for every natural number n. For the M¨obius function µ (see, e.g., [21]) for every natural number n it is valid the following equality   1 ω(n)+1 µ(n) = (−1) . , CF(n)

On Modifications and Extensions of the Arithmetic Functions ...

133

where for every integer number n function ω(n) is the number of the prime divisors of n and [x] is the integer part of real number x ≥ 0. Really, if n is a prime number, then   2 1 = 1 = µ(n); (−1) 1 if n = p1 p2 ...ps, where p1 , p2 , ..., ps are prime numbers, then ω(n) = s and     1 s+1 s+1 1 (−1) = (−1) = (−1)s+1 = µ(n); CF(p1 p2 ...ps) 1 if there exists such a prime p that n = ps m, where s, m are natural numbers, m does not divide by p and s ≥ 2, then     1 1 ω(n)+1 ω(n)+1 = (−1) = 0 = µ(n), (−1) CF(ps m) s p .CF(m) because at least s p > 1. It can be easily seen that IF(n) = n if and only if CF(n) = 1 if and only if |µ(n)| = 1.

3.4. Restrictive Factor: Definition, Properties and Problems Here, we juxtapose to the natural number n the natural number k

RF(n) = ∏ pαi i −1 i=1

which we call Restrictive factor introduced in [110, 111] It can be easily seen that if for every i (1 ≤ i ≤ k) αi = 1, then RF(n) = 1. On the other hand, if there is at least one αi > 1, then n > RF(n) > 1. It can be easily seen that RF is a multiplicative function and k

k

k

k

k

i=1

i=1

i=1

2αi −1 i i −2 RF(∏ p2α > ∏ p2α = (∏ pαi i −1 )2 = (RF(∏ pαi i ))2 . i ) = ∏ i = 1pi i i=1

134

J´ozsef S´andor and Krassimir Atanassov Moreover, let n = k.l, m = k.s for (k, l) = (k, s) = (l, s) = 1. Then

RF(m, n) = RF(k2 ).RF(l).RF(s) > RF(k)2 .RF(l).RF(s) = RF(m).RF(n). k

Also, RF(n) = 1 iff n = ∏ pi . In particular, RF(p) = 1 for each prime i=1

number p. RF is not a monotonic function. Its first 40 values are given in Table 3.4.1. Table 3.4.1 n 1 2 3 4 5 6 7 8 9 10

RF(n) 1 1 1 2 1 1 1 4 3 1

n 11 12 13 14 15 16 17 18 19 20

RF(n) 1 2 1 1 1 8 1 3 1 2

n 21 22 23 24 25 26 27 28 29 30

RF(n) 1 1 1 4 5 1 9 2 1 1

n 31 32 33 34 35 36 37 38 39 40

RF(n) 1 16 1 1 1 6 1 1 1 4

For each natural number n: RF(n) > ϕ(n). Theorem 3.4.1. For every natural number n, A(n) ≡ n2 − ϕ(n).σ(n) − RF(n) ≥ 0. Proof. Let d(n) = 1, i.e., n is a prime number. Then, n2 − ϕ(n).σ(n) − RF(n) = n2 − (n − 1).(n + 1) − 1 = 0. Let us suppose that for every natural number n, if 1 ≤ d(n) ≤ k, then the assertion is valid. We will prove that if conditions d(n0 ) = k + 1 are valid for the natural number n0 , then the assertion of the Theorem 3.4.1 is valid for n0, too. However, n0 = n.p, where d(n) = k and p is a prime number. Two cases are possible for p. Case 1: p 6∈ set(n). Then A(n0) = n2 p2 − ϕ(n.p).σ(n.p) − RF(n).RF(p)

On Modifications and Extensions of the Arithmetic Functions ...

135

= n2 p2 − ϕ(n).σ(n)(p2 − 1) − RF (n) > p2 .(n2 − ϕ(n).σ(n) − RF(n)) ≥ 0.

Case 2: p ∈ set(n). Then n0 = m.pa , where d(m) ≤ k − 1, a ≥ 1 and A(n0) = m2 p2a+2 − ϕ(m.pa+1 ).σ(m.pa+1) − RF(m).RF(pa+1 ) = m2 p2a+2 − ϕ(m).σ(m).pa .(pa+2 − 1) − RF(m).RF(pa+1 ) > p2a+2 (m2 − ϕ(m).σ(m) − RF(m)) geq0.

Therefore, n2 − ϕ(n).σ(n) ≥ RF(n). Analogously, it can be proved that ϕ(n) + σ(n) − 2n ≥ RF(n). Finally, we must mention, that the functions from the Sections 3.2, 3.3 and 3.4 are object of research in [108, 109, 110, 111].

3.5. On an Arithmetic Function, Related to Operation “Derivative” In [8, 112] a new arithmetic function and its modifications have been introduced. Here we discuss some of the properties of this function. For the natural number n we define the function k

α

α

i−1 αi −1 i+1 δ(n) = ∑ αi pα1 1 ...pi−1 pi pi+1 ...pαk k .

(3.5.1)

i=1

It is similar to the operation “derivative”, but it is used only for natural numbers. We will show some of its properties. Obviously, if p is a prime number, then from the definition it follows that δ(p) = 1. From (3.5.1) it follows that for every natural number n k

δ(n) = ∑ αi i=1

n , pi

136

J´ozsef S´andor and Krassimir Atanassov

hence, k

δ(n) = n. ∑ i = 1 i=1

αi . pi

Theorem 3.5.1 [8]. For every two natural numbers m and n, it holds that (a) δ(m.n) = δ(m).n + m.δ(n); (b) if mn is a natural number, then δ

m n

=

δ(m).n − m.δ(n) ; n2

(c) δ(mn ) = n.mn−1 .δ(m). ! !

Proof. Let

k

∏ pαi

m=

k+l

i=1

k+l

β pi i

∏

n=

i=k+1

pαi i ,

∏

.

i

i=k+1

!

k+l+s

β pi i

∏

.

i=k+l+1

!

,

where k, l, s ≥ 1 and α1 , ..., αk+l, βk+1, ..., βk+l+s ≥ 1 are natural numbers, p1 , ..., pk+l+s are different prime numbers. Then, for (a) we have !! ! ! ! k

k+l

∏ pαi i .

δ(m.n) = δ

∏

i=1

k

∏ pαi

=δ

i

i=1

k

=

∑

k+l

.

∏

α +βi

pi i

i=k+1

k

(α j + β j ).

k+l+s

k

∑

β j.

j=k+l+1 k

=

∑

j=1

∏ pαi i=1

j=k+1

+

!

k+l+s

.

k+l

∏

∏ pαi i=1

i

!

i

pi i

∏

β

pi i

i=k+l+1 α +βi

pi i

!

!! k+l+s

∏

.

β

pi i

i=k+l+1

α j−1 α j −1 α j+1 αk+l αk+1 .pk+1 ...p j−1 p j p j+1 ...pk+l .

!

k+l+s

∏

β pi i

i=k+l+1 k+l+s

.

β

∏

.

i=k+l+1

i=k+1

!

k+l+s

β

pi i

i=k+1

α j−1 α j −1 α j+1 α j pα1 1 ...p j−1 p j p j+1 ...pαk k .

k+l

∑

∏

i=k+1

!

j=1

+

k+l

pαi i .

∏

i=k+l+1

β pi i

!

α −1 α j+1 αk+l+s p j+1 ...pk+l+s

α

α

j−1 k+l+1 .pk+1+1 ...p j−1 pj j

α j−1 α j −1 α j+1 α j pα1 1 ...p j−1 p j p j+1 ...pαk k .

k+l

∏

i=k+1

α +β pi i i

!

!

k+l+s

.

∏

i=k+l+1

β pi i

!

On Modifications and Extensions of the Arithmetic Functions ... 137 ! ! k+l

k

∑

+

α j.

k+l

∑

k

β j.

k+l+s

∑

∏

pαi i

i=1

j=k+1

+

α

α

k

β j.

∏

pαi i

i=1

j=k+l+1

k+l

= l+s

∑

∏

∑

k+l

!

β j.

∏

∏

i=k+l+1 k+l+s

∏

.

β pi i

i=k+l+1

!

α

α

!

k+l+s

.

∏

β pi i

i=k+1

!

β pi i

!

α −1 α j+1 αk+l+s p j+1 ...pk+l+s

j−1 k+l+1 .pk+l+1 ...p j−1 pj j

k+l+s

∏

β pi i

i=k+1

i=1

j=k+1

k+l+s

α j−1 α j −1 α j+1 αk+1 αk+l .pk+1 ...p j−1 p j p j+1 ...pk+l .

α j−1 α j −1 α j+1 p j p j+1 ...pαk k . α j pα1 1 ...p j−1

pαi i

β

pi i

i=k+l+1

!

j=1

+

k+l+s

α −1 α j+1 αk+l p j+1 ...pk+l .

j−1 k+1 ...p j−1 pj j .pk+1

i

i=1

j=k+1

+

∏ pαi

α

α

!

α −1 α j+1 αk+l+s p j+1 ...pk+l+s

j−1 k+l+1 .pk+l+1 ...p j−1 pj j

= δ(m).n + m.δ(n). (b) and (c) are proved by the same manner. From Theorem 3.5.1 (a) it directly follows that for every natural numbers n ≥ 2 and m1 , m2 , ..., mn : ! k

δ

k

∏ mi i=1

= ∑ m1 . . . ..mi−1 .δ(mi).mi+1 . . . ..mn . i=1

It directly holds that k

δ(n) = n if and only if

αi

∑ pi = 1

i=1

and δ(p p ) = p p . It is well-known, that for k = ω(n) > 1 : k

αi

∑ pi 6= 1,

i=1

because the denominators of the sum are different. Therefore, δ(n) = n if and only if n = p p

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J´ozsef S´andor and Krassimir Atanassov

for some prime number p. The first 100 values of δ(n) are given in Table 3.5.1. k

Following the above idea, in [112] for the two natural numbers n = ∏ pαi i , i=1

(where k, α1 , α2 , ..., αk ≥ 1 are natural numbers, p1 , p2 , ..., pk are different prime numbers) and s ≥ 1, the function δs is defined as follows:  1, if s > k        αi1 −1 αi −1 αi1 +1 αi .pi1 .pi1 +1 . ∑ αi1 . · · · .αis .p1 1 . · · · .pi1 −1 δs (n) = (i1 ,...,is )∈S       αi αis −1 αs −1 αis +1  if s ≤ k .pis .pis +1 . · · · .pik k , · · · .pis−1

where

S = {(i, i, ..., i) | 1 ≤ i1 < i2 < ... < is ≤ k} ⊂ N s and N is the set of all natural numbers. Let n and s have the above forms everywhere. From the above definition, the validity of the following assertion follows directly. Theorem 3.5.2. ([112]) For the natural numbers n and s:  1, if s > k   δs (n) = αi1 .αi2 .··· .αis  if s ≤ k  n. ∑ pi .pi .··· .pis , (i1 ,...,is )∈S

1

2

Obviously, if p1 , p2 , ..., pt are prime numbers and t ≤ s, then, δs (p1 .p2 . · · · .pt ) = 1.

When s = 1, we obtain function δ. Now, we shall introduce some other properties of function δ. We shall start with the following Lemma 3.5.1. There are infinitely many natural numbers such that: (a) δ(n) < n and δ(n) < ϕ(n), (b) δ(n) = n,

On Modifications and Extensions of the Arithmetic Functions ...

139

Table 3.5.1. n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

δ(n) 1 1 1 4 1 5 1 12 6 7 1 16 1 9 8 32 1 21 1 24 10 13 1 44 10

n 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

δ(n) 15 27 32 1 31 1 80 14 19 12 60 1 21 16 68 1 41 1 48 39 25 1 112 14 45

n 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75

δ(n) 20 56 1 81 16 92 22 31 1 92 1 33 51 192 18 61 1 72 26 59 1 156 1 39 55

n 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

δ(n) 80 18 71 1 176 108 43 1 124 22 45 32 130 1 123 20 96 34 49 24 272 1 77 75 140

(c) δ(n) > n, (d) δ(n) > ψ(n) and δ(n) > σ(n). As we noted above, for each prime number n the two inequalities of (a) are valid and each n = p p satisfies equality (b) for some prime number p. Obviously, if n = pm for some prime number p and for some natural number

140

J´ozsef S´andor and Krassimir Atanassov

m > p, then n satisfies inequality (c), while, if m > 2p, then δ(pm ) = m.pm−1 > 2.pm >

pm+1 − 1 = σ(pm ) > ψ(pm ), p−1

henve, (d) is valid, too. The following stronger assertion is valid, also. Theorem 3.5.3. For each natural number k there are infinitely many natural numbers n such that: (a) δ(n) = kn, (b) δ(n) ≥ k.σ(n). Proof. (a) Let k be a fixed natural number, and let p be a fixed prime number. Let n = pkp . Then, δ(n) = δ(pkp ) = kp.pkp−1 = k.pkp = kn. (b) Let p be a prime number, let k and m > 2p be natural numbers and let n = pkm . Then, δ(n) = kmpkm−1 > 2kpkm > k.

pkm+1 − 1 = kσ(pkm) = kσ(n). p−1

Theorem 3.5.4. For every natural number n: B(n).δ(n) > n.(d(n) − 1),

(3.5.2)

where function B is defined in the Section 3.1. Proof. We prove this by induction about ω(n). Let n = p be a prime number. Then (3.5.2) is obviously valid. Let us assume that for each n such that d(n) = k the assertion is valid. We shall prove it for the case when n0 = n.p, i.e., when d(n0 ) = k + 1. Then d(n0 ) = d(n) + 1, B(n0) = B(np) = B(n) + p, δ(n0 ) = δ(np) = δ(n).p + n.

On Modifications and Extensions of the Arithmetic Functions ...

141

and B(n0).δ(n0 ) − n0 .(d(n0) − 1)

= B(n.p).δ(n.p) − n.p.(d(n.p) − 1)

= (B(n) + p).(δ(n).p + n) − n.p.d(n)

= p.(B(n).δ(n) − n.d(n)) + p2.δ(n) + n.B(n) = p2 .δ(n) + n.B(n) > 0. The so defined function is not an exact analogue of the operation “derivative”. For example, there are natural numbers m, n for which δ(m + n) 6= δ(m) + δ(n), e.g., 10 = δ(12 + 13) 6= δ(12) + δ(13) = 17. There are natural numbers such that δ(δ(n)) > n, and others, for which δ(δ(n)) < n. For example, δ(δ(8)) = δ(12) = 16 > 8, δ(δ(10)) = δ(7) = 1 < 10. There are natural numbers n for which the sequence n, δ(n), δ(δ(n)), ... goes to 1. Let δs (n) = δ(...(δ(n)...). | {z } s

times

The following Open Problem 3.5.1. is interesting: Are there natural numbers n such that the sequence {δs (n)}∞ s=1 is strictly increasing? Let us define i(n) = {m|n = δ(m)}. Therefore, i(1) = {2, 3, 5, 7, ..} = P ,

where P is the set of the prime numbers. From the above table we see that / i(2) = i(3) = 0. Now, we see that if Goldbach’s hypothesis is valid, i.e., if for each even number 2s there are at least two prime numbers p and q, so that 2s = p + q, then / because δ(p.q) = p + q = 2s. Therefore, if there is an even number i(2s) 6= 0, / then Goldbach’s hypothesis will not be valid for this 2s for which i(2s) = 0, number 2s.

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J´ozsef S´andor and Krassimir Atanassov

3.6. On an Arithmetic Function Related to Function σ Having in mind that function σ has the obvious representation: k

σ(n) = ∏(pαi i + pαi i −1 + ... + 1), i=1

Here, following [113], we introduce a new function that is somehow dual to function σ. It will have the form: k

χ(n) = ∏(pαi i − pαi i −1 + ... + (−1)αi−1 pi + (−1)αi ) i=1

for n ≥ 2. Therefore, for each prime number p: χ(p) = p − 1. It is appropriate to define that χ(1) = 1. In Table 3.6.1, the values of functions χ and σ for the first 50 natural numbers are given. It is easily proved the validity of the following assertions. Proposition 3.6.1. For every natural number n ≥ 2 n < χ(n) ≤ n − 1. 2 Proposition 3.6.2. For every natural number n = p.q, where p, q are prime numbers and p > q ≥ 2, the equality σ(n) =

q+1 .χ(n) q−1

holds. Function χ is multiplicative. Of course, for every natural number n: ϕ(n) ≤ χ(n) < n < ψ(n) ≤ σ(n). Having in mind the results from Section 1.2, we prove

On Modifications and Extensions of the Arithmetic Functions ...

143

Table 3.6.1. n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

χ(n) 1 1 2 3 4 4 6 5 7 6 10 8 12 8 12 11 16

σ(n) 1 3 4 7 6 12 8 15 13 18 12 28 14 24 24 31 18

n 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

χ(n) 13 18 12 16 12 22 16 21 14 20 18 28 22 30 31 24 18

σ(n) 39 20 42 32 36 24 60 31 42 40 56 30 72 32 63 48 54

n 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

χ(n) 32 25 36 20 28 24 40 32 42 30 36 24 46 32 43 31

σ(n) 48 91 38 60 56 90 42 96 44 84 78 72 48 124 57 93

Theorem 3.6.2. For every natural number n: a) ϕ(n) + σ(n) ≥ ψ(n) + χ(n), b) n + ϕ(n) ≥ 2χ(n), c) χ(n) + ψ(n) ≥ 2n,

d) ϕ(n)σ(n) ≥ χ(n)ψ(n) > nn > ψ(n)χ(n) ≥ σ(n)ϕ(n).

Proof. a) Let ω(n) = 1. Then (a) is obviously valid. Let us assume that the assertion is valid for every natural numbers n and k, such that d(n) ≤ k. Let n = n.p, where d(n) = k and p is a prime number. For p there are two cases again. Case 1: p 6∈ set(n). Then, ϕ(n) + σ(n) − ψ(n) − χ(n)

144

J´ozsef S´andor and Krassimir Atanassov = ϕ(n).(p − 1) + σ(n).(p + 1) − ψ(n).(p + 1) − χ(n).(p − 1) = p.(ϕ(n) + σ(n) − ψ(n) − χ(n)) + σ(n) − ψ(n) + χ(n) − ϕ(n) ≥ 0.

Case 2: p ∈ set(n). Then n = m.pa for some natural numbers m, p ≥ 1 and n = m.pa+1 . = ϕ(n).p + σ(n).

pa+2 − 1

pa+1 − 1

ϕ(n) + σ(n) − ψ(n) − χ(n) − ψ(n).p − χ(n).

pa+1 − pa + ... + (−1)a p + (−1)a+1 pa − pa−1 + ... + (−1)a−1 p + (−1)a

= p.(ϕ(n) + σ(n) − ψ(n) − χ(n)) + σ(n). −χ(n).

p−1 pa+1 − 1

(−1)a+1 pa − pa−1 + ... + (−1)a−1 p + (−1)a = σ(m).

p − 1 pa+1 − 1 . pa+1 − 1 p − 1

−χ(m).(pa − pa−1 + ... + (−1)a−1 p + (−1)a ).

(−1)a+1 pa − pa−1 + ... + (−1)a−1 p + (−1)a

= σ(m) − χ(m).(−1)a+1 ≥ σ(m) − χ(m) > 0.

b) Let ω(n) = 1. Then the inequality is obviously valid. Let us assume that the assertion is valid for every natural numbers n and k, such that d(n) ≤ k. Let n = n.p, where d(n) = k and p is a prime number. Case 1: p 6∈ set(n). Then, n + ϕ(n) − 2χ(n) = np + ϕ(n)(p − 1) − 2χ(n)(p − 1) > 0. Case 2: p ∈ set(n). Then n = m.pa for some natural numbers m, p ≥ 1 and n = m.pa+1 . n + ϕ(n) − 2χ(n) = m.pa+1 + ϕ(n).pa .(p − 1) − 2χ(m).(pa+1 − pa + ... + (−1)a+1) = pa+1 .(m + ϕ(m) − 2χ(m)) − ϕ(m).pa + 2χ(m).(pa + ... + (−1)a) > pa−1 .(2χ(n).(p − 1) − ϕ(n).p) > 0,

because 2(p − 1) ≥ p and χ(n) ≥ ϕ(n). Inequalities (c) and (d) are proved by analogy. The results from this Section are published in [106]

On Modifications and Extensions of the Arithmetic Functions ...

145

Table 3.7.1. n 1 2 3 4 5 6 7 8 9 10

EF(n) 1 4 9 8 25 36 49 16 27 100

n 11 12 13 14 15 16 17 18 19 20

EF(n) 121 72 169 196 225 32 289 108 361 200

n 21 22 23 24 25 26 27 28 29 30

EF(n) 441 484 529 144 125 676 81 392 841 900

n 31 32 33 34 35 36 37 38 39 40

EF(n) 961 64 1089 1156 1225 216 1369 1444 15321 400

3.7. Extension Factor: Definition, Properties and Problems k

In the present Section, for each natural number n = ∏ pαi i , where k, α1 , α2 , i=1

..., αk ≥ 1 are natural numbers and p1 , p2 , ..., pk are different prime numbers, we will introduce a new arithmetic function, in some sense, opposite to the restrictive factor. Here, following [114, 115], we juxtapose to the natural number n the (natural) number k

EF(n) = ∏ pαi i +1 i=1

that we call Extension Factor. Hence, EF(n) = n.γ(n). The first 40 values of EF are given in Table 3.7.1. If (m, n) = 1, where for the natural numbers m, n, (m, n) is the Greatest Common Divisor (GCD), then EF(m.n) = EF(m).EF(n),

146

J´ozsef S´andor and Krassimir Atanassov

i.e., EF is a multiplicative function, k

EF(n) = ∏ EF(pαi i ), i=1

and

k

k

k

i=1

i=1

i=1

2 i EF(n) = EF(∏ pαi i ) = ∏ pαi i +1 ≤ ∏ p2α i =n .

On the other hand, it can be seen that if for every i (1 ≤ i ≤ k) αi = 1, then EF(n) = n2 . Therefore, for each prime number p: EF(p) = p2 . From the definitions of functions RF and EF it follows the basic identity EF(n).RF(n) = n2 .

(3.7.1)

Therefore, EF(n) = n2 if and only if RF(n) = 1, i.e., when n = γ(n), so when n is a squarefree number. Theorem 3.7.1. For every two natural numbers m and n: EF(m).EF(n) = EF(m.n).γ((m.n)). Proof. Let (m, n) = r ≥ 1 and let m = s.r, n = t.r. Then EF(m).EF(n) = EF(s.r).EF(t.r) = (s.r.γ(s.r)).(t.r.γ(t.r)) = s.r2 .t.γ(s).γ(r)2.γ(t) = (s.r2.t.γ(s).γ(r).γ(t)).γ(r) = EF(m.n).γ(r) = EF(m.n).γ((m.n)). Theorem 3.7.1 follows also from the definitions, and the following property of the function γ: γ(n).γ(m) = γ(mn).γ((m, n)). Theorem 3.7.2. For every natural number n: RF(EF(n)) = n ≥ EF(RF(n)).

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Proof. Let n be the natural number with the above form. Let for each real number x   1, if x > 0 sg(x) = .  0, if x ≤ 0

Then

k

RF(EF(n)) = RF(∏ pαi i +1 ) i=1

k

k

i=1

i=1

α .sg (αi −1)

= ∏ pαi i = n ≥ ∏ pi i

(so, we eliminate the prime numbers with power 1) k

= EF(∏ pαi i −1 ) = EF(RF(n)). i=1

Another Proof of Theorem 2. follows from: γ(n.γ(n)) = n

(3.7.2)



(3.7.3)

and γ

n γ(n)



≤ n.

(3.7.2) follows from the fact that n and γ(n) have the same prime factors, n while (3.7.3) from the fact that the prime factors of γ(n) are among the prime factors of n . There is equality in (3.7.3) only when n > 1 is squarefull number (i.e., when from each prime power divisor pa of n one has a ≥ 2). Thus one has γ(RF(n)) ≤ γ(n) and γ(EF(n)) = n and the result follows. It could be mentioned that there is equality in Theorem 3.7.2 only when n is squarefull. Theorem 3.7.3. For every natural number n:

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(a) ϕ(EF(n)) = ϕ(n).γ(n), (b) ψ(EF(n)) = ψ(n).γ(n), (c) σ(EF(n)) ≥ σ(n).γ(n). Proof. Let n be a natural number. Then k

k

i=1

i=1

ϕ(EF(n)) = ϕ(∏ pαi i +1 ) = ∏ pαi i (pi − 1) = ϕ(n).γ(n), i.e., (a) is valid. (b) is proved analogously, while the proof of (c) is the following. k

k pαi i +2 − 1 pαi +2 − 1 = σ(n). ∏ αi i +1 ≥ σ(n).γ(n). −1 i=1 pi − 1 i=1 p1 k

σ(EF(n)) = σ(∏ pαi i +1 ) = ∏ i=1

Another proof of inequality (c) of Theorem 3.7.3 is based on the known inequality σ(a.b) ≥ a.σ(b), with equality only for a = 1. Let a = γ(n), b = n, and the result follows. When a = n, b = γ(n), one obtains another inequality: σ(EF(n)) ≥ n.σ(γ(n)) = n.(p1 + 1)...(pk + 1), where p1 , ..., pk are the distinct prime factors of n. Since (p1 + 1)...(pk + 1) =

ψ(n) , RF(n)

we get the inequality: nψ(n) . RF(n) Another result of this type is the following σ(EF(n)) ≥

Theorem 3.7.4. For every natural number n: σ(EF(n)) ≤

σ(n).ψ(n) . RF(n)

Proof. Applying the known inequality σ(ab) ≤ σ(a).σ(b) for a = n and b = γ(n), and letting the distinct prime factors of n to be p1 , ..., pk, remark that one has ψ(n) σ(γ(n)) = (p1 + 1)...(pr + 1) = . RF(n)

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The result follows by the definitions. Theorem 3.7.5. For every natural number n: EF(n) > σ(n). Proof. Let n be a natural number. Then k

k

i=1

i=1

EF(n) = ∏ pαi i +1 > ∏

pαi i +1 − 1 = σ(n). pi − 1

The inequality of Theorem 3.7.5 can be improved when n is odd. Theorem 3.7.6. When n > 1 is odd number, then EF(n) > σ(n) + n. Proof. Apply the known inequality σ(n).φ(n) < n2 (see e.g., [21, 24]). Thus, p1 ...pk n2 n σ(n) < φ(n) . Since φ(n) = (p1 −1)...(p , and EF(n) − n = n(p1 ...pk − 1), it will k −1) be sufficient to prove that: p1 ...pk ≤ p1 ...pk − 1. (p1 − 1)...(pk − 1) Put pi − 1 = xi . Since n is odd, one has xi ≥ 2 for all i = 1, 2, ..., k. We have to prove the inequality x1 ...xk ≤ (x1 + 1)...(xk + 1)(x1 ...xk − 1), or x1 ...xk + (x1 + 1)...(xk + 1) ≤ x1 ...xk (x1 + 1)...(xk + 1). Put x1 ...xk = a, (x1 + 1)...(xk + 1) = b. Then we have to prove that a + b ≤ a.b, or, this can be written also as (a − 1)(b − 1) ≥ 1. This is true, as a − 1 ≥ x1 − 1 ≥ 1, and b ≥ x1 + 1 ≥ 3. The inequality is strict. Now, we will formulate and prove the following common refinement of the last two theorems.

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Theorem 3.7.7. a) For any n > 1 one has σ(n) < n(ω(n) + 1) ≤ EF(n)

(3.7.4)

b) For any n > 1 odd one has σ(n) < n(ω(n) + 1) ≤ EF(n) − n,

(3.7.5)

where ω(n) denotes the number of distinct prime factors of n. Proof. The first inequalities of both a) and b), namely σ(n) < (ω(n) + 1) appeared for the first time in paper [49] from 1989. Now, to prove the second inequality of (3.7.4), remark that γ(n) = p1 ...pk ≥ 2k , where p1 , ..., pk are the prime divisors of n, and k = ω(n). Now, 2k ≥ k + 1 holds true for any k ≥ 1. Thus (3.7.4) follows, as EF(n) = n.γ(n). For the proof of second inequality of b), remark that when n > 1 is odd, then γ(n) ≥ 3k , as p1 , ..., pk ≥ 3. Now, the inequality 3k ≥ k + 2 for k ≥ 1 follows at once, e.g., by mathematical induction. This proves γ(n) ≥ k + 2, so (3.7.5) follows. Theorem 3.7.8. For every natural number n: EF(n) + RF(n) ≥ 2n, with equality only for n = 1. √ Proof. This follows from the classical inequality x + y ≥= 2 xy applied for x = EF(n), y = RF(n), and using the basic identity (3.7.1). Another simple related inequality is the following. Theorem 3.7.9. For every natural number n: n2 ≥

EF(n) ≥ 4ω(n), RF(n)

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where ω(n) is the number of distinct prime factors of n. Proof. There is equality on the right only when n has a single prime factor, i.e., when ω(n) = 1, and on the left, when n is squarefree number. This follows from EF(n) = (γ(n))2. RF(n) Now, from γ(n) ≤ n, the left side inequality follows. For the right side, remark that γ(n) ≥ 2k as any prime divisor is higher or equal to 2. Theorem 3.7.10. For every natural number n: B(EF(n)) = B(n) + B(γ(n)). Proof. Let n be a natural number. Then k

k

k

k

i=1

i=1

i=1

i=1

B(EF(n)) = B(∏ pαi i +1 ) = ∑ (αi + 1).pi = ∑ (αi ).pi + ∑ pi = B(n) + B(γ(n)). Theorem 3.7.11. For every natural number n: δ(EF(n)) = δ(n)γ(n) + nδ(γ(n)). Proof. Let n be a natural number. Then k

δ(EF(n)) = δ(∏ pαi i +1 ) = i=1

k

α +1

= ∑ αi p1 1 i=1

α

i−1 ...pi−1

k

α

∑ (αi + 1)pα1 1 +1...pi−1i−1

+1 αi αi+1 +1 α +1 pi pi+1 ...pk k

i=1

+1 αi αi+1 +1 α +1 pi pi+1 ...pk k +

k

α

∑ pα1 1 +1...pi−1i−1

+1 αi αi+1 +1 α +1 pi pi+1 ...pk k

i=1

= δ(n)γ(n) + nδ(γ(n)). Theorem 3.7.12. If p is a prime number and s < p is a natural number, then EF(s) < EF(p). Proof. Let s < p be an arbitrary natural number. Then EF(s) = s.γ(s) ≤ s2 < p2 = EF(p).

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Corollary 3.7.1. Let n > 1 be squarefree, written as n = ∏ pi . Then i=1

EF(ϕ(n)) ≤ EF(n). Proof. We obtain sequentially that k

k

EF(ϕ(n)) = EF(ϕ(∏ pi )) = EF(∏(pi − 1)) i=1

k

i=1

k

≤ ∏ EF(pi − 1) ≤ ∏ EF(pi ) = EF(n). i=1

i=1

Theorem 3.7.13. For infinitely many n one has EF(ϕ(n)) < EF(n), and for infinitely many m one has EF(ϕ(m)) > EF(m). Proof. When n = 2k for arbitrary natural number k, we obtain for k = 1: EF(ϕ(n)) = EF(ϕ(2)) = EF(1) = 1 < 4 = EF(2) = EF(n), and for k ≥ 2 we obtain: EF(ϕ(n)) = EF(ϕ(2k )) = EF(2k−1) = 2k < 2k+1 = EF(2k) = EF(n), but, when n = 3k , for k = 1: EF(ϕ(n)) = EF(ϕ(3)) = EF(2) = 4 < 9 = EF(3) = EF(n), and for k ≥ 2 we obtain: EF(ϕ(n)) = EF(ϕ(3k )) = EF(2.3k−1) = 4.3k > 3k+1 = EF(3k) = EF(n). More generally, we can prove the following: Theorem 3.7.14. has

For any odd prime p, and k ≥ 2, for m = pk , one EF(ϕ(m)) > EF(m).

Proof. First, we see that for k = 1 from Theorem 3.7.12: EF(ϕ(m)) = EF(ϕ(p)) = EF(p − 1) < EF(p) = EF(m),

(3.7.6)

On Modifications and Extensions of the Arithmetic Functions ...

153

and for k ≥ 2 we obtain: ϕ(m) = pk−1 .(p − 1), so EF(ϕ(m)) = pk .EF(p − 1). Now, EF(p − 1) = (p − 1).γ(p − 1) ≥ 2.(p − 1), as γ(p − 1) ≥ 2, because p ≥ 3. Now, as 2(p − 1) > p, inequality (3.7.6) follows. Theorem 3.7.15. Let k, s ≥ 1 and p ≥ 3 be an odd prime, satisfying EF(p − 1) ≤ 2p. Let n = 2k .ps . Then one has E(ϕ(n)) < EF(n). Proof. One has ϕ(n) = 2k−1 .ps−1 .(p − 1), so using the inequality EF(u.v) ≤ EF(u).EF(v), we can write EF(ϕ(n)) ≤ 2k .ps .EF(p − 1) ≤ 2k+1 .ps+1 = EF(2k .ps ) = EF(n), by using the assumption EF(p − 1) ≤ 2p. The cases k = 1 and/or s = 1 are checked as above. So, the inequality (3.7.6) holds. We must mention that examples of odd primes p, for which EF(p − 1) ≤ 2p are the following: p = 3, 5, 17, and generally any Fermat prime p = 2r + 1. On the other hand, in the general case, between EF(p) and, e.g., EF(p + 1) there is not a fixed relation, because, for example EF(8) = 8 × 2 = 16 < 49 = 72 = EF(7), while EF(13) = 132 = 169 < 196 = 14 × 14 = EF(14). For this case, the following assertion is valid. Theorem 3.7.16. If p is a prime number and p + 1 is squarefree , then EF(p) < EF(p + 1); If p ≥ 3 is a prime number and p + 1 is not squarefree , then EF(p) > EF(p + 1).

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Proof. First suppose that for the prime number p, p + 1 is squarefree. If p = 2, then EF(2) = 4 < 9 = EF(3) = EF(2 + 1). If p ≥ 3, then

EF(p + 1) = (p + 1)2 > p2 = EF(p),

i.e., the first case is valid. If p + 1 is not squarefree , then k

p + 1 = ∏ pαi i , i=1

where k, α1 , α2 , ..., αk ≥ 1 are natural numbers and p1 , p2 , ..., pk are different prime numbers and there is at least one i for which αi > 1. Let α1 = 2, p1 = 2 for α2 = · · · = αk = 1. Then EF(p + 1) = (p + 1)γ(p + 1) =

(p + 1)2 < p2 = EF(p), 2

for p ≥ 3. Obviously, if there is more than one αi > 1, or if the smallest pi > 2, then the inequality will be more powerful. Corollary 3.7.2. For every prime number p ≥ 3 with p + 1 squarefree, EF(ψ(n)) > EF(n), while, otherwise , we have EF(ψ(n)) < EF(n). From Theorem 3.7.11 it follows that for each prime number p: EF(p) > EF(p − 1), but there is not a fixed relation between EF(q) and EF(p − 1), where q is the greatest prime number smaller than p, because, for example for p = 7, q = 5: EF(5) = 25 < 36 = EF(6), while for p = 17, q = 13: EF(16) = 32 < 169 = EF(13).

On Modifications and Extensions of the Arithmetic Functions ...

155

It can be directly seen that for n = 1, 2: EF(n)n = nEF(n) , because EF(1)1 = 11 = 1 = 11 = 1EF (1) and EF(2)2 = 42 = 16 = 24 = 2EF (2). Theorem 3.7.17. For the natural number n ≥ 3: EF(n)n < nEF(n) . The proof follows directly from D. Mitrinovi´c’s inequality (n + r)r < nn+r for the natural numbers r and n ≥ 3 [74]. Now, we will discuss some relations between the function EF and some other arithmetic functions. Obviously, µ(EF(n)) = 0 for each natural number n > 1. It is seen directly that for each natural number n: k k EF(n) 1+sg(αi −1) sg(α −1) = ∏ pi = γ(n). ∏ pi i . RF(n) i=1 i=1

Now, we prove the following assertions. Theorem 3.7.18. For every natural number n: IF(EF(n)) < IF(n). Proof. Let n be a fixed natural number. Then k

k

1 α +1

IF(EF(n)) = IF(∏ pαi i +1 ) = ∏ pi i i=1

i=1

k

1 α

< ∏ pi i = IF(n).

Theorem 3.7.19. For every natural number n: CF(EF(n)) > CF(n).

i=1

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J´ozsef S´andor and Krassimir Atanassov

Proof. Let n be a fixed natural number. Then k

CF(EF(n)) = CF

∏ i=1

pαi i +1

!

k

= ∏(αi + 1) pi i=1

k

> ∏ pαi i +1 = CF(n). i=1

We check directly that EF(1)RF(1) = 11 = 1 = 11 = RF(1)EF(1) , EF(2)RF(2) = 41 = 4 > 14 = RF(2)EF(2) , EF(4)RF(4) = 82 = 64 < 256 = 28 = RF(4)EF(4) . More generally, the following assertion holds. Theorem 3.7.20. For every squarefree natural number n > 1 one has: EF(n)RF(n) > RF(n)EF(n); and if n is not squarefree , then EF(n)RF(n) < RF(n)EF(n) . Proof. In the first case, we get EF(n)RF(n) > 1 = RF(n)EF(n) . In the second one, we will discuss the simplest case, when n = 2k for k ≥ 2. The case k = 2 was checked above, so, let k ≥ 3. Then: k

k

RF(n)EF(n) − EF(n)RF(n) = RF(2k )EF(2 ) − EF(2k )RF(2 )

= 2(k−1)2

k+1

 2k+1  2k−1 = 2k−1 − 2k+1

− 2(k+1)2

k−1

= 24(k−1)2

k−1

− 2(k+1)2

k−1

> 0,

because 4(k − 1) − (k + 1) = 3k − 5 > 0 for each natural number k ≥ 1. Obviously, in all other cases for n the inequality will be more powerful.

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157

Theorem 3.7.21. For every natural number n not squarefree and with γ(n) > 6, it follows that RF(n)EF(n) > EF(n)n . Proof. Let the natural number n not squarefree be given, and let for brevity r = γ(n) > 6. We check sequentially that  n nr − (nr)n > 0 RF(n)EF(n) − EF (n)n = r

if and only if

nnr − (nr)n .rnr = nnr − rn+nr .nn = nnr − rn(r+1).nn > 0 if and only if nn(r−1) − rn(r+1) > 0. Since n is not squarefree, it follows that n ≥ 2r. Let us assume that n = 2r. Then (2r)2r(r−1) − r2r(r+1) = 22r(r−1).r2r(r−1) − r2r(r+1) = 4r(r−1).r2r(r−1) − r2r(r+1) = r2r(r−1).(4r(r−1) − r4r ) > 0,

that is obvious valid for r > 6. Theorem 3.7.22. For every odd not squarefree number n one has: RF(n)EF(n) > EF(n)n . Proof. Let the odd number n be not squarefree . Then, as above, we must prove that nn(r−1) − rn(r+1) > 0. Now, clearly n ≥ 3r. Let us assume that n = 3r. Then (3r)3r(r−1) − r3r(r+1) = 33r(r−1).r3r(r−1) − r3r(r+1) = r3r(r−1).(33r(r−1). − r6r ) > 0,

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because 33r(r−1). − r6r > 0 for r ≥ 3 that is valid, because n ≥ 3. Corollary 3.7.3. For every squarefree natural number n nEF (n) > EF(n)n > EF(n)RF(n) > RF(n)EF(n); If n is not squarefree , and n is even, and γ(n) > 6, or n is an odd number, then nEF (n) > RF(n)EF(n) > EF(n)n > EF(n)RF(n). Finally, we will discuss the relations between arithmetic functions EF and π. In Table 3.7.2 we give the values for the first 20 natural numbers Table 3.7.2. n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

π(n) 0 1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 7 7 8 8

γ(n) 1 2 3 2 5 6 7 2 3 10 11 6 13 13 15 2 17 6 19 10

π(γ(n)) 0 1 2 1 3 3 4 1 2 4 5 3 6 6 6 1 7 3 8 4

EF(n) 1 4 9 8 25 36 49 16 27 100 121 72 169 196 225 32 289 108 361 200

π(EF(n)) 0 2 4 4 9 11 15 6 9 25 31 20 39 44 48 11 61 30 72 46

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159

We will formulate the following Theorem 3.7.23. For each natural number n ≥ 8: π(EF(n)) ≥ π(n).π(γ(n)). Proof. In 1998, in [116], L. Panaitopol proved the inequality π(x).π(y) > π(xy) for any integers x, y ≥ 2, excepting the following pairs: (x, y) = (5, 7), (7, 5), (7, 7). Now, put x = n, y = γ(n) in Panaitopol’s inequality, to prove Theorem 3.7.23. In fact, from the above, a slightly stronger form can be deduced, namely, it holds true for any n ≥ 2 distinct from 7. Theorem 3.7.24. For each natural number n ≥ 2: π(EF(n)) ≥ π(n) + π(γ(n)). Namely, in 1934 in [117], H. Ishikawa proved the inequality π(xy) ≥ π(x) + π(y) for any integers x, y ≥ 2. Now put x = n, y = γ(n) in Ishikawa’s inequality, to prove Theorem 3.7.24. We can also state two results similar to Theorems 3.7.23 and 3.7.24, but involving the function RF, namely Theorem 3.7.25. For each natural number n ≥ 2: π(RF(n)) ≤ π(n) − π(γ(n)) and π(RF(n)) ≤

π(n) . π(γ(n))

(3.7.7)

(3.7.8)

Proof. Apply the Ishikawa inequality π(xy) ≥ π(x) + π(y) for x, y ≥ 2 for x = π(n) γ(n) ≥ 2 and y = ≥ 1. Then we get the inequality (3.7.7). π(γ(n))

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We will remark, that (3.7.7) holds also for x, y ≥ 1, since π(1) = 0. Now, for the proof of (3.7.7) apply the Panaitopol  inequality π(xy) ≥ π(n) π(n) π(x)π(y) for x = γ(n) and y = . Now the pairs γ(n), should π(γ(n)) π(γ(n)) be distinct from (5,7), (7,5) and (7,7). This is possible only if we do not have π(n) γ(n) = 7 and = 7, i.e., when n = 49. Therefore, (3.7.8) follows. π(γ(n))

3.8. Extensions of Restrictive and Extension Factors This section is based on our paper [118].

3.8.1.

First Round of Generalizations

Let

k

n = ∏ pαi i , i=1

where k, α1 , α2 , ..., αk ≥ 1, are the prime factorization of n > 1. Define r

i +1 EFs (n) = ∏ psα i

i=1

and

r

i −1 , RFs (n) = ∏ psα i

i=1

where s ∈ R , and R is the set of real numbers. Then, clearly EF1 (n) = EF(n), RF1 (n) = RF(n) and EFs (n).RFs(n) = n2s . √ So, using the inequality x + y ≥ 2 xy, from (3.8.1) we get We have also

(3.8.1)

EFs (n) + RFs (n) ≥ 2ns.

(3.8.2)

EFs (n) = ns.γ(n) . ns RFs (n) = γ(n)

(3.8.3)

161

On Modifications and Extensions of the Arithmetic Functions ... We get

where s ≥ 1 and for 0 < s ≤ 1. Similarly, as

EFs (n) ≥ EF(n) , RFs (n) ≥ RF(n)

(3.8.4)

EFs (n) ≤ EF(n) , RFs (n) ≤ RF(n)

(3.8.5)

for s ≥ 1, we get by (3.8.3) that

  n s ns ≥ γ(n) γ(n)

EFs (n) ≥ (EF(n))s , RFs (n) ≥ (RF(n))s

(3.8.6)

for s ≥ 1. For 0 < s ≤ 1, the reverse inequalities hold true. Now, we prove Theorem 3.8.1. Let Js denote the Jordan totient function. Then we have for n > 1: RFs (n) ≤ (γ(n))s−1(ns − Js (n)), (3.8.7) for s > 0. Proof. We have

s

JS (n) = n

s

∏ i=1



 1 1− s . pi

Now, first we prove that s

∏ i=1



1 1− s pi



≤ 1−

1 s

∏ psi

i=1

or equivalently, s

s

∏(psi − 1) ≤ ∏ psi − 1. i=1

Put

psi − 1

i=1

= xi for i = 1, 2, ..., r. Then we have to prove that s

s

∏ xi ≤ ∏(xi + 1) − 1, i=1

i=1

(3.8.8)

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or

s

s

∏(xi + 1) ≥ ∏ xi + 1. i=1

i=1

s

This holds true, as xi > 0 by pi ≥ 2 > 1 for s > 0. For r = 1 we have an equality. Now, by (3.8.8) we can write Js (n) ≤ nk −

ns 1 1 . = ns − RFs (n). γ(n) (γ(n))s−1 (γ(n))s−1

and equality (3.8.7) follows. Obviously, for s = 1 we get from (3.8.7): RF(n) ≤ n − ϕ(n)

(3.8.9)

for n > 1. Theorem 3.8.2. For s ≥ 1 we have EFs (n) > σs (n)

(3.8.10)

EFs (n) > σs (n) + ns ,

(3.8.11)

for n > 1. When n ≥ 3 is odd, then

where σs (n) denotes the sum of s-th powers of divisors of n. Proof. As

s(αi +1)

r

σs (n) = ∏ i=1

pi

−1

ps1 − 1

,

for the proof of (3.8.10) it will be sufficient to show that psa+1 >

ps(a+1) − 1 . ps − 1

Now, (3.8.12) is equivalent to psa+s+1 − psa+1 − psa+s > −1

(3.8.12)

On Modifications and Extensions of the Arithmetic Functions ...

163

that is valid, because ps − ps−1 = ps−1 (p − 1) ≥ p − 1 ≥ 1 by s ≥ 1. For the proof of (3.8.11) we will use the following well-known inequality (see, e.g., [110]) for s ≥ 1: σs (n).Js(n) < n2s. Thus, we get n2s < ns .(γ(n))s − ns . Js (n) The right inequality is equivalent to σs (n)
0, α1 , . . ., αr > 0 and α1 + · · · + αr = 1, α1 a1

1 ≤ aα1 1 . . .aαr r ≤ αa a1 + . . . αr ar . + · · · + αarr

(3.8.15)

We will mention that this is the classical Weighted Harmonic Mean – Geometric Mean – Arithmetic Mean Inequality. Theorem 3.8.3. For n > 1 we have   s.B(n) + β(n) k EFs (n) ≤ , k

(3.8.16)

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where k = s.Ω(n) + ω(n) and s > 0 and  m s.B(n) − β(n) RFs (n) ≤ , m

(3.8.17)

where m = s.Ω(n) − ω(n) and s ≥ 1. Proof. Apply the right-hand side of inequality (3.8.15) to a1 = p1 , . . ., ar = pr and sa1 + 1 sar + 1 α1 = , . . ., αr = . k k Then, clearly, α1 + . . .αr = α1 =

sΩ(n) + ω(n) s(a1 + . . .ar ) + r = α1 = = 1. k k

After elementary computations, we get (3.8.16). In the same manner, apply the right-hand side of (3.8.15) to a1 = p1 , . . ., ar = pr and α1 =

sa1 − 1 sar − 1 , . . ., αr = . k k

Then α1 + . . .αr = α1 =

s(a1 + . . . ar ) + r sΩ(n) − ω(n) = α1 = k k

and from α1 , . . ., αr > 0 by sai − 1 ≥ ai − 1 ≥ 1 > 0; inequality (3.8.17) follows, as well. In that follows, we will introduce the following new arithmetic functions: let 1 1 β∗ (n) = +···+ p1 pr and B∗ (n) = Theorem 3.8.4. For n > 1 we have  EFs (n) ≥

a1 ar +···+ . p1 pr

k ∗ sB (n) + β∗ (n)

k

(3.8.18)

On Modifications and Extensions of the Arithmetic Functions ... for s > 0; and RFs (n) ≥



k ∗ sB (n) − β∗ (n)

k

165

(3.8.19)

for s ≥ 1, where k and m are defined as in Theorem 3.8.3. Proof. Use the left-hand side of inequality (3.8.15) to a1 = p1 , . . ., ar = pr and α1 =

sa1 + 1 sar + 1 , . . ., αr = k k

and use the new arithmetic functions β∗ and B∗ . So, inequality (3.8.18) follows. Inequality (3.8.19) follows in the same manner. From (3.8.15), by letting α1 = · · · = αr = 1r , we get   1 1 (a1 + · · · + ar ). +···+ ≥ r2 a1 ar

(3.8.20)

so we get the relation β(n).β∗(n) ≥ (ω(n))2 .

(3.8.21)

We shall prove the similar inequality B(n).B∗(n) ≥ (Ω(n))2.

(3.8.22)

For this purpose, apply the classical Cauchy-Bunyakowski inequality (see [119]) !2 ! ! r

r

∑ xiyi

i=1

to xi =

√

ai pi , yi =

q

≤

∑ x2i

i=1

r

.

∑ y2i

i=1

ai pi . As xi yi = ai , by the given definitions, inequality (3.8.22)

√ follows. By x + y ≥ 2 xy, clearly from (3.8.21) and (3.8.22), we get: β(n) + β∗ (n) ≥ 2ω(n),

B(n) + B∗ (n) ≥ 2Ω(n).

.

(3.8.23)

166

J´ozsef S´andor and Krassimir Atanassov

3.8.2.

Second Round of Generalizations

A second generalization of EF and RF will be given by r

EF (s) (n) = ∏ pai i+s i=1

and

r

RF (s)(n) = ∏ pai i−s , i=1

(1)

where s ∈ R . Then clearly EF (n) = EF(n) and RF (1) (n) = RF(n). Now, EF (s) (n).RF (s)(n) = n2 . (3.8.24) Thus, we have the inequality similar to (3.8.2): EF (s) (n) + RF (s) (n) = 2n. We have also

EF (s) (n) = n(γ(n))s RF (s) (n) =

n (γ(n))s

.

(3.8.25)

(3.8.26)

From (3.8.26) and (3.8.3) it is clear that EFs (n) ≥ EF (s) (n)

(3.8.27)

for s ≥ 1 with an equality only when s = 1, and RFs (n) ≥ RF (s) (n)

(3.8.28)

for s ≥ 1 with an equality only when s = 1. For 0 < s ≤ 1, the inequalities in (3.8.27) and (3.8.28) are reversed. By (3.8.26), we get that RF (s) (n) ≤ RF(n) ≤ n − ϕ(n).

(3.8.29)

Now, we shall introduce an extension of the function σ. Put r

σ(s) (n) = ∏ i=1

pai i+s − 1 pi − 1

(3.8.30)

On Modifications and Extensions of the Arithmetic Functions ...

167

for n > 1. Clearly, we have σ(1) (n) = σ(n). As s(ai + 1) ≥ ai + s for s ≥ 1, we get that σs (n) ≥ σ(s) (n),

(3.8.31)

where σs (n) is the sum of the s-th powers of the divisors of n. Theorem 3.8.5. For s ≥ 1 we have for n > 1: n < σ(s) (n) < EF(s)(n).

(3.8.32)

Proof. The following double inequality can be directly proved: pa
1, we get the classical inequalities (see [110]): 6 ϕ(n).σ(n) < < 1. 2 π n2

(3.8.35)

Corollary 3.8.1. For s ≥ 1 one has EF s−1 (n) ϕ(n)σ(s) (n) < < EF (s−1)(n). ζ(s + 1) n

(3.8.36)

Using now Lemma 3.8.1, we can obtain results similar to these stated in Theorems 3.8.3 and 3.8.4: Theorem 3.8.7. For n > 1 we have   B(n) + sβ(n) Ω(n)+sω(n) (s) , EF (n) ≤ Ω(n) + sω(n) (s)

EF (n) ≥



Ω(n) + sω(n) B(n) + sβ(n)

Ω(n)+sω(n)

,

(3.8.37)

(3.8.38)

On Modifications and Extensions of the Arithmetic Functions ... 169   B(n) − sβ(n) Ω(n)−sω(n) (s) , (3.8.39) RF (n) ≤ Ω(n) − sω(n)  Ω(n)−sω(n) Ω(n) − sω(n) (s) EF (n) ≥ . (3.8.40) B(n) − sβ(n)

Proof. For the proof of (3.8.37), apply the right-hand side of inequality (3.8.15) to a1 = p1 , . . ., ar = pr and α1 = a1t+s , . . ., αr = art+s , where t = Ω(n) + sω(n). Then, Ω(n) + sω(n) α1 + . . . αr = = 1. t So, after elementary computations, we get (3.8.37). For the inequality (3.8.38), apply the left-hand side of inequality (3.8.15), and use the new arithmetic functions β∗ and B∗ (see the proof of Theorem 3.8.4). Inequalities (3.8.39) and (3.8.40) can be proved in the same manner, and we omit the details.

We now state an auxiliary result, which is essentially due to Minkowski [119]: Lemma 3.8.2. Let A, B ≥ 0. Then we have: r

∏(Ai + Bi ) i=1

! 1r

r

≥

∏ Ai i=1

! 1r

r

+

∏ Bi i=1

! 1r

.

(3.8.41)

! 1r

.

(3.8.42)

If Ai ≥ Bi (i = 1, . . ., r), then r

∏(Ai − Bi ) i=1

! 1r

r

≤

∏ Ai i=1

! 1r

r

−

∏ Bi i=1

Proof. (3.8.41) is well-known. For the proof of (3.8.42), for each i (i = 1, . . ., r) put: Ai := Ai − Bi and Bi := Bi in stead of Ai and Bi in (3.8.41). Then we get from (3.8.41) the inequality (3.8.42). Theorem 3.8.8. From any s ∈ R we have 

EF (s) (n)



1 ω(n)

  1 1 ω(n) + EF (s−1)(n) ≤ ((γ(n))s.ψ(n)) ω(n) ,

(3.8.43)

170

J´ozsef S´andor and Krassimir Atanassov 

EF (s) (n)



1 ω(n)

  1 1 ω(n) − EF (s−1)(n) ≥ ((γ(n))s.ϕ(n)) ω(n) .

(3.8.44)

Proof. Let Ai = pai i +s , Bi = pai i +s−1 in (3.8.41). Then

Ai + Bi = pai i +s−1 (pi + 1) = psi .pai i−1 (pi + 1). r

As ψ(n) = ∏ pai i −1 (pi + 1), by definitions, we get the desired inequality i=1

(3.8.43). Inequality (3.8.44) can be deduced in the same manner from (3.8.42). Theorem 3.8.9. From any s ∈ R we have EF(s−1)(n).ψ(n) EF(s−1) (n).ϕ(n) for n > 1.







1 ω(n)

1 ω(n)

≥ EF(s) (n) ≥ EF(s) (n)







1 ω(n)

1 ω(n)

s

+ n ω(n) ,

(3.8.45)

s

+ n ω(n) ,

(3.8.46)

i +1 i Proof. Apply (41) to Ai = psa , Bi = psa i i . Now,

a (s−1)+1

ai −1 i .(pi + 1).pi i Ai + Bi = psa i .(pi + 1) = pi

.

So, by the given definition, inequality (3.8.45) follows from (3.8.41). The similar proof applies to (3.8.46), and we omit the details. We can mention that when Ai > 0, Bi > 0 hold true for any s ∈ R and inequality Ai ≥ Bi is equivalent to pi ≥ 1, so, we can assume again that s can take any real natural value. Theorem 3.8.10. For s > −1 we have   1 ω(n) EF (s) (n) ≥

r

∏(pi − 1) i=1

!

1 ω(n)

1

. (σs (n)) ω(n) + 1.

(3.8.47)

a +s

Proof. Apply inequality (3.8.42) of Lemma 3.8.2 to Ai = Then r

∏(Ai − Bi ) = σ(s) (n), i=1

pi i pi −1

and Bi =

1 pi −1 .

171

On Modifications and Extensions of the Arithmetic Functions ... r

∏ Ai = i=1

ES(s) (n) r

,

∏ (p1 − 1)

i=1 r

∏ Bi = i=1

1 r

,

∏ (p1 − 1)

i=1

and after elementary transformations, we get inequality (3.8.47). We will mention that it is immediate that r

∏(p1 − 1) = i=1

γ(n).ϕ(n) , n

(3.8.48)

so (3.8.47) can be written also in terms of the arithmetic functions γ and ϕ.

3.8.3.

Additive Analogues

As β(n) is an additive analogue of γ(n) and B(n) – of the identity function n, respectively, one can introduce the additive analogues of the functions EF and RF. More generally, let us denote (s)

r

RF+ (n) = ∑ pai i −s, (s)

i=1 r

EF+ (n) = ∑ pai i +s ,

(3.8.49)

i=1

and similarly, r

i −1 RF+,s (n) = ∑ psa , i

i=1 r

i +1 EF+,s (n) = ∑ psa . i

(3.8.50)

i=1

They are generalizations of the additive functions: r

RF+ (n) = ∑ pai i −1 , i=1 r

EF+ (n) = ∑ pai i +1 .

(3.8.51)

i=1

(s)

Here, the respective conditions for the s-argument of RFs (n) and RF+ are valid, as above.

172

J´ozsef S´andor and Krassimir Atanassov

We will study first the arithmetic functions (3.8.51), as these have not been studied in the literature up to now. First, we prove Theorem 3.8.11. For n > 1, 1

RF+ (n) ≥ ω(n)(RF(n)) ω(n) , 1

EF+ (n) ≥ ω(n)(EF(n)) ω(n) ,  1  B (n) − RF+ (n) ≥ ϕ(n), ω(n)   1 B (n) + RF+ (n) ≥ ψ(n), ω(n)   EF+ (n) − RF+ (n) ϕ(n)ψ(n) ≥ . ω(n) RF(n)

(3.8.52) (3.8.53) (3.8.54) (3.8.55) (3.8.56)

Proof. Inequality (3.8.52) follows by applying the arithmetic-geometric mean inequality s ! 1 r

∑ xi ≥ r

i=1

r

∏ xi

,

(3.8.57)

i=1

for xi = pai i −1 (i = 1, . . ., r), r = ω(n). For (53) put xi = pai i +1 ; for (54) remark r

that pa − pa−1 = pa−1 (p − 1) and ∏ (pai i − pai i −1 ) = ϕ(n). Let xi = pai i − pai i −1 r

in (3.8.57). As ∑ i=1

pai i

i=1 r

= B(n) and ∑ pai i −1 = RF+(n), (3.8.54) follows. i=1

Apply (3.8.57) for xi = pai i + pai i −1 to deduce (3.8.55). Finally, as pai i +1 − pai i −1 = pai i −1 (pi − 1)(pi + 1), we get r

∏(pai +1 − pai −1) = i

i=1

i

ϕ(n)ψ(n) RF(n)

and (3.8.56) follows by applying (3.8.57) to xi = pai i +1 − pai i −1 . Theorem 3.8.12. For n > 1 we have (B1 (n))2 ≤ RF+ (n).EF+(n),

(3.8.58)

On Modifications and Extensions of the Arithmetic Functions ... 173   2 m , (3.8.59) (RF+ (n))2 ≤ ω(n).(RF(n)) ω(n) + (ω(n) − 1).RF+ γ(n)   2 m 2 . (3.8.60) (EF+ (n)) ≤ ω(n).(EF(n)) ω(n) + (ω(n) − 1).EF+ γ(n) Proof. For the proof of (3.8.58) q apply the q classical Cauchy-Bunyakowski inai −1 equality (see [119]) for xi = pi , yi = pai i+1 . Then, the inequality (3.8.58) follows. For the proof of (3.8.59) and (3.8.60), we will use the following inequality due to T. Popoviciu and V. Cˆırtraje (see [120]). If I ⊆ R is an interval and f : I → R is a convex function, and a1 , . . .ar ∈ I for r > 2, then  r    ∑ ai  r  i=1 ai + a j r 2   . (3.8.61) ∑ f (ai) + r − 2 . f  r  ≥ r − 2 . ∑ f 2 i=1 1≤i< j≤r Put f (x) = ex in (3.8.61) and then, let ai = 2 logxi for xi > 0. As   !2 r r 1 ∑ xi x j = r − 2  ∑ xi − ∑ x2i  , 1≤i< j≤r i=1 i=1

after some transformations, we get from (3.8.61): r

(r − 1) ∑ x2i + r i=1

s 1 r

r

r

∏ x2i ≥ ∑ xi i=1

i=1

!2

.

Now, apply first the inequality (3.8.62) for xi = pai i−1 . As r n2 i −1 = ∏ p2a , i γ(n) i=1

we get that r

r

n2

∑ x2i = ∑ pai −1 = RF+ ( γ(n) ), i

i=1

i=1

(3.8.62)

174

J´ozsef S´andor and Krassimir Atanassov

and (3.8.59) follows. In the same manner, apply (3.8.62) to xi = pai i +1 . As EF+ ( inequality (3.8.60) follows.

r n2 i +2 ) = ∑ p2a , i γ(n) i=1

Chapter 4

Arithmetic Functions of Other Types 4.1. A Digital Arithmetic Function The digital function, discussed here has a long history. The interest of K. Atanassov in this topic was generated in the beginning of 1970s, when he was a school boy in the secondary school, but his first paper on this topic [121] was published 15 years later and the second one – [122] after another 20 years. Here, the most interesting results over this function are given. The digital arithmetical function is defined and its properties are described, following [121, 122] (see, also [123, 124]). Everywhere here and below we use the natural number n of the following form k

n = ∑ ai .10k−i ≡ a1 a2 ...ak, i=1

where ai is a natural number and 0 ≤ ai ≤ 9 (1 ≤ i ≤ k). We define a function denoted by ϕ by:  if n = 0  0, k ϕ(n) =  ∑ ai , if n > 0 i=1

We use the decimal numeral system everywhere hereafter.

176

J´ozsef S´andor and Krassimir Atanassov

Lemma 4.1.1. Function ϕ satisfies the scheme:  ϕ(n) + 1 , if ak 6= 9        ϕ(n) − 9r + 1 , if ak = ak−1 = ... = ak−r+1 = 9 ϕ(n + 1) = where 1 ≤ r ≤ k and ak−r 6= 9        1 , if ak = ak−1 = ... = a1 = 9

Proof. Let ak 6= 9. Then,

ϕ(n + 1) = ϕ(a1 a2 ...ak + 1) = ϕ(a1 a2 ...a0k), where a0k = ak + 1 ≤ 9. Hence, ϕ(n + 1) =

k−1

k

i=1

i=1

∑ ai + a0k = ∑ ai + 1 = ϕ(n) + 1.

If ak = ak−1 = ... = ak−r+1 = 9 and ak−r 6= 9 for some r (1 ≤ r ≤ k), then ϕ(n + 1) = ϕ(a1 a2 ...ak−r−1a0k−r |{z} 0...0), r

where a0k−r = ak−r + 1 ≤ 9. Hence, k−r−1

ϕ(n + 1) =

∑

i=1

k−r

k

i=1

i=1

ai + a0k−r = ∑ ai + 1 = ∑ ai − 9r + 1 = ϕ(n) − 9r + 1.

If r = k, then ϕ(n + 1) = ϕ(9...9 |{z}) = 1. |{z} +1) = ϕ(1 0...0 k

k

Now, we define a sequence of functions ϕ0 , ϕ1 , ϕ2 , ..., where for each natural number l: ϕ0 (n) = n, ϕl+1 = ϕ(ϕl (n)). Obviously, for every l ∈ N the set of the natural numbers, ϕl : N → N . Since for k > 1 k

k

ϕ(n) = ∑ ai < ∑ ai .10k−i = n. i=1

i=1

Arithmetic Functions of Other Types

177

Then for every n ∈ N there exists l ∈ N such that ϕl (n) = ϕl+1 (n) ∈ ∆ ≡ {0, 1, 2, ..., 9}. Let the function ψ be defined by ψ(n) = ϕl (n), where ϕl+1 (n) = ϕl (n). Hence, ψ : N → N . The following assertions are easy to prove. Theorem 4.1.1. Function ψ satisfies the scheme: ψ(0) = 0 ψ(n + 1) = ψ(ψ(n) + 1). Theorem 4.1.2. For every two natural numbers m and n: (a) ψ(m + n) = ψ(ψ(m) + ψ(n)), (b) ψ(m.n) = ψ(ψ(m).ψ(n)) = ψ(m.ψ(n)) = ψ(ψ(m).n), (c) ψ(mn ) = ψ(ψ(m)n ), (d) ψ(n + 9) = ψ(n), (e) ψ(9n) = 9. Theorem 4.1.3. For every natural number n: ψ(n) ≡ ϕ(n) ≡ n(mod 9). Let us define the function Mod9 : N → {0, 1, ..., 8} such that Mod9 (n) ≡ n(mod 9). Then from Theorem 4.1.3 it can be directly seen that there is a remarkable resemblance between functions ψ and Mod9 . However, the difference is also essential: ψ : N → {0, 1, ..., 9} and for example, ψ(9) = 9 6= 0 = Mod9 (9).

178

J´ozsef S´andor and Krassimir Atanassov Let the sequence of natural numbersa1 , a2 , ... be given and let ci = ψ(ai ) (i = 1, 2, ...).

Hence, we deduce the sequence c1 , c2 , ... from the former sequence. If k and l exist such that l ≥ 0, ci+l = ck+i+l = c2k+i+l = ... for 1 ≤ i ≤ k, then we will say that [cl+1 , cl+2 , ..., cl+k] is the base of the sequence a1 , a2 , ... with length of k and with respect to function ψ. For example, the Fibonacci sequence {Fi }∞ i=0 , for which F0 = 0, F1 = 1, Fn+2 = Fn+1 + Fn (n ≥ 0) has a base of length 24 with respect to the function ψ and it is the following: [1, 1, 2, 3, 5, 8,4, 3, 7,1, 8,9, 8, 8,7, 6, 4,1, 5, 6,2, 8,1, 9]. The Lucas sequence {Li }∞ i=0 , for which L2 = 2, L1 = 1, Ln+2 = Ln+1 + Ln (n ≥ 0) also has a base of length 24 with respect to the function ψ and it is the following: [2, 1, 3, 4, 7, 2,9, 2, 2,4, 6,1, 7, 8,6, 5, 2,7, 9, 7,7, 5,3, 8]. Even the Lucas–Lehmer sequence {li }∞ i=0 , for which l1 = 4, ln+1 = ln2 − 2 (n ≥ 0), has a base of length 1 and it is the following [5] with respect to the function ψ. We can construct the following table for the natural number k ≥ 0. n

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

ψ(Fn6k+2 )

1

1

4

9

7

1

7

9

4

1

1

9

1

1

4

9

7

1

7

9

4

1

1

9

ψ(Fn6k+3 )

1

1

8

9

8

8

1

9

1

1

8

9

8

8

1

9

1

1

8

9

8

8

1

9

ψ(Fn6k+4 )

1

1

7

9

4

1

4

9

7

1

1

9

1

1

7

9

4

1

4

9

7

1

1

9

ψ(Fn6k+5 )

1

1

5

9

2

8

7

9

4

1

8

9

8

8

4

9

7

1

2

9

5

8

1

9

ψ(Fn6k+6 )

1

1

1

9

1

1

1

9

1

1

1

9

1

1

1

9

1

1

1

9

1

1

1

9

ψ(Fn6k+7 )

1

1

2

9

5

8

4

9

7

1

8

9

8

8

7

9

4

1

5

9

2

8

1

9

Arithmetic Functions of Other Types

179

The Pell sequence {Pi }∞ i=0 , for which P1 = P2 = 2, Pn+2 = 2Pn+1 + Pn (n ≥ 0) has a base of length 24 with respect to the function ψ, and it is the following: [1, 2, 5, 3, 2, 7,7, 3, 4,2, 8,9, 8, 7,4, 6, 7,2, 2, 6,5, 7,1, 9]. The Tribonacci sequence {Ti }∞ i=0 , for which T1 = T2 = T3 = 1, Tn+3 = Tn+2 + Tn+1 + Tn (n ≥ 0) has a base of length 39 with respect to the function ψ and it is the following: [1, 1, 1, 3, 5,9, 8, 4,3, 6, 4,4, 5,4, 4, 4,3, 2, 9,5, 7, 3,6, 7,7, 2, 7,7, 7, 3,8, 9,2, 1, 3, 6, 1, 1, 8]. The Tribonacci sequences {Ti }∞ i=0 , for which T1 = 0, T2 = T3 = 1, Tn+3 = Tn+2 + Tn+1 + Tn (n ≥ 0) or T1 = T2 = 0, T3 = 1, Tn+3 = Tn+2 + Tn+1 + Tn (n ≥ 0) have both bases of length 39 with respect to the function ψ, which are identical, namely the following: [1, 1, 2, 4, 7,4, 6, 8,9, 5, 4,9, 9,4, 4, 8,7, 1, 7,6, 5, 9,2, 7,9, 9, 7,7, 5, 1,4, 1,6, 2, 9, 8, 1, 9, 9]. The Padovan sequence {Pi }∞ i=0 , for which P1 = P2 = P3 = 2, Pn+3 = Pn+1 + Pn (n ≥ 0) has a base of length 39 with respect to the function ψ, and it is the following: [1, 1, 1, 2, 2,3, 4, 5,7, 9, 3,1, 1,4, 2, 5,6, 7, 2,4, 9, 6,4, 6,1, 1, 7,2, 8, 9,1, 8,1, 9, 9, 1, 9, 1, 9].

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The Tetrabonacci sequence {Ui}∞ i=0 , for which a) U1 = U2 = U3 = 0,U4 = 1, or U1 = U2 = 0,U3 = U4 = 1, and Un+4 = Un+3 + Un+2 +Un+1 +Un (n ≥ 0) has a base of length 78 (= 39 × 2) with respect to the function ψ, and it is the following: [1, 1, 2, 4, 8, 6,2, 2, 9,1, 5, 8,5, 1,1, 6, 4,3, 5, 9,3, 2, 1,6, 3,3, 4, 7, 8, 4, 5, 6, 5,2, 9, 4,2, 8,5, 1, 7,3, 7, 9,8, 9, 6,5, 1,3, 6, 6, 7, 4, 5, 4, 2, 6, 8,2, 9, 7,8, 8,5, 1, 4,9, 1, 6,2, 9, 9,8, 1,9, 9, 9] b) U1 = 0,U2 = U3 = U4 = 1, Un+4 = Un+3 + Un+2 + Un+1 + Un (n ≥ 0) has a base of length 78 with respect to the function ψ, and it is the following: [1, 1, 1, 3, 6, 2,3, 5, 7,8, 5, 7,9, 2,5, 5, 3,6, 1, 6,7, 2, 7,4, 2,6, 1, 4, 4, 6, 6, 2, 9,5, 4, 2,2, 4,3, 2, 2,2, 9, 6,1, 9, 7,5, 4,7, 5, 3, 1, 7, 7, 9, 6, 2, 6,5, 1, 5,8, 1,6, 2, 8,8, 6, 6,1, 3, 7,8, 1,1, 8, 9] c) U1 = U2 = U3 = U4 = 1, Un+4 = Un+3 +Un+2 +Un+1 +Un (n ≥ 0) has a base of length 78 with respect to the function ψ, and it is the following: [1, 1, 1, 1, 4, 7,4, 7, 4,4, 1, 7,7, 1,7, 4, 1,4, 7, 7,1, 1, 7,7, 7,4, 7, 7, 7, 7, 1, 4, 1,4, 1, 1,7, 4,4, 7, 4,1, 7, 1,4, 4, 7,7, 4,4, 4, 1, 4, 4, 4, 4, 7, 1, 7,1, 7, 7,4, 1,1, 4, 1,7, 4, 7,1, 1, 4,4, 1,1, 1, 7] The Pentabonacci sequence {Vi }∞ i=0 , for which a) V1 = V2 = V3 = V4 = 0,V5 = 1, or V1 = V2 = V3 = 0,V4 = V5 = 1, and Vn+5 = Vn+4 +Vn+3 +Vn+2 +Vn+1 +Vn (n ≥ 0) has a base of length 312 (= 39 × 8) with respect to the function ψ, and it is the following: [1, 1, 2, 4, 8, 7,4, 7, 3,2, 5, 3,2, 6,9, 7, 9,6, 1, 5,1, 4, 8,1, 1,6, 2, 9, 1, 1, 1, 5, 8,7, 4, 7,4, 3,7, 7, 1,4, 4, 5,3, 8, 6,8, 3,1, 8, 8, 1, 3, 3, 5, 2, 5, 9,6, 9, 4,6, 7,5, 4, 8,3, 9, 2,8, 3, 7,2, 4,6, 4, 5, 3, 4, 4, 2, 9, 4, 5,6, 8, 5,1, 7,9, 3, 7,9, 8, 9,9, 6, 5,1, 3,6, 3, 9, 4, 7, 2, 7, 2, 4, 4,1, 9, 2,2, 9,5, 9, 9,7, 3, 6,7, 5, 1,4, 5,4, 1, 6, 2, 9, 4, 4, 7, 8, 5,1, 7, 1,4, 9,4, 7, 7,4, 4, 8,3, 8, 9,5, 6,4, 5, 2,

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4, 3, 9, 5, 5, 8, 3,3, 6, 7,9, 1,8, 4, 2,6, 3, 5,2, 9, 7,8, 4,3, 4, 8, 9, 1, 7, 2, 9, 1, 2,3, 8, 5,1, 1,9, 6, 4,3, 5, 9,9, 3, 2,1, 6,3, 6, 9, 7, 4, 2, 1, 5, 1, 4,4, 6, 2,8, 6,8, 3, 9,7, 6, 6,4, 5, 1,4, 2,7, 1, 6, 2, 9, 7, 7, 4, 2, 2,4, 1, 4,4, 6,1, 7, 4,4, 4, 2,3, 8, 3,2, 9,7, 2, 5, 7, 3, 6, 5, 8, 2, 6,9, 3, 1,3, 4,2, 4, 5,9, 6, 8,5, 6, 7,5, 4,9, 4, 2, 6, 7, 1, 2, 9, 7, 8,9, 8, 5,1, 4,9, 9, 1,6, 2, 9,9, 9, 8,1, 9,9, 9, 9] b) V1 = V2 = 0,V3 = V4 = V5 = 1, Vn+5 = Vn+4 +Vn+3 +Vn+2 +Vn+1 +Vn (n ≥ 0) has a base of length 312 with respect to the function ψ, and it is the following: [1, 1, 1, 3, 6, 3,5, 9, 8,4, 2, 1,6, 3,7, 1, 9,8, 1, 8,9, 8, 7,6, 2,5, 1, 3, 8, 1, 9, 4, 7,2, 5, 9,9, 5,3, 4, 3,6, 3, 1,8, 3, 3,9, 6,2, 5, 7, 2, 4, 2, 2, 8, 9, 7,1, 9, 7,6, 3,8, 6, 3,8, 1, 8,8, 1, 8,8, 6,4, 9, 8, 8, 8, 1, 7, 5, 2, 5,2, 3, 8,2, 2,8, 5, 7,6, 1, 9,1, 6, 5,4, 7,5, 9, 3, 1, 7, 7, 9, 9, 6, 2,6, 5, 1,2, 7,3, 9, 4,7, 3, 8,4, 8, 3,8, 4,9, 5, 2, 1, 3, 2, 4, 3, 4, 7,2, 2, 9,6, 8,9, 7, 3,6, 6, 4,8, 9, 6,6, 6,8, 8, 7, 8, 1, 5, 2, 5, 3, 7,4, 3, 4,3, 3,8, 3, 3,2, 1, 8,8, 4, 5,8, 6,4, 9, 5, 5, 2, 7, 1, 2, 8, 2,2, 6, 2,2, 5,8, 5, 4,6, 1, 6,4, 3, 2,7, 4,2, 9, 6, 1, 4, 4, 6, 3, 9, 8,3, 2, 7,2, 4,9, 6, 1,4, 6, 8,7, 8, 6,8, 1,3, 8, 8, 1, 3, 5, 7, 6, 4, 7,2, 8, 9,3, 2,6, 1, 3,6, 9, 7,8, 6, 9,3, 6,5, 2, 7, 5, 7, 8, 2, 2, 6, 7,7, 6, 1,9, 3,8, 9, 3,5, 1, 8,8, 7, 2,8, 6,4, 9, 2, 2, 5, 4, 4, 8, 5, 8,2, 9, 5,2, 8,8, 5, 1,6, 1, 3,7, 9, 8,1, 1,8, 9, 9] c) V1 = 0,V2 = V3 = V4 = V5 = 1, Vn+5 = Vn+4 +Vn+3 +Vn+2 +Vn+1 +Vn (n ≥ 0) has a base of length 312 with respect to the function ψ, and it is the following: [1, 1, 1, 1, 4, 8,6, 2, 3,5, 6, 4,2, 2,1, 6, 6,8, 5, 8,6, 6, 6,4, 3,7, 8, 1, 5, 6, 9, 2, 5,9, 4, 2,4, 6,7, 5, 6,1, 7, 8,9, 4, 2,3, 8,8, 7, 1, 9, 6, 4, 9, 2, 3, 6,6, 8, 7,3, 3,9, 3, 7,7, 2, 1,2, 1, 4,1, 9,8, 5, 9, 5, 9, 9, 1, 6, 3, 1,2, 4, 7,8, 4,7, 3, 2,6, 4, 4,1, 8, 5,4, 4,4, 7, 6,

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J´ozsef S´andor and Krassimir Atanassov 7, 1, 7, 1, 4, 2, 6,2, 6, 2,9, 7,8, 5, 4,6, 3, 8,8, 2, 9,3, 3,7, 6, 1, 2, 1, 8, 9, 3, 5, 8,6, 4, 8,4, 3,7, 8, 3,7, 1, 8,9, 1, 8,9, 8,8, 7, 4, 9, 9, 1, 3, 8, 3, 6,3, 5, 7,6, 9,3, 3, 1,4, 2, 4,5, 7, 4,4, 6,8, 2, 6, 8, 3, 9, 1, 9, 3, 7,2, 4, 7,5, 7,7, 3, 2,6, 7, 7,7, 2, 2,7, 7,7, 7, 3, 4, 1, 4, 1, 4, 5, 6,2, 9, 8,3, 1,5, 8, 7,6, 9, 8,2, 5, 3,9, 9,1, 9, 4, 5, 1, 2, 3, 6, 8, 2,3, 4, 5,4, 9,7, 2, 9,4, 4, 8,9, 7, 5,6, 8,8, 7, 7, 9, 3, 7, 6, 5, 3, 6,9, 2, 7,9, 6,6, 3, 4,1, 2, 7,8, 4, 4,7, 3,8, 8, 3, 2, 6, 9, 1, 3, 3, 4,2, 4, 7,2, 1,7, 3, 2,6, 1, 1,4, 5, 8,1, 1,1, 7, 9]

d) V1 = V2 = V3 = V4 = V5 = 1, Vn+5 = Vn+4 +Vn+3 +Vn+2 +Vn+1 +Vn (n ≥ 0) has a base of length 312 with respect to the function ψ, and it is the following: [1, 1, 1, 1, 1, 5,9, 8, 6,2, 3, 1,2, 5,4, 6, 9,8, 5, 5,6, 6, 3,7, 9,4, 2, 7, 2, 6, 3, 2, 2,6, 1, 5,7, 3,4, 2, 3,1, 4, 5,6, 1, 8,6, 8,2, 7, 4, 9, 3, 7, 3, 8, 3, 6,9, 2, 1,3, 3,9, 9, 7,4, 5, 7,5, 1, 4,4, 3,8, 2, 3, 2, 9, 6, 4, 6, 9, 7,5, 4, 4,2, 4,1, 6, 8,3, 4, 4,7, 8, 8,4, 4,4, 1, 3, 7, 1, 7, 1, 1, 8, 9,8, 9, 8,6, 4,8, 8, 7,6, 6, 8,8, 8, 9,3, 9,1, 3, 7, 5, 7, 5, 9, 6, 5, 5,3, 1, 2,7, 9,4, 5, 9,7, 7, 5,6, 7, 5,3, 8,2, 7, 7, 9, 6, 4, 6, 5, 3, 6,6, 8, 1,6, 9,3, 9, 1,1, 5, 1,8, 7, 4,7, 9,8, 8, 9, 5, 3, 6, 4, 9, 9, 4,5, 4, 4,8, 7,1, 6, 8,3, 7, 7,4, 2, 5,7, 7,7, 1, 9, 4, 1, 4, 1, 1, 2, 9,8, 3, 5,9, 7,5, 2, 1,6, 3, 8,2, 2, 3,9, 6,4, 6, 1, 8, 7, 8, 3, 9, 8, 8,9, 1, 8,7, 6,4, 8, 6,4, 1, 5,6, 4, 2,9, 8,2, 7, 1, 9, 9, 1, 9, 2, 3, 6,3, 5, 1,9, 6,6, 9, 4,7, 5, 4,2, 4, 4,1, 6,8, 5, 6, 8, 6, 6, 4, 3, 9, 1,5, 4, 4,5, 1,1, 6, 8,3, 1, 1,1, 5, 2,1, 1,1, 1, 6] The non-homogenous sequence {Gi}∞ i=0 , for which G1 = G2 = 1, Gn+2 = Gn+1 + Gn + 1 (n ≥ 0)

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has a base of length 24 with respect to the function ψ, and it is the following: [1, 1, 3, 5, 9, 6,7, 5, 4,1, 6,8, 6, 6,4, 2, 7,1, 9, 2,3, 6,1, 8]. The generating function for the Euler’s polynomials En (x) is defined by ∞ ext tn = ∑ En (x). e + 1 n=0 n! t

for |t| < π, from which we get that the Euler’s numbers En are defined for each n ∈ N as follows:   1 n En = 2 En . 2 A. Shannon, J. Clarke and K. Atanassov proved in [125] that the sequence {En }∞ i=1 has a base of length 6 and it is the following [1, 5, 7, 8, 4, 2]. The n-th Jacobsthal number (n ≥ 0) is defined by Jn =

2n − (−1)n . 3

The first ten Jacobsthal numbers are: n 0 1 2 3 4 5 6 7 8 9 Jn 0 1 1 3 5 11 21 43 85 171 It can be checked directly that for each n ≥ 0: Jn+1 = 2.Jn − (−1)n+1 . The base of the sequence {Jn } is [7, 4, 9, 8, 8,6, 4] (with length 7). The proofs of these facts are easy and can be produced, e.g., by induction. The same is valid for the next facts, as well. The k-th triangular number tk is defined by the formula tk =

k(k + 1) 2

and it has a base of length 9, namely: [1, 3, 6, 1, 6, 3, 1,9, 9].

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The following property of the triangular numbers is valid for each two natural numbers m and r such that m ≥ 1 and 1 ≤ r ≤ 9: ψ(t9m+r ) = ψ(tr ). The k-th n-gonal number pnk is defined by the formula (see, e.g., [126]) pnk = k +

k(k − 1) .(n − 2) 2

and it has a base with respect to k (i.e., for a fixed n) with length of 9 with the form [1, ψ(n), ψ(3n − 1), ψ(6n − 8), ψ(10n − 15), ψ(15n − 24),ψ(21n − 35), ψ(28n − 48), ψ(36n − 63)] and a base with respect to n ≥ 3 (i.e., for a fixed k) with length of 9 with the form   3   2   2

k −k 3k − k) 5k − 3k) 2 2 ,ψ k ,ψ , ψ 2k − k , ψ , [ψ 2 2 2  2   2 

7k − 5k 9k − 7k 2 2 ψ 3k − 2k , ψ , ψ 4k − 3k , ψ ]. 2 2

For these numbers the following properties are valid. Theorem 4.1.4.

a) For each three natural numbers m, n and r, such that m, n ≥ 1 and 1 ≤ r ≤ 9, it holds that ψ(p9m+r ) = ψ(prn ). n b) For each three natural numbers m, k and r, such that m, k ≥ 1 and 3 ≤ r ≤ 11 it holds that ψ(pk9m+r ) = ψ(pkr ). The k-th n-pyramidal number Pkn (n ≥ 3, k ≥ 1) is defined by the formula Pkn =

k(k + 1) (3 + (k − 1)(n + 2)) 6

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and it has a base with respect to k of length 27. For example, the tetrahedral numbers (n = 3) have the base: [1, 8, 8, 6, 7, 7, 2,6, 6,7, 5, 5,3, 4, 4,8, 3, 3,4, 2,2, 9, 1,1, 5, 9,9] and the following base with respect to n ≥ 3 with length of 9     2   k(k + 1)(2k + 1) k (k + 1) k(k + 1)(k + 2) [ψ ,ψ ,ψ , 6 6 2       k(k + 1)(4k − 1) k(k + 1)(5k − 2) k(k + 1)(2k − 1) ψ ,ψ ,ψ , 6 6 2       k(k + 1)(7k − 4) k(k + 1)(8k − 5) k(k + 1)(3k − 2) ψ ,ψ ,ψ ]. 6 6 6 Vassia Atanassova remarked in 2001 that the 8-digit numbers 188677266, 755344833 and 422911599 obtained by the digits of the base, satisfy the equalities: ψ(188677266) = ψ(755344833) = ψ(422911599) = 6, ψ(188) = ψ(755) = ψ(422) = 8, ψ(677) = ψ(344) = ψ(911) = 2, ψ(266) = ψ(833) = ψ(599) = 6. For these numbers the following properties are valid Theorem 4.1.5. a) For each three natural numbers m, n and r, such that m, n ≥ 1 and 1 ≤ r ≤ 27, ψ(Pn27m+r ) = ψ(Pnr ); b) For each three natural numbers m, k and r, such that m, k ≥ 1 and 3 ≤ r ≤ 11, k ψ(P9m+r ) = ψ(Prk ). Following [127] we can give some other bases generated by the ψ-function, using formulas from [126] for the n-th member of the respective sequence: pentagonal number. Pen = 12 n(3n − 1)

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hexagonal number. Hen = n(2n − 1)

truncated tetrahedral number. T tetn = 16 n(23n2 − 27n + 10) square pyramid number. Pyrn = 16 n(n + 1)(2n + 1) octahedral number. Octn = 31 n(2n2 + 1) stella octangula number. Steln = n(2n2 − 1)

centred cube number. Ccubn = (2n − 1)(n2 − n + 1)

truncated octahedral number. Toctn = 16n3 − 33n2 + 24n − 6

rhombic dodecahedral number. Rhon = (2n − 1)(2n2 − 2n + 1) pentatope number. Ptopn =

1 24 n(n + 1)(n + 2)(n + 3).

Now, for each sequence from above we shall give its length and basis: Pen Hen T tetn Pyrn Octn Steln Ccubn Toctn Rhon Ptopn

9 9 27 27 27 9 3 3 9 27

[1, 5, 3, 4, 8, 6, 7,2, 9] [1, 6, 6, 1, 9, 3, 1,3, 9] [1, 7, 5, 9, 6, 1, 8,5, 6, 7,4, 2,6, 3, 7,5, 2, 3,4, 1, 8,3, 9,4, 2, 8,9] [1, 5, 5, 3, 1, 1, 5,6, 6, 7,2, 2,9, 7, 7,2, 3, 3,4, 8, 8,7, 4,4, 8, 9,9] [1, 6, 1, 8, 4, 2, 6,2, 3, 4,9, 4,2, 7, 5,9, 5, 6,7, 3, 7,5, 1,8, 3, 8,9] [1, 5, 6, 7, 2, 3, 4,8, 9] [1, 9, 8] [1, 2, 3] [1, 6, 2, 4, 9, 5, 7,3, 8] [1, 5, 6, 8, 7, 9, 3,6, 9, 4,2, 6,2, 4, 9,6, 3, 9,7, 8, 6,5, 1,9, 9, 9,9]

It can be directly checked that the bases of sequences {nk }∞ k=1 for n = 1, 2, ..., 9 are those introduced in the following table. n 1 2 3 4 5 6 7 8 9

base of a sequence {nk }∞ i=1 1 2,4,8,7,5,1 9 4,7,1 5,7,8,4,2,1 9 7,4,1 8,1 9

base length 1 6 1 3 6 1 3 2 1

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187

On the other hand, sequence {nn }∞ n=1 has a base (of length 9) in the form [1, 4, 9, 1, 2, 9, 7,1, 9], and the sequence {kn!}∞ n=1 has a base (of length 1) in the form   [1] , if k 6= 3m for some natural number m 

[9] , if k = 3m for some natural number m

In [128] the following results, related to the concept “factorial”, were published. The concepts of n!! is well-known. Let us define the new factorial n!!! by: n!!! = 1.2.4.5.7.8.10.11...n, where n=

  n 

, if n = 3s + 1 or n = 3s + 2

n − 1 , if n = 3s + 3,

for some natural number s. Therefore 3 does not divide n!!!. We shall prove that sequence {ψ(n!!!)}∞ n=1 has a base of length 12 with respect to the function ψ and it is [1, 2, 8, 4, 1,8, 8, 7,1, 5, 8,1]. V. Atanassova mentioned that the 4-digit numbers 1284, 1887 and 1581, obtained by the digits of the base, satisfy the equalities: ψ(1284) = ψ(1887) = ψ(1581) = 6. Really, the validity of the assertion for the first 12 natural numbers with the above mentioned forms, i.e., the numbers 1, 2, 4, 5, 7, 8,10,11, 13, 14, 16, 17, is directly checked. Let us assume that the assertion is valid for the numbers (18k + 1)!!!, (18k + 2)!!!, (18k + 4)!!!, (18k + 5)!!!, (18k + 7)!!!, (18k + 8)!!!,

188

J´ozsef S´andor and Krassimir Atanassov (18k + 10)!!!, (18k + 11)!!!, (18k + 13)!!!, (18k + 14)!!!, (18k + 16)!!!, (18k + 17)!!!.

Then, ψ((18k + 19)!!!) = ψ((18k + 17)!!!.(18k + 19)) = ψ(ψ(18k + 17)!!!.ψ(18k + 19)) = ψ(1.1) = 1; ψ((18k + 20)!!!) = ψ((18k + 19)!!!.(18k + 20)) = ψ(ψ(18k + 19)!!!.ψ(18k + 20)) = ψ(1.2) = 2; ψ((18k + 22)!!!) = ψ((18k + 20)!!!.(18k + 22)) = ψ(ψ(18k + 20)!!!.ψ(18k + 22)) = ψ(2.4) = 8, etc., with which the assertion is proved. Having in mind that every natural number different from 0 can be represented in exactly one of the forms 3k + 1, 3k + 2 and 3k + 3, for the natural number n = 3k + m, where m ∈ {1, 2, 3} and k ≥ 1 being a natural number, we can define:  1.4...(3k + 1), if n = 3k + 1 and m = 1      n!m = 2.5...(3k + 2), if n = 3k + 2 and m = 2      3.6...(3k + 3), if n = 3k + 3 and m = 3

As above, we can prove that: • for the natural number n of the form 3k + 1, sequence {ψ(n!1 )}∞ n=1 has a base of length 3 with respect to function ψ and it is [1, ψ(3k + 1), 1];

• for the natural number n of the form 3k + 2, sequence {ψ(n!1 )}∞ n=1 has a base of length 6 with respect to function ψ and it is [2, ψ(6k + 4), 8, 7, ψ(3k + 5), 1];

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Arithmetic Functions of Other Types

• for the natural number n of the form 3k + 3, sequence {ψ(n!1 )}∞ n=1 has a base of length 1 with respect to function ψ. Only the first member of the sequence is 3, but in general, it is [9]. One problem is introduced and solved in [129] by Iseki Kiyoshi and Nakakura Manabu. In the above terms it can be represented in the following form: Find all (integer) solutions of the equation ϕ(nm ) = n, where m and n are natural numbers. Based on the above definitions, we shall formulate and solve two assertions, connected to the above one. Problem 4.1.1. Find all (integer) solutions of the equation ψ(nm) = ψ(n).

(4.1.1)

Problem 4.1.2. Find all (integer) solutions of the equation ψ(nm) = ψ(m).

(4.1.2)

As we saw above, the bases of the sequences {nk }∞ k=1 for n ∈ N are the following for the first 18 values. 1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8 9

1 4 9 7 7 9 4 1 9

1 8 9 1 8 9 1 8 9

1 7 9 4 4 9 7 1 9

1 5 9 7 2 9 4 8 9

1 1 9 1 1 9 1 1 9

1 2 9 4 5 9 7 8 9

1 4 9 7 7 9 4 1 9

1 8 9 1 8 9 1 8 9

1 7 9 4 4 9 7 1 9

1 5 9 7 2 9 4 8 9

1 1 9 1 1 9 1 1 9

1 2 9 4 5 9 7 8 9

1 4 9 7 7 9 4 1 9

1 8 9 1 8 9 1 8 9

1 7 9 4 4 9 7 1 9

Using this table we obtain all solutions of the above two problems. For Problem 4.1.1 they are:

1 5 9 7 2 9 4 8 9

1 1 9 1 1 9 1 1 9

190

J´ozsef S´andor and Krassimir Atanassov form of n 9l + 1 9l + 2 9l + 3 9l + 4 9l + 5 9l + 6 9l + 7 9l + 8 9l + 9

ψ(n) 1 2 3 4 5 6 7 8 9

form of m k 6k + 1 if n = 3, then m = 1; no other solutions 3k + 1 6k + 1 if n = 6, then m = 1; no other solutions 3k + 1 2k + 1 k

(for arbitrary natural numbers k and l). If equation (4.1.1) is substituted by the equation: ψ(nm ) = n, then the solution is rather similar to the solution in the above table for l = 0. Obviously, other values of n cannot be solutions of the problem, because 1 ≤ ψ(nm ) ≤ 9.

(4.1.3)

It can be directly seen that the solutions given in [129] are special cases of the solutions introduced in the above table. The solutions of Problem 4.1.2 are obtained in a similar way. From the table of the bases for different values of m (= 18l + 1, 18l + 2, ..., 18l + 18; l being a natural number), i.e., for different ψ(m) (= 1, 2, ..., 9, 1, 2,..., 9) are obtained those numbers of the elements put down by repeating of the bases of ψ(nm ), which coincide with the corresponding values of ψ(m). In this way we obtain the following solutions of (4.1.2): m 18l + 1 18l + 2 18l + 3 18l + 4 18l + 5 18l + 6 18l + 7 18l + 8 18l + 9 18l + 10 18l + 11 18l + 12 18l + 13 18l + 14 18l + 15 18l + 16 18l + 17 18l + 18

form of ψ(m) 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

form of n (for every natural number l) 9l + 1 no solutions no solutions 9k + 4, 9l + 5 9k + 2 no solutions 9k + 7 no solutions 9l + 3, 9l + 6, 9l + 9 9l + 1, 9l + 8 9l + 5 no solutions 9k + 4 no solutions no solutions 9k + 2, 9l + 7 9l + 8 9l + 3, 9l + 6, 9l + 9

191

Arithmetic Functions of Other Types (for every natural number k). As in the first problem, if we substitute (4.1.2) by the equation ψ(nm ) = m,

according to (4.1.3), we shall obtain that the solutions coincide with those from the first nine rows of the above table. In [130], V. Atanassova determined all natural numbers k and n, for which ψ(n)k = n.

(4.1.4)

Obviously, the natural number n must be the k-th power of some natural number m, for which 0 ≤ m ≤ 9. The solutions of (4.1.4) for 2 ≤ k ≤ 7 are the following: ψ(81)2 = 81, ψ(512)3 = 512, ψ(729)3 = 729, ψ(256)4 = 256, ψ(2401)4 = 2401, ψ(6561)4 = 6561, ψ(32768)5 = 32768, ψ(59049)5 = 59049, ψ(531441)6 = 531441, ψ(128)7 = 128, ψ(16384)7 = 16384, ψ(78125)7 = 78125, ψ(823543)7 = 823543, ψ(2097152)7 = 2097152, ψ(4782969)7 = 4782969. She proved the following Theorem 4.1.6 [130] All solutions hk, ni of (4.1.3) are: (a) h1, 1i, h1, 3i, h1, 6i; (b) hl, 0i for every natural number l; (c) h3.l + 1, 43l+1i and h3.l + 1, 73.7l+1i for every natural number l, if ψ(n) = 4 and ψ(n) = 7, respectively; (d) h6.l + 1, 56l+1 i for every natural number l, if ψ(n) = 5; (e) h2.l + 1, 82l+1 i for every natural number l, if ψ(n) = 8; (f) hl, 9l i for every natural number l, if ψ(n) = 9.

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J´ozsef S´andor and Krassimir Atanassov

The following open problem of E. Avanessov is introduced in [131]: Find all natural numbers x, for which there exist natural numbers n such that x = n + ϕ(n). Using the definition of function ψ we can formulate a problem similar to the above one, following [132]. Find all natural numbers x, for which there exist natural numbers n such that x = n + ψ(n). (4.1.5) We shall prove the following Theorem 4.1.7. only

For all natural number

x ≥ 10

there exists one and

one natural number n such that (4.1.5) is valid. Proof. Let x = 18.k + 2.r, where k, r are natural numbers and 0 ≤ r ≤ 8. Then for the natural number n = x − r : x − r + ψ(x − r) = 18.k + r + ψ(18.k + r) = 18.k + r + r = 18.k + 2.r = x. Let x = 18.k + 2.r + 9, where k, r are natural numbers and 1 ≤ r ≤ 9. Then for the natural number n = x − r : x − r + ψ(x − r) = 18.k + r + 9 + ψ(18.k + r + 9) = 18.k + 2.r + 9 = x. Therefore, for every natural number x there exists at least one solution of (4.1.5). Now, we shall prove the uniqueness of the solution. Let x = m + ψ(m) = n + ψ(n), i.e., let x have at least two different representations in the form of (4.1.5). Then, m − n = ψ(n) − ψ(m). Let m > n. From ψ(n) − ψ(m) ≤ 8 it follows that, if m = n + k, then 1 ≤ k ≤ 8. Therefore, m + ψ(m) = n + k + ψ(n + k) = n + ψ(n),

193

Arithmetic Functions of Other Types i.e. ψ(n + k) = ψ(n) − k.

(4.1.6)

Let g(n, k) = ψ(n + k) − ψ(n) + k. It directly can be seen that g(n + 9.s, k) = ψ(n + 9.s + k) − ψ(n + 9.s) + k = ψ(n + k) − ψ(n) + k. Hence, we can consider only the values of g(n, k) for 0 ≤ n ≤ 9. They are the following:

k n 0 1 2 3 4 5 6 7 8 9

1

2

3

4

5

2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2 −5 −7 −5

6 6 6 6 6 6 6 −3 −3 −3

8 8 8 8 8 8 −1 −1 −1 −1

10 10 10 10 10 1 1 1 1 1

6

7

8

12 14 16 12 14 16 12 14 7 12 5 7 3 5 7 3 5 7 3 5 7 3 5 7 3 5 7 3 5 7

Therefore, for every two natural numbers n, k (1 ≤ k ≤ 8) : g(n, k) 6= 0, i.e., (4.1.6) has no solution. Hence, the representation (4.1.5) of x is unique.

4.2. On an Inequality of Klamkin, Its Arithmetic Applications, Modifications and Extension In 1974 M. S. Klamkin [133] proved the following result: Let x be a nonnegative real number, and m, n integers with m ≥ n ≥ 1. Then (m + n)(1 + xm ) ≥ 2n

1 − xm+n . 1 − xn

(4.2.1)

194

J´ozsef S´andor and Krassimir Atanassov We note that for x = 1, the right-hand side of (4.2.1) is understood as lim , x→1

when the inequality becomes an equality. Also, for x = 0 (4.2.1) becomes m + n ≥ 2n, which is true. For m = n there is equality in (4.2.1). In fact, it can be shown that for all real numbers m > n > 0, and all x > 0, (4.2.1) holds true with strict inequality (see the solutions of (1) in [134]). Assume now that x = a ≥ 1, m = p, n = q, where p ≥ q ≥ 0 are real numbers. Then, since (1 + a p )(1 − aq ) = a p − aq + 1 − a p+q , after some transformations (4.2.1) becomes equivalent to (p − q)(a p+q − 1) ≥ (p + q)(a p − aq ).

(4.2.2)

In the case of p − q ≤ 1, a weaker result than (4.2.2) appears in the famous monograph by D. S. Mitronovi´c [135] (3.6.26, page 276). For certain arithmetical applications of Klamkin’s inequality, see [19]. In what follows we will point out some surprising connections of inequality (4.2.2) (i.e., in fact (4.2.1)) with certain special means of two arguments. Also, a new application of (4.2.1) will be given. m−n Let m, n > 0 and put p + q = m, p − q = n. Then p = m+n 2 , q = 2 and (4.2.2) gives am − 1 m (m−n)/2 > a . (4.2.3) an − 1 n By letting a = xy (x > y > 0), relation (4.2.3) may be written also as 

xm − yn n · xn − yn m

1/(m−n)

>

√

xy.

(4.2.4)

If n = 1, the expression on the left-hand side of (4.2.4) is called as the “Stolarsky mean” of x and y. Put   m x − ym 1 1/(m−1) · . S(m) = S(x, y, m) = x−y m

It is not difficult to see that S can be defined also for all real numbers m 6∈ {0, 1}, while for m = 0, and m = 1, S can be defined by the limits lim S(x, y, m) =

m→0

x−y lnx − ln y

195

Arithmetic Functions of Other Types and for y 6= x,

1 lim S(x, y, m) = m→1 e



yy xx

1/(y−x)

,

the definition of S can be extended to all real numbers m. Let for y 6= x, x−y , L(x, y) = ln x − ln y   1 yy 1/(y−x) , I(x, y) = e xx and L(x, x) = I(x, x) = x. These means are known as the “logarithmic” and “identric means” of x and y (see e.g. [136] for their properties). Stolarsky [137] has proved that S is a strictly increasing function of m. Therefore S(−1) < S(0) < S(1) < S(2), giving √

xy < L(x, y) < I(x, y)
0, we get √

xy < L(x, y)
n, √ xy < (L(xm−n , ym−n ))1/(m−n) < x



xm − ym n · xn − yn m

1/(m−n)

.

f (p+q)

(4.2.7)

Proof. Put f (x) = a x−1 (x > 0), where a > 1; and let ϕ(p) = f (p−q) (p > q > 0), where q is fixed. We first show that ϕ is a strictly increasing function. Since f 0 (p + q) f (p − q) − f 0 (p − q) f (p + q) ϕ0 (p) = , f 2 (p − q)

196

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it will be sufficient to prove that follow, if

f0 f

f 0 (p+q) f (p+q)

>

f 0 (p−q) f (p−q) .

= g is an increasing function. By g0 (t) = ( f 0 (t)/ f (t))0 =

Since p + q > p − q, this will

f 00 (t) f (t) − ( f 0(t))2 , f 2 (t)

it will be sufficient to show that f is strictly log-convex (i.e., ln f is strictly convex). Lemma 4.2.1. The function f is strictly log-convex. Proof. After certain simple computations (which we omit here), it follows that tat lnt − (at − 1) f 0 (t) = , t2 f 00 (t) =

t 2 at ln2 a − 2tat ln a + 2at − 2 , t3

and

a2t − 2at − t 2 at ln2 a + 1 f 00 (t) f (t) − ( f 0(t))2 = t4 √ √ (at − 10 at ln at )(at − 1 + at lnat ) = . t4 √ √ √ Put at = h. Then h − 1 − h lnh > 0, since h−1 ln h > h by L(h, 1) > h (lefthand side of (4.2.5)). This proves the log-convexity property of f for a > 1. Since ϕ is strictly increasing, one can write ϕ(p) >

lim

p→q,p>q

f (p + q) a2q − 1 = . f (p − q) 2q lna

Write p + q = m, p − q = n, a = xy , and the right-hand side of (4.2.7) follows. For the left-hand side of (4.2.7) observe that again by the left-hand side of (4.2.5) one has p L(xm−n , ym−n ) > xm−n ym−n = (xy)(m−n)/2, which implies the desired inequality.

Arithmetic Functions of Other Types

197

Remark 4.2.1. ϕ being strictly increasing, it follows also that a p+q − 1 p − q · = a2q , p→∞ a p−q − 1 p + q

ϕ(p) < lim i.e.

(p − q)(a p+q − 1) ≤ (p + q)a2q (a p−q − 1),

(4.2.8)

which is complementary to (4.2.2). A divisor d of N is called unitary divisor of the positive integer N > 1, if (d, Nd ) = 1. For k ≥ 0, let σk (N), respectively σ∗k (N), denote the sum of k-th powers of divisors, respectively unitary divisors of N. Remark that σ0 (N) = d(N), σ∗0 (N) = d ∗ (N) are the numbers of these divisors of N. It is well-known that (see e.g. [23, 24]) if the prime factorization of N is r

N = ∏ pai i i=1

(pi distinct primes, ai ≥ 1 integers), then σk (N) = ∏ri=1

k(ai +1)

pi

−1 , pki −1

d(N) = ∏ri=1 (ai + 1), (4.2.9)

∗ r ω(N) i σ∗k (N) = ∏ri=1 (pka ), i + 1), d (N) = 2 (= 2

where ω(N) = r denotes the number of distinct prime divisors of N. Write now (4.2.1) (see [134]) in the form 2n

xm+n − 1 xm+n − 1 ≤ (m + n)(1 + xm ) ≤ 2m n , n x −1 x −1

(4.2.10)

where x > 1, m ≥ n ≥ 1. Put n = k, m = kai , x = pi (i = 1, 2, . . ., r). Writing (4.2.10), after termwise multiplication, we get 2ω(N) σk (N) ≤ d(N)σ∗k (N) ≤ 2ω(N)β(N)σk (N),

(4.2.11)

where β(N) = ∏ri=1 ai (for this, and the other functions, too, see e.g., [21, 24]). The left-hand side of (4.2.11) appears also in [19]. Now, remarking that r

r

i=1

i=1

2ω(N) β(N) = ∏(2ai ) ≤ ∏ 2ai = 2Ω(N),

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J´ozsef S´andor and Krassimir Atanassov

where Ω(N) denotes the total number of prime factors of N (using the classical inequality 2a−1 ≥ a for all a ≥ 1), relation (4.2.11) also implies 2ω(N) ≤

d(N)σ∗k (N) ≤ 2Ω(N). σk (N)

(4.2.12)

We say that the normal order of magnitude of the arithmetical function f (n) is the function g(n), if for each ε > 0, the inequality | f (n) − g(n)| < εg(n) holds true for almost all positive integers n. Theorem 4.2.2. The normal order of magnitude of log(d(N)σ∗k (N)/σk (N)) is (log2)(loglogN). Proof. Let P be a property in the set of positive integers and let set a p (n) = 1 if n has the property P; a p (n) = 0, otherwise. Let A p (x) = ∑ a p (n). n≤x

If A p (x) ∼ x (x → ∞) we say that the property P holds for almost all natural numbers. By a well-known result of Hardy and Ramanujan (see e.g., [21, 23, 135]), the normal order of magnitude of ω(N) and Ω(N) is loglogN. By (4.2.12) we can write (1 − ε)(loglogN) < ω(N) ≤

1 logd(N)σ∗k (N)/σk (N) log2

≤ Ω(N) < (1 + ε) lglgN for almost all N, so Theorem 4.2.2 follows. Now, following [138]. we will introduce two modifications of Klamkin’s inequality. We will modify (4.2.1) to two different forms. Theorem 4.2.3. Let x ≥ 0, m ≥ k ≥ n ≥ 1 and 2k ≥ m + n. Then (m + k + n)(1 + xm )(1 + xk ) ≥ 3n

1 − xm+k+n . 1 − xn

(4.2.13)

Arithmetic Functions of Other Types

199

Proof. Let x > 1. Then, (4.2.13) has the equivalent forms (m + k + n)(1 + xm )(1 + xk ) ≥ 3n

xm+k+n − 1 xn − 1

(4.2.14)

and (m + k + n)(xm + 1)(xk + 1)(xn − 1) ≥ 3n(xm+k+n − 1).

(4.2.15)

Now, having in mind (4.2.1), in its form (m + n)(xm + 1) ≥ 2n

xm+n − 1 xn − 1

(4.2.16)

for (4.2.15) we obtain sequentially (m + k + n)(xm + 1)(xk + 1)(xn − 1) − 3n(xm+k+n − 1) =

=

m+k+n (m + n)(xm + 1)(xn − 1)(xk + 1) − 3n(xm+k+n − 1) m+n m+k+n .2n(xm+n − 1)(xk + 1) − 3n(xm+k+n − 1) ≥ m+n

n (2(m + k + n)(xm+k+n + xm+n − xk − 1) − 3(m + n)(xm+k+n − 1)) m+n n = ((2k − m − n)(xm+k+n − 1) + 2(m + k + n)(xm+n − xk )) m+n

(from xm+n ≥ xk )

n (2k − m − n)(xm+k+n − 1) m+n (from 2k ≥ m + n and x > 1) ≥ 0. ≥

If x = 1, then both sides of (4.2.15) are equal to 0. If x = 0, then (4.2.14) is transformed to the inequality m + k + n ≥ 3n, that is true. Let below 0 < x < 1. Then, (4.2.13) has the form (m + k + n)(1 + xm )(1 + xk )(1 − xn ) ≥ 3n(1 − xm+k+n)

(4.2.17)

200

J´ozsef S´andor and Krassimir Atanassov Having in mind (4.2.1), for (4.2.17) we obtain sequentially (m + k + n)(1 + xm )(1 + xk )(1 − xn ) − 3n(1 − xm+k+n ) =

=

m+k+n (m + n)(1 + xm )(1 − xn )(1 + xk ) − 3n(1 − xm+k+n ) m+n m+k+n ≥ .2n(1 − xm+n )(1 + xk ) − 3n(1 − xm+k+n ) m+n

n (2(m + k + n)(1 − xm+k+n − xm+n + xk ) − 3(m + n)(1 − xm+k+n )) m+n n = ((2k − m − n)(1 − xm+k+n ) + 2(m + k + n)(xk − xm+n )) m+n

(from xm+n ≤ xk )

n (2k − m − n)(1 − xm+k+n ) m+n (from 2k ≥ m + n and x < 1) ≥ 0. ≥

Therefore, in both cases (4.2.13) is valid. Theorem 4.2.4. Let x ≥ 0, k ≥ m ≥ n ≥ 1 and m + n ≥ k. Then (4.2.13) is valid. Proof. Let x > 1. We will use again (4.2.15) (as an equivalent form of (4.2.13)) and (4.2.16): (k + m + n)(xk + 1)(xm + 1)(xn − 1) − 3n(xm+k+n − 1) =

=

k+m+n (m + n)(xk + 1)(xn − 1)(xm + 1) − 3n(xk+m+n − 1) m+n k+m+n ≥ .2n(xm+n − 1)(xk + 1) − 3n(xk+m+n − 1) m+n

n (2(m + k + n)(xk+m+n + xm+n − xk − 1) − 3(m + n)(xk+m+n − 1)) m+n n = ((2k − m − n)(xk+m+n − 1) + 2(k + m + n)(xm+n − xk )) m+n

(from xm+n ≥ xk )

≥

n (2k − m − n)(xm+k+n − 1) m+n

201

Arithmetic Functions of Other Types (from k ≥ m ≥ n and x > 1)

≥ 0.

The rest of the cases are checked by analogy. Now, following [139], we will formulate an extension of inequality (4.2.1). Let m1 , m2 , ... be an infinite increasing sequence of real numbers. We will define recurrently for the natural number s ≥ 2 the infinite sequence α2 , α3 , ... by α2 = 1,  if ms+1 ≥ m1 + m2 + ... + ms  αs , αs+1 =  if ms+1 ≤ m1 + m2 + ... + ms αs . m1 +mm2s+1 +...+ms , Let x > 1 be a real number. Firstly, we will prove that

(m1 + m2 + ... + ms )(xm1 − 1)(xm2 + 1)...(xms + 1) ≥ 2s−1 αs m1 (xm1 +m2 +...+ms − 1).

(4.2.18)

If s = 2, m1 = n, m2 = m, then the assertion is valid, because Klamkin’s inequality (4.2.1) for x > 1 has the equivalent form (m + n)(xm + 1)(xn − 1) ≥ 2n(xm+n − 1).

(4.2.19)

Let us assume that (4.2.18) is valid for some natural number s ≥ 2. Then, for the real numbers m1 ≤ m2 ≤ ... ≤ ms+1 there are two cases. Case 1: ms+1 ≥ m1 + m2 + ... + ms . Then, we obtain (m1 + m2 + ... + ms+1 )(xm1 − 1)(xm2 + 1)...(xms+1 + 1) =

m1 + m2 + ... + ms+1 (m1 + m2 + ... + ms )(xm1 − 1)(xm2 + 1)...(xms+1 + 1) m1 + m2 + ... + ms

(by induction) ≥

m1 + m2 + ... + ms+1 s−1 2 αsm1 (xm1 +m2 +...+ms − 1)(xms+1 + 1) m1 + m2 + ... + ms

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J´ozsef S´andor and Krassimir Atanassov

(because αs+1 = αs ) =

m1 2s−1 αs+1 (m1 + m2 + ... + ms+1 ) m1 + m2 + ... + ms .(xm1 +m2 +...+ms − 1)(xms+1 + 1)

(from (4.2.19), for the two numbers m1 + m2 + ... + ms and ms+1 ) m1 2s−1 αs+1 .(2(m1 + m2 + ... + ms )(xm1 +m2 +...+ms+1 − 1)) m1 + m2 + ... + ms

≥

= 2s αs+1 m1 (xm1 +m2 +...+ms+1 − 1). Case 2: ms+1 ≤ m1 + m2 + ... + ms . Then, (xm1 +m2 +...+ms − 1)(xms+1 + 1) − (xm1 +m2 +...+ms + 1)(xms+1 − 1) = (xm1 +m2 +...+ms+1 − xms+1 + xm1 +m2 +...+ms − 1) −(xm1 +m2 +...+ms+1 + xms+1 − xm1 +m2 +...+ms − 1) = 2(xm1 +m2 +...+ms − xms+1 ) ≥ 0.

Therefore, (xm1 +m2 +...+ms − 1)(xms+1 + 1) ≥ (xm1 +m2 +...+ms + 1)(xms+1 − 1)

(4.2.20)

and (m1 + m2 + ... + ms+1 )(xm1 − 1)(xm2 + 1)...(xms+1 + 1)

=

m1 + m2 + ... + ms+1 (m1 + m2 + ... + ms )(xm1 − 1)(xm2 + 1)...(xms+1 + 1) m1 + m2 + ... + ms

(by induction) ≥

m1 + m2 + ... + ms+1 s−1 2 αsm1 (xm1 +m2 +...+ms − 1)(xms+1 + 1) m1 + m2 + ... + ms

(from (4.2.20)) ≥

m1 + m2 + ... + ms+1 s−1 2 αsm1 (xm1 +m2 +...+ms + 1)(xms+1 − 1) m1 + m2 + ... + ms =

m1 2s−1 αs (m1 + m2 + ... + ms+1 ) m1 + m2 + ... + ms

Arithmetic Functions of Other Types

203

.(xm1 +m2 +...+ms + 1)(xms+1 − 1) (from (4.2.19) for the two numbers m1 + m2 + ... + ms and ms+1 ) ≥

m1 2s−1 αs.(2ms+1 (xm1 +m2 +...+ms+1 − 1)) m1 + m2 + ... + ms =

ms+1 2s αs .m1 (xm1 +m2 +...+ms+1 − 1) m1 + m2 + ... + ms = 2sαs+1 m1 (xm1 ,m2 ,...,ms+1 − 1).

Therefore, (4.2.18) is proved. If x ∈ [0, 1), then we can prove analogously to the above, that (m1 + m2 + ... + ms )(1 − xm1 )(1 + xm2 )...(1 + xms ) ≥ 2s−1 αs m1 (1 − xm1 +m2 +...+ms ).

(4.2.21)

In x = 1, then (4.2.18) and (4.2.21) obtain the form of equality. If x = 0, then (4.2.21) obtains the form (m1 + m2 + ... + ms ) ≥ 2s−1 αsm1 .

(4.2.22)

The validity of (4.2.22) can be proved directly as an independent inequality. From the validity of (4.2.18), (4.2.21) and (4.2.22) we can formulate the following Theorem 4.2.5. Let s ≥ 2 be a natural number, x ≥ 0 and m1 ≤ m2 ≤ ... ≤ ms be real numbers. Then, the inequality (m1 + m2 + ... + ms )(1 + xm2 )...(1 + xms ) ≥ 2s−1 αsm1

1 − xm1 +m2 +...+ms 1 − xm1

(4.2.23)

holds. It is obvious that (4.2.23) is an extension of (4.2.1). Now, we will discuss two interesting particular cases. Let the real numbers m1 ≤ m2 ≤ ... ≤ ms for every i (2 ≤ i ≤ s) satisfy one of the two following condifions. Condition 1. mi+1 ≤ m1 + m2 + ... + mi ,

204

J´ozsef S´andor and Krassimir Atanassov

Then (4.2.23) obtains the form (m1 + m2 + ... + ms )(1 + xm2 )...(1 + xms ) ≥ 2s−1

m3 m4 ms 1−xm1 +m2 +...+ms ... m1 ; m1 + m2 m1 + m2 + m3 m1 + m2 + ... + ms−1 1 − xm1

Condition 2.

mi+1 ≥ m1 + m2 + ... + mi , Then (4.2.23) obtains the form (m1 + m2 + ... + ms )(1 + xm2 )...(1 + xms ) ≥ 2s−1 m1

1 − xm1 +m2 +...+ms . 1 − xm1

4.3. Some Representations Related to Arithmetic Function “Factorial” In the present Section, we describe different (well-known and new) representations of function “factorial” n! for the natural number n and comment them. The following denotations are used everywhere below: • [x] is the integer part of the real nonnegative number x; •

ord p m is the largest exponent of power p which divides the natural number m, where p is a prime number, i.e.,ord p m = k if and only if pk divides m, but pk+1 does not (another notation for ordp m is e p (m));

• Λ marks Mangoldt’s function; µ marks M¨obius function; • P is the set of all primes; (m)

• Sn denotes Stirling numbers of first kind;   n is used for binomial coefficients. • k

205

Arithmetic Functions of Other Types First, we give a short list of formulas for ord p n!: [log p n] 

∑

ordp n! =

k=1

[ np ]

ord p n! =

∑ k=1

h

 n ; pk

(4.3.1)

ni ; k

(4.3.2)



(4.3.3)

logp

[ np ] [ mpn ]

 n ; ord p n! = − ∑ ∑ µ(kp) kmp m=1 k=1 ord p n! = 2ord p and especially for ord2 n!:

hni

1 1 ! + [logp n] − 2 2 2

1 [log2 n] ∑ (−1) 2 k=1

ord2 n! =

h

n 2k

i

∑

(−1)

n pk

i

,

(4.3.4)

(4.3.5)



(4.3.6)

k.

k=1

∑

k=1

h

hni 1 ; − [log2 n] + 2 2 2

[log2 n]

ord2 n! =

[log p n]

 n + 2k . 2k+1

Identity (4.3.1) is known as Legendre’s formula for ord p n!. Identity (4.3.2) holds true from (4.3.1) and from the identity [log2 n] 

∑

k=1

 [ np ] h n ni = log . ∑ p k pk k=1

(4.3.7)

Identity (4.3.7) is a particular case of the following identity proved by M. Vassilev (Missana) in [140]: [Ψ−1 (n)] 

∑

k=1



n = Ψ(k)

h

n Ψ(1)

∑ k=1

i

h

Ψ−1

 n i k

,

(4.3.8)

under the assumption that Ψ(x) is an increasing function on [1, ∞) with Ψ(1) > 0, where Ψ−1 is the inverse function of Ψ. Putting in (4.3.8) Ψ(m) = pm and using that Ψ−1 (s) = logp s, we get (4.3.7).

206

J´ozsef S´andor and Krassimir Atanassov Identity (4.3.3) follows from the relation [ np ]



 n − ∑ µ(kp) = [logp n], kp k=1

(4.3.9)

proved by M. Vassilev in [141], and from (4.3.2). We need the representation (−1)[x] = 1 − 2[x] + 4[ 2x ], which is valid for any real x ≥ 0, in order to prove (4.3.4) and after that to use (4.3.1). (4.3.5) is a corollary from (4.3.4) when p = 2, while formulas (4.3.6) and (4.3.9) are proposed by Ivan Mladenov. The proof of (4.3.6) is based on the equality       x x x + 2k − k+1 = , 2k 2 2k+1 which is valid for any real x ≥ 1. Second, we give a short list of formulas for n!. Using the obvious formula n! =

pord p n! =

∏

p∈P , p≤n

∏ pord n!, p

(4.3.10)

p∈P

i.e., the product is finite, because for p > n : prd p n! = 0, we may use each one of formulas (4.3.1), (4.3.2) and (4.3.3) to represent n!. Other representations are given below. n

n! =

(m)

∑ (−1)n−mSn

(4.3.11)

;

m=0

n n n! = exp( ∑ Ψ( )), k k=1

(4.3.12) [x]

where Ψ denotes Chebyshev’s function, i.e., Ψ(x) = ∑ Λ(n); n=1 n

n! = exp

∑ Λ(k). k=1



hn i



k

 kn   n! =   n , k

!

;

(4.3.13)

(4.3.14)

207

Arithmetic Functions of Other Types for k ≥ 2.nn+2;



n k

j.

"

n

n! =

∑ (−1)k

k=1



(x − k)n ,

(4.3.15)

for every complex number x; 

p[log p n]+1 − 1 n! = ∏  + p−1 p∈P ,p≤n

[log p n]

∑

j=1

# " #! n − p[logp n] n − p[logp n] . − p j+1 pj

(4.3.16)

Identities (4.3.11) and (4.3.12) are well-known (see, e.g., [66, 142]). For the proof of (4.3.13) it is necessary firstly to note that ∑ Λ(d) = lnk, where d runs d|k

all divisors of k. Therefore, n

∑ ∑ Λ(d) = ∑ ln k = ln n!.

k=1 d|k

(4.3.17)

k=1

[ kn ]

n

Using the well-known identity ∑ f (k) ∑ g(d) = ∑nk=1 g(k) ∑ f (ks) in the k=1

case f ≡ 1 and g ≡ Λ, we obtain n

s=1

d|k

n

∑ ∑ Λ(d) = ∑ Λ(k). k=1 d|k

k=1

hn i k

.

(4.3.18)

Then (4.3.17) and (4.3.18) yield (4.3.13). Identity (4.3.14) is contained in [143], and identity (4.3.15) is well-known, while the proof of the particular case x = 0 is given in [144]. Identity (4.3.16) can be proved as by induction, as well as directly from the construction from [145], where the 66-th, 67-th, and 68-th Smarandache’s problems (see [146, 147]) were solved. Note that in [146, 147] there are other problems, related to the function “factorial”, too. Another representation of n! is based on (4.3.10), (4.3.4) and the following form of the Chebyshev’s function (see [148]): Ψ(n) =

∑ [logp n]. ln p = ln ∏ p[log

p∈P

p∈P

i.e., eΨ(n) =

∏ p[log

p∈P

p n]

.

p n]

,

208

J´ozsef S´andor and Krassimir Atanassov It has the form: n! =

∏ pord n! = ∏ p p

p∈P

2ord

!+ 12 [log p n]− 21

[ ]

n p 2

[log p n]

∑ (−1)



n pk



k=1

p∈P

∏ (pord p [ 2 ]! )2 . n

=

p∈P

v u u t

r

∏ p[log p n] p∈P

[log p n]

∏ p

∑ (−1)



n pk



k=1

p∈P

√ ∏ ([ n2 ]!)2 . eΨ(n) p∈P =v  , u [log p n] n u k ∑ (−1) p t ∏ p k=1 p∈P

i.e., we received the formula

√ ∏ ([ n2 ]!)2 . eΨ(n) p∈P n! = v  , u [log p n] n u k ∑ (−1) p t ∏ p k=1 p∈P

which admits an interesting interpretation. Namely, it is known that eΨ(n) is the Least Common Multiple LCM of the natural numbers less than n. Therefore, p ∏ ([ n2 ]!)2 . LCM(1, 2, ...,n) p∈P v n! = .   u n [log p n] u k ∑ (−1) p t ∏ p k=1 p∈P

Another form of the same formula is the following:

LCM(1, 2, ..., n) =

(n!)2

[log p n]

∏p

. ([ 2n ]!)4 p∈P

∑ (−1) k=1



n pk



.

Arithmetic Functions of Other Types

209

Of course, the product on the right-hand side of each of the above three formulas is finite and it is restricted up to p ≤ n, because, if p > n, then [logp n] = [log p n]

0 and the sum ∑ vanishes. k=1

4.4. Some Representations Concerning the Product of Divisors of n Let us denote by d(n) the number of all divisors of n. It is well-known (see, e.g., [149]) that p (4.4.1) Pd (n) = nd(n)

and of course, we have

Pd (n) . (4.4.2) n But (4.4.1) is not a good formula for Pd (n), because it depends on function d and to express d(n) we need the prime number factorization of n. Below, we give other representations of Pd (n) and pd (n), which do not use the prime number factorization of n. pd (n) =

Proposition 4.4.1. For n ≥ 1 the representation n

n

Pd (n) = ∏ k[ k ]−[

n−1 k ]

(4.4.3)

k=1

holds. Proof. We have hni





  1,

n−1 θ(n, k) ≡ − =  k k

Therefore,

n

∏ k[

n n−1 k ]−[ k ]

k=1

and Proposition 4.4.1 is proved.

if k is a divisor of n (4.4.4)

0,

otherwise

= ∏ k ≡ Pd (n) k/n

210

J´ozsef S´andor and Krassimir Atanassov Here and further the symbols

∏ • and ∑ • k/n

k/n

mean the product and the sum, respectively, of all divisors of n. The following assertion is obtained as a corollary of (4.4.2) and (4.4.3). Proposition 4.4.2. For n ≥ 1 the representation n−1

n

pd (n) = ∏ k[ k ]−[

n−1 k ]

(4.4.5)

k=1

holds. For n = 1 we have pd (1) = 1. Proposition 4.4.3. For n ≥ 1 the representation h i n ! n k Pd (n) = ∏  n−1  k=1

k

(4.4.6)

!

holds, where here and hereafter we assume that 0! = 1.

Proof. Obviously, we have that h i  n n !  k, h k i = n−1 !  1, k Hence

n

n

!

if k is a divisor of n . otherwise n

k  = ∏ = ∏ k ≡ Pd (n), ∏  n−1 k !

k=1

k

k/n

k/n

since, if k describes all divisors of n, then n describes them, too. k Now (4.4.2) and (4.4.6) yield. Proposition 4.4.4. For n ≥ 2 the representation h i n ! n k i pd (n) = ∏ h n − 1 ! k=2 k

(4.4.7)

211

Arithmetic Functions of Other Types holds. Another type of representation of pd (n) is the following Proposition 4.4.5. For n ≥ 3 the representation n−2

pd (n) = ∏ (k!)θ(n,k)−θ(n,k+1)

(4.4.8)

k=1

holds, where θ(n, k) is given by (4.4.4). Proof. Let r(n, k) = θ(n, k) − θ(n, k + 1). The assertion holds from the fact, that  1, if k is a divisor of n and k + 1 is not a divisor of n      r(n, k) = −1, if k is not a divisor of n and k + 1 is a divisor of n      0, otherwise We are ready to prove the following interesting theorem.

Theorem 4.4.1. For n ≥ 2 the identity n

hni n−1 n n ! = ∏ k ∏ (k!)[ k ]−[ k+1 ] k=2 k=1

(4.4.9)

holds. Proof. By induction. For n = 2 (4.4.9) is true. Let us assume, that (4.4.9) holds for some n ≥ 2. Then we must prove that  n+1  n n+1 n+1 n+1 (4.4.10) ∏ k ! = ∏ (k!)[ k ]−[ k+1 ] k=2 k=1 holds, too. Dividing (4.4.10) by (4.4.9) we obtain h i n + 1 ! n−1 n k ∏ h n i = ∏ (k!)r(n+1,k). k=2 ! k=1 k

(4.4.11)

212

J´ozsef S´andor and Krassimir Atanassov Since, for k = n + 1

and for k = n

 

 n+1 !=1 k

   n+1 n+1 − = 0, k k+1

then (4.4.10) is true, if and only if (4.4.11) is true. Therefore, we must prove (4.4.11) for proving of the Theorem 4.4.1. From (4.4.7), the left-hand side of (4.4.11) is equal to pd (n + 1). From (4.4.8), the right side of (4.4.11) is equal to pd (n + 1), too. Therefore, (4.4.11) is true.

4.5. A Note on Certain Euler–Mascheroni Type Sequences Let (an ) be a sequence of strictly positive real numbers, and construct the new sequence (xn ) defined for n ∈ N by n

xn =

1

∑ ak − log an .

(4.5.1)

k=1

For ak = k (k = 1, 2, ...) one obtains n

xn =

1

∑ k − log n,

k=1

which gives the well-known Euler sequence (or Euler–Mascheroni sequence), having as limit the Euler–Mascheroni constant γ (see [150]). In his book [151] (see p. 42) K. Kashihara posed the problems of convergence or divergence of sequence (xn ) given by (4.5.1) for the particular cases ak = pk , the k-th prime; as well as ak = S(k), the Smarandache function value. We will prove the following theorem. Theorem 4.5.1. The sequence (x1n ) given by n

x1n =

1

∑ pk − log pn

k=1

Arithmetic Functions of Other Types

213

is divergent, being unbounded from below. The sequence (x2n ) given by n

x2n =

1

∑ S(k) − logS(n)

k=1

is divergent, being unbounded from above. Proof. An old result of P. Chebyshev (see e.g., [24]) states that 1

∑ p = log logx + B + O(1),

(4.5.2)

p≤x

where p denotes a prime. This means that ∑ p≤x

1 p

is a convergent sequence. Re-

marking that x1n

! 1 ∑ − log log pn + log log pn − log pn , p≤pn p

=

and by loglog pn − log pn = log logpnpn , since logpnpn → 0 as n → ∞, we get that x1n → −∞ as n → ∞. This proves the first part of the theorem. For the second part, put n = m! Then, since S(n) = min{k ≥ 1 : n|k!}, we have S(n) = m, and x2n =

m

1

∑ k − log m +

k=1

1 , S(k) k 0. Then c 1 1 . < , 2 log pn pn+1 − pn

so bn =

pn+1 − pn 2 < = K. log pn c

This means that the sequence of general terms (bn ) is bounded above. On the other hand, a well-known theorem by E. Westzynthius (see [24], p. 256) states that lim supbn = +∞, i.e., the sequence (bn ) is unbounded. This shows n→∞ that (4.6.1) cannot hold. For the proof of convergence of (xn ) given by (4.6.2), we apply the result

∑

p≤x

pα ∼

x1+α as x → ∞ (α ≥ 0) (1 + α) logx

(4.6.4)

due to T. Sal´at and S. Zn´am (see [24], p. 257). We note that for α = 1, relation (4.6.4) was discovered first by E. Landau. Now, letting α = 1, respectively, α = 2 in (4.6.4), we can write

∑

p≤pn

p∼

p2n as x → ∞ 2 logx

(4.6.5)

216

J´ozsef S´andor and Krassimir Atanassov

and

∑

p≤pn

p2 ∼

p3n as x → ∞ 3 logx

(4.6.6)

Thus, xn =

∑

p

p≤pn

2

!

p4n 3 logx
. 3 . pn ∑ p≤pn p .4 log2 pn

!

4 log pn . . . 3 pn

By (4.6.5) and (4.6.6), the limit of term on the right-hand side of the last equality is 1. Since 43 . logpnpn → 0, we get lim xn = 0. This finishes the proof that n→∞ (4.6.2) is convergent. We must mention that an extension of (4.6.4) is due to M. Kalecki [153]: Let f : (0, +∞) → R be an arbitrary function having the following properties: (a) f (x) > 0; (b) f (x) is a non-decreasing function; f (nx) n→∞ f (x)

(c) for each n > 0, ϕ(n) = lim

exists.

Put s = logϕ(e). Then

∑ f (p) ∼

p≤x

f (x).x 1 . as x → ∞. logx s + 1

(4.6.7)

For f (x) = xα (α ≥ 0) we get ϕ(n) = nα , so s = α and relation (4.6.4) is reobtained. We note that for α = 0, relation (4.6.4) implies the “Prime Number Theorem” (see [24]): x π(x) ∼ as x → ∞. logx By letting f (x) = (g(x))α , where g satisfies conditions (a) - (c), a general sequence of terms n

∑ (g(pi ))α i=1 xn =  α n ∑ g(pi ) i=1

may be studied (via (4.6.7)) in a similar manner and by this reason, we omit the details.

Arithmetic Functions of Other Types

217

4.7. A Generalization of J´ozsef S´andor and Florian Luca’s Theorem In [154] Florian Luca proved a conjecture of J´ozsef S´andor that for every natural number k ≥ 5: (pk+1 − 1)! divides (pk − 2)!pk !, where the prime numbers are denoted by: p1 = 2, p2 = 3, p3 = 5, ... Krassimir Atanassov generalized this result, proving in [155] that for k ≥ 5: (pk+1 − 1)! divides pk−1 !pk ! (of course, for every natural number k ≥ 2: pk+1 − 2 ≥ pk ). This assertion will be valid if we show that Zk ≡ (pk + 1)(pk + 2)...(pk+1 − 1) is a divisor of pk−1!. Of course, none of the numbers pk + 1, pk + 2, ..., pk+1 − 1 is prime. Let us represent each of the members of the product Zk in the form pk + i = qi .ri,

(4.7.1)

where 1 ≤ i ≤ dk ≡ pk+1 − pk − 1, qi ≥ 2 is the smallest prime divisor of pk + i and ri is the greatest proper divisor, that is not obligatory a prime number. From Lemma 1 in [1], it follows that for every natural number n ≥ 59: 2pn >

4 16 pn > pn+1 > pn+2 , 9 3

i.e.,

pn+2 . 2 It is directly seen, that the same is valid for prime numbers pn with n ≥ 3. Therefore, each of the divisors of numbers pk + 1, pk + 2, ..., pk+1 − 1 is met in pk−1 !. The number of the members of the product Zk is pk+1 − pk − 1, but from the above inequality it follows that pn >

pk+1 − pk
((2n)!)2.((2n + 1)2.(2n + 2)2 − (3n + 1).(3n + 2).(3n + 3)) = ((2n)!)2.(16n4 + 21n3 − 2n2 − 9n − 2) > 0.

From (4.7.3) it follows that G is less than (2n)!. Now, we must show that each divisor of G is a divisor of (2n)!, too. It is directly checked that the Lemma holds for each natural number n such that 5 ≤ n ≤ 9. Hence, let n ≥ 10. Let X = {δ(2n + i)|1 ≤ i ≤ n} − {1}. Therefore, X is the set of the natural numbers that are not prime and lie between 2n + 1 and 3n. It is easy to see that between these two numbers there is at least one prime number. Indeed, if pm = 2k − 1 (in this case pm is the greatest prime number less than 2k + 1) and pm+1 = 3k + 1 (in this case pm+1 is the least prime number greater than 3k), then from (4.7.2) we obtain: pm + 1 pm > pm+1 − pm = k + 2 = + 2, 3 2

Arithmetic Functions of Other Types

219

i.e., 2pm > 3pm + 15, which is impossible. Hence, between 2n + 1 and 3n there is at least one prime number and if we put that pk = 2n − 1, then pk+1 < 3n. We must show that the power of each of the prime multipliers of Zk is less than the power with which it is met in pk−1 !. Let the prime number p be a divisor of natural number x. Then, ord p x = y iff py divides x, but py+1 does not divide it. We must prove that n

ord p ∏ δ(2n + i) ≤ ord p ((2n)!)2 = 2ordp (2n)!.

(4.7.4)

i=1

It is well-known that ordp x! =

[log p x] 

∑

k=1

 x . pk

Let s = [logp n] and let everywhere below p ≤ 2n − 1. If s = 0, then p > n and therefore, p is met exactly once among the multipliers of (2n)! and it is met among the multipliers of G once if n < p ≤ 23 n, or never if 32 n < p ≤ 2n − 1, i.e., (4.7.4) is valid. We must note that for each real number x ≥ 0, if 0 ≤ x < if if if

1 3 1 2 2 3

≤x< ≤x
0. By condition n ≥ 10. Let p = 2. Then for s = [log2 n] ≥ 3: n = a0 .2s + a1 .2s−1 + ... + as , where a0 = 1, 0 ≤ ai ≤ 1 for 1 ≤ i ≤ s, 3n 3(a0 .2s + a1 .2s−1 + ... + as ) 3(2s + 2s−1 + ... + 1) = ≤ 2s+2 2s+2 2s+2

220

J´ozsef S´andor and Krassimir Atanassov =

i.e.

3(2s+1 − 1) < 2, 2s+2 

 3n ≤ 1, 2s+2

and s + 2 ≥ [log2 3n] ≥ [log2 2n] = s + 1 and

n

A2 ≡ 2ord2 (2n)! − ord2 ∏ δ(2n + i) ≥ 2ord2 (2n)! − ord2 (3n)! i=1

 [log2 3n]   3n 2n − ∑ =2 ∑ k 2 2k k=1 k=1  s+2   s+1  2n 3n ≥2∑ k −∑ k 2 k=1 2 k=1      s+1   2n 3n 3n = ∑ 2. k − k − s+2 2 2 2 k=1      s+1 2(a0 .2s + a1 .2s−1 + ... + as) 3.(a0 .2s + a1 .2s−1 + ... + as) ≥ ∑ 2. − −1 2k 2k k=1 [log2 2n] 

s+1

=

∑ (4.(a0.2s−k + a1.2s−1−k + ... + as−k) − 3.(a0.2s−k + a1.2s−1−k + ... + as−k))

k=1

 s+1   2(as−k+1.2k−1 + as−k+2 .2k−2 + ... + as) + ∑ 2. 2k k=1   3.(as−k+1.2k−1 + as−k+2.2k−2 + ... + as) − −1 2k

(from (4.7.5) and a0 ≥ 1) s+1

≥ ( ∑ a0 .2s−k + a1 .2s−1−k + ... + as−k ) − (s + 1) − 1 k=1

s+1

≥ ( ∑ 2s−k ) − (s + 2) k=1

s

≥ 2 − 1 − (s + 2) = 2s − (s + 3) > 0.

Arithmetic Functions of Other Types

221

Let below p ≥ 3. Then, for s = [logp n] ≥ 1. We have two cases: s = 1 and s ≥ 2. Let s = 1. Therefore, n < p2 , i.e., n = a.p + b for 1 ≤ a ≤ p − 1 and 0 ≤ b ≤ p − 1. From n ≥ 10 it follows that p ≥ 5. Now, if a ≥ 3, then       2(ap + b) 3(ap + b) 3(ap + b) − − Ap = 2 p p p2       2b b 3(ap + b) = a+2 − − ≥ a − 3 ≥ 0. p p p2 If a = 1, then p + b ≥ 10 and hence p ≥ 7. Therefore,       2(p + b) 3(p + b) 3(p + b) Ap = 2 − − p p p2       2b b 3(p + b) = 1+2 − − ≥ 1 − 1 = 0. p p p2 If a = 2, then from n ≥ 10 it follows that p ≥ 5. Let p = 5. Then   3(ap + b) =1 p2 and

   2b 3b Ap = 2 + 2 − − 1 ≥ 0. p p 

If p > 5, i.e., p ≥ 7, then  and

 3(ap + b) ≤1 p2



   2b 3b Ap = a + 2 − ≥ 2 − 1 − 1 = 0. p p

Let s ≥ 2. Then n = a0 .ps + a1 .ps−1 + ... + as, where 1 ≤ a0 ≤ p − 1, 0 ≤ ai ≤ p − 1 for 1 ≤ i ≤ s, 3(a0 .ps + a1 .ps−1 + ... + as ) 3(p − 1)(ps + ps−1 + ... + 1) 3n = ≤ ps+1 ps+1 ps+1

222

J´ozsef S´andor and Krassimir Atanassov =

i.e.

3(ps+1 − 1) < 3, ps+1 

 3n ≤ 2, ps+1

and s + 1 ≥ [logp 3n] ≥ [logp 2n] ≥ s and n

A p ≡ 2ordp (2n)! − ord p ∏ δ(2n + i) ≥ 2ordp (2n)! − ord p (3n)! i=1

 [log p 3n]   3n 2n − ∑ =2 ∑ k p pk k=1 k=1  s+1   s  2n 3n ≥2∑ k −∑ k k=1 p k=1 p        s 2n 3n 3n = ∑ 2. k − k − s+1 p p p k=1      s 2(a0 .ps + a1 .ps−1 + ... + as ) 3.(a0.ps + a1 .ps−1 + ... + as ) ≥ ∑ 2. − −2 pk pk k=1 [log p 2n] 

s

= ∑ (4.(a0.ps−k +a1 .ps−1−k +...+as−k)−3.(a0.ps−k +a1 .ps−1−k +...+as−k)) k=1

   2(as−k+1.pk−1 + as−k+2 .pk−2 + ... + as ) + ∑ 2. pk k=1   3.(as−k+1.pk−1 + as−k+2 .pk−2 + ... + as ) −2 − pk s

(from (4.7.5) and a0 ≥ 1) s

≥ ( ∑ a0 .ps−k + a1 .ps−1−k + ... + as−k ) − s − 2 k=1

s

≥ ( ∑ ps−k ) − (s + 2) k=1

Arithmetic Functions of Other Types = Therefore,

223

ps − 1 − (s + 2) ≥ ps−1 − (s + 2). p−1 A p ≥ ps−1 − (s + 2).

(4.7.6)

When s ≥ 3, or s = 2, and p ≥ 5, then, obviously, A p ≥ 0. When s = 2, and p = 3 the inequality (4.7.6) is not valid, because above we used a very powerful inequality, while in this case: n = 9a + 3b + c, where 1 ≤ a ≤ 2, 0 ≤ b, c ≤ 2 and, as above:     18a + 6b + 2c 18a + 6b + 2c +2 − (9a + 3b + c) Ap = 2 3 9     9a + 3b + c 9a + 3b + c − − 3 9       2c 6b + 2c = 2 6a + 2b + 2 + 2 2a + − (9a + 3b + c) − (3a + b) − a 3 9 ≥ 3a − c ≥ 3 − c > 0. Therefore, in all cases A p ≥ 0. Thus, the Lemma 4.7.1 is proved. Obviously, the validity of the generalized J´ozsef S´andor and Florian Luca’s result follows from the Lemma 4.7.1. The present result is published in [155].

References [1] Atanassov, K., J. S´andor. On some modifications of the ϕ and σ functions, Comptes Rendus de l’Academie bulgare des Sciences [Reports of the Bulgarian Academy of Sciences]. Vol. 42 (1989), 55-58. [2] Mullin A., Problems. Bulletin of Number Theory and Related Topics, Vol. VI (1982), No. 3, 218-219. [3] Atanassov K. On one problem of A. Mullin. Bulletin of Number Theory and Related Topics. Vol. VIII (1984), No. 3, 1-5. [4] Atanassov K. Short proof of a hypothesis of A. Mullin. Bulletin of Number Theory and Related Topics. Vol. IX (1985), No. 2, 9-11. [5] Atanassov K. Some properties of ϕ and σ functions. Comptes Rendus de l’Academie bulgare des Sciences [Reports of the Bulgarian Academy of Sciences]. Tome 40 (1987), No. 4, 35-38. [6] Atanassov K., S. Danchev. A modification of A. Mullin’s inequality. Notes on Number Theory and Discrete Mathematics. Vol. 3 (1997), No. 1, 58-60. [7] Dwight, H.B. Tables of Integrals and Other Mathematical Data, The MacMillan Co., New York, 1961. [8] Atanassov K. New integer functions, related to ϕ and σ functions. Bull. of Number Theory and Related Topics. Vol. XI (1987), No. 1, 3-26. [9] Atanassov K. Remarks on ϕ, σ and ψ functions. Mathematical Forum. Vol. 3 (2001), No. 2, 53-54.

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About the Authors J´ozsef S´andor, PhD Professor, Department of Mathematics Babes¸-Bolyai University, Cluj, Romania Email: [email protected] Prof. J´ozsef S´andor was born on 19.11.1956 in Farcd (Forteni), Jud. Harghita, Romania. He studied in Department of Mathematics and Physics in Babes¸-Bolyai University of Cluj, Romania, and defended his PhD dissertation in 1998 in the subject of generalized convexity and its applications. His main fields of interests are essentially in Number theory, Mathematical analysis, Special functions, Classical geometry and History of mathematics. He published more than 1000 scientific papers and 500 methodicalscientific papers as well as 12 books in various fields of mathematics. His best known books are Handbook of Number Theory I, II, published by Springer (1996, 2004, 2006). He is the editor or associate editor of more than 20 international journals. His main interests in Number theory are arithmetical functions in algebra, analysis and geometry; asymptotics of arithmetic functions; inequalities in number theory; irrationality or transcendence of series and special numbers; special functions related to number theory and history of number theory and related fields. Krassimir Atanassov, PhD Professor, Corresponding Member of the Bulgarian Academy of Sciences Department of Bioinformatics and Mathematical Modelling Institute of Biophysics and Biomedical Engineering Bulgarian Academy of Sciences, Sofia, Bulgaria Email: [email protected] Prof. Krassimir Atanassov was born on 23.03.1954 in Burgas, Bulgaria. He studied in Department of Mathematics in Sofia University and defended dissertations for PhD in 1986, Doctor of Technical Sciences in 1997, and Doctor of Mathematical Sciences in 2000. Since 1998 he is a Full Professor in the Department of Bioinformatics and Mathematical Modelling of Institute of Biophysics and Biomedical Engineering of

238

About the Authors

the Bulgarian Academy of Sciences, since 2012 a Corresponding Member of Bulgarian Academy of Sciences, since 2013 fellow of IFSA, since 2018 Senior of IEEE. He has more than 700 papers in journals, 300 conference reports and 35 monographs, and is member of the Editorial Boards of more than 20 international journals. His research interests are in the areas of fuzzy sets, where he introduced the concept of intuitionistic fuzzy sets, and investigated its basic properties and applications, mathematical modelling by generalized nets and their applications in artificial intelligence, systems theory, medicine, economics, etc., and number theory. In the area of Number Theory his research interests include generalizations the Fibonacci sequence, properties of arithmetic functions, formulation and solution to extremal and combinatorial problems.

Index (α, β)-superunitary perfect number, 89 σl , viii (m, k)-perfect number, 61 βs , vii +-lovely-pair numbers, 58 ϕ (Euler’s totient function), viii +-perfect number, 58 d, viii 2ω -perferct number, 86 ϕ∗ , 13 [G, η]-perfect number, 91 σ∗ , 13 ord p , 196 d ∗ , 16 f − n-hyperperfect number, 106 EF (s) , 159 g-additive perfect number, 78 EFs , 154 g-multiplicative perfect number, 78 RF (s) , 159 (s) k-balanced number, 109 RF+ , 164 k-multiplicatively perfect number (kRFs , 154 m-perfect number), 68 RF+,s , 164 n-gonal number, 176 arithmetic function ψ, 169 n-hyperperfect number, 106 Atanassova, V., 177, 179, 183 Avanesov, E., 184 Alaoglu, L., 46 almost perfect number, 89 base of a sequence, 170 amicable numbers, 53 bi-unitary divisor, 95 arithmetic function bi-unitary hatmonic number, 96 Ω, vii, ix Bi-unitary hyperperfect number, 106 ϕs (Jordan’s totient function), viii bi-unitary perfect number, 95 λ (Liouville’s function), ix Bunyakowski, V., 159, 166 µ (M¨obius function), ix Cˆırtraje, V., 166 γ, vii Cauchy, A.-L., 159, 166 ω, vii Cauchy-Bunyakowski inequality, 159, π, ix 166 ψ (Dedekind’s function), viii centred cube number, 178 σ, vii Chebyshev, P.L., 198, 199, 204

240

Index

Clarke, J., 175 Cohen, G.L., 30, 61, 94, 95 converse factor, 126

Jacobsthal, E., 175 Jordan totient function Js , 155 Jordan, C., viii, 73, 155

Dedekind, R., viii, 22, 60 Deng, M., 94 digital arithmetical function ϕ, 167

Kannan, V., 22 Kanold, H.J., 29, 40, 53, 55, 57, 59, 66, 95 Kashihara, K., 204, 205 Kiyoshi, K., 181 Klamkin, M.S., 186, 190, 193

Erd¨os, P., 31, 46 Euclid, 29, 53, 69, 71 Euler’s polynomials, 175 Euler, L., viii, 18, 19, 22, 26, 27, 29, 43, 53, 59, 69, 71, 79, 117, 161, 175, 204–206 Euler–Mascheroni sequence, 204 exponential divisor (e-divisor), 105 extension factor, 139 factorial n m, 180 Fermat, P., 41–45, 147 Fibonacci sequence, 170 Fibonacci, L., 170 Ford, K., 30

Landau, E., 207 least common multiple, ix Legendre, A.-M., 197 Lehmer, D.N., 170 Liouville, J., ix Lord, G., 95 lovely-pair numbers, 53 Luca, F., 30, 208, 214 Lucas sequence, 170 Lucas, E., 170 Lucas–Lehmer sequence, 170

M¨obius, A., ix, 80, 124, 127, 196 Makowski, A., 30, 31, 48, 49 Gato, T., 95 Manabu, N., 181 greatest common divisor, ix Mascheroni, L., 204, 205 Grytczuk, A., 30 Mersenne, M., 29, 38, 39, 41, 42, 46, Gupta, R., 30 49, 53, 54, 69, 70 Mertens, F., 206 Hagis, P., 95, 106 Mills, H., 94, 97 Hardy, G.H., 71, 190 Minkowski, H., 162 harmonic divisor number (Ore numMitrinovi´c, D., 149 ber), 92 Mitronovi´c, D., 186 hexagonal number, 178 Mladenov, I., 198 modified f − n-hyperperfect number, irrational factor, 123 106 Jacobsthal number, 175 modified hyperperfect number, 106

Index Mullin, A., 1–3 multiplicatively perfect number (mperfect number), 66 multiplicatively super-perfect number (m-super-perfect number), 66

241

Schinzel, A., 30, 31, 48, 49 set function set, vii Shannon, A.G., 175 Shibata, S., 95 Sitaramaiah, V., 30 Sivaramakrishnan, R., 28 Nageswara Rao, K., 95 Smarandache, F., 199, 204, 205 Nicolas, J.L., 72 Soril, R.M., 94 non-homogenous sequence sequence, square pyramid number, 178 175 Srikanth, R., 22 stella octangula number, 178 octahedral number, 178 Subbarao, M.V., 30, 92, 94, 95, 109 Okeya, K., 95 Omega-perfect number (ΩF -perfect super lovely-pair numbers, 53 super perfect number, 53 number), 78 super-perfect numbers, 54 Ore, O., 92, 96, 97 superunitary perfect number, 89 Suryanarana, D., 95 Padovan sequence, 172 Suryanarayana, D., 29, 40, 53, 55, 57, Padovan, R., 172 59, 66 Panaitolop, L., 152, 153 Pe, J.L., 78 te Riele, H.J.J., 61 Pell sequence, 171 Tetrabonacci sequence, 172 Pell, J., 171 triangular number, 176 Pentabonacci sequence, 173 Tribonacci sequence, 171 pentagonal number, 178 truncated octahedral number, 178 pentatope number, 178 truncated tetrahedral number, 178 perfect number, 53 Pomerance, C., 30, 94 Vassilev-Missana, M., 77, 197 Popoviciu, T., 166 Venkataraman, C., 28 Vitek, V., 31 Ramanujan, S., 71, 190 restrictive factor, 127 rhombic dodecahedral number, 178 Riemann zeta-function, 77 Riemann, B., 77, 111, 161 Robin, G., 72 Ruzsa, I.Z., 53, 58, 59 Sal´at, T., 207

Wall, C., 17, 37, 95, 96, 100, 107 Warren, L.J., 95 Weierstrass inequality, 40 Weierstrass, K., 18, 40, 45 Westzynthius, E., 206 Wiegert, S., 72 Wojtowicz, M., 30 Zn´am, S., 207