Research On Smarandache Problems In Number Theory. II (Proceedings)

Page 1

ZHANG WENPENG, LI JUNZHUANG, LIU DUANSEN (EDITORS) RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY (VOL. II) PROCEEDINGS OF THE FIRST NORTHWEST CONFERENCE ON NUMBER THEORY

Hexis Phoenix, AZ 2005

EDITORS:

ZHANG WENPENG DEPARTMENT OF MATHEMATICS NORTHWEST UNIVERSITY XI’AN, SHAANXI, P.R.CHINA LI JUNZHUANG, LIU DUANSEN INSTITUTE OF MATHEMATICS SHANGLUO TEACHER’S COLLEGE SHANGLUO, SHAANXI, P.R.CHINA

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY (VOL. II) PROCEEDINGS OF THE FIRST NORTHWEST CONFERENCE ON NUMBER THEORY

Hexis Phoenix, AZ 2005

This book can be ordered in a paper bound reprint from: Books on Demand ProQuest Information & Learning (University of Microfilm International) 300 N. Zeeb Road P.O. Box 1346, Ann Arbor MI 48106-1346, USA Tel.: 1-800-521-0600 (Customer Service) http://wwwlib.umi.com/bod/search/basic

Peer Reviewers: Shigeru Kanemitsu, Graduate School of Advanced Technology, University of Kinki, 11-6 Kayanomori, Iizuka 820-8555, Japan. Zhai Wenguang, Department of Mathematics, Shandong Teachers' University, Jinan, Shandong, P. R. China. Guo Jinbao, College of Mathematics and Computer Science, Yanan University, Yanan, Shaanxi, P. R. China.

Copyright 2005 by Hexis and Zhang Wenpeng, Li Junzhuang, Liu Duansen, and all authors for their own articles

Many books can be downloaded from the following E-Library of Science: http://www.gallup.unm.edu/~smarandache/eBooks-otherformats.htm

ISBN: 1-931233-99-3 Standard Address Number: 297-5092 Printed in the United States of America

Contents

Dedication Preface

v xi

On the Smarandache m-th power residues Zhang Wenpeng

1

An equation involving Euler’s function Yi Yuan

5

On the Function ϕ(n) and δk (n) Xu Zhefeng

9

On the mean value of the Dirichlet’s divisor function in some special sets Xue Xifeng

13

A number theoretic function and its mean value Ren Ganglian

19

On the additive k -th power complements Ding Liping

23

On the Smarandache-Riemann zeta sequence Li Jie

29

On the properties of the hexagon-numbers Liu Yanni

33

On the mean value of the k -th power part residue function Ma Jinping

37

On the integer part of the m-th root and the k -th power free number Li Zhanhu1,2

41

A Formula For Smarandache LCM Ratio Sequence Wang Ting

45

On the m-th power complements sequence Yang Hai, Fu Ruiqin

47

On the Smarandache ceil function and the Dirichlet divisor function Ren Dongmei

51

viii

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

On a dual function of the Smarandache ceil function Lu Yaming

55

On the largest m-th power not exceeding n Liu Duansen and Li Junzhuang

59

On the Smarandache back concatenated odd sequences Li Junzhuang and Wang Nianliang

63

On the additive hexagon numbers complements Li chao and Yang Cundian

71

On the mean value of a new arithmetical function Yang Cundian and Liu Duansen

75

On the mean value of a new arithmetical function Zhao Jian and Li Chao

79

On the second class pseudo-multiples of 5 sequences Gao Nan

83

A class of Dirichlet series and its identities Lou Yuanbing

87

An arithmetical function and the perfect k -th power numbers Yang Qianli and Yang Mingshun

91

An arithmetical function and its mean value formula Ren Zhibin and Zhao Xiaopeng

95

On the square complements function of n! Fu Ruiqin, Yang Hai

99

On the Smarandache function Wang Yongxing

103

Mean value of the k -power complement sequences Feng Zhiyu

107

On the m-th power free number sequences Chen Guohui

111

A number theoretic function and its mean value Du Jianli

115

On the additive k -th power part residue function Zhang Shengsheng

119

An arithmetical function and the k -th power complements Huang Wei

123

On the triangle number part residue of a positive integer Ji Yongqiang

127

Contents

ix

An arithmetical function and the k -full number sequences Zhao Jiantang

131

On the hybrid mean value of some special sequences Li Yansheng and Gao Li

135

On the mean value of a new arithmetical function Ma Junqing

143

This book is dedicated to Professor Florentin Smarandache, who listed many new and unsolved problems in number theory.

Preface Arithmetic is where numbers run across your mind looking for the answer. Arithmetic is like numbers spinning in your head faster and faster until you blow up with the answer. KABOOM!! Then you sit back down and begin the next problem. Alexander Nathanson Mathematics is often referred to as a servant of science and number theory as a queen of mathematics, by which we are to understand that number theory should contribute to the development of science by continuously supplying challenging problems and illustrative examples of its own discipline in anticipation of later practical use in the real world. The research on Smarandache Problems plays a key role in the development of number theory. Therefore, many mathematicians show their interest in the Smarandache problems and they conduct much research on them. Under such circumstances, we published the book <RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY>, Vol. I, in September, 2004. That book stimulated more Chinese mathematicians to pay attention to Smarandache conjectures, open and solved problems in number theory. The First Northwest Number Theory Conference was held in Shangluo Teacher's College, China, in March 2005. One of the sessions was dedicated to the Smarandache problems. In that session, several professors gave a talk on Smarandache problems and many participants lectured on Smarandache problems both extensively and intensively. This book includes 34 papers, most of which were written by participants of the above mentioned conference. All these papers are original and have been refereed. The themes of these papers range from the mean value or hybrid mean value of Smarandache type functions, the mean value of some famous number theoretic functions acting on the Smarandache sequences, to the convergence property of some infinite series involving the Smarandache type sequences. We sincerely thank all the authors and the referees for their important contributions. Thanks are also due to Dr. Xu Zhefeng for his effort of making files of LaTeX style. The last, but not the least, thanks are due to the teachers and students of Shangluo Teacher's College in China for their great help in our successful conference. August 10, 2005 Zhang Wenpeng, Li Junzhuang, Liu Duansen

ON THE SMARANDACHE M -TH POWER RESIDUES∗ Zhang Wenpeng Department of Mathematics, Northwest University, Xi’an, Shaanxi, P.R.China wpzhang@nwu.edu.cn

Abstract

For any positive integer n, let φ(n) be the Euler function, and am (n) denotes the Smarandache m-th power residues function of n. The main purpose of this paper is using the elementary method to study the number of the solutions of the equation φ(n) = am (n), and give all solutions for this equation.

Keywords:

Arithmetical function; Equation; Solutions.

§1. Introduction For any positive integer n, let n = pα1 1 pα2 2 · · · pαk k denotes the factorization of n into prime powers. The Smarandache m-th power residues function am (n) are defined as am (n) = pβ1 1 pβ2 2 · · · pβk k ,

βi = min{m − 1, αi }, i = 1, 2. · · · , k.

In problem 65 of [1], Professor F.Smarandache asked us to study the properties of this function. Let φ(n) denotes the Euler function. That is, φ(n) denotes the number of all positive integers not exceeding n which are relatively prime to n. It is clear that φ(n) and am (n) both are multiplicative functions. In this paper, we shall use the elementary method to study the solutions of the equation involving these two functions, and give all solutions for it. That is, we shall prove the following: Theorem. Let m be a fixed integer with m ≥ 2. Then the equation φ(n) = am (n) have m + 1 solutions, namely n = 1, 2m , 2α 3m ,

α = 1, 2, · · · , m − 1.

§2. Proof of the theorem In this section, we shall complete the proof of the theorem. Let n = pα1 1 pα2 2 · · · pαk k denotes the factorization of n into prime powers, then from the definitions of φ(n) and am (n) we have am (n) = pβ1 1 pβ2 2 · · · pβk k , ∗ This

βi = min{m − 1, αi }, i = 1, 2. · · · , k

work is supported by N.S.F. of P.R.China (10271093, 60472068)

(1)

2

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

and φ(n) = pα1 1 −1 (p1 − 1)pα2 2 −1 (p2 − 1) · · · pαk k −1 (pk − 1).

(2)

It is clear that n = 1 is a solution of the equation φ(n) = am (n). If n > 1, then we will discuss the problem in four cases: (i) If ∃ αi > m, then we have αi − 1 ≥ m, max {βj } ≤ m − 1. Combining 1≤j≤k

(1) and (2), we know that φ(n) 6= am (n). That is, there is no any solution satisfied φ(n) = am (n) in this case; (ii) If α1 = α2 = · · · = αk = m, then we have am (n) = pm−1 pm−1 · · · pm−1 1 2 k and φ(n) = pm−1 (p1 − 1)pm−1 (p2 − 1) · · · pm−1 (pk − 1). 1 2 k It is clear that only n = 2m is a solution of the equation in this case; (iii) If max {αi } < m, then from (1) and (2), and noting that βk = αk > 1≤i≤k

αk − 1, we know that φ(n) 6= am (n) for all n in this case; (iv) If ∃ i, j such that αi = m and αj < m, then from (1), (2) and equation φ(n) = am (n), we get p1 = 2. If not, then φ(n) is an even number, but am (n) is an odd number. Noting that if αk < m, then φ(n) 6= am (n), so we have αk = m and φ(n) = 2α1 −1 pα2 2 −1 (p2 − 1) · · · pm−1 (pk − 1) k α2 α1 m = am (2 )am (p2 ) · · · am (pk ) = am (n). If α1 = m, then n = 2m is the case of (ii). Hence, α1 < m. Now from the equation, we have pα2 2 −1 (p2 − 1) · · · pm−1 (pk − 1) = 2am (pα2 2 ) · · · am (pm k ). k

(3)

Noting that 2|(pi − 1) if i > 1, from (3) we can deduce that only one term with the form pi − 1 in the left side of (3). That is, n = 2α pm . From φ(2α pm ) = am (2α pm ) ⇐⇒ pm−1 (p − 1) = 2pm−1 , we get n = 2α 3m , 1 ≤ α < m. That is, n = 2α 3m , (1 ≤ α < m) are all the solutions of φ(n) = am (n) in this case. Now combining the above four cases we may immediately get all m + 1 solutions of equation φ(n) = am (n), namely n = 1, 2m , 2α 3m ,

α = 1, 2. · · · , m − 1.

This completes the proof of Theorem.

On the Smarandache m-th power residues1

3

References [1] F. Smarandache. Only Problems, Not Solutions. Chicago: Xiquan Publishing House, 1993, pp54. [2] Tom M. Apostol. Introduction to Analytic Number Theory. New York: Springer-Verlag, 1976.

AN EQUATION INVOLVING EULER’S FUNCTION∗ Yi Yuan Research Center for Basic Science , Xi’an Jiaotong University, Xi’an, Shaanxi, P.R.China yuanyi@mail.xjtu.edu.cn

Abstract

In this paper, we use the elementary methods to study the number of the solutions of an equation involving the Euler’s function φ(n), and give an interesting identity.

Keywords:

Multiplicative function; Arithmetical property; Identity.

§1. Introduction For any positive integer n ≥ 1, the Euler function φ(n) is defined to be the number of all positive integers not exceeding n which are relatively prime to n. We also define J(n) as the number of all primitive Dirichlet’s characters mod n. Let A denotes the set of all positive integers n satisfying the equation φ2 (n) = nJ(n). In this paper, we using the elementary methods to study the convergent properties of a new Dirichlet’s series involing the solutions of the equation φ2 (n) = nJ(n), and give an interesting identity for it. That is, we shall prove the following conclusion: Theorem. For any real number s > 12 , we have the identity ∞ X 1 n=1 n∈A

ns

=

ζ(2s)ζ(3s) , ζ(6s)

where ζ(s) is the Riemann zeta-function. Taking s = 1 and 2, and note that ζ(2) = π 2 /6, ζ(4) = π 4 /90, ζ(6) = 6 π /945, ζ(12) = 691π 12 /638512875, we may immidiately deduce the following identities: ∞ X 1 n=1 n∈A

∗ This

315 = 4 ζ(3) and n 2π

∞ X 1 n=1 n∈A

work is supported by N.S.F. of P.R.China (10271093)

n2

=

15015 1 . 1382 π 2

6

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

§2. Proof of the theorem In this section, we will complete the proof of the theorem. First note that both φ(n) and J(n) are multiplicative functions, and if n = pα , then J(p) = p − 2, φ(p) = p − 1, J(pα ) = pα−2 (p − 1)2 and φ(pα ) = pα−1 (p − 1) for all positive integer α > 1 and prime p. So we will discuss the solutions of the equation φ2 (n) = nJ(n) into three cases. (a) It is clear that n = 1 is a solution of the equation φ2 (n) = nJ(n). (b) If n > 1, let n = pα1 1 pα2 2 · · · pαs s denotes the prime powers decomposition of n with all αi ≥ 2 (i = 1, 2, · · · · · · , k), then we have J(n) = pα1 1 −2 (p1 − 1)2 · · · pαk k −2 (pk − 1)2 and 2

φ2 (n) = pα1 1 −1 (p1 − 1)pα2 2 −1 (p2 − 1) · · · pαk k −1 (pk − 1)

.

For this case, n is also a solution of the equation φ2 (n) = nJ(n). αr+1 (c) If n = p1 p2 · · · pr pr+1 · · · pαk k with p1 < p2 < · · · < pr and αj > 1, j = r + 1, r + 2, · · · , k, then from (b) and the definition of φ(n) and J(n) we have φ2 (n) = nJ(n) if and only if φ2 (p1 p2 · · · pr ) = p1 p2 · · · pr J(p1 p2 · · · pr ) or (p1 − 1)2 (p2 − 1)2 · · · (pr − 1)2 = p1 p2 · · · pr (p1 − 2)(p2 − 2) · · · (pr − 2). It is clear that pr can not divide (p1 −1)2 (p2 −1)2 · · · (pr −1)2 . So the equation φ2 (n) = nJ(n) has no solution in this case. Combining the above three cases we may immediately obtain the set of all solutions of the equation φ2 (n) = nJ(n) is n = pα1 1 pα2 2 · · · pαs s with all αi > 1 and 1. That is, A is the set of all square-full numbers and 1. Now we define the arithmetical function a(n) as follows:

a(n) =

n ∈ A, otherwise.

1, if 0, if

For any real number s > 0, it is clear that ∞ X 1 n=1 n∈A ∞ X 1

ns

<

∞ X 1 n=1

ns

,

is convergent if s > 1. So for s ≥ 2, from the Euler product ns formula (see [2]),we have

and

n=1

An equation involving Euler’s function1 ∞ X 1 n=1

ns

7 !

=

a(p2 ) a(p3 ) 1 + 2s + 3s + · · · p p

Y p

=

Y p

=

Y p

= =

Y

=

Y

p

p

! 1 ps )

p2s − ps p3s + 1 ps (p2s − 1) p6s − 1 ps (p2s − 1)(p3s − 1) 1−

Y p

=

1 1 + 2s p (1 −

Y p2s − ps + 1 p

=

1 1 + 3s + · · · 2s p p

1+

(1 −

1 p6s

1 )(1 p2s

−

1 ) p3s

ζ(2s)ζ(3s) , ζ(6s)

where ζ(s) is the Riemann zeta-function, and

Y

denotes the product over all

p

primes. This completes the proof of the theorem.

References [1] F.Smaradache. Only problems, not solutions, Xiquan Publishing House, Chicago, 1993. [2] Tom M. Apostol. Introduction to Analytic Number Theory. New York: Springer-Verlag, 1976.

ON THE FUNCTION ϕ(N ) AND δK (N ) ∗ Xu Zhefeng Department of Mathematics, Northwest University, Xi’an, Shaanxi, P.R.China zfxu@nwu.edu.cn

Abstract

The main purpose of this paper is using the elementary method to study the divisibility of δk (n) by ϕ(n), and find all the n such that ϕ(n) | δk (n).

Keywords:

Euler function; Arithmetical function; Divisibility.

§1. Introduction and main results For a fixed positive integer k and any positive integer n, we define a new arithmetic function as following: δk (n) = max{d | d | n, (d, k) = 1} If n ≥ 1 the Euler function ϕ(n) is defined to be the number of positive integers not exceeding n which are relatively prime to n; thus, ϕ(n) =

n X 0

1

k=1

where n.

n X 0

indicates that the sum is extended over those k relatively prime to

k=1

In this paper, we shall study the divisibility of δk (n) by ϕ(n), and find all the n such that ϕ(n) | δk (n). In fact, we shall prove the following result: Theorem. 0, α, β ∈ N .

ϕ(n) | δk (n) if and only if n = 2α 3β , where α > 0, β ≥

§2. Proof of the theorem In this section, we will complete the proof of the theorem. First we need the following Lemma. ∗ This

work is supported by N.S.F. of P.R.China (60472068)

10

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

Lemma For n ≥ 1, we have ϕ(n) = n

Y p|n

1 . 1− p

Proof. See Theorem 2.4 in reference [2]. Now we use the above lemma to complete the proof of Theorem. We will discuss it in two cases. I. (n, k) = 1, then δk (n) = n. Let n = pα1 1 pα2 2 · · · pαs s is the factorization of n into prime powers. From the Lemma, we can write ϕ(n) = pα1 1 −1 (p1 − 1)pα2 2 −1 (p2 − 1) · · · pαs s −1 (ps − 1), If ϕ(n) | δk (n), then pα1 1 −1 (p1 − 1)pα2 2 −1 (p2 − 1) · · · pαs s −1 (ps − 1) | n, that is (p1 − 1)(p2 − 1) · · · (ps − 1) | p1 p2 · · · ps , from the above formula, we are sure of i) p1 = 2, if not, (p1 − 1)(p2 − 1) · · · (ps − 1) is even but p1 p2 · · · ps odd and ii) s ≤ 2, since 2|(pi − 1) (i = 2, 3, · · · , s) but p2 p3 · · · ps can not be divided by 2. If s = 1, then n = 2α , α ≥ 1. If s = 2, then n = 2α 3β , α ≥ 1, β > 1. Thus, we can obtain that n = 2α 3β , α ≥ 1, β ≥ 0 such that ϕ(n) | δk (n) when (n, k) = 1. II. (n, k) 6= 1, we can write n = n1 · n2 , where (n1 , k) = 1 and (n1 , n2 ) = 1, then δk (n) = n1 , and ϕ(n) = ϕ(n1 )ϕ(n2 ), If ϕ(n) | δk (n), that means ϕ(n1 )ϕ(n2 ) | n1 , that is ϕ(n1 ) | n1

(1)

ϕ(n2 ) | n1

(2)

n1 = 2α1 3β1 ,

(3)

from (1), we can get

On the Function ϕ(n) and δk (n) 1

11

where α1 ≥ 1, β1 ≥ 0, α1 , β1 ∈ N . combining (3)and(4), we can easily get ϕ(n2 ) | 2α1 3β1 , this means ϕ(n2 ) = 1. Otherwise (n1 , n2 ) 6= 1. So n2 = 1 Altogether, whether (n, k) = 1 or not, we can obtain n = 2α 3β , α ≥ 1, β ≥ 0 such that ϕ(n) | δk (n). This completes the proof of the theorem.

References [1] Ibstedt. Surfinig On the Ocean of Numbers- A Few Smarandache Notions and Similar Topics, Erhus University Press, New Mexico. [2] T.M.Apostol. Introduction to Analytic Number Theory, Springer-verlag, New York, 1976. [3] F.Mertens. Ein Beitrag Zur Analytischen Zahlentheorie, Crelle’s Journal, 78(1874), 46-62.

ON THE MEAN VALUE OF THE DIRICHLET’S DIVISOR FUNCTION IN SOME SPECIAL SETS∗ Xue Xifeng Department of Mathematics, Northwest University, Xi’an, Shaanxi, P.R.China

Abstract

The main purpose of this paper is using the analytic methods to study the mean value properties of the Dirichlet’s divisor function in some special sets, and give several interesting asymptotic formulae for them.

Keywords:

Divisor function; Mean value; Asymptotic formula.

§1. Introduction For any positive integer n and k ≥ 2, the k-th power complement function bk (n) of n is the smallest positive integer such that nbk (n) is a perfect k-th power. In problem 29 of reference [1], Professor F.Smarandache asked us to study the properties of this function. About this problem, some people had studied it before, and obtained some interesting conclusions, see references [4] and [5]. In this paper, we define two new sets B = {n ∈ N, bk (n) | n} and C = {n ∈ N, n | bk (n)}. Then we use the analytic methods to study the mean value properties of the Dirichlet’s divisor function d(n) acting on these two special sets, and obtain two interesting asymptotic formulae for them. That is, we shall prove the following : Theorem 1 . For any real number x ≥ 1, we have the asymptotic formula 1

1 1 mx m d(n) = m+1 R p m · f (log x) + +O x 2m +ε , ζ (2) n≤x X

n∈B

where

1

R pm

=

Y

1 +

p

−

∗ This

1

1

pm (p m − 1)(m + 1) + p m

1

1

(p + 1)m+1 (p m − 1)2

pm − 1

2 P

m+1 i=2

m+1 i 1

(p + 1)m+1 (p m − 1)2

work is supported by N.S.F. of P.R.China (60472068).

pm+1−i 

,

14

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

f (y) is a polynomial of y with degree m = [ k+1 2 ], and ε is any fixed positive number. Theorem 2 . For any real number x ≥ 1, we have the asymptotic formula X

x log x Y (l + 1)(p − 1) 1− ζ(l + 1) p pl+2 − p

d(n) =

n≤x n∈C

where l =

h i k 2

1

+ Ax + +O(x 2 +ε ),

, A is a constant, and ε is any fixed positive number.

§2. Proof of the theorems In this section, we shall complete the proof of the theorems. In fact, let n = pα1 1 pα2 2 · · · pαs s denotes the factorization of n into prime powers. Then it is clear that bk (n) = bk (pα1 1 pα2 2 · · · pαs s ) = bk (pα1 1 )bk (pα2 2 ) · · · bk (pαs s ). That is, bk (n) is a multiplicative function. So we first study the problem in the case n = pα . (1) If α ≥ k, then from the definition of bk (n) we know that bk (n) | n. Therefore n ∈ B. (2) If α ≤ k, then bk (n) = pk−α . So combining (1), (2) and the definition k of B and C, we deduce that n ∈ B if α ≥ [ k+1 2 ], and n ∈ C if α ≤ [ 2 ]. Now we prove the theorem 1 and theorem 2 respectively. First let f (s) =

X d(n) n∈B

ns

.

Then from the definition of bk (n) and B, the properties of the Dirichlet’s divisor function and the Euler product formula [2], we have !

f (s) =

d(pm ) d(pm+1 ) 1 + ms + (m+1)s + · · · p p

Y p

=

Y

1+

m+1 m+2 + (m+1)s + · · · pms p

1+

m+1 ps m+1 + + pms pms (ps − 1) pms (ps − 1)2

p

=

Y p

2

=

ζ m+1 (ms) Y pm s ((ps − 1)(m + 1) + ps ) 1 + ζ m+1 (2ms) p (pms + 1)m+1 (ps − 1)2 (ps − 1)2 −

m+1 X

 m+1 i

p

m(m+1−i)s

i=2

(pms + 1)m+1 (ps − 1)2

   ,  

(1)

On the mean value of the Dirichlet’s divisor function in some special sets1

15

where Îś(s) is the Riemann zeta-function and m = [ k+1 2 ]. Obviously, we have

∞

X d(n)

1

≤

Ďƒ

n Ďƒâˆ’1− n=1

|d(n)| ≤ n,

1 , m

1 is the real part of s. Therefore by Perron formula [3], with where Ďƒ > 1 − m 3 2 s0 = 0, b = m , T = x 2m , we have

X

d(n) =

n≤x n∈B

1 2Ď€i

2 +iT m

Z

2 −iT m

1 Îś m+1 (ms) xs +Îľ 2m ), R(s) ds + O(x Îś m+1 (2ms) s

where   Y  R(s) = 1 +  p 

2 pm s ((ps

− 1)(m + 1) +

ps )

−

(ps

−

1)2

(pms + 1)m+1 (ps −

m+1 X

 m+1 i

p

m(m+1−i)s

i=2 1)2

To estimate the main term 1 2Ď€i

Z

2 +iT m 2 −iT m

Îś m+1 (ms) xs R(s) ds, Îś m+1 (2ms) s

we move the integral line from s =

2 m

Âą iT to s =

1 2m

Âą iT , then the function

Îś m+1 (ms) xs R(s) Îś m (2ms) s have one m + 1 order pole point at s =

1 m

with residue

(m) 1 R(s)xs lim (ms − 1)m+1 Îś m+1 (ms) m+1 s→1 m! Îś (2ms)s (m) R(s)xs 1 m lim (0 ) (ms − 1)m+1 Îś m+1 (ms) + s→1 m! Îś m+1 (2ms)s (m−1) R(s)xs 0 1 m m+1 m+1 + lim (1 ) (ms − 1) Îś (ms) + ¡¡¡ s→1 m! Îś m+1 (2ms)s

=

1 m R(s)xs (m ) (ms − 1)m+1 Îś m+1 (ms) m+1 s→1 m! Îś (2ms)s

(m)

+ lim 1

=

1 mx m 1 R f (log x) + O x 2m + , m+1 Îś (2) m

where f (y) is a polynomial of y with degree k, and Îľ is any fixed positive number.

   .  

16

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

So that we can get 1 2Ï€i mx

=

2 +iT m

Z

2 −iT m 1 m

ζ m+1 (2)

R

Z

+

1 +iT 2m 2 +iT m

1 −iT 2m

Z

+

1 +iT 2m

2 −iT m

Z

+

!

1 −iT 2m

ζ m+1 (ms)xs R(s)ds ζ m+1 (2ms)s

1 1 f (log x) + O(x 2m + ). m

It is easy to estimate

1

1 Z 2m +iT ζ m+1 (ms)xs 1

R(s)ds

x 2m +ε ,

2Ï€i 2 +iT ζ m+1 (2ms)s m

1 Z

2Ï€i

and

1 Z

2Ï€i

2 −iT m 1 −iT 2m 1 −iT 2m

1 +iT 2m

1 ζ m+1 (ms)xs

R(s)ds

x 2m +ε , m+1

ζ (2ms)s

1 ζ m+1 (ms)xs

R(s)ds

x 2m +ε . m+1

ζ (2ms)s

Therefore, we have 1

1 1 mx m d(n) = m+1 R p m · f (log x) + +O x 2m +ε , ζ (2) n≤x X

n∈B

where 

R p

1 m

 pm (p Y  = 1 +  p 

1 m

− 1)(m + 1) + p

1 m

− p

1

X d(n) n∈C

ns

.

Then from the Euler product formula [3] we have X d(n)

=

Y p

−1

(p + 1)m+1 (p m − 1)2

g(s) =

n∈C

2 m+1 X i=2

This completes the proof of Theorem 1. For any integer k ≥ 2 and any real number s > 1, let

g(s) =

1 m

ns d(p) d(p2 ) d(pl ) 1 + s + 2s + · · · + ms p p p

!

 m+1 i

pm+1−i 

  .  

On the mean value of the Dirichlet’s divisor function in some special sets2

=

Y p

=

Y p

2 3 l+1 1 + s + 2s + · · · + ls p p p 1 1 − p1s 

= ζ(s)

Y  p

!

1−

1

1−

1 ps

l+1 − (l+1)s  p

ζ 2 (s) Y (l + 1)(ps − 1) , 1− ζ((l + 1)s) p p(l+2)s − ps

=

1 1 1 1 l+1 1 + s + 2s + 3s + · · · + ls − (l+1)s p p p p p

p(l+1)s

17

where l = [ k2 ]. So by Perron formula [3] and the methods of proving Theorem 1 we can easily get Theorem 2.

References [1] F. Smarandache. Only Problems, Not solutions. Chicago: Xiquan Publishing House, 1993. [2] Apostol T M. Introduction to Analytic Number Theory. New York: Springer- Verlag, 1976. [3] Pan Chengdong and Pan Chengbiao. Foundation of Analytic Number Theory. Beijing: Science press, 1997. [4] Zhu Weiyi. On the k-power complement and k-power free number sequence. Smarandache Notions Journal, 2004, 14: 66-69. [5] Liu Hongyan and Lou Yuanbing. A Note on the 29-th Smarandache’s problem. Smarandache Notions Journal, 2004, 14: 156-158.

A NUMBER THEORETIC FUNCTION AND ITS MEAN VALUE∗ Ren Ganglian Department of Mathematics, Northwest University, Xi’an, Shaanxi, P.R.China

Abstract

Let p and q be two distinct primes, epq (n) denote the largest exponent of power pq which divides n. In this paper, we study the properties of theX sequence epq (n). epq (n), and give an interesting asymptotic formula for the mean value n≤x

Keywords:

Largest exponent; Asymptotic formula; Mean value.

§1. Introduction Let p and q be two distinct primes, epq (n) denote the largest exponent of power pq which divides n. In problem 68 of [1], Professor F. Smarandache asked us to study the properties of the sequence ep (n). About this problem, some people had studied it, and obtained a series of interesting results (See references [4] ). In this paper, we use the analytic method to study the properties ofX the sequence epq (n), and give a sharp asymptotic formula for its mean epq (n). That is, we shall prove the following: value n≤x

Theorem. Let p and q be two distinct primes, then for any real number x ≥ 1, we have the asymptotic formula X x epq (n) = + O x1/2+ε , pq − 1 n≤x where ε is any fixed positive number.

§2. Proof of the theorem In this section, we shall complete the proof of Theorem. For any complex s, we define the function f (s) =

∞ X epq (n) n=1

∗ This

ns

work is supported by N.S.F. of P.R.China (60472068).

.

20

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

Note that the definition of epq (n), and applying the Euler product formula (See Theorem 11.6 of [2]), we may get ∞ X ∞ X ∞ X

f (s) =

Îą (pÎą+t q Îą n

Îą=0 t=1 n1 =1 (n1 ,pq)=1 ∞ X ∞ X ∞ X

+

∞ X

+

Îą

(pÎą q Îą+t n1 )s =1

Îą=0 t=1 n1 (n1 ,pq)=1 ∞ X

Îą (pq)Îąs Îą=0

=

s 1)

t=1

∞ X

1 pts

∞ X

Îą Îąs (pq) Îą=0

+

t=1

Îą Îą q Îą n )s (p 1 =1

Îą=0 n1 (n1 ,pq)=1

1 n1 s

n1 =1 (n1 ,pq)=1 ∞ ∞ X X

1 q ts

∞ X ∞ X

n1 =1 (n1 ,pq)=1

∞ X 1 Îą + n1 s Îą=0 (pq)Îąs

∞ X n1 =1 (n1 ,pq)=1

1 n1 s

Îś(s) , (pq)s − 1

=

where Îś(s) is the Riemann zeta-function. Obviously, we have

∞

X epq (n)

1

,

≤ Ďƒ

n Ďƒâˆ’1 n=1

epq (n) ≤ logpq n ≤ ln n

where Ďƒ is the real part of s. Therefore by Parron’s formula ( See reference [3]) we can get X epq (n) n≤x

ns0

1 = 2Ď€i

Z

Îś(s + s0 ) xs ds + O (pq)s+s0 − 1 s

b+iT

b−iT

+O x1âˆ’Ďƒ0 H(2x) min 1,

log x T

xb B(b + Ďƒ0 ) T

+ O xâˆ’Ďƒ0 H(N ) min 1,

!

x kxk

where N is the nearest integer to x, kxk = |x − N |. Taking s0 = 0, b = 3 1 2 , H(x) = ln x, B(Ďƒ) = Ďƒâˆ’1 , we have 1 epq (n) = 2Ď€i n≤x X

Z

3 +iT 2 3 −iT 2

3

Îś(s) xs x 2 +Îľ ds + O( ). (pq)s − 1 s T

To estimate the main term 1 2Ď€i

Z

3 +iT 2 3 −iT 2

Îś(s) xs ds, (pq)s − 1 s

A number theoretic function and its mean value1

we move the integral line from s = function g(s) =

3 2

21

± iT to s =

Z

3 +iT 2 3 −iT 2

Z

+

1 +iT 2 3 +iT 2

Z

+

1 −iT 2 1 +iT 2

± iT . This time, the

ζ(s) xs (pq)s − 1 s

has a simple pole point at s = 1, and the residue is 1 2πi

1 2

Z

+

3 −iT 2

!

1 −iT 2

x pq−1 .

So we have

ζ(s) xs x ds = . (pq)s − 1 s pq − 1

Taking T = x, and note that

1

2πi

Z

1 +iT 2 3 +iT 2

3 2

Z

T

+

3 −iT 2 1 −iT 2

!

ζ(s) xs

ds

(pq)s − 1 s

3

1 x 2

ζ(σ + iT )

dσ

(pq)σ − 1 T

1 2 3 +ε 2

x

Z

1

= x 2 +ε

and

1 Z T

1 Z 12 −iT 1 1 ζ(s) xs

1 x2

|ζ( ds + it) |dt x 2 +ε ,

1 s

2πi 1 +iT (pq) − 1 s

2 0 (pq) 2 − 1 t 2

we may immediately get the asymptotic formula X n≤x

epq (n) =

x + O x1/2+ε . pq − 1

This completes the proof of Theorem.

References [1] F. Smarandache. Only Problems, Not solutions. Chicago: Xiquan Publishing House, 1993. [2] Apostol T M. Introduction to Analytic Number Theory. New York: Springer- Verlag, 1976. [3] Pan Chengdong and Pan Chengbiao. Foundation of Analytic Number Theory. Beijing: Science press, 1997. [4] Zhang Wenpeng. Reseach on Smarandache problems in number theory. Hexis Publishing House, 2004.

ON THE ADDITIVE K -TH POWER COMPLEMENTS∗ Ding Liping Department of Mathematics, Northwest University, Xi’an, Shaanxi, P.R.China

Abstract

The main purpose of this paper is using the elementary and analytic methods to study the mean value properties of the additive k-th power complements, and give some interesting asymptotic formulae for it.

Keywords:

Additive k-th power complements; Mean value; Asymptotic formula.

§1. Introduction For any positive integer n, the Smarandache k-th power complements bk (n) is the smallest positive integer such that nbk (n) is a complete k-th power, see problem 29 of [1]. Similar to the Smarandache k-th power complements, the additive k-th power complements ak (n) is defined as follows: ak (n) is the smallest nonnegative integer such that ak (n) + n is a perfect k-th power. For example, if k = 2, we have the additive square complements sequence {a2 (n)} (n = 1, 2, · · ·) as follows: a2 (1) = 0, a2 (2) = 2, a2 (3) = 1, a2 (4) = 0, a2 (5) = 4, a2 (6) = 3, a2 (7) = 2, a2 (8) = 1, a2 (9) = 0, · · ·. About this problem, many authors have studied it before, and obtained some interesting results. For example, Z.F. Xu [4] studied the mean value properties of the additive k-th power complements, and gave the following: Proposition . For any real number x ≥ 3 and fixed positive integer k ≥ 2, we have the asymptotic formula: X

ak (n) =

n≤x

1 2 k2 x2− k + O x2− k . 4k − 2

For any fixed positive integer m, the definition of the arithmetical function δm (n) is

δm (n) =

max{d ∈ N

| d|n, (d, m) = 1}, if 0, if

n 6= 0, n = 0.

In this paper, we shall use the elementary and analytic methods to study the mean value properties of the new arithmetical function δm (ak (n)), and give ∗ This

work is supported by N.S.F. of P.R.China (60472068).

24

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

an interesting asymptotic formula for it. That is, we shall prove the following conclusion: Theorem. For any real number x ≼ 3 and positive integer m, we have the asymptotic formula: X

δm (ak (n)) =

n≤x

where

Q

1 Y 2 p k2 x2− k + O x2− k , 2(2k − 1) p+1 p|m

if k > 3,

denotes the product over all prime divisors p of m, and is any fixed

p|m

positive number. Taking k = 2, 3 in our Theorem, we may immediately deduce the following:

Corollary. For any real number x ≼ 1, we have the asymptotic formulae 5 2 3Y p δm (a2 (n)) = x 2 + O x 4 + 3 p|m p + 1 n≤x X

and X

δm (a3 (n)) =

n≤x

4 9 5Y p x3 + O x 3 + . 10 p|m p + 1

§2. Some lemmas To complete the proof of the theorem, we need following Lemmas: Lemma 1. For any real number x > 1 and positive integer m, we have the asymptotic formula X

δm (n) =

n≤x

3 x2 Y p + O x 2 + , 2 p|m p + 1

where is any fixed positive number. Proof. Let s = Ďƒ+it be a complex number and f (s) =

∞ P n=1

δm (n) ns .

Note that

δm (n) n, so it is clear that f (s) is a Dirichlet series absolutely convergent for Re(s)> 2, by the Euler product formula [2] and the definition of δm (n) we get f (s) =

∞ X δm (n) n=1

ns

!

=

Y p

δm (p) δm (p2 ) δm (p2n ) 1+ + + ¡ ¡ ¡ + + ¡¡¡ ps p2s pns

!

=

Y p|m

δm (p) δm (p2 ) δm (p2n ) 1+ + + ¡ ¡ ¡ + + ¡¡¡ ps p2s pns

On the additive k -th power complements1

25 !

δm (p) δm (p2 ) δm (p2n ) 1+ + + ¡¡¡ + + ¡¡¡ s 2s p p pns

Y

Ă—

p†m

=

Y p|m

1 1 1 1 + s + 2s + ¡ ¡ ¡ + ns + ¡ ¡ ¡ p p p

!

p p2 p2n 1 + s + 2s + ¡ ¡ ¡ + ns + ¡ ¡ ¡ p p p

Y

Ă—

p†m

=

1 1 − p1s

Y p|m

= Îś(s − 1)

Y

!

!

1

Y

1 ps−1

1−

p†m s p −

p , s p −1

p|m

where Îś(s) is the Riemann zeta-function, and

Q

(1) denotes the product over all

p

primes. From (1) and Perron’s formula [3], we have 1 δm (n) = 2Ď€i n≤x X

5 +iT 2

Z

5 −iT 2

Îś(s − 1)

Y ps − p xs

where is any fixed positive number. Now we move the integral line in (2) from s = time, the function Îś(s − 1)

ps −p

Q

with residue

¡

ps −1

p|m

¡

ps − 1

p|m

xs

5 2

s

5

ds + O

Âą iT to s =

x 2 + T 3 2

!

, (2)

Âą iT . This

has a simple pole point at s = 2

s

x2 Y p . 2 p|m p + 1

(3)

Hence, we have 1 2Ď€i =

x2 2

Z

5 −iT 2

Z

+

3 −iT 2

5 +iT 2 5 −iT 2

Z

+

3 +iT 2 5 +iT 2

Z

+

3 −iT 2

!

3 +iT 2

Îś(s − 1)

Y ps − p xs p|m

ps − 1

¡

p . p + 1 p|m Y

Z

5 −iT 2 3 −iT 2

Z

+

ds

(4)

We can easily get the estimate

1

2Ď€i

s

3 +iT 2 5 +iT 2

!

Îś(s − 1)

5

x 2 +

¡ ds s

T

Y ps − p xs p|m

ps − 1

(5)

and

s Z 3 −iT s Y

1

3 2 p −p x

x 2 + . Îś(s − 1) ¡ ds

2Ď€i 3

s−1 p s +iT

2 p|m

(6)

26

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

Taking T = x, combining (2), (4), (5) and (6) we deduce that X

δm (n) =

n≤x

3 x2 Y p + O(x 2 + ). 2 p|m p + 1

(7)

This completes the proof of Lemma 1. Lemma 2. For any real number x ≼ 3 and any nonnegative arithmetical function f (n) with f (0) = 0, we have the asymptotic formula: h X

f (ak (n)) =

n≤x

1

i





x k −1

X

X

t=1

n≤g(t)

  

X

f (n) + O  n≤g

h

  

f (n) , 1

xk

i

where [x] denotes the greatest integer not exceeding x and g(t) =

k−1 X i=1

i k

ti .

Proof. See reference [4].

§3. Proof of the theorem In this section, we will complete the proof of the theorem. From the definition of δm (ak (n)), Lemma 1 and Lemma 2, we have X

δm (ak (n))

n≤x

h

=

h

=

1

i



x k −1

X

X

t=1

n≤g(t)

1

n≤g

h

  

δm (n) 1

xk

i

i

X



1 k 2 t2k−2 Y p  + O(t2k−3 ) + O x1− k ln x 2 p+1 p|m h

=

X

δm (n) + O 

x k −1 

t=1

=

  



k2 Y p 2 p|m p + 1

1

if

i

x k −1

X

2

t2k−2 + O(x2− k )

t=1

Y p 2 1 k2 x2− k + O(x2− k ). 2(2k − 1) p|m p + 1

For the cases of k = 2, 3, we can also prove the results by Lemma 2. For example, X n≤x

δm (a2 (n))

k>3

On the additive k -th power complements2 h

=

h

=

1

i

X

X

t=1

n≤2t

  

δm (n) + O  n≤g

X h

  

δm (n) 1

x2

i

i

x 2 −1 

X t=1

=





x 2 −1

1

27



1 3 p 2t2 + O(t 2 + ) + O x1− 2 ln x p+1 p|m Y

3 5 2x 2 Y p + O(x 4 + ). 3 p|m p + 1

This completes the proof of the theorem.

References [1] F.Smarandache. Only problems, not solutions, Xiquan Publishing House, Chicago, 1993. [2] Tom M.Apostol. Introduction to Analytic Number Theory. New York: Springer-Verlag, 1976. [3] Pan Chengdong and Pan Chengbiao. Foundation of Analytic Number Theory. Beijing: Science Press, 1997. [4] Xu Zhefeng. On the additive k-th power complements. Research on Smarandache Problems in Number Theory, Hexis 13-16, 2004.

ON THE SMARANDACHE-RIEMANN ZETA SEQUENCE Li Jie Department of Mathematics, Northwest University, Xi’an, Shaanxi, P.R.China jli0815@126.com

Abstract

The main purpose of this paper is using the elementary method to study the properties of the Smarandache-Riemann zeta sequence, and solve a conjecture posed by Murthy.

Keywords:

Smarandache-Riemann zeta sequence; Murthy’s conjecture; Elementary method.

§1. Introduction For any positive integer n, we let a(2n, 2) denotes the sum of the base 2 digits of 2n. That is, if 2n = a1 2α1 + a2 2α2 + . . . + as 2αs with αs > αs−1 > . . . > α1 ≥ 0, where ai = 0 or 1, i = 1, 2, . . . , s, then a(2n, 2) =

s X

ai .

i=1

For any complex number s with Re(s) > 1, we define Riemann-zeta function as ∞ X 1 ζ(s) = . ns n=1 For any positive integer n, let Tn be a number such that ζ(2n) =

π 2n , Tn

where π is ratio of the circumference of a circle to its diameter. Then the sequence T = {Tn }∞ n=1 is called the Smarandache-Riemann zeta sequence. In [1], Murthy believed that Tn is a sequence of integers. Simultaneous, he proposed the following: Conjecture. No two terms of Tn are relatively prime. In this paper, we shall prove the following conclusion: Theorem. There exists infinite positive integers n such that Tn is not an integer. From this Theorem we know that the Murthy’s conjecture is not correct, because there exists infinite positive integers n such that Tn is not an integer.

30

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

§2. Some simple lemmas Before the proof of the theorem, some simple lemmas will be useful. Lemma 1. If ord(2, (2n)!) < 2n − 2, then Tn is not an integer, where ord(2, (2n)!) denotes the order of prime 2 in (2n)!. Proof. See reference [2]. Lemma 2. For any positive integer n ≥ 1, we have the identity: α2 (2n) ≡

+∞ X i=1

2n = 2n − a(2n, 2), 2i

where [x] denotes the greatest integer not exceeding x. Proof. From the properties of [x] we know that

2n 2i

=

a1 2α1 + a2 2α2 + . . . + as 2αs 2i

 s X   aj 2αj −i , if =   j=k

0,

αk−1 < i ≤ αk ;

if i > αs .

So from this formula we have

α2 (2n) ≡

+∞ X i=1

= =

+∞ X a1 2α1 + a2 2α2 + · · · + as 2αs 2n = 2i 2i i=1

αj s X X j=1 k=1 s X

aj ·

j=1

aj 2αj −k = 2αj

s X

aj (1 + 2 + 22 + · · · + 2αj −1 )

j=1 s X

−1 (aj 2αj − aj ) = 2−1 j=1

= 2n − a(2n, 2) This completes the proof of Lemma 2.

§3. Proof of the theorem In this section, we shall complete the proof of the theorem. From Lemma 2 and the Theorem 3.14 of [3] we know that ord(2, (2n)!) = α2 (2n) ≡

+∞ X i=1

2n = 2n − a(2n, 2). 2i

Now it is clear that ord(2, (2n)!) < 2n − 2 if and only if 2n − a(2n, 2) < 2n − 2.

On the Smarandache-Riemann zeta sequence

31

That is a(2n, 2) ≥ 3. In fact, there exists infinite positive integers n such that a(2n, 2) ≥ 3. For example, taking n = 2α1 + 2α2 + 2α3 + 2α4 with α4 > α3 > α2 > α1 ≥ 0, then a(2n, 2) ≥ 3. For these n, from Lemma 1 we know that Tn is not an integer. Since there are infinite positive integers α1 , α2 , α3 and α4 , it means that there exists infinite positive integers n such that a(2n, 2) ≥ 3. Therefore, there exists infinite positive integers n such that Tn is not an integer. This completes the proof of Theorem.

References [1] A.Murthy, Some more conjectures on primes and divisors, Smarandache Notions J.12(2001), pp. 311. [2] Le Maohua, The Smarandache-Riemann zeta sequence, Smarandache Notions J.14(2004), pp. 346. [3] Tom M. Apostol, Introduction to Analytic Number Theory, New York: Springer-Verlag, 1976.

ON THE PROPERTIES OF THE HEXAGON-NUMBERS Liu Yanni Department of Mathematics, Northwest University, Xi’an, Shaanxi, P.R.China

Abstract

In this paper, we shall use the elementary method to study the properties of the hexagon-numbers, and give an interesting identity involving the hexagonnumbers.

Keywords:

Hexagon-number; Arithmetical property; Identity.

§1. Introduction For any positive integer m, if n = m(2m−1), then we call such an integer n as hexagon-number (see reference [1]). The hexagon-numbers 1, 6, 15, 28, · · · are closely related to the hexagons. Moreover, the hexagon-numbers are the partial sums of the terms in the arithmetic progression 1, 5, 9, 13, · · · , 4n + 1, · · · . For any positive integer n, let m be the largest positive integer that satisfy the inequality m(2m − 1) ≤ n < (m + 1)(2m + 1). Now we define a(n) = m(2m − 1), and call a(n) as the hexagon-number part of n. In this paper, we shall use the elementary method to study the convergent ∞ X 1 properties of the Dirichlet’s series f (s) = , and give an interesting s (n) a n=1 identity for the case s = 2. That is, we shall prove the following: Theorem. gent, and

For any real number s > 1, the infinity series f (s) is converf (2) =

∞ X n=1

1 a2 (n)

5 = π 2 − 4 ln 2. 3

§2. Proof of Theorem In this section, we shall complete the proof of Theorem. First from the definition of a(n) we know that there exists (m + 1)(2m + 1) − m(2m − 1) = 4m + 1

34

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

solutions for the equation a(n) = m(2m − 1). So we can easy deduce that f (s) =

∞ X

∞ X 1 = as (n) m=1 n=1

∞ X n=1 a(n)=m(2m−1)

∞ X 1 4m + 1 = . s s a (n) m=1 m (2m − 1)s

Thus we may conclude that f (s) is convergent if 2s − 1 > 1. That is, s > 1. Now we shall use the elementary method to calculate the exact value of f (2). Applying the above identity we have f (2) = =

∞ X

4m + 1 2 (2m − 1)2 m m=1 ∞ X 1 m=1 ∞ X

m2

+

∞ X

∞ X 12 8 − 2 (2m − 1) m(2m − 1) m=1 m=1

1 + 12 = m2 m=1

∞ X 1

∞ X

1 − 2 m 4m2 m=1 m=1

!

−

∞ X

8 m(2m − 1) m=1

∞ X 8 1 = ζ(2) + 12 ζ(2) − ζ(2) − 4 m(2m − 1) m=1

= 10ζ(2) −

∞ X

8 m(2m − 1) m=1 ∞ X

∞ X 1 1 = 10ζ(2) − 4 − 2m − 1 2m m=1 m=1

!

1 1 1 1 1 + − + − + ··· , 2 3 4 5 6

= 10ζ(2) − 4 1 −

where ζ(s) is the Riemann zeta-function. From the Taylor’s expansion (see reference [2]) for ln(1 + x) we know that ln(1 + x) = x −

x2 x3 x4 xn + − + ··· + + ···. 2 3 4 n

Using this identity with x = 1 we may immediately get ln 2 = 1 −

1 1 1 1 1 + − + − + ···. 2 3 4 5 6

Combing all the above, and note that ζ(2) =

π2 6

we may obtain

5 f (2) = 10ζ(2) − 4 ln 2 = π 2 − 4 ln 2. 3 This completes the proof of Theorem.

On the properties of the hexagon-numbers

35

References [1] Tom M. Apostol, Introduction to Analytic Number Theory, New York: Springer-Verlag, 1976. [2] Tom M. Apostol, Mathematical Analysis (2nd Edition), Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills, Ont., 1974.

ON THE MEAN VALUE OF THE K -TH POWER PART RESIDUE FUNCTION Ma Jinping Department of Mathematics, Northwest University, Xi’an, Shaanxi, P.R.China

Abstract

Similar to the Smarandache k-th power complements, we define the k-th power part residue function fk (n) is the smallest nonnegative integer such that n − fk (n) is a perfect k-th power. The main purpose of this paper is using the 1 elementary methods to study the mean value properties of fk (n)+1 , and give an interesting asymptotic formula for it.

Keywords:

k-th power part residue function; Mean value; Asymptotic formula.

§1. Introduction and results For any positive integer n, the Smarandache k-th power complements bk (n) is the smallest positive integer such that nbk (n) is a perfect k-th power (see problem 29 of [1]). Similar to the Smarandache k-th power complements, Xu Zhefeng in [2] defined the additive k-th power complements ak (n) as follows: ak (n) is the smallest nonnegative integer such that n + ak (n) is a perfect k-th power. As a generalization of [2], we will define the k-th power part residue function fk (n) as the smallest nonnegative integer such that n − fk (n) is a perfect k-th power. For example, if k = 2, we have the square part residue sequence {f2 (n)} (n = 1, 2, . . .) as following: 0, 1, 2, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4,5, 6, 0, 1, 2, . . . . Meanwhile, let p be a prime, ep (n) denotes the largest exponent of power p which divides n. About the relations between ep (n) and fk (n), it seems that none had studied them before, at least we couldn’t find any reference about it. In this paper, we use the elementary methods to study the mean value prop1 and ep (fk (n)), and obtain two sharper asymptotic formulae erties of fk (n)+1 for them. That is, we will prove the following conclusions: Theorem 1.

For any real number x ≥ 3, we have the asymptotic formula

1 1 k−1 1 = x k ln x + (ln x + γ − k + 1)x k + O(ln x), f (n) + 1 k n≤x k

X

where γ is the Euler constant.

38

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

Theorem 2.

For any real number x ≥ 3, we have the asymptotic formula X

1 1 k x+O x1− k p−1 p−1

ep (fk (n)) =

n≤x

.

§2. Some Lemmas To complete the proof of the above theorems, we need following several Lemmas. Lemma 1. For any real number x ≥ 1, we have X

ep (n) =

n≤x

1 x + O(ln2 x). p−1

Proof. See reference [3]. Lemma 2. Let h(n) be a nonnegative arithmetical function with h(0) = 0. Then, for any real number x ≥ 1 we have the asymptotic formula: X

h(fk (n)) =

 X

h(n) + O 

t=1 n≤g(t)

n≤x

where g(t) =

M −1 X

k−1 X i=1

 X

h(n) ,

n≤g(M )

!

1 k i t and M = [x k ], [x] denotes the greatest integer not i

exceeding x. Proof. For any real number x ≥ 1, let M be a fixed positive integer such that M k ≤ x < (M + 1)k . Noting that if n pass through the integers in the interval [tk , (t + 1)k ), then fk (n) pass through all integers in the interval [0, (t + 1)k − tk ), so we can deduce that X

h(fk (n)) =

M −1 X

X

t=1 tk ≤n<(t+1)k

n≤x

=

M −1 X

X

t=1 n≤g(t)

=

M −1 X

h(n) +

=

h(fk (n))

M k ≤n≤x

X

h(n)

0≤n<x−M k

 X

 X

h(n) + O 

t=1 n≤g(t) M −1 X

X

h(fk (n)) +

 X

t=1 n≤g(t)

h(n)

0≤n≤(M +1)k −M k

h(n) + O 

 X n≤g(M )

h(n) .

On the mean value of the k -th power part residue function

39

This proves the Lemma 2. Lemma 3. For any real number x > 1, we have the asymptotic formula X 1 n≤x

n

1 , x

= ln x + γ + O

where γ is the Euler constant. Proof. See reference [4]. Lemma 4 For any real number x ≥ 3, we have the asymptotic formula X

fk (n) =

n≤x

1 2 k2 x2− k + O x2− k . 2(2k − 1) 1

Proof. Let h(n) = n and M = [x k ], then from Lemma 2 and Euler summation formula (see reference [4]) we obtain X

fk (n) =

M −1 X

 X

n+O

t=1 n≤g(t)

n≤x

 X

n

n≤g(M )

1 [x k

=

]−1 2 1 X 2 2k−2 k t + O(x2− k ) 2 t=1

=

1 2 k2 x2− k + O(x2− k ). 2(2k − 1)

This proves Lemma 4.

§3. Proof of the theorems In this section, we shall complete the proof of Theorems. First we prove 1 Theorem 1. Let h(n) = n and M = [x k ], then from Lemma 2 and Lemma 3 we obtain X 1 f (n) + 1 n≤x k =

M −1 X t=1

=

X 1  1 +O n+1 n+1 n≤g(t) n≤g(M ) X

M −1 X

ln(kt

k−1

t=1

1 ) + ln 1 + O t

1 +γ+O g(t)

= (k − 1) ln((M − 1)!) + (ln k + γ)(M − 1) + O(ln x)

1

1

1

= (k − 1) ([x k ] − 1) ln([x k ] − 1) − ([x k ] − 1) 1

=

+(ln k + γ)([x k ] − 1) + O(ln x) 1 k−1 1 x k ln x + (ln x + γ − k + 1)x k + O(ln x). k

+ O(ln x)

40

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

This completes the proof of Theorem 1. The proof of Theorem 2. Note that the definition of ep (n), from Lemma 1 and Lemma 4 we have X

ep (fk (n))

n≤x

=

M X

X

ep (fk (n)) +

t=1 (t−1)k ≤n<tk k

=

t=1

=

t=1

 X

ep (j) + O 

j=0

M X

ep (fk (n))

M k ≤n<x

k

M (t+1) X X−t

X

ep (fk (n))

M k ≤n<(M +1)k

1 (t + 1)k − tk + O ln2 ((t + 1)k − tk ) p−1 

+O 

X

ep (n)

0≤n<(M +1)k −M k

1 g(M ) (M + 1)k − 1 + O(ln2 (kM k − 1)) + O p−1 p−1 1 k Mk + O M k−1 . p−1 p−1

= =

1

Taking M = [x k ], we easily get X n≤x

1 1 k x+O x1− k p−1 p−1

ep (fk (n)) =

.

This completes the proof of Theorem 2.

References [1] F.Smarandache. Only Problems, not Solutions, Xiquan Publishing House, Chicago, 1993. [2] Xu Zhefeng. On the Additive k-power Complements. Research on Smarandache Problems in Number Theory, Hexis 13-16, 2004. [3] Lv Chuan. A Number Theoretic Function and its Mean Value. Research on Smarandache Problems in Number Theory, Hexis 33-36, 2004. [4] Tom M. Apostol. Introduction to Analytic Number theory. New York: Springer-Verlag, 1976. [5] Pan Chengdong and Pan Chengbiao. Foundation of Analytic Number Theory. Beijing: Science Press, 1997, P98.

ON THE INTEGER PART OF THE M -TH ROOT AND THE K -TH POWER FREE NUMBER Li Zhanhu1,2 1. Department of Mathematics, Northwest University, Xi’an, Shaanxi, P.R.China 2. Department of Mathematics, Xianyang Teacher’s College, Xianyang, Shaanxi, P.R.China

Abstract

The main purpose of this paper is using the elementary methods to study the mean value properties of a special arithmetical function involving the integer part of the m-th root of an integer and the k-th power free numbers, and give an interesting asymptotic formula for it.

Keywords:

Integer part of the m-th root; k-th power free numbers; Asymptotic formula.

§1. Introduction and results Let m and k are two fixed positive integers with k ≥ 2. For any positive integer n, we define arithmetical function bm (n) as the integer part of the m1 th root of n. That is, bm (n) = [n m ], where [x] denotes the greatest integer ≤ x. For example, b2 (1) = 1, b2 (2) = 1, b2 (3) = 1, b2 (4) = 2, b2 (5) = 2, b2 (6) = 2, b2 (7) = 2, b2 (8) = 2, b2 (9) = 3, · · · · · ·. In reference [1], Professor F.Smarandache asked us to study the properties of the sequences {bk (n)}. About this problem, I do not know whether there exists any progress. But none study the mean value properties of {bk (n)} over the k-th power free numbers, here we call a positive integer n as k-th power free number, if pk † n for any prime p. For convenience, we let Ak denotes the set of all k-th power free numbers. In this paper, we shall use the elementary methods to study the mean value properties of bm (n) over the set Ak , and give an interesting formula for it. That is, we shall prove the following result: Theorem. Let m and k are two fixed positive integers with k ≥ 2. Then for any real number x ≥ 1, we have the asymptotic formula X n≤x n∈Ak

bm (n) =

m+1 1 m x m + O (x) , ζ(k) m + 1

where ζ(k) is the Riemann zeta-function. From this theorem we may immediately deduce the following Corollaries:

42

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

Corollary 1. For any real number x ≥ 1, we have the asymptotic formula X

b2 (n) =

n≤x n∈A2

4 3 x 2 + O (x) π2

X

and

b3 (n) =

n≤x n∈A2

9 4 x 3 + O (x) . 2π 2

Corollary 2. For any real number x ≥ 1, we have the asymptotic formula X

b4 (n) =

n≤x n∈A4

64 5 x 4 + O (x) π4

X

and

b5 (n) =

n≤x n∈A4

75 6 x 5 + O (x) . π4

§2. Some Lemmas Before completing the proof of the theorem, we need the following lemmas. Lemma 1. Let m and k are two fixed positive integers with k ≥ 2. Then for any real number x ≥ 1, we have the asymptotic formula X n≤x

1

nm =

X µ(n)

m+1 m x m + O(x) and m+1

n≤x

nk

=

1 + O x−k+1 , ζ(k)

where µ(n) is the M¨ obius function, and ζ(k) is the Riemann zeta-function. Proof. Using the Euler’s summation formula (see Theorem 3.1 of [2]) we can easily deduce these conclusions. Lemma 2. Let a(n) denotes the character function of the k-th power free numbers (That is, if n is a k-th power free number, then a(n) = 1; if n is not a k-th power free number, then a(n) = 0). Then for any positive integer n, we have the identity X µ(d), a(n) = dk |n

where µ(d) is the M o¨bius function. Proof. From the properties of M o¨bius function we know that for any positive integer n, X 1 . µ(d) = n d|n Let uk denotes the greatest k-th power divisor of n. That is, n = uk v, where v is a k-th power free number. It is clear that n is a k-th power free number if and only if u = 1. In this case, X

µ(d) =

dk |n

X

µ(d) =

dk |uk

X

µ(d) = µ(1) = 1 = a(n).

d|u

If u > 1, then from the above formula we have X dk |n

µ(d) =

X dk |uk

µ(d) =

X d|u

1 = 0 = a(n). u

µ(d) =

On the integer part of the m-th root and the k -th power free number

43

This proves Lemma 2.

§3. Proof of the theorem In this section, we shall complete the proof of the theorem. For any real number x ≥ 1 and positive integer k ≥ 2, from Lemma 1 and Lemma 2 and the definition of bm (n) we have X

bm (n)

n≤x n∈Ak

=

X

a(n)bm (n)

n≤x

=

XX

h

1

µ(d) n m

i

n≤x dk |n

=

1

h

X

µ(d) (dk t) m

i

dk t≤x

=

X

µ(d)d

k m

dk ≤x

=

X

X

t

1 m

+O

t≤x/dk

µ(d)d

dk ≤x

k m

X

X

dk ≤x t≤x/dk

m+1 m 1 x m k(m+1) + O m+1 d m

x dk

1 !! m

+ O x

X 1 dk ≤x

dk

m+1 X µ(d) 1 X m 1 + O (x) x m + O x m k m+1 d k k

d ≤x

=

=

1

d ≤x

m+1 m 1 x m + O (x) . ζ(k) m + 1

This completes the proof of Theorem. Now the Corollaries follows from ζ(2) = π 2 /6 and ζ(4) = π 4 /90.

References [1] Smarandache F. Only problems, not Solutions. Chicago: Xiquan Publ. House, 1993. [2] Tom M.Apostol. Introduction to Analytic Number Theory. New York: Springer-Verlag, 1976. [3] Zhang Wenpeng. Research on Smarandache problems in number theory. Hexis Publishing House, 2004.

 

A FORMULA FOR SMARANDACHE LCM RATIO SEQUENCE Wang Ting Department of Mathematics, Xi’an Science and Arts College, Xi’an, Shaanxi, P.R.China

Abstract

In this paper, a reduction formula for Smarandache LCM ratio sequences SLR(5) is given.

Keywords:

Smarandache LCM ratio sequences; Reduction formula.

§1. Introduction Let (x1 , x2 , ..., xt ) and [x1 , x2 , ..., xt ] denote the greatest common divisor and the least common multiple of any postive integers x1 , x2 , ..., xt respectively. Let r be a positive integer with r > 1. For any positive integer n, let [n, n + 1, ..., n + r − 1] , T (r, n) = [1, 2, ...r] then the sequences SLR(r) = {T (r, n)}∞ is called Smarandache LCM ratio sequences of degree r. In reference [1], Maohua Le studied the properties of SLR(r), and gave two reduction formulas for SLR(3) and SLR(4). In this paper, we study the calculating problem of SLR(5), and prove the following conclusion. Theorem. For any postive integer n, we have the calculating formula :

T (5, n) =

                

1 1440 n(n + 1)(n + 2)(n + 3)(n + 4), 1 120 n(n + 1)(n + 2)(n + 3)(n + 4), 1 720 n(n + 1)(n + 2)(n + 3)(n + 4), 1 360 n(n + 1)(n + 2)(n + 3)(n + 4), 1 480 n(n + 1)(n + 2)(n + 3)(n + 4), 1 240 n(n + 1)(n + 2)(n + 3)(n + 4),

if n ≡ 0, 8 mod 12; if n ≡ 1, 7 mod 12; if n ≡ 2, 6 mod 12; if n ≡ 3, 5, 9, 11 mod 12; if n ≡ 4 mod 12; if n ≡ 10 mod 12.

§2. Proof of the theorem To complete the proof of Theorem, we need following several simple Lemmas. Lemma 1. For any postive integer a and b, we have (a, b)[a, b] = ab.

46

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

Lemma 2. For any postive integer s with s < t, we have (x1 , x2 , ..., xt ) = ((x1 , ..., xs ), (xs+1 , ..., xt )) and [x1 , x2 , ..., xt ] = [[x1 , ..., xs ], [xs+1 , ..., xt ]]. Lemma 3. For any postive integer n, we have (

T (4, n) =

1 24 n(n 1 72 n(n

+ 1)(n + 2)(n + 3), if n ≡ 1, 2 mod 3; + 1)(n + 2)(n + 3), if n ≡ 0 mod 3.

The proof of Lemma 1 and Lemma 2 can be found in [3], Lemma 3 is proved in reference [1]. Now we use these Lemmas to complete the proof of our Theorem. In fact, for any positive integer n, from the properties of the least common multiple of any positive integers we know that [n, n + 1, n + 2, n + 3, n + 4] = [[n, n + 1, n + 2, n + 3], n + 4] [n, n + 1, n + 2, n + 3](n + 4) . = ([n, n + 1, n + 2, n + 3], n + 4) Note that [1, 2, 3, 4, 5] = 60, [1, 2, 3, 4] = 12 and   4,     1,   

if n ≡ 0, 4 mod 12; if n ≡ 1, 3, 5, 7, 9 mod 12; 6, if n ≡ 2 mod 12; ([n, n + 1, n + 2, n + 3], n + 4) =  2, if n ≡ 6, 10 mod 12;     12, if n ≡ 8 mod 12;    3, if n ≡ 11 mod 12, Combining Lemma 3, we and easily get the theorem.

References [1] Maohua Le, Two formulas for smarandache LCM ratio sequences, Smarandache Notions Journal, Vol.14, 2004, 183-185. [2] F.Smarandache, Only problems, not Solutions, Xiquan Publ. House, Chicago, 1993. [3] Tom M.Apostol, Introduction to Analytic Number Theory, SpringerVerlag, New York, Heidelberg Berlin, 1976.

ON THE M -TH POWER COMPLEMENTS SEQUENCE∗ Yang Hai Research Center for Basic Science, Xi’an Jiaotong University, Xi’an, Shaanxi, P.R.China xjtu88yh@163.com

Fu Ruiqin School of Science, Xi’an Shiyou University, Xi’an, Shaanxi, P.R.China xd2008frq@163.com

Abstract

The main purpose of this paper is using the elementary and analytic methods to study the asymptotic properties of the m-th power complements sequence, and give several interesting asymptotic formulae for it.

Keywords:

M -th power complements; Riemann zeta-function; Asymptotic formula.

§1. Introduction and results For any positive integer n ≼ 2, let bm (n) denote the m-th power complements sequence. That is, bm (n) denotes the smallest positive integer such that nbm (n) is a complete m-th power. In problem 29 of the reference [1], Professor F.Smarandache asked us to study the properties of this sequence. About this problem, some authors had studied it before, and obtained some interesting results, see reference [4] and [5]. The main purpose of this paper is using the elementary and analytic methods to study the asymptotic properties of the mth power complements sequence, and give an interesting asymptotic formula for it. That is, we shall prove the following: Theorem . For any real number x > 1 and any fixed positive integers m and k, we have the asymptotic formula

X

1

δ (b (n)) n≤x k m =

∗ This

x2 Y pm (pm − pm−1 + 1) Y pm+1 2Îś(m) p†k (p2m − 1) (p + 1)(pm − 1) p|k

work is supported by N.S.F. of P.R.China (10271093)

!

3

+ O x 2 + ,

48

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

where δk (n) defined as following: δk (n) = max{d ∈ N Q

| d|n,

(d, k) = 1},

denotes the product over all prime numbers.

p

From this theorem, we may immediately get the following: Corollary 1. Let a(n) be the square complements sequence, then for any real number x > 1 and any fixed positive integer k, we have the asymptotic formula 3x2 Y p4 − p3 + p2 Y p3 1 = 2 δ (a(n)) Ď€ p†k (p4 − 1) p|k (p + 1)(p2 − 1) n≤x k

!

X

3

+ O x 2 + .

Corollary 2. Let b(n) be the cubic complements sequence, then for any real number x > 1 and any fixed positive integer k, we have the asymptotic formula !

3 1 x2 Y p6 − p5 + p3 Y p4 = +O x 2 + . 6 − 1) 3 − 1) δ (b(n)) 2Îś(3) (p (p + 1)(p k n≤x p†k p|k X

§2. Proof of the theorem In this section, we will complete the proof of Theorem. Let s = Ďƒ + it be a complex number and f (s) =

∞ P n=1

1 δk (bm (n)) ns

. Note that

1 δk (bm (n))

1, so it is clear that f (s) is a Dirichlet series absolutely convergent for Re(s)> 1. By the Euler product formula [2] and the multiplicative property of δk (bm (n)) we get f (s) =

∞ X n=1

1 δk (bm (n)) ns



=

Y

1 +

p



=

Y p

1 +

(1) 1

δk (bm (p)) ps 1 δk (pm−1 ) ps

+

1 δk (bm (p2 )) p2s

+ ¡¡¡ +

1 δk (1) pms

!

=

Y p†k

=

+ ¡¡¡ +

+

1 δk (bm (pm )) pms

1 δk (pm−1 ) p(m+1)s

Y pm(s−1)−1 1 1 + s−1 ms (p − 1)(p − 1) p|k 1 − p1s



+ ¡ ¡ ¡

+ ¡¡¡ +

1 δk (1) p2ms

!

Îś(s − 1) Y (ps−1 − 1)p(m−1)(s−1) p(m−1)(s−1) + Îś(m(s − 1)) p†k (pms − 1) pm(s−1) − 1

!



+ ¡ ¡ ¡

On the m-th power complements sequence1

49 !

×

Y p|k

ps (pm(s−1) − p(m−1)(s−1) ) , (ps − 1)(pm(s−1) − 1)

where ζ(s) is the Riemann zeta-function and

Q

(2) denotes the product over all

p

prime numbers. From (1) and Perron’s formula [3], we have 1

X

(3)

δ (b (n)) n≤x k m =

1 2πi ×

Z

5 +iT 2

ζ(s − 1) Y (ps−1 − 1)p(m−1)(s−1) p(m−1)(s−1) + ζ(m(s − 1)) p†k (pms − 1) (pm(s−1) − 1)

5 −iT 2

Y p|k

ps (pm(s−1) − p(m−1)(s−1) ) (ps − 1)(pm(s−1) − 1)

!

3

x 2 + T

xs · ds + O s

where is any fixed positive number. Now we move the integral line in (2) from s = time, the function

5 2

!

± iT to s =

,

3 2

(4)

± iT . This

ζ(s − 1) Y (ps−1 − 1)p(m−1)(s−1) p(m−1)(s−1) + ζ(m(s − 1)) p†k (pms − 1) (pm(s−1) − 1) ×

Y p|k

ps (pm(s−1) − p(m−1)(s−1) ) (ps − 1)(pm(s−1) − 1)

!

·

!

!

xs s

has a simple pole point at s = 2 with residue !

x2 Y pm (pm − pm−1 + 1) Y pm+1 . 2ζ(m) p†k (p2m − 1) (p + 1)(pm − 1) p|k

(5)

Hence, we have 1 2πi ×

Y p†k

×

Y p|k

Z

5 −iT 2 3 −iT 2

Z

+

5 +iT 2 5 −iT 2

Z

+

3 +iT 2 5 +iT 2

Z

+

3 −iT 2

!

3 +iT 2

(ps−1 − 1)p(m−1)(s−1) p(m−1)(s−1) + (pms − 1) (pm(s−1) − 1) ps (pm(s−1) − p(m−1)(s−1) ) (ps − 1)(pm(s−1) − 1)

!

·

ζ(s − 1) ζ(m(s − 1)) !

(6)

xs ds s !

=

pm+1 x2 Y pm (pm − pm−1 + 1) Y . m − 1) 2ζ(m) p†k (p2m − 1) (p + 1)(p p|k

(7)

50

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

We can easily get the estimate

5

1 Z 25 −iT x 2 + xs

f (s) ¡ ds ,

2Ď€i 3 −iT s

T 2

(8)

5

1 Z 23 +iT x 2 + xs

f (s) ¡ ds ,

2Ď€i 5 +iT s

T 2

(9)

1 Z 32 −iT 3 xs

f (s) ¡ ds x 2 + .

2Ď€i 3 +iT s

(10)

and

2

Taking T = x, combining (2), (4), (5), (6) and (7) we may deduce that 1 δ (b (n)) n≤x k m X

=

pm+1 x2 Y pm (pm − pm−1 + 1) Y 2Îś(m) p†k (p2m − 1) (p + 1)(pm − 1) p|k

!

3

+ O x 2 + .

This completes the proof of Theorem.

References [1] Smarandache F. Only problems, not Solutions. Chicago: Xiquan Publ. House, 1993. [2] Tom M.Apostol. Introduction to Analytic Number Theory. New York: Springer-Verlag, 1976. [3] Pan Chengdong and Pan Chengbiao. Foundation of Analytic Number Theory. Beijing: Science Press, 1997. [4] Zhu Weiyi. On the k-power complements and k-power free number sequence. Smarandache Notions Journal, 2004, 14: 66-69. [5] Liu Hongyan and Lou Yuanbing. A note on the 29-th Smarandache’s problem. Smarandache Notions Journal, 2004, 14: 156-158.

ON THE SMARANDACHE CEIL FUNCTION AND THE DIRICHLET DIVISOR FUNCTION Ren Dongmei Research Center for Basic Science, Xi’an Jiaotong University, Xi’an, Shaanxi, P.R.China

Abstract

The main purpose of this paper is using the elementary methods to study the mean value properties of the composite function involving Dirichlet divisor function and Smarandache ceil function, and give an interesting asymptotic formula for it.

Keywords:

Smarandache Ceil function; Dirichlet divisor function; Mean value; Asymptotic formula.

§1. Introduction For a fixed positive integer k and any positive integer n, the Smarandache ceil function Sk (n) is defined as following: Sk (n) = min{m ∈ N : n|mk }. This function was introduced by Professor Smarandeche (see reference [1]). About this function, many scholars studied its properties (see reference [2] and [3]). In reference [2], Ibstedt presented the following property: (∀a, b ∈ N )(a, b) = 1 =⇒ Sk (ab) = Sk (a)Sk (b). That is, Sk (n) is a multiplicative function. In this paper, we use elementary methods to study the mean value properties of the composite function involving d(n) and Sk (n), and give an interesting asymptotic formula for it. That is, we shall prove the following: Theorem. Let k be a given positive integer with k ≥ 2. Then for any real number x ≥ 1, we have the asymptotic formula: X n≤x

1 6ζ(k)x ln x Y 1− k 2 π p + pk−1 p

d(Sk (n)) =

1

+ Cx + O(x 2 +ε ).

where C is a computable constant, and ε is any fixed positive number. Taking k = 2 and k = 4 in Theorem 1, noting that: ζ(2) =

π2 , 6

ζ(4) =

π4 , 90

52

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

we may immediately deduce the following: Corollary. For any real number x ≥ 1, we have the asymptotic formula: X

d(S2 (n)) = x ln x

Y

1−

p

n≤x

1 p2 + p

π 2 x ln x Y 1 1− 4 15 p + p3 p

X

d(S4 (n)) =

n≤x

1

+ C1 x + O(x 2 +ε ),

1

+ C2 x + O(x 2 +ε ),

where C1 , C2 are computable constants. If k = 1, then function Sk (n) turns into the identical transformation and the composite function d(Sk (n)) turns into d(n). We have the following asymptotic formula: X

d(S1 (n)) =

n≤x

X

1

d(n) = x ln x + (2γ − 1)x + O(x 2 ),

n≤x

where γ is the Euler’s Constant.

§2. Proof of the theorem In this section, we shall complete the proof of the theorem. Let ∞ X d(Sk (n))

f (s) =

n=1

ns

.

From the Euler product formula (see reference [5]) and the multiplicative property of Sk (n), we have !

f (s) =

d(Sk (p)) d(Sk (p2 )) 1+ + + ··· ps p2s

Y p

!

=

d(p) d(p) d(p2 ) d(p2 ) 1 + s + · · · + ks + (k+1)s + · · · + 2ks + · · · p p p p

Y p

=

Y

2 2 3 + · · · + ks + (k+1)s + · · · ps p p

1+

p

=

Y

1 +

1

p

= ζ(s)

1 pks − p1s

1−

Y p

×

1 1 1 1 + s × 1 + ks + 2ks + · · · p p p

= ζ(s)ζ(ks)

Y p

1 1 1 − ks + s p p

ζ(s) Y 1 = ζ(s)ζ(ks) 1 − ks , ζ(2s) p p + p(k−1)s

2 3 + (k+1)s + · · ·  s p p

53

On the Smarandache ceil function and the Dirichlet divisor function

where ζ(s) is the Riemann zeta-function. It is obvious that

∞

X d(Sk (n))

1

,

≤ σ

n σ−1 n=1

|d(Sk (n))| ≤ n,

where σ > 1 is the real part of s. By Perron formula (see reference [4]), let s0 = 0, b = 32 , T = x, then we have 1 d(Sk (n)) = 2πi n≤x X

3 +iT 2

Z

1 ζ 2 (s)ζ(ks) xs R(s) ds + O(x 2 +ε ), ζ(2s) s

3 −iT 2

1 , and ε is any fixed positive number. where R(s) = 1 − ks p + p(k−1)s p Now we calculate the term

Y

1 2πi

Z

3 +iT 2

ζ 2 (s)ζ(ks) xs R(s) ds. ζ(2s) s

3 −iT 2

3 2

We move the integral line from

1 2

± iT to

ζ 2 (s)ζ(ks) ζ(2s)

± iT . Then the function

R(s)

xs s

has a second order pole at s = 1 with residue 2ζ

lim (s − 1)

2 (s)ζ(ks)

ζ(2s)

s→1

2ζ

(s − 1)

= lim

s→1

2ζ

+(s − 1)

xs R(s) s

2 (s)ζ(ks)

ζ(2s)

2 (s)ζ(ks)

ζ(2s)

!0

R(s)

xs s

sxs ln x − xs R(s) s2

6ζ(k)x ln(x) Y 1 1− k π2 p + pk−1 p

=

!0

!

+ Cx,

where C is a computable constant. So we can obtain 1 2πi =

Z

3 +iT 2 3 −iT 2

6ζ(k)x ln(x) π2

1 +iT 2

Z

+

+

3 +iT 2

Y

1−

p

1 −iT 2

Z

pk

Z

+

1 +iT 2

1 + pk−1

3 −iT 2 1 −iT 2

!

ζ 2 (s)ζ(ks) xs R(s) ds ζ(2s) s

+ Cx.

Taking T = x, we have the estimate

1

2πi 3

<<

Z

1 +iT 2 3 +iT 2

1 x 2 +ε = x 2 +ε T

Z

+

3 −iT 2 1 −iT 2

!

ζ 2 (s)ζ(ks) xs

R(s) ds

ζ(2s) s

54

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

and 1 | 2πi

Z

1 −iT 2 1 +iT 2

1 ζ 2 (s)ζ(ks) xs R(s) ds| << x 2 +ε . ζ(2s) s

So we may immediately get the asymptotic formula X

1 6ζ(k)x ln(x) Y 1− k 2 π p + pk−1 p

d(Sk (n)) =

n≤x

1

+ Cx + O(x 2 +ε ).

This completes the proof of Theorem.

References [1] F. Smarandache. Only Problems, Not solutions. Chicago: Xiquan Publishing House, 1993. [2] Ibstedt. Surfining On the Ocean of Number-A New Smarandache Notions and Similar Topics. New Mexico: Erhus University Press. [3] Sabin Tabirca and Tatiana Tairca. Some new results concerning the Smarandache ceil function, Smarandache Notions, book series Vol. 13, 2002, 30-36. [4] Pan Chengdong and Pan Chengbiao. Foundation of Analytic Number Theory. Beijing: Science Press, 1999. [5] Tom M, Apostol. Introduction to Analytic Number Theory. New York: Springer-Verlag, 1976.

ON A DUAL FUNCTION OF THE SMARANDACHE CEIL FUNCTION Lu Yaming Research Center for Basic Science, Xi’an Jiaotong University, Xi’an, Shaanxi, P.R.China

Abstract

Let n and k are any two positive integers, Sk (n) denotes the largest positive integer x satisfying xk |n. The main purpose of this paper is using the elementary methods to study the mean value properties of the arithmetical function d(Sk (n)), where d(n) is the Dirichlet divisor function, and give an interesting asymptotic formula for it.

Keywords:

Arithmetical function; Mean value; Asymptotic formula.

§1. Introduction and results Let k be a fixed positive integer. For any positive integer n, the Smarandache ceil function of order k is denoted by Sk (n) and has the following definition: Sk (n) = min{x ∈ N : n|xk }. This arithmetical function is a multiplicative function, and has many interesting properties, so it had be studied by many people, see reference [2]. Now we introduce a dual function of Sk (n) as follows: Sk (n) = max{x ∈ N : xk |n}. It is clear that Sk (n) is also a multiplicative function, but about its arithmetical properties, we know very little at present. In this paper, we use the elementary methods to study the mean value properties of d(Sk (n)), where d(n) is the Dirichlet divisor function, and obtain a sharper asymptotic formula for it. That is, we will prove the following: Theorem. Let k ≥ 2 be a fixed integer. Then for any real number x > 1, we have the asymptotic formula X

1

d(S1 (n)) = x ln x + (2γ − 1)x + O x 3

n≤x

and X n≤x

1 1 1 x k + O x k+1 , k

d(Sk (n)) = ζ(k)x + ζ

56

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

where γ is the Euler constant, and ζ(n) is the Riemann zeta-function. From this Theorem we may immediately deduce the following Corollary 1. For any real number x > 1, we have the asymptotic formula X

d(S2 (n)) =

n≤x

π2 x+ζ 6

1 1 1 x2 + O x3 . 2

Corollary 2. For any real number x > 1, we have the asymptotic formula π4 d(S4 (n)) = x+ζ 90 n≤x X

1 1 1 x4 + O x5 . 4

§2. Proof of the theorem In this section, we will give the proof of the theorem. The following Lemma is necessary. Lemma. If x ≥ 1 and s ≥ 0, s 6= 1, we have X 1 n≤x

ns

=

x1−s + ζ(s) + O(x−s ), 1−s

where

∞ t − [t] 1 −s dt. ζ(s) = 1 + s−1 ts+1 1 Proof. This Lemma can be easily proved by using the Euler summation formula, see Theorem 3.2(b) of [3]. Now we come to the proof of the theorem. It is obvious that S1 (n) = n, and this deduces the first part of the theorem immediately by the classical result on Dirichlet divisor problem, see reference [4]. Now assume k ≥ 2, we have

Z

X

d(Sk (n)) =

X X

1,

n≤x d|Sk (n)

n≤x

from the definition of Sk (n) we know that d|Sk (n) ⇔ dk |n, hence X

d(Sk (n)) =

n≤x

XX n≤x dk |n

1

1=

X

1.

dk l≤x

Let δ = x k+1 , applying the above formula and Theorem 3.17 of [3] we have X n≤x

d(Sk (n)) =

X dk l≤x

1

57

On a dual function of the Smarandache ceil function X

=

X

1+

1≤dk ≤δ k 1≤l≤x/dk

X

=

X

X x

=

dk

1≤d≤δ

+

X

X

X

X

1−

1≤l≤δ 1≤dk ≤x/l

1+

1≤d≤δ 1≤l≤x/dk

X

1

1≤dk ≤δ k

X

1

1≤l≤δ

1 − [δ]2

1≤l≤δ 1≤dk ≤x/l

X

" # 1/k

1≤l≤δ

x l

− (δ 2 − 2δ{δ} + {δ}2 ).

Using the above Lemma we get X

d(Sk (n))

n≤x

= x

X 1≤d≤δ

X 1 1 1/k + x + O(δ) − (δ 2 + O(δ)) 1/k dk l 1≤l≤δ !

δ 1−k + ζ(k) + O(δ −k ) + x1/k = x 1−k

1

δ 1− k +ζ 1 − k1

1 k

+ O(δ

− k1

!

)

−δ 2 + O(δ). Notice that x = δ k+1 , we have X

=

d(Sk (n))

n≤x δ2

1−k

+ ζ(k)x +

δ2 +ζ 1 − k1

1 1 x k − δ 2 + O(δ) k

1 1 x k + O(δ) k 1 1 1 = ζ(k)x + ζ x k + O(x k+1 ). k

= ζ(k)x + ζ

This completes the proof of Theorem.

References [1] F. Smarandache. Only Problems, Not solutions. Chicago: Xiquan Publishing House, 1993. [2] Sabin Tabirca and Tatiana Tairca. Some new results concerning the Smarandache ceil function, Smarandache Notions, book series Vol. 13, 2002, 30-36. [3] Tom M, Apostol. Introduction to Analytic Number Theory. New York: Springer-Verlag. 1976 [4] Pan Chengdong and Pan Chengbiao. Foundation of Analytic Number Theory. Beijing: Science Press, 1999.

ON THE LARGEST M -TH POWER NOT EXCEEDING N∗ Liu Duansen and Li Junzhuang Institute of Mathematics, Shangluo Teacher’s College, Shangluo, Shaanxi, P.R.China

Abstract

In this paper, some elementary methods are used to study the properties of the largest m-th power not exceeding n, and give an identity about it.

Keywords:

The largest m-th power not exceding n ; Dirichlet series; Identity.

§1. Introduction and results Let n is a positive integer. It is clear that there exists an integer k such that k m ≤ n < (k + 1)m . Now we define bm (n) = k m . That is, bm (n) is the largest m-th power not exceeding n. In problems 40 and 41 of reference [1], Professor F.Smarandache asked us to study the properties of the sequences {b2 (n)} and {b3 (n)}. About these problems, some people had studied them, and obtained some interesting results, see references [2] and [3]. In this paper, we using the elementary methods to study the convergent properties of the Dirichlet series f (s) = ∞ X 1 , and give an interesting identity. That is, we shall prove the fols b (n) n=1 m lowing: Theorem. Let m be a fixed positive integer. Then for any real number s > 1, the Dirichlet series f (s) is convergent and 1 2 m−1 f (s) = Cm ζ(ms−m+1)+Cm ζ(ms−m+2)+· · ·+Cm ζ(ms−1)+ζ(ms), m! n = where Cm n!(m−n)! , and ζ(s) is the Riemann zeta-function. From this theorem we may immediately deduce the following:

∗ This

work is supported by N.S.F. of P.R.China(10271093) and the Education Department Foundation of Shannxi Province(04JK132)

60

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

Corollary 1. Taking m = 2 and s = 3/2 or m = s = 2 in the above Theorem, then we have the identities ∞ X n=1

1 b

3 22

(n)

=

π2 + ζ(3) and 3

∞ X

1 π4 = 2ζ(3) + . 2 90 n=1 b2 (n)

Corollary 2. Taking m = 3 and s = 2 or m = 2 and s = 3 in the above theorem, we have the identities ∞ X

1 π4 π6 = + 3ζ(5) + 2 30 945 n=1 b3 (n)

and

∞ X

π6 1 = 2ζ(5) + . 3 945 n=1 b2 (n)

§2. Proof of the theorem In this section, we shall complete the proof of the theorem. For any positive integer n, let bm (n) = k m . It is clear that there are exactly (k + 1)m − k m integer n such that bm (n) = k m . So we may get f (s) = =

∞ X (k + 1)m − k m 1 = bs (n) k=1 k ms n=1 m ∞ X

∞ 1 k m−1 + C 2 k m−2 + · · · + C m−1 k + 1 X Cm m m k=1

k ms

.

From the integral criterion, we know that f (s) is convergent if ms−(m−1) > ∞ X 1 1. That is, s > 1. If s > 1, note that ζ(s) = , we have s n n=1 1 2 m−1 f (s) = Cm ζ(ms − m + 1) + Cm ζ(ms − m + 2) + · · · + Cm ζ(ms − 1) + ζ(ms).

This completes the proof of Theorem. It’s easy to compute that 1 2 m−1 f (2) = Cm ζ(m + 1) + Cm ζ(m + 2) + · · · + Cm ζ(2m − 1) + ζ(2m), 1 2 m−1 f (3) = Cm ζ(2m + 1) + Cm ζ(2m + 2) + · · · + Cm ζ(3m − 1) + ζ(3m).

Now the corollaries follows from ζ(2) = π 2 /6, ζ(4) = π 4 /90, ζ(6) = (see reference [4]) and the above theorem.

π 6 /945

References [1] F. Smarandache. Only Problems, Not solutions. Chicago: Xiquan Publishing House, 1993.

On the largest m-th power not exceeding n1

61

[2] Zhang Wenpeng. On the cube part sequence of a positive integer. Journal of Xian Yang Teacher’s College, 2003, 18: 5-7. [3] Zheng Jianfeng. On the inferior and superior k-th power part of a positive integer and divisor function. Smarandache Notions Journal, 2004, 14: 88-91. [4] Tom M. Apostol. Introduction to Analytic Number Theory. New York: Springer-Verlag, 1976.

ON THE SMARANDACHE BACK CONCATENATED ODD SEQUENCES ∗ Li Junzhuang and Wang Nianliang Institute of Mathematics, Shangluo Teacher’s College Shangluo, Shaanxi, P.R.China

Abstract

The main purpose of this paper is to study the arithmetical properties of the Smarandache back concatenated odd sequences, and give several simple properties involving the recursion formula and exact expressions for the general term.

Keywords:

The Smarandache back concatenated odd sequence; Arithmetical properties; Recursion formula.

§1. Introduction and main results The famous Smarandache back concatenated odd sequence {bn } is defined as following 1, 31, 531, 7531, 97531, 1197531, 131197531, 15131197531, 1715131197531, · · · . In problem 3 of reference [1], Professor Mihaly Bencze and Lucian Tutescu asked us to study the arithmetical properties about this sequence. It is interesting for us to study this problem. But it’s a pity that none had studied it before. At least we haven’t seen such a paper yet. In this paper, we shall use the elementary methods to study the arithmetical properties of the Smarandache back concatenated odd sequences, and give several simple properties involving the recursion formula, exact expressions for the general term, and so on. That is, we shall prove the following: Theorem 1. Let n ≥ 2 be any positive integer with (2n − 3) has k digits. Then for the Smarandache back concatenated odd sequences {bn }, we have the following recursion formula bn = bn−1 + (2n − 1) × 10

10−10k 18

+k×(n−1)

,

where b1 = 1.

∗ This

work is supported by N.S.F. of P.R.China (10271093)and the Education Department Foundation of Shannxi Province(04JK132)

64

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

Theorem 2. Let (2n − 3) has k digits. Then the bn−1 -th term in the k Smarandache back concatenated odd sequences {bn } has 10−10 +k ×(n −1) 18 digits. Theorem 3.

Let Sk(m,n) =

n X

(2i − 1) × 10k(i−1) , then we have the

i=m

following exact expression for the general term. That is, −(k−1)

z }| {

bn = 1+S1(2,6) +10−5 ×S2(7,51) +10−55 ×S3(52,501) +· · ·+10−5 · · · 5 ×Sk(m,n) .

Theorem 4. Let S50 denotes the summation for the first 50 terms in the Smarandache back concatenated odd sequences {bn }, then we have S50 =

11 × 45 × 106 − 201 × 50 −81 × 106 + 970 + 9 92 95 6 −4 × 10 + 4 × 100 99 × 10 − 13 × 44 × 107 + + 93 99 95 × 1095 + 73 × 109 −4 × 1095 + 4 × 1011 + + . 992 993

§2. Some lemmas Lemma 1. Let k, m, n are positive integers with m ≤ n, then we have Sk(m,n) =

(2m − 1) × 10k(m−1) 10km [1 − 10k(n−m) ] (2n − 1) × 10kn +2× + . 1 − 10k (1 − 10k )2 1 − 10k

Proof. From the definition of Sk(m,n) , we have Sk(m,n) =

n X

(2i − 1) × 10k(i−1)

i=m

= (2m − 1) × 10k(m−1) + (2m + 1) × 10km + · · · + (2n − 1) × 10k(n−1) and 10k Sk(m,n) = (2m−1)×10km +(2m+1)×10k(m+1) +· · ·+(2n−1)×10kn . Thus we can get (1 − 10k )Sk(m,n) = (2m − 1) × 10k(m−1) + 2 × [10km + 10k(m+1)

On the Smarandache back concatenated odd sequences 1

65

+ · · · + 10k(n−1) ] − (2n − 1) × 10kn = (2m − 1) × 10k(m−1) + 2 ×

10km [1 − 10k(n−m) ] 1 − 10k

−(2n − 1) × 10kn . So we have Sk(m,n) =

10km [1 − 10k(n−m) ] (2n − 1) × 10kn (2m − 1) × 10k(m−1) +2× + . 1 − 10k (1 − 10k )2 1 − 10k

This proves Lemma 1. Lemma 2.

0 Let Sk(n,m) =

n X

i2 × 10−ki . Then for any positive integers

i=m

k, m, n, we have 0 Sk(m,n) =

m2 × 10−km (2m + 1) × 10−k×(m+1) + 1 − 10−k (1 − 10−k )2 +2 × −

10−k(m+2) [1 − 10−k(n−m−1) ] (1 − 10−k )3

(2n − 1) × 10−k×(n+1) n2 × 10−k(n+1) − . (1 − 10−k )2 1 − 10−k

0 Proof. From the definition of Sk(m,n) , we have 0 Sk(m,n)

= m2 × 10−k×m + (m + 1)2 × 10−k×(m+1) + (m + 2)2 × 10−k×(m+2) + · · · + [m + (n − m)]2 × 10−k[m+(n−m)] and 0 10−k Sk(m,n) = m2 × 10−k×(m+1) + (m + 1)2 × 10−k×(m+2)

+ · · · + [m + (n − m)]2 × 10−k[m+(n−m)+1] . Thus we can get 0 (1 − 10−k )Sk(m,n)

= m2 × 10−k×m + (2m + 1) × 10−k×(m+1) + (2m + 3) × 10−k×(m+2) + · · · + {2[m + (n − m)] − 1} × 10−k[m+(n−m)] −[m + (n − m)]2 × 10−k[m+(n−m)]+1 .

66

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

So we have n X 0 Sk(m,n) =

=

(2i − 1) × 10−k×i

n2 × 10−k(n+1) × i=m+1 + + 1 − 10−k 1 − 10−k 1 − 10−k 2 −km −k×(m+1) m × 10 (2m + 1) × 10 + −k 1 − 10 (1 − 10−k )2

m2

10−km

+2 × −

10−k(m+2) [1 − 10−k(n−m−1) ] (1 − 10−k )3

(2n − 1) × 10−k×(n+1) n2 × 10−k(n+1) − . (1 − 10−k )2 1 − 10−k

This proves Lemma 2. Lemma 3.

00 Let Sk(n,m) =

n X

i × 10−ki . Then for any positive integers

i=m

k, m, n, we have 00 Sk(m,n) =

m2 × 10−km 10−k(m+1) [1 − 10−k(n−m) ] n × 10−k(n+1) + − . 1 − 10−k (1 − 10−k )2 (1 − 10−k )

00 Proof. From the definition of Sk(m,n) , we have 00 Sk(m,n) = m × 10−km + (m + 1) × 10−k(m+1)

+ · · · + [m + (n − m)] × 10−k[m+(n−m)] and 00 10−k Sk(m,n) = m × 10−k(m+1) + (m + 1) × 10−k(m+2)

+ · · · + [m + (n − m)] × 10−k[m+(n−m)+1] . Thus we can get 00 (1 − 10−k )Sk(m,n) = m × 10−km + 10−k(m+1) + · · · + 10−k[m+(n−m)]

−[m + (n − m)] × 10−k[m+(n−m)+1] . So we have 00 Sk(m,n) =

m × 10−km 10−k(m+1) [1 − 10−k(n−m) ] n × 10−k(n+1) + × − . 1 − 10−k (1 − 10−k )2 (1 − 10−k )

This proves Lemma 3.

On the Smarandache back concatenated odd sequences 2

67

§3. Proof of the theorems In this section, we shall use the above lemmas to complete the proof of the theorems. First we use mathematical induction to prove Theorem 1. (i) If n = 2, 3, it is obvious that Theorem 1 is true. (ii) Assuming that Theorem 1 is true for n = m. That is, the following recursion formula bm = bm−1 + (2m − 1) × 10

10−10k 18

+k×(m−1)

is true for the Smarandache back concatenated odd sequence {bn }. Comparing the difference between bm and bm−1 , we know that bm−1 has 10−10k + k × (m − 1) digits. If n = m + 1, we can divide it into two cases: 18 k (i) If 2m − 1 still has k digits, then we know that bm has [ 10−10 +k× 18 (m − 1) + k] digits. Comparing the difference between bm+1 and bm , we may immediately deduce that 10−10k bm+1 − bm = 10 18 +k×(m−1)+k . 2m + 1

That is, bm+1 = bm + (2m + 1) × 10

10−10k 18

+k×m

. k

+k× (ii) If 2m − 1 has k + 1 digits, then we know that bm has [ 10−10 18 (m − 1) + k + 1] digits. Recall that (2m − 3) has k digits and (2m − 1) has k + 1 digits, which exist only in the case that 2m − 3 = 10k − 1, 2m − 1 = 10k + 1. That is, m =

10k 2

+ 1. Therefore, we have

10 − 10k + k × (m − 1) + (k + 1) = 18 =

10 − 10k 10k +k× + (k + 1) 18 2 10 − 10k+1 + (k + 1) × m 18

Comparing the difference between bm+1 and bm , we can deduce that 10−10k 10−10k+1 bm+1 − bm = 10 18 +k×(m−1)+(k+1) = 10 18 +(k+1)×m . 2m + 1

That is, bm+1 = bm + (2m + 1) × 10

10−10k+1 +(k+1)×m 18

.

Combining (i) and (ii), Theorem 1 is true for any positive integer n. This completes the proof of Theorem 1. By using the result of Theorem 1, and note that the difference between bn and bn−1 , we can immediately get the result of Theorem 2.

68

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

By Theorem 1, when (2n − 3) has k digits, let (2m − 3) be the least positive k−1 integer which has k odd digits (here m = 2 when k = 1, and m = 10 2 + 2). Then we have bn = 1 + (2 × 2 − 1) × 10

10−101 +1×(2−1) 18

+(2 × 3 − 1) × 10

10−101 +1×(3−1) 18

+(2 × 4 − 1) × 10

10−101 +1×(4−1) 18

+ (2 × 5 − 1) × 10

10−101 +1×(5−1) 18

+(2 × 6 − 1) × 10

10−101 +1×(6−1) 18

+ (2 × 7 − 1) × 10

10−102 +2×(7−1) 18

+(2 × 8 − 1) × 10

10−102 +2×(8−1) 18

+ · · · + (2 × 51 − 1) × 10

10−102 +2×(51−1) 18

+(2 × 52 − 1) × 10

10−103 +3×(52−1) 18

+(2 × 53 − 1) × 10

10−103 +3×(53−1) 18

+ · · · + (2 × m − 1) × 10 + · · · + (2 × n − 1) × 10 +10

10−102 18

+ · · · + 10 +1 + 10

10−10k 18

× S2(7,51) + 10 10−10k 18

10−101

+ · · · + 10

10−10k 18

18

18

= 1 + S1(2,6) + 10

+k×(n−1)

10−103 18

+ 1 + 10

10−101 18

× S1(2,6)

× S3(52,501)

× Sk(m,n)

× S1(2,6) + 10

10−10k

+k×(m−1)

10−102 18

× S2(7,51) + 10

10−103 18

× S3(52,501)

× Sk(m,n)

−5

× S2(7,51) + 10−55 × S3(52,501)

−(k−1)

z }| {

+ · · · + 10−5 · · · 5 × Sk(m,n) . Applying the result of Lemma 1, we can deduce Theorem 3. Now we come to prove Theorem 4. By using the result of Theorem 1, we have S50 = b1 + b2 + b3 + · · · + b49 + b50 = 50 + 49 × (2 × 2 − 1) × 10

10−101 +1×(2−1) 18

+48 × (2 × 3 − 1) × 10

10−101 +1×(3−1) 18

+47 × (2 × 4 − 1) × 10

10−101 +1×(4−1) 18

+46 × (2 × 5 − 1) × 10

10−101 +1×(5−1) 18

+45 × (2 × 6 − 1) × 10

10−101 +1×(6−1) 18

On the Smarandache back concatenated odd sequences 3

69

+44 × (2 × 7 − 1) × 10

10−102 +2×(7−1) 18

+43 × (2 × 8 − 1) × 10

10−102 +2×(8−1) 18

+ · · · + 3 × (2 × 48 − 1) × 10

10−102 +2×(48−1) 18

+2 × (2 × 49 − 1) × 10

10−102 +2×(49−1) 18

+1 × (2 × 50 − 1) × 10

10−102 +2×(50−1) 18

= 50 × [2 × (51 − 50) − 1] × 10

10−101 +1×(50−50) 18

+49 × [2 × (51 − 49) − 1] × 10

10−101 +1×(50−49) 18

+48 × [2 × (51 − 48) − 1] × 10

10−101 +1×(50−48) 18

+47 × [2 × (50 − 47) − 1] × 10

10−101 +1×(50−47) 18

+ · · · + 45 × [2 × (51 − 45) − 1] × 10 +44 × [2 × (50 − 44) − 1] × 10

10−102 +2×(50−44) 18

+ · · · + 2 × [2 × (51 − 2) − 1] × 10 +1 × [2 × (51 − 1) − 1] × 10 = 101 × [50 × 10

+44 × 10

+1 × 10

10−101 +1×50−1×45 18

10−102 +2×50−2×2 18

10−102 +2×50−2×1 18

−2 × [502 × 10 +492 × 10 +442 × 10

10−101 +1×50−1×49 18 10−101 +1×50−1×45 18

10−102 +2×50−2×44 18

+ · · · + 22 × 10

= 101 ×

]

10−101 +1×50−1×50 18

+ · · · + 452 × 10

+12 × 10

+ 49 × 10

10−102 +2×50−2×44 18

+ · · · + 2 × 10

10−102 +2×50−2×2 18

10−102 +2×50−2×1 18

50 X

i × 10

i=45 44 X

+101 ×

i=1

]

10−101 +1×50−1×i 18

i × 10

10−102 +2×(50−2) 18

10−102 +2×(50−1) 18

10−101 +1×50−1×50 18

+ · · · + 45 × 10

10−101 +1×(51−45) 18

10−102 +2×50−2×i 18

10−101 +1×50−1×49 18

70

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

−2 × −2 ×

50 X

10−101 +1×50−1×i 18

i2 × 10

i=45 44 X

i2 × 10

10−102 +2×50−2×i 18

i=1 50

= 101 × 10

×

50 X

−i

i × 10

95

+ 101 × 10

i=45

−2 × 1050 ×

50 X

i2 × 10−i − 2 × 1095 ×

i=45

×

44 X

i × 10−2×i

i=1 44 X

i2 × 10−2×i .

i=1

Applying the results of Lemma 2 and Lemma 3, we can get S50 =

11 × 45 × 106 − 201 × 50 −81 × 106 + 970 + 9 92 6 95 −4 × 10 + 4 × 100 99 × 10 − 13 × 44 × 107 + + 93 99 95 × 1095 + 73 × 109 −4 × 1095 + 4 × 1011 + . + 992 993

This completes the proof of Theorem 4.

References [1] Mihaly Bencze and Lucian Tutescu (Eds.). Some notions and questions in number theory. Vol 2. http://www.gallup.unm.edu/ smarandache/SNAQINT2.TXT. [2] F.Smarandache, Only problems, not Solutions, Xiquan Publ. House, Chicago, 1993. [3] Tom M.Apostol, Introduction to Analytic Number Theory, SpringerVerlag, New York, Heidelberg Berlin, 1976.

ON THE ADDITIVE HEXAGON NUMBERS COMPLEMENTS∗ Li chao and Yang Cundian Institute of Mathematics, Shangluo Teacher’s College, Shangluo, Shaanxi, P.R.China

Abstract

In this paper, similar to the Smarandache k-th power complements, we defined the hexagon numbers complements. Using the elementary method, we studied the mean value properties of the additive hexagon numbers complements, and obtained some interesting asymptotic formulae for it.

Keywords:

Additive hexagon numbers complements; Mean value; Asymptotic formula.

§1. Introduction and results Let n be a positive integer. If there exists a positive integer m such that n = m(2m − 1), then we call n as a hexagon number. For any positive integer n, the Smarandache k-th power complements bk (n) is the smallest positive integer such that nbk (n) is complete k-th power, see problem 29 of [1]. Similar to the Smarandache k-th power complements, we define the additive hexagon numbers complements a(n) as follows: a(n) is the smallest nonnegative integer such that a(n) + n is a hexagon number. For example, if n = 1, 2, · · · 15, we have the additive hexagon number sequences {a(n)} (n = 1, 2, · · · 15) as follows: 0, 4, 3, 2, 1, 0, 8, 7, 6, 5, 4, 3, 2, 1, 0. In this paper, we study the mean value properties of the composite arithmetic function d(a(n)) (where d(n) is the Dirichlet divisor function), and give some interesting asymptotic formulae for it. That is, we shall prove the following conclusions: Theorem 1. For any real number x ≥ 3, we have the asymptotic formula: √ X 2 2 3 a(n) = x 2 + O(x). 3 n≤x Theorem 2. For any real number x ≥ 3, we have the asymptotic formula: 1 d(a(n)) = x log x + 2 n≤x X

∗ This

2 3 1 log 2 + (2γ − 1) − x + O(x 3 ), 2 2

work is supported by N.S.F. of P.R.China(10271093) and the Education Department Foundation of Shannxi Province(04JK132)

72

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

where γ is the Euler constant.

§2. Some Lemmas Before the proof of the theorems, some Lemmas will be useful. Lemma 1. For any real number x ≥ 3, we have the asymptotic formula: X

1

d(x) = x ln x + (2γ − 1)x + O x 3 ,

n≤x

where γ is the Euler constant. Proof. See reference [2]. Lemma 2. For any real number x ≥ 3 and any nonnegative arithmetical function f (n) with f (0) = 0, we have the asymptotic formula: X

f (a(n)) =

1 x2 2

   f (i) + O   i≤4m 

X X m=1

n≤x

   f (i) ,  1  x2

X i≤4

2

where [x] denotes the greatest integer less than or equal to x. Proof. For any real number x ≥ 1, let M be a fixed positive integer such that M (2M − 1) ≤ x < (M + 1)(2M + 1). Noting that if n pass through the integers in the interval [m(2m − 1), (m + 1)(2m + 1)) , then a(n) pass through the integers in the interval [0, 4m] and f (0) = 0, we can write X

X

f (a(n)) =

n≤x

n≤M (2M −1)

=

X

f (a(n)) +

M X X

f (a(n))

M (2M −1)<n≤x

 X

f (i) + O 

m=1 i≤4m

f (i) ,

i≤x−M (2M −1)

Since x − M (2M − 1) < (M + 1)(2M + 1) − M (2M − 1) = 4M + 1 and 1

M=

x2 2

, we have X n≤x

f (a(n)) =

1 x2 2

   f (i) + O   i≤4m 

X X m=1

   f (i) .  1  x2

X i≤4

2

On the additive hexagon numbers complements1

73

This proves Lemma 2. Note: This Lemma is very useful. Because if we have the mean value formula of X f (n), then from this Lemma, we can easily get the mean value formula of f (a(n)). n≤x

§3. Proof of the theorems In this section, we will complete the proof of the theorems. First, we prove Theorem 1. From the definition of a(n), and the Euler summation formula (see [3]), we have X

X

a(n) =

n≤x

n≤M (2M −1)

=

M X X

=

X

i

i≤x−M (2M −1)

(4M )2 2

2m(4m + 1) + O

m=1

= =

a(n)

M (2M −1)<n≤x

i+

m=1 i≤4m M X

X

a(n) +

!

4 M (M + 1)(2M + 1) + O(M 2 ) 3√ 2 2 3 x 2 + O(x). 3

This proves Theorem 1. Now we prove Theorem 2. From Lemma 2, Lemma 1 and the Abel’s identity (see [3]), we have X

d(a(n))

n≤x

=

d(i) + O 

m=1 i≤4m

=

M X

M X X

X

d(i)

i≤4M 1

4m log 4m + (2γ − 1)4m + O (4m) 3

m=1

1

+O 4M log 4M + (2γ − 1)4M + O (4M ) 3 = (8 log 2 + 4(2γ − 1))

X m≤M

+O 

m+4

X

m log m

m≤M

 X

1 3

m  + O (4M log 4M )

m≤M

= (8 log 2 + 4(2γ − 1))

1 2 M + O(M ) 2

74

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II 4 1 2 1 +4 M log M − (M 2 − 1) + O(M log M ) + O(M 3 ) 2 4

4

= 2M 2 log M + (4 log 2 + 2(2γ − 1) − 1)M 2 + O(M 3 ) 2 1 3 1 x + O(x 3 ). = x log x + log 2 + (2γ − 1) − 2 2 2 This completes the proof of the theorems.

References [1] F.Smaradache. Only problems, not solutions, Xiquan Publishing House, Chicago, 1993. [2] H.Iwaniec and C.J.Mozzochi. On the divisor and circle problems. Journal of Number theory, Vol. 28(1988), 60-93. [3] Tom M. Apostol. Introduction to Analytic Number Theory, SpringerVerlag, New York, 1976, pp54 and pp77.

ON THE MEAN VALUE OF A NEW ARITHMETICAL FUNCTION∗ Yang Cundian and Liu Duansen Institute of Mathematics, Shangluo Teacher’s College, Shangluo, Shaanxi, P.R.China zzz@zzz.com

Abstract

The main purpose of this paper is using the elementary methods to study the mean value properties of a new arithmetical function σ(S(n)), and give an interesting mean value formula for it.

Keywords:

Smarandache function; Mean value; Asymptotic formula.

§1. Introduction and results For any positive integer n, the Smarandache function S(n) is defined as follows: S(n) = min{m ∈ N : n|m!}. This function was introduced by American-Romanian number theorist Professor F.Smarandache, see reference [1]. About its arithmetical properties, many scholars had studied it, and obtained some interesting conclusions, see reference [2] and [3]. In this paper, we shall use the elementary methods to study the mean X value properties of a new arithmetical function σ(S(n)), where σ(n) = d, and d|n

give an interesting mean value formula for it. That is, we shall prove the following: Theorem. For any real number x ≥ 3, we have the asymptotic formula π 2 x2 σ(S(n)) = +O 12 ln x n≤x X

∗ This

!

x2 . ln2 x

work is supported by N.S.F. of P.R.China(10271093) and the Education Department Foundation of Shannxi Province(04JK132)

76

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

§2. Some simple lemmas To complete the proof of Theorem, we need some simple Lemmas. For convenience, we denotes the greatest prime divisor of n by p(n). Then we have √ Lemma 1. If n is a square free number or p(n) > n, then S(n) = p(n). Proof. If n be a square-free number, let n = p1 p2 · · · pr p(n), then pi < p(n) for i = 1, 2, · · · , r. Thus pi |p(n)!,

i = 1, 2, · · · , r.

So n|p(n)! , but p(n)†(p(n) − 1)! and n†(p(n) − 1)!. In this case, we have S(n) = p(n);√ If p(n) > n, then p2 (n)†n. Let n = pα1 1 pα2 2 · · · pαr r p(n), so we have √ pα1 1 pα2 2 · · · pαr r < n < p(n). It is clear that pαi i |p(n)!,

i = 1, 2, · · · , r.

So n|(p(n))!, but p(n)†(p(n) − 1)!, in this case, we can also deduce that S(n) = p(n). This proves Lemma 1. Lemma 2. Let p be a prime, then we have the asymptotic formula X √

x≤p≤x

x2 p= +O 2 ln x

!

x2 . ln2 x

Proof. Let π(x) denotes the number of the primes up to x. Noting that (see [4]) x x π(x) = +O , ln x ln2 x from the Abel’s summation formula [5], we have X √

Z x √ √ p = π(x)x − π( x) x − √ π(t)dt x

x≤p≤x

This proves Lemma 2.

=

x2 x2 − +O ln x 2 ln x

=

x2 +O 2 ln x

x2 ln2 x !

x2 . ln2 x

!

On the mean value of a new arithmetical function1

77

§3. Proof of the theorem In this section, we shall complete the proof of Theorem. First we define the sets A and B as following: √ √ A = {n|n ≤ x, p(n) ≤ n} and B = {n|n ≤ x, p(n) > n}. Note that S(n) p(n) ln n and σ(n) n ln ln n, we have the estimate X

σ(S(n))

n∈A

X√

X

S(n) ln ln (S(n))

n∈A 3

n ln n ln ln n x 2 ln x ln ln x.

(1)

n≤x

Then using the above Lemmas we may immediately get X

σ(S(n)) =

X

σ(p(n)) =

n≤x √ p(n)> n

n∈B

X

X

√ √ x n≤ x n≤p≤ n

σ(p)

=

=

X

√ √ x n≤ x x≤p≤ n

X n≤ln2 x

=

X

(p + 1) + O 

X

√ √ √ n≤ x n≤p≤ x

x2 + 2n2 ln nx

π 2 x2 +O 12 ln x

 X

X √ ln2 x≤n≤ x

p

3 x2 2 x + O(x ln x) 2 2n ln n

!

x2 . ln2 x

(2)

Combining (1) and (2) we obtain X

σ(S(n)) =

n≤x

X

σ(S(n)) +

n∈A

=

π 2 x2 +O 12 ln x

X

σ(S(n))

n∈B

!

x2 . ln2 x

This completes the proof of Theorem.

References [1] F. Smarandache. Only problems, Not solutions. Chicago: Xiquan Publ. House, 1993. [2] Li Hailong and Zhao Xiaopeng. Research on Smarandache Problems In Number Theory. Hexis, 2004, 119-122. [3] Mark Farris and Patrick Mitchell. Bounding the Smarandache function. Smarandache Notions Journal, Vol 13(2002), 37-42. [4] Tom M.Apostol. Introduction to Analytic Number Theory. SpringerVerlag: New York, 1976.

ON THE MEAN VALUE OF A NEW ARITHMETICAL FUNCTION∗ Zhao Jian and Li Chao Institute of Mathematics, Shangluo Teacher’s College, Shangluo, Shaanxi, P.R.China zzz@zzz.com

Abstract

The main purpose of this paper is using the elementary methods to study the convergent properties of a new Dirichlet’s series involving the triangular numbers, and give an interesting identity for it.

Keywords:

Dirichlet’s series; Convergence; Identity.

§1. Introduction and results For any positive integer m, it is clear that m(m + 1)/2 is a positive integer, and we call it as the triangular number, because there are close relations between these numbers and the geometry. Now for any positive integer n, we define arithmetical function c(n) as follows: c(n) = max {m(m + 1)/2 : m(m + 1)/2 ≤ n, m ∈ N } . That is, c(n) is the greatest triangular number ≤ n. From the definition of c(n) we can easily deduce that c(1) = 1, c(2) = 1, c(3) = 3, c(4) = 3, c(5) = 3, c(6) = 6, c(7) = 6, c(8) = 6, c(9) = 6, c(10) = 10, · · · · · ·. About this function, it seems that none had studied it before, even we do not know its arithmetical properties. In this paper, we introduce a new Dirichlet’s series f (s) involving the sequences {c(n)}, i.e., f (s) =

∞ X

1 . s (n) c n=1

Then we using the elementary methods to study the convergent properties of f (s), and obtain an interesting identity. That is, we shall prove the following result: Theorem Let s be any positive real number. Then the Dirichlet’s series f (s) is convergent if and only if s > 1. Especially for s = 2, 3 and 4, we have ∗ This

work is supported by N.S.F. of P.R.China(10271093) and the Education Department Foundation of Shannxi Province(04JK132)

80

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

the identities: ∞ X

2π 2 1 = − 4; c2 (n) 3 n=1 ∞ X

1 = 8ζ(3) − 4π 2 + 32; 3 (n) c n=1 ∞ X

1 π4 160 2 = − 48ζ(3) + π − 384; 4 c (n) 5 3 n=1 π 4 88 2 + π − 192. 5 3 where ζ(k) is the Riemann zeta-function. 6f (3) + f (4) =

§2. Proof of the theorem In this section, we will complete the proof of the theorem. It is easy to see ≤ n < (m+1)(m+2) , then c(n) = m(m+1) . So the same numthat if m(m+1) 2 2 2 m(m+1) (m+1)(m+2) m(m+1) bers repeated − = m + 1 times in the sequences 2 2 2 {c(n)}. Hence, we can write f (s) = = =

∞ X

1 s (n) c n=1 ∞ X

m+1

m(m+1) s ) m=1 ( 2 ∞ X 2s m=1

ms (m + 1)s−1

.

It is clear that f (s) is convergent if s > 1, divergent if s ≤ 1. Specially if s = 2, we can write f (2) = 4 = 4

∞ X

1 m2 (m

m=1 ∞ X m=1

+ 1)

1 1 1 − + 2 m m m+1

= 4ζ(2) − 4. Using the same method, we can also obtain f (3) = 8ζ(3) − 24ζ(2) + 32; f (4) = 16ζ(4) − 48ζ(3) + 320ζ(2) − 384. Now the theorem follows from the identities (see reference [2]) ζ(2) = π 2 /6 and ζ(4) = π 4 /90. This complete the proof of the theorem.

On the mean value of a new arithmetical function1

81

Note: In fact, for any positive integer s ≼ 2, using our methods we can express f (s) as the Riemann zeta-function.

References [1] Smarandache F. Only problems, not Solutions. Chicago: Xiquan Publ. House, 1993. [2] Tom M.Apostol. Introduction to Analytic Number Theory. New York: Springer-Verlag, 1976.

ON THE SECOND CLASS PSEUDO-MULTIPLES OF 5 SEQUENCES Gao Nan School of Science, Xi’an Shiyou University, Xi’an, Shaanxi, P.R.China

Abstract

The main purpose of this paper is to study the mean value properties of the second class pseudo-multiples of 5 sequences, and give an interesting asymptotic formula for it.

Keywords:

Second class pseudo-multiples of 5 sequences; Mean value; Asymptotic formula.

§1. Introduction and results A positive integer is called pseudo-multiple of 5 if some permutation of its digits is a multiple of 5, including the identity permutation. In reference [1], Professor F.Smarandache asked us to study the properties of the pseudomultiple of 5 sequences. About this problem, Wang Xiaoying [2] had studied it, and obtained some interesting results. Now, we define the second class pseudo-multiples of 5 numbers as following: A positive integer is called the second class pseudo-multiple of 5 if it is not a multiple of 5, but its some permutation of its digits is a multiple of 5. For convenience, let A denotes the set of all pseudo-multiple of 5 sequences, and B denotes the set of all second class pseudo-multiple of 5 sequences. In this paper, we shall use the elementary methods to study the mean value properties of the second class pseudo-multiple of 5 sequences, and obtain an interesting asymptotic formula for it. That is, we shall prove the following:

Theorem. For any real number x ≥ 1, we have the asymptotic formula ln 5 16 d(n) = x ln x + 2γ − 1 + 25 2 n∈B X

ln 8

+ O x ln 10 +ε ,

n≤x

where d(n) is the Dirichlet’s divisor function, γ is the Euler constant, and ε denotes any fixed positive number.

84

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

§2. Some Lemmas To complete the proof of the above Theorem, we need the following two Lemmas: Lemma 1. Let q be a fixed prime or q = 1. Then for any real number x ≼ 1, we have the asymptotic formula 1 ln q x ln x + 2Îł − 1 + d(qn) = 2 − q 2q −1 n≤x

X

1

+ O x 2 + ,

where Îł is the Euler constant, and denotes any fixed positive number. Proof. If q = 1, then from Theorem 3.3 of [3] we know that Lemma 1 is ∞ X d(qn) . correct. Now for any prime q and real number s > 1, let f (s) = ns n=1 Note that d(n) is a multiplicative function, so by the Euler product formula [3] we have

f (s) =

∞ X d(qn) n=1

=

ns

=

∞ X 2+Îą Îą=0

q sÎą

∞ X

∞ X

Îą=0

n1 =1 (n1 ,q)=1

∞ X n1 =1 (n1 ,q)=1

d(q Îą+1 n1 ) q sÎą ns1

d(n1 ) ns1 



1   1 =  2  1 + 1 1 − qs 1 − qs !

Ă—

Y p (p,q)=1

d(p) d(p2 ) d(pk ) 1 + s + 2s + ¡ ¡ ¡ + ks + ¡ ¡ ¡ p p p





1 1 1  2 + 2  Îś (s) 1 − s q 1 1 − q1s 1 − qs

=  

2

1 = Îś (s) 2 − s . q 2

Then from this identity and the Perron’s formula [4] we may immediately deduce the asymptotic formula X n≤x

d(qn) = 2 −

1 ln q x ln x + 2Îł − 1 + q 2p − 1

This completes the proof of Lemma 1.

1

+ O x 2 + .

On the second class pseudo-multiples of 5 sequences

Lemma 2.

85

For any real number x ≥ 1, we have the asymptotic formula X

ln 8

d(n) = x ln x + (2γ − 1)x + O x ln 10 +ε .

n∈A n≤x

Proof. For any real number x ≥ 1, it is clear that there exists a nonnegative integer k such that 10k ≤ x < 10k+1 . That is, k ≤ log x < k + 1. According to the definition of A we know that the largest number of digits (≤ x) not in A is 8k+1 . In fact, there are 8 one digit, they are 1, 2, 3, 4, 6, 7, 8, 9; There are 82 two digits ; · · · · · ·; The number of k digits are 8k . Since ln 8

8k ≤ 8log x = x ln 10 , we have X

1 ≤ 8 + 8 2 + 83 + . . . + 8k ≤

n6∈A n≤x

Note that d(n) nε and X

ln 8 ln 10

d(n) =

> 12 , applying Lemma 1 with q = 1 we have

X

d(n) −

X

d(n)

n6∈A n≤x

n≤x

n∈A n≤x

64 64 ln 8 8k+2 ≤ 8k ≤ x ln 10 . 7 7 7

=

X

 X   d(n) + O  xε  n6∈A n≤x

n≤x

=

X

ln 8

d(n) + O x ln 10 +ε

n≤x

ln 8

= x ln x + (2γ − 1)x + O x ln 10 +ε . This proves Lemma 2.

§3. Proof of the theorem In this section, we complete the proof of Theorem. From the definition of set A and set B, we know the relationship between them: A − B={multiples of 5}. Combining the above Lemmas we may immediately get X

d(n) =

n∈B n≤x

X

d(n) −

X

d(5n)

5n≤x

n∈A n≤x

ln 8

= x ln x + (2γ − 1)x + O x ln 10 +ε

9 ln 5 x ln x − ln 5 + 2γ − 1 + 25 9 ln 8 16 ln 5 x ln x + 2γ − 1 + + O x ln 10 +ε . 25 2

−

=

86

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

This completes the proof of Theorem. Notes: In fact, we may use the similar method to study other arithmetical functions on the pseudo-multiples of 5 sequences and the second class pseudomultiples of 5 sequences.

References [1] F.Smarandache. Only problems, not solutions, Xiquan Publishing House, Chicago, 1993. [2] Wang Xiaoying. On the Smarandache Pseudo-multiples of 5 Sequence, Hexis, 2004, 17-19. [3] Tom M. Apostol. Introduction to Analytic Number Theory, New York, 1976. [4] Pan Chenddong and Pan Chengbiao. Elements of the Analytic Number Theory, Science Press, Beijing, 1991.

A CLASS OF DIRICHLET SERIES AND ITS IDENTITIES Lou Yuanbing College of Science, Tibet University, Lhasa, Tibet, P.R.China zzz@zzz.com

Abstract

In this paper, we using the elementary methods to study the convergent properties of one class Dirichlet series involving a special sequences, and give an interesting identity for it.

Keywords:

Convergent property; Dirichlet series; Identity.

§1. Introduction and results For any positive integer n and m ≥ 2, we define the m-th power complement bm (n) is the smallest positive integer such that nbm (n) is a complete m-th power, see problem 29 of [1]. Now for any positive integer k, we also define an arithmetic function δk (n) as follows:

δk (n) =

max{d ∈ N

| d|n, (d, k) = 1}, if 0, if

n 6= 0, n = 0.

Let A denotes the set of all positive integers n such that the equation δk (n) = bm (n). That is, A = {n : n ∈ N, δk (n) = bm (n)}. In this paper, we using the elementary methods to study the convergent properties of the Dirichlet series involving the set A, and give an interesting identity for it. That is, we shall prove the following conclusion: Theorem. Let m be a positive even number. Then for any real number s > 1 and positive integer k, we have the identity: 3 Y ζ m p 2 ms 2s , = ns ζ(ms) p|k (pms − 1) p 12 ms − 1

∞ X 1 n=1 n∈A

where ζ(s) is the Riemann zeta-function, and

Q

denotes the product over all

p

primes. From this Theorem we may immediately deduce the following corollaries:

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RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

Corollary 1. Let B = {n : n ∈ N, δk (n) = b2 (n)}, then we have ∞ X 1 n=1 n∈B

n2

=

15 Y p6 . π 2 p|k (p4 − 1)(p2 − 1)

Corollary 2. Let C = {n : n ∈ N, δk (n) = b4 (n)}, then we have ∞ X 1 n=1 n∈C

n2

=

∞ X 1 n=1 n∈B

=

n4

105 Y p12 . π 4 p|k (p8 − 1)(p4 − 1)

Corollary 3. Let C = {n : n ∈ N, δk (n) = b4 (n)}, then we have ∞ X 1 n=1 n∈C

n3

=

p18 675675 1 Y . 691 π 6 p|k (p12 − 1)(p6 − 1)

§2. Proof of the theorem In this section, we will complete the proof of the theorem. For any real number s > 0, it is clear that ∞ X 1 n=1 n∈A

and

∞ X 1 n=1

ns

ns

<

∞ X 1 n=1

is convergent if s > 1, thus

ns

,

∞ X 1

ns

is also convergent if s > 1.

n=1 n∈A α1 α2 p1 p2 · · · pαs s

Now we find the set A. Let n = denotes the factorization of n into prime powers. First from the definitions of δk (n) and bm (n) we konw that δk (n) and bm (n) both are multiplicative functions. So in order to find A, we only discuss the problem in case n = pα . If n = pα and (p, k) = 1, then we α have δk (pα ) = pα ; bm (pα ) = pm−α , if 1 ≤ α ≤ m; bm (pα ) = pm+m[ m ]−α , if α > m and α 6= m; bm (pα ) = 1, if α = rm, where r is any positive integer, and [x] denotes the greatest integer ≤ x. So in this case δk (pα ) = bm (pα ) if and only if α = m 2. α If n = p and (p, k) 6= 1, then δk (pα ) = 1, so in this case the equation δk (pα ) = bm (pα ) has solution if and only if n = prm , r = 0, 1, 2, · · ·. Now from the Euler product formula (see [2]) and the definition of A, we have ∞ X 1 n=1 n∈A

ns

=

Y p†k

1+

1

!

Y

m

p2s

p|k

1+

1 pms

+

1 p2ms

+

1 p3ms

+ ···

89

A class of Dirichlet series and its identities

=

Y

1+

p

1

! Y

m

p2s

p|k

Y 1 1 1 + ms 1 1 − pms p|k p2

!−1

3 Y ζ m p 2 ms 2s , ζ(ms) p|k (pms − 1) p 12 ms + 1

=

where ζ(s) is the Riemann zeta-function, and

Q

denotes the product over all

p

primes. This completes the proof of Theorem. The Corollaries follows from ζ(2) = π 2 /6, ζ(4) = π 4 /90, ζ(6) = π 8 /945, 691π 8 ζ(8) = π 8 /9450 and ζ(12) = 638512875 .

References [1] F.Smarandache. Only problems, not solutions, Xiquan Publishing House, Chicago, 1993. [2] Tom M.Apostol. Introduction to Analytic Number Theory. New York: Springer-Verlag, 1976.

AN ARITHMETICAL FUNCTION AND THE PERFECT K -TH POWER NUMBERS Yang Qianli and Yang Mingshun Department of Mathematics, Weinan Teacher’s College, Weinan, Shaanxi, P.R.China

Abstract

For any primes p and q with (p, q) = 1, the arithmetical function epq (n) defined as the largest exponent of power pq which divides n. In this paper, we use the elementary methods to study the mean value properties of epq (n) acting on the perfect k-th power number sequences, and give an interesting asymptotic formula for it.

Keywords:

the perfect k-th power number; Asymptotic formula; Mean value.

§1. Introduction For any prime p, let ep (n) denotes the largest exponent of power p which divides n. In problem 68 of reference [1], Professor F.Smarandache asked us to study the properties of this arithmetical function. About this problem, many scholars showed great interest in it, and obtained some interesting results, see references [2] and [3]. Similarly, we will define arithmetical function epq (n) as follows: for any two primes p and q with (p, q) = 1, let epq (n) denotes the largest exponent of power pq which divides n. That is, epq (n) = max{Îą : (pq)Îą | n, Îą ∈ N + }. According to [1], a number n is called a perfect k-th power number if it satisfied k|Îą for all pÎą kn, where pÎą kn denotes pÎą |n, but pÎą+1 †n. Let A denotes the set of all the perfect k-th power numbers. It seems that no one knows the relations between these two arithmetical functions before. The main purpose of this paper is using the elementary methods to study the mean value properties of epq (n) acting on the set A, and give an interesting asymptotic formula for it. That is, we shall prove the following: Theorem. Let p and q are two primes with (p, q) = 1, then for any real number x ≼ 1, we have the asymptotic formula X n≤x n∈A

1

1

epq (n) = Cp,q kx k + x 2k + ,

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RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

where Cp,q =

∞ (p − 1)(q − 1) X n n pq (pq) n=1

is a computable positive constant, and denotes any fixed positive number.

§2. Proof of the theorem In this section, we shall complete the proof of the theorem. First we define arithmetical function a(n) as follows:

1, if n is a perfect k-th power number; 0, otherwise .

a(n) =

In order to complete the proof of Theorem, we need the following: Lemma. For any real number x ≼ 1, we have the asymptotic formula 1

X

a(n) = x k

n≤x (n,pq)=1

1 (p − 1)(q − 1) + O x 2k + . pq

Proof. Let

∞ X

f (s) =

n=1 (n,pq)=1

a(n) . ns

where Re(s) > 1. From the Euler product formula [4] and the multiplicative properties of a(n), we have !

a(P k ) a(P 2k ) + + ¡¡¡ 1+ P ks P 2ks

Y

f (s) =

P (P,pq)=1

=

Y

1 1 1 + 2ks + ¡ ¡ ¡ Ă— 1 − ks ks P P p

1+

P

= Îś(ks) 1 −

1 pks

Ă— 1−

Ă— 1−

1 q ks

1 , q ks

where Îś(s) is the Riemann zeta-function, and

Y

denotes the product over all

P

primes. Now by Perron formula [5] with s0 = 0, b = and B(Ďƒ) = Ďƒâˆ’1 1 , we have

1 k

1

+ log1 x , T = x 2k , H(x) = x

k

X n≤x (n,pq)=1

a(n) =

1 2Ď€i

Z

b+iT

Îś(ks) b−iT

1 (pks − 1)(q ks − 1) xs ds + O(x 2k +Îľ ). ks (pq) s

An arithmetical function and the perfect k -th power numbers

93 1 k

To estimate the main term, we move the integral line from b = 1 a = 2k + log1 x Therefore, 1 2Ď€i

Z

b+iT

a+iT

Z

a−iT

Z

!

+

+

+

a−iT

a+iT

b+iT

b−iT

b−iT

Z

+

1 log x

to

xs xs 1 f (s) ds = Res f (s) , . s s k

Note that lim Îś(s)(s − 1) = 1, we may immediately get s→1

1 xs 1 Res f (s) , = xk s k

1 1− p

1 1− . q

Now from the estimate

1

2Ď€i

Z

a+iT

Z

a−iT

+ b+iT

Z

b−iT

!

1 xs

f (s) ds x 2k + ,

s

+ a+iT

a−iT

we can easily get 1

X

a(n) = x k

n≤x (n,pq)=1

1 (p − 1)(q − 1) + O x 2k + . pq

This proves the lemma. Now we prove the theorem. From the properties of geometrical series and the definition of epq (n), combining the lemma we have X

epq (n)

n≤x n∈A

=

X ι≤logpq x k|ι

=

X

Îą

(n,pq)=1

X

a(n)

x n≤ (pq) ι

kÎą

log x ι≤ kpq

x (pq)kÎą

1

k

(p − 1)(q − 1) +O pq

 1

= kx k

∞ X (p − 1)(q − 1)  n  −  pq (pq)n n=1

x (pq)kÎą

1

X Îą>

logpq x k

∞ X

(p − 1)(q − 1)  n −  pq (pq)n n=1

1

+O x 2k +

1 + 2k

!!

 1 ι   + O x 2k + (pq)ι 



= kx k

∞ X

1 h

(pq)

logpq x k

i Îą=1

Îą+

h log

pq

k Îą (pq)

x

i   

94

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II 1 1 (p − 1)(q − 1) ap,q + O(x− k log x) + O x 2k + pq 1 1 (p − 1)(q − 1) ap,q kx k + x 2k + , pq 1

= kx k = where

ap,q =

∞ X

n (pq)n n=1

is a computable positive constant. This completes the proof of Theorem.

References [1] F.Smarandache. Only problems, not Solutions. Xiquan Publ.House, Chicago, 1993. [2] Lv Chuan. A number theoretic function and its mean value, Research on Smarandache problems in number theory (collected papers). Hexis, 2004, 33-35. [3] Zhang Wenpeng. An arithmetic function and the primitive number of power p, Research on Smarandache problems in number theory (collected papers). Hexis, 2004, 1-4. [4] T.M.Apostol. Introduction to Analytic Number Theory. Springer-Verlag, 1976. [5] Pan Chengdong and Pan Chengbiao. Elemens of the analytic number Theory. Beijing, Science Press. 1991.

AN ARITHMETICAL FUNCTION AND ITS MEAN VALUE FORMULA Ren Zhibin and Zhao Xiaopeng Department of Mathematics, Weinan Teacher’s College, Weinan, Shaanxi, P.R.China

Abstract

In this paper, we use the elementary methods to study the mean value properties of an arithmetical function, and give an interesting asymptotic formula for it.

Keywords:

Arithmetical function; Mean value; Asymptotic formula.

§1. Introduction Let k be a fixed positive integer. For any positive integer n, we define the arithmetical function bk (n) as follows: (

bk (n) = max m|

m X

)

ik ≤ n, n ∈ N

.

i=1

That is, bk (n) is the greatest positive integer m such that

m X

ik ≤ n. For

i=1

example, b2 (1) = 1, b2 (2) = 1, b2 (3) = 1, b2 (4) = 1, b2 (5) = 2, · · · · · ·. In fact, from the definition of bk (n) we know that if 1 ≤ n < (1 + 2k ), then b(n) = 1; if 1 + 2k ≤ n < 1 + 2k + 3k , then b(n) = 2; · · · · · ·, if

m X

ik ≤

i=1

n<

m+1 X

ik , then b(n) = m. So the same positive integer m repeated (m + 1)k

i=1

times in the sequence {bk (n)} (n = 1, 2, 3, 4, · · ·). About this arithmetical function, it seems that none had studied it, at least we have not see any related references. In this paper, we shall use the elementary methods to study the mean value properties of bk (n), and give an interesting asymptotic formula for it. That is, we shall prove the following: Theorem. For any real number x > 1, we have the asymptotic formula k+2

(k + 1) k+1 k+2 bk (n) = x k+1 + O(x). k+2 n≤x X

From the Theorem, we may immediately deduce the following two Corollaries

96

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

Corollary 1. For any real number x > 1, we have the asymptotic formula 3

22 3 b1 (n) = x 2 + O(x). 3 n≤x X

Corollary 2. For any real number x > 1, we have the asymptotic formula 4

33 4 b2 (n) = x 3 + O(x). 4 n≤x X

§2. Proof of the theorem In this section, we shall complete the proof of the theorem. First we define fk (n) =

n X

ik . For any real number x > 1, it is clear that there exists one and

i=1

only one positive integer N such that fk (N ) ≤ x < fk (N + 1). For any fixed positive integer k, from the Euler’s summation formula (see Theorem 3.1 of [2]) we have fk (N ) =

N X

ik =

i=1

N k+1 (N + 1)k+1 +O Nk = + O N k = fk (N + 1). k+1 k+1

So that from the inequality fk (N ) ≤ x < fk (N + 1) we have x=

N k+1 + O Nk k+1

or

1

1

(k + 1) k+1 x k+1 = N + O(1).

Then from the Euler’s summation formula and the definition of bk (n), we have X

bk (n) =

N −1 X

X

n=1 fk (n)≤j<fk (n+1)

n≤x

=

N −1 X

X

bk (j) +

bk (n)

fk (N )≤n<x

(n + 1)k n + O(N k+1 ) =

N X

nk (n − 1) + O(N k+1 )

n=2

n=1

=

N k+2 N k+1 N k+2 − + O(N k+1 ) = + O(N k+1 ) k+2 k+1 k+2

=

k+2 (k + 1) k+1 k+1 x + O(x). k+2

k+2

This completes the proof of the theorem.

References [1] F. Smarandache. Only Problems, Not solutions. Chicago: Xiquan Publishing House, 1993.

An arithmetical function and its mean value formula

97

[2] Tom M.Apostol. Introduction to Analytic Number Theory. New York: Springer-Verlag, 1976. [3] Zhang Wenpeng. Research on Smarandache problems in number theory. Hexis Publishing House, 2004.

ON THE SQUARE COMPLEMENTS FUNCTION OF N! Fu Ruiqin School of Science, Xi’an Shiyou University, Xi’an, Shaanxi, P.R.China xd2008frq@163.com

Yang Hai Research Center for Basic Science , Xi’an Jiaotong University, Xi’an, Shaanxi, P.R.China xjtu88yh@163.com

Abstract

For any positive integer n, let b(n) denote the square complements function of n. That is, b(n) denotes the smallest positive integer such that n · b(n) is a perfect square number. In this paper, we use the elementary method to study the asymptotic properties of b(n!), and give an interesting asymptotic formula for ln(b(n!)).

Keywords:

Square complements function; Standard factorization; Asymptotic formula.

§1. Introduction For any positive integer n, let b(n) denote the square complements function of n. That is, b(n) denotes the smallest positive integer k such that nk is a perfect square. For example, b(1) = 1, b(2) = 2, b(3) = 3, b(4) = 1, b(5) = 5, b(6) = 6, b(7) = 7, b(8) = 2, · · ·. In problem 27 of [1], Professor F. Smarandache ask us to study the properties of b(n). About this problem, some authors had studied it before, and obtained some interesting results, see references [5] and [6]. In this paper, we use the elementary method to study the asymptotic properties of the square complements function b(n!) of n, and give an interesting asymptotic formula for ln(b(n!)). That is, we shall prove the following: Theorem . For any positive integer n, we have the asymptotic formula 3

ln b(n!) = n ln 2 + O n exp

−A ln 5 n 1

(ln ln n) 5 where A > 0 is a constant.

!!

,

100

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

§2. Two simple lemmas To complete the proof of the theorem, we need the following two simple lemmas: Lemma 1. Let n! = pα1 1 pα2 2 · · · pαk k denotes the factorization of n into prime powers. Then we have the calculate formula b(n!) = b(pα1 1 ) · b(pα2 2 ) · · · · · b(pαk k ) ord(p1 )

= p1

ord(p2 )

· p2

ord(pk )

· · · · · p2

,

where the ord function is defined as:

ord(pi ) =

1 , 0 ,

if αi is odd, if αi is even.

Proof. See reference [2]. Lemma 2. For any real number x ≥ 2, we have the asymptotic formula θ(x) =

X

ln p

p≤x 3

= x + O x exp

−A ln 5 x 1

!!

.

(ln ln x) 5 Proof. See references [3] or [4].

§3. Proof of the theorem In this section, we shall complete the proof of Theorem. First from Lemma 1 we have ln b(n!) = ln b(pα1 1 ) · b(pα2 2 ) · · · · · b(pαk k )

=

X

ln p

p≤n 2†ord(p)

=

X

ln p +

n <p≤n 2

X n <p≤ n 4 3

ln p +

X

ln p + · · · + O(1).

n <p≤ n 6 5

Let n be a positive integer large enough, if a prime factor p of n! in the interval ( n2 , n], then the power of p is 1 in the standard factorization of n!. Similarly, if a prime factor p of n! in the interval ( n3 , n2 ], then the power of p is 2 in the standard factorization of n!; If a prime factor p of n! in the interval ( n4 , n3 ], then the power of p is 3 in the standard factorization of n!, · · · . On the other hand, note that the identity 1−

1 1 1 + − + · · · = ln 2, 2 3 4

On the square complements function of n!

101

then from Lemma 2 we have ln b(n!) n = θ (n) − θ 2

n +θ 3

n −θ 4

n +θ 5

n 6

−θ 3

1 1 1 −A ln 5 n = n 1 − + − + · · · + O n exp 1 2 3 4 (ln ln n) 5

3

= n ln 2 + O n exp

−A ln 5 n 1

+ · · · + O(1)

!!

!!

.

(ln ln n) 5 This completes the proof of Theorem.

References [1] Smarandache F. Only problems, not Solutions. Chicago: Xiquan Publ. House, 1993. [2] Felice Russo. An introduction to the Smarandache square complements. Smarandache Notions Journal, 2002, 13:160-172. [3] Pan Chengdong and Pan Chengbiao. Foundation of Analytic Number Theory. Beijing: Science Press, 1997. [4] J.B.Rosser and L.Schoenfeld. Approximates formulas for some functions of prime numbers. Illinois J. Math. 1962, 6: 64-94. [5] Liu H.Y. and Gou S. On a problem of F. Smarandache, Journal of Yanan University (Natural Science Edition) 2001, 20: 5-6. [6] Zhang H.L. and Wang Y. On the mean value of divisor function for square complements, Basic Science Journal of Textile University 2002,15: 4446.

ON THE SMARANDACHE FUNCTION Wang Yongxing Department of Mathematics, Weinan Teacher’s college, Weinan, Shaanxi, P.R.China

Abstract

For any positive integer n, let S(n) denotes the Smarandache function. In this paper, we study the mean value properties of S(n) , and give an interesting asn ymptotic formula for it.

Keywords:

Smarandache function; Mean value; Asymptotic formula.

§1. Introduction For any positive integer n, let S(n) denotes the Smarandache function. That is, S(n) is the smallest positive integer m such that n|m!. From the definition of S(n), we can easily deduce that if n = pα1 1 pα2 2 · · · pαk k is the prime powers factorization of n, then S(n) = max {f (pαi i )}. 1≤i≤k

About the arithmetical properties of S(n), many scholars had studied it before (see reference [1]). The main purpose of this paper is to study the mean value properties of S(n) n , and obtain an interesting asymptotic formula for it. That is, we shall prove the following: Theorem.

For any real number x ≥ 2, we have the asymptotic formula X S(n) n≤x

n

π2 x x = +O . 6 ln x ln2 x

§2. Some lemmas To complete the proof of Theorem, we need the following several Lemmas: Lemma 1. For any positive integer n, if n has √ the prime powers factorization n = pα1 1 pα2 2 · · · pαr r P (n) such that P (n) > n, then we have the identity S(n) = P (n). where P (n) denotes the greatest prime divisor of n. Proof. From the prime powers factorization of n, we may immediately get √ pα1 1 pα2 2 · · · pαr r < n.

104

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

Then we have pαi i |P (n)! i = 1, 2, · · · r. Thus we can easily obtain n|P (n)!. But P (n) † (P (n) − 1)!, so we have S(n) = P (n). This completes the proof of Lemma 1. Lemma 2. For any real number x ≥ 2, we have the asymptotic formula !

π 2 x2 +O S(n) = 12 ln x n≤x

x2 . ln2 x

X

Proof. First we define two sets A and B as following: √ A = {n|n ≤ x, P (n) ≤ n} and B = {n|n ≤ x, P (n) >

√

n}.

Using the Euler’s summation formula (see reference [2]), we may get X

S(n)

n∈A

=

X√ n≤x Z x√

n ln n Z

t ln tdt +

1

x

√ √ (t − [t])( t ln t)0 dt + x ln x(x − [x])

1

3 2

x ln x. Similarly, from the Abel’s summation forumula we also have X

S(n) =

n∈B

X

P (n)

n≤x √ P (n)> n

=

X

X

p

√

x n≤ x n≤p≤ n

=

=

X

X

√ √ x n≤ x x≤p≤ n

X √

n≤ x

p+O

x x π n n

X √

n≤ x

−

√

X √

x

x n≤p≤ n x

!

Z 3 √ n xπ( x) − √ π(s)ds + O x 2 ln x , x

where π(x) denotes all the numbers of primes which is not exceeding x. Note that x x π(x) = +O . ln x ln2 x

105

On the Smarandache function

Using the above asymptotic formula, we have

√

x x π n n

√

X

p =

x x≤p≤ n

−

x

Z √ n xπ( x) − √ π(s)ds x

1 x2 1 x √ +O − 2 2 n ln x/n 2 ln x

=

x 2√ ln x

+O

!

x2 n2 ln2 x/n

!

!

+O

x x2 − . √ 2 2 n2 ln x/n ln x

Considering the following X √

n≤ x

1 1 = ζ(2) + O( ). n2 x

Hence we have x2

X √

n≤ x

n2 ln nx

=

n2 ln nx

2

n≤ln x

=

π2

x2

X

x2

6 ln x

+O

+O

√

2

ln x≤n≤ x

x2

x2

n2 ln nx

X

!

ln2 x

and x2 √ n2 ln2 n≤ x

!

X

x n

=O

x2 . ln2 x

Combining all the above, we may immediately deduce that X

X

S(n) =

n≤x

S(n) +

n∈A

π 2 x2 +O 12 ln x

=

X

S(n)

n∈B

!

x2 . ln2 x

This completes the proof of Lemma 2.

§3. Proof of the theorem In this section, we shall complete the proof of Theorem. First applying the Abel’s summation, and note that the results of Lemma 1 and Lemma 2, we may have X S(n) n≤x

n

=

1 X S(n) + x n≤x

Z 1

x

1 X S(t) dt t2 n≤t

106

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

=

1 x

π 2 x2 +O 12 ln x

=

!!

π2 x x +O + 12 ln x ln2 x π2 x x . +O 6 ln x ln2 x

=

x2 ln2 x

Z

+ 1

Z 1

x

x

1 t2

π 2 t2 +O 12 ln t

π2 dt + O 12 ln t

Z 1

x

t2 ln2 t

1 dt ln2 t

!!

This completes the proof of Theorem.

References [1] Li Hailong and Zhao Xiaopeng. On the Smarandache function and the k-th roots of a positive integer. Hexis, 119-122, 2004. [2] Tom M A. Introduction to Analytic Number Theory. New York: SpringerVerlag, 1976.

dt

MEAN VALUE OF THE K -POWER COMPLEMENT SEQUENCES Feng Zhiyu Fengxiang Teacher’s School, Fengxiang, Shaanxi, P.R.China

Abstract

For any prime p ≥ 3 and any fixed integer k ≥ 2, let ep (n) denote the largest exponent of power p which divides n, a(n, k) denotes the k-power complement number of n. In this paper, we study the properties of the sequence ep (a(n, k)), and give an interesting asymptotic formula for it.

Keywords:

Largest exponent; Mean value; Asymptotic formula.

§1. Introduction Let p ≥ 3 be a prime, ep (n) denote the largest exponent of power p which divides n. For any fixed integer k ≥ 2, let a(n, k) denote k-power complement sequence of n. That is, a(n, k) is the smallest positive integer such that na(n, k) is a perfect k-power. In problem 29 of reference [1], professor F. Smarandach asked us to study the properties of this sequences. About this problem, some people had studied it before, and made some progress, see reference [2]. The main purpose of this paper is using the analytic method to study the properties of the sequence ep (a(n, k)), and give an interesting asymptotic formula for its mean value. That is, we shall prove the following: Theorem. Let p ≥ 3 be a prime and k ≥ 2 a fixed positive integer. Then for any real number x ≥ 1, we have the asymptotic formula X

ep (a(n, k)) =

n≤x

(k − 1)pk − kpk−1 + 1 x + O x1/2+ε , k (p − 1)(p − 1)

where ε denotes any fixed positive number. From this Theorem we may immediately deduce the following two corollaries: Corollary 1. Let p ≥ 3 be a prime. Then for any real number x ≥ 1, we have the asymptotic formula X n≤x

ep (a(n, 2)) =

1 x + O x1/2+ε . p−1

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RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

Corollary 2. Let p ≼ 3 be a prime. Then for any real number x ≼ 1, we have X 2p + 1 x + O x1/2+ξ . ep (a(n, 3)) = 2 p +p+1 n≤x

§2. Proof of the theorem In this section, we shall complete the proof of Theorem. In fact, for any complex number s with Re(s) > 1 , we define the Dirichlet series f (s) =

∞ X ep (a(n, k)) n=1

ns

.

From the definitions of ep (n) and a(n, k), and applying the Euler product formula (See reference [4]) we have f (s) =

p(Îąk+β)s Îą=0 β=1 ∞ k−1 X X

= k

∞ X

1

1

1

1 (n1 )s

n1 =1 (n1 ,p)=1



∞ X X β 1 k−1 1 − (n1 )s Îą=0 pÎąks β=1 pβs

s − pks k 1 p  Îś(s) 1 − 1 1 ps 1 − pks 1 − ps



1

∞ X n1 =1 (n1 ,p)=1

1 (n1 )s 

1

s − pks 1 k − 1 p − Îś(s) 1 1 − pks 1 − pks 1 − ps

p(k−1)s − 1 [(k − 1)ps − k]

= =

n1 =1 (n1 ,p)=1

p(ιk+β)s ι=0 β=1 

=

∞ X

∞ k−1 X k−β X

(pks

−

1) (ps

1)pks

− 1)

Îś(s) +

k−1 Îś(s) pks − 1

kp(k−1)s

(k − − +1 Îś(s), ks s (p − 1)(p − 1)

where Μ(s) is the Riemann zeta-function. Obviously, we have ep (a(n, k)) ≤ k logp n ≤ k ln n

∞

X k ep (a(n, k))

,

≤ Ďƒ

n Ďƒâˆ’1 n=1

where Ďƒ is the real part of s. Therefore by Parron’s formula (See reference [3]) we can get X ep (a(n, k))

ns0

n≤x

1âˆ’Ďƒ0

+O x

1 = 2Ď€i

Z

b+iT

b−iT

xs f (s + s0 ) ds + O s

ln x H(2x) min 1, T

âˆ’Ďƒ0

+O x

xb B(b + Ďƒ0 ) T

!

x H(N ) min 1, kxk

Mean value of the k -power complement sequences

109

where N is the nearest integer to x, kxk = |x − N |. k Taking s0 = 0, b = 32 , H(x) = k ln x, B(σ) = σ−1 , we have 1 ep (a(n, k)) = 2πi n≤x X

where

3 +iT 2

Z

3 −iT 2

3

xs x 2 +ε ζ(s)R(s) ds + O( ), s T

(k − 1)pks − kp(k−1)s + 1 . (pks − 1)(ps − 1)

R(s) = To estimate the main term 1 2πi

3 +iT 2

Z

3 −iT 2

we move the integral line from s = This time, the function

ζ(s)R(s)

3 2

xs ds, s

± iT to s =

g(s) = ζ(s)R(s)

1 2

± iT .

xs s

has a simple pole point at s = 1, and the residue is (k − 1)pk − kpk−1 + 1 x. (pk − 1)(p − 1) So we have 1 2πi

3 +iT 2

Z

3 −iT 2

1 +iT 2

Z

+

3 +iT 2

Z

+

1 −iT 2 1 +iT 2

Z

+

3 −iT 2

!

ζ(s)R(s)

1 −iT 2

xs ds s

(k − 1)pk − kpk−1 + 1 x

=

.

(pk − 1)(p − 1)

Note that the estimate

1

2πi

Z

1 +iT 2 3 +iT 2

Z

+

3 −iT 2 1 −iT 2

Z

+

1 −iT 2

!

1 +iT 2

(k − 1)pks − kp(k−1)s + 1 xs

ζ(s) ds

(pks − 1)(ps − 1) s

3

x 2 +ε . T Taking T = x, from the above formula we may immediately get the asymptotic formula X n≤x

ep (a(n, k)) =

(k − 1)pk − kpk−1 + 1 x (pk

− 1)(p − 1)

This completes the proof of Theorem.

+ O x1/2+ε .

110

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

References [1] F. Smarandache. Only Problems, Not solutions. Chicago: Xiquan Publishing House, 1993. [2] Zhang Wenpeng. Reseach on Smarandache problems in number theory. Hexis Publishing House, 2004. [3] Pan Chengdong and Pan Chengbiao. Foundation of Analytic Number Theory. Beijing: Science press, 1997. [4] Apostol T M. Introduction to Analytic Number Theory. New York: Springer- Verlag, 1976.

ON THE M -TH POWER FREE NUMBER SEQUENCES

Chen Guohui Department of Mathematics, Hainan Normal University, Haikou, Hainan, P.R.China

Abstract

In this paper, we using the elementary methods to study the arithmetical properties of the m-th power free number sequences, and give some interesting identities for it.

Keywords:

m-free number sequence; Dirichlet’s series; Identity.

§1. Introduction For any positive integer n and m with m ≥ 2, we define the m-th power free number function am (n) of n as follows: If n = pα1 1 pα2 2 · · · pαs s is the prime powers decomposition of n, then the m-th power free number function of n is the function: am (n) = pβ1 1 pβ2 2 · · · pβs s , where βi = αi , if αi ≤ m − 1, and βi = 0, if αi ≥ m. For any positive integer k, we also define the arithmetical function δk (n) as follows:

δk (n) =

max{d ∈ N

| d|n, (d, k) = 1}, if 0, if

n 6= 0, n = 0.

Let A denotes the set of all the positive integers n such that the equation am (n) = δk (n). That is, A = {n ∈ N, am (n) = δk (n)}. In this paper, we using the elementary methods to study the convergent properties of the Dirichlet series involving the set A, and give an interesting identity. That is, we shall prove the following conclusion: Theorem. Let m ≥ 2 be a positive integer. Then for any real number s > 1, we have the identity: ∞ X 1 n=1 n∈A

ns

=

ζ(s) Y pms − p(m−1)s + 1 , ζ(ms) p|k pms − 1

where ζ(s) is the Riemann zeta-function. From this theorem we may immediately deduce the following:

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RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

Corollary . Let B = {n ∈ N, a2 (n) = δk (n)} and C = {n ∈ N, a3 (n) = δk (n)}, then we have the identities: ∞ X 1

n2

n=1 n∈B

and

∞ X 1 n=1 n∈C

n2

=

15 Y p4 − p2 + 1 π 2 p|k p4 − 1

305 1 Y p6 − p4 + 1 . 2 π 4 p|k p6 − 1

=

§2. Proof of the theorem In this section, we will complete the proof of the theorem. First, we define the arithmetical function a(n) as follows:

a(n) =

n ∈ A, otherwise.

1, if 0, if

For any real number s > 0, it is clear that ∞ X 1 n=1 n∈A

and

∞ X 1 n=1

ns

ns

=

∞ X a(n) n=1

ns

is convergent if s > 1, thus

<

∞ X 1 n=1

∞ X 1 n=1 n∈A

ns

ns

,

is also convergent if s > 1.

Now let n = pα1 1 pα2 2 · · · pαs s denotes the factorization n into prime powers. Then from the definition of am (n) and δk (n), we know that if αj ≥ m for α some j, then am (pj j ) = 1. So only when αi ≤ m − 1 and (pi , k) = 1 for all i, the equation am (n) = δk (n) has solution. If (pj , k) 6= 1 for some j, then the α α equation am (pj j ) = δk (pj j ) has solution if and only if αj ≥ m. So from the Euler product formula (see [1]), we have ∞ X 1 n=1 n∈A

ns

!

=

a(p) a(p2 ) a(pm−1 ) 1 + s + 2s + · · · + (m−1)s + · · · p p p

Y p

=

a(p) a(p2 ) a(pm−1 ) 1 + s + 2s + · · · + (m−1)s p p p

Y p†k

!

!

×

Y p|k

=

Y p†k

a(p) a(p2 ) a(p3 ) 1 + ms + (m+1)s + (m+2)s + · · · p p p

1 1 1 1 + s + 2s + · · · + (m−1)s p p p

On the m-th power free number sequences

×

Y p|k

=

1+

1 pms

+

113 1 p(m+1)s

+

1 p(m+2)s

+ ···

ζ(s) Y pms − p(m−1)s + 1 , ζ(ms) p|k pms − 1

where ζ(s) is the Riemann zeta-function and

Q

denotes the product over all

p

primes. This completes the proof of Theorem.

References [1] Tom M.Apostol. Introduction to Analytic Number Theory. New York: Springer-Verlag, 1976. [2] F.Smarandache. Only problems, not solutions, Xiquan Publishing House, Chicago, 1993.

A NUMBER THEORETIC FUNCTION AND ITS MEAN VALUE Du Jianli School of Science, Xi’an Shiyou University, Xi’an, Shaanxi, P.R.China

Abstract

Let p and q are two distinct primes, epq (n) denotes the largest exponent of power X n epq epq (t). In this paper, we t

pq which divides n, and let bpq (n) =

t|n

using the analytic methods to study the mean vlaue properties of the bpq (n), and give an interesting asymptotic formula for it. Keywords:

Largest exponent; Mean value; Asymptotic formula.

§1. Introduction Let p and q are two distinct primes, epq (n) denotes the largest exponent of power pq which divides n. In problem 68 of [3], professor F.Smarandache asked us to study the properties of the sequence ep (n). About this problem, some people had studied it, and obtained a series interesting results, see references [2]. In this paper, we define arithmetical function bpq (n) = X n epq (t), then we using the analytic methods to study the mean epq t t|n value properties of bpq (n), and give give a sharp asymptotic formula for it. That is, we shall prove the following: Theorem. Let p and q are two distinct primes, then for any real number x ≥ 1, we have the asymptotic formula X n≤x

bpq (n) =

1 1 − 2γ − pq + 2pqγ − 2pq ln(pq) x ln x + x + O x 2 +ε , (pq − 1)2 (pq − 1)3

where ε is any fixed positive number, and γ is the Euler constant.

§2. Proof of the theorem In this section, we shall complete the proof of Theorem. For any complex s, we define the Dirichlet’s series f (s) =

∞ X epq (n) n=1

ns

and

g(s) =

∞ X bpq (n) n=1

ns

.

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RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

It is clear that g(s) = f 2 (s). Îś(s) (pq)s −1 , where Îś 2 (s) g(s) = ((pq) s −1)2 .

Ren Ganglian has given (in a paper to appear ) f (s) = Îś(s) is the Riemann zeta-function. So we can obtain that Obviously, we have

∞

X b (n)

1 pq

,

≤ Ďƒ

n Ďƒâˆ’2 n=1

bpq (n) ≤ logpq n ≤ n ln n

where Ďƒ(≼ 2) is the real part of s. Therefore, by Parron’s formula ( See reference [4]) we can get X bpq (n) n≤x

ns0

1 = 2Ď€i

Z

b+iT

b−iT

xs Îś (s + s0 )R(s + s0 ) ds + O s 2

xb B(b + Ďƒ0 ) T

x log x +O xâˆ’Ďƒ0 H(N ) min 1, +O x H(2x) min 1, T kxk where N be the nearest integer to x, kxk = |x − N |. Let

1âˆ’Ďƒ0

R(s) =

bpq (n) =

n≤x

1 2Ď€i

Z

5 +iT 2 5 −iT 2

,

1 . − 1)2

((pq)s

Taking s0 = 0, b = 52 , H(x) = x ln x, B(Ďƒ) = X

!

1 Ďƒâˆ’2 ,

we have 3

Îś 2 (s)R(s)

x 2 +Îľ xs ds + O( ). s T

To estimate the main term 1 2Ď€i

Z

5 +iT 2 5 −iT 2

Îś 2 (s)R(s)

xs ds, s

we move the integral line from s = 25 Âą iT to s = 12 Âą iT . This time, the function xs Îś 2 (s)R(s) s has a second order pole point at s = 1, and the residue is x ln x 1 − 2Îł − pq + 2pqÎł − 2pq ln(pq) + x, 2 (pq − 1) (pq − 1)3 where Îł is Euler constant. So we have 1 2Ď€i

Z

5 +iT 2 5 −iT 2

=

Z

+

1 +iT 2 5 +iT 2

Z

+

1 −iT 2 1 +iT 2

Z

+

5 −iT 2 1 −iT 2

!

Îś 2 (s)R(s)

xs ds s

x ln x 1 − 2Îł − pq + 2pqÎł − 2pq ln(pq) + x. 2 (pq − 1) (pq − 1)3

117

A number theoretic function and its mean value

Taking T = x2 , we can easily obtain

1

2πi

and

Z

1 +iT 2 5 +iT 2

Z

+

5 −iT 2 1 −iT 2

!

1 xs

ζ (s)R(s) ds x 2 +ε s

2

1 Z 21 −iT 1 xs

2 ζ (s)R(s) ds x 2 +ε .

2πi 1 +iT s

2

So from the above formula, we may immediately get the asymptotic formula X

bpq (n) =

n≤x

x ln x 1 − 2γ − pq + 2pqγ − 2pq ln(pq) 1/2+ε + x+O x . (pq − 1)2 (pq − 1)3

This completes the proof of Theorem.

References [1] Apostol T M. Introduction to Analytic Number Theory. New York: Springer- Verlag, 1976. [2] Zhang Wenpeng. Reseach on Smarandache problems in number theory. Hexis Publishing House, 2004. [3] F. Smarandache. Only Problems, Not solutions. Chicago: Xiquan Publishing House, 1993. [4] Pan Chengdong and Pan Chengbiao. Foundation of Analytic Number Theory. Beijing: Science press, 1997.

ON THE ADDITIVE K -TH POWER PART RESIDUE FUNCTION Zhang Shengsheng Shaanxi Technical College of Finance and Economics, Xianyang, Shaanxi, P.R.China

Abstract

Similar to the Smarandache k-th power complements, we define the additive kth power part residue fk (n) of n is the smallest nonnegative integer such that n − fk (n) is a perfect k-th power. The main purpose of this paper is using the elementary methods to study the mean value properties of fk (n) and d(fk (n)), and give two interesting asymptotic formulae for them.

Keywords:

Additive k-th power part residue function; Mean value; Asymptotic formula.

§1. Introduction and results For any positive integer n, the Smarandache k-th power complements bk (n) is the smallest positive integer bk (n) such that nbk (n) is a complete k-th power (see problem 29 of [1]). Similar to the Smarandache k-th power complements, Xu Zhefeng in [2] defined the additive k-th power complements ak (n) as follows: ak (n) is the smallest nonnegative integer such that n + ak (n) is a complete k-th power. Similarly, we will define the additive k-th power part residue fk (n) as following: for any positive integer n, fk (n) = min{r|0 ≤ r = n − mk , m ∈ N }. For example, if k = 2, we have the additive square part residue sequences {f2 (n)} (n = 1, 2, . . .) as following: 0, 1, 2, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, . . . . In this paper, we use the elementary methods to study the mean value properties of fk (n) and d(fk (n)) (where d(n) is the Dirichlet divisor function), and obtain some interesting asymptotic formulae for them. That is, we will prove the following conclusions: Theorem 1. For any real number x ≥ 3 and integer k ≥ 2, we have the asymptotic formula X n≤x

fk (n) =

1 2 k2 x2− k + O x2− k . 2(2k − 1)

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RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

Theorem 2. For any real number x ≥ 3 and integer k ≥ 2, we also have the asymptotic formula X

1 1 1 x ln x + 2γ + ln k − 2 + x + O x1− k ln x , k k

d(fk (n)) = 1 −

n≤x

where γ is the Euler constant.

§2. Some Lemmas To complete the proof of the theorems, we need following two Lemmas. First we have Lemma 1. For any real number x ≥ 3, we have the asymptotic formula X

1

d(n) = x ln x + (2γ − 1)x + O(x 2 ),

n≤x

where γ is the Euler constant. Proof. See reference [3]. Lemma 2. For any real number x ≥ 3 and any nonnegative arithmetical function h(n) with h(0) = 0, we have the asymptotic formula X

h(fk (n)) =

 X

h(n) + O 

t=1 n≤g(t)

n≤x

where g(t) =

M −1 X

k−1 X i=1

 X

h(n) ,

x≤g(M )

!

1 k i t and M = [x k ], [x] denotes the greatest integer not i

exceeding x. Proof. For any real number x ≥ 1, let M be a fixed positive integer such that M k ≤ x < (M + 1)k . Noting that if n pass through the integers in the interval [tk , (t + 1)k ), then fk (n) pass through the integers in the interval [0, (t+1)k −tk −1] and h(0) = 0, we can deduce that X

h(fk (n)) =

M −1 X

X

t=1 tk ≤n<(t+1)k

n≤x

=

M −1 X

X

t=1 n≤g(t)

=

M −1 X

h(n) +

=

h(fk (n))

M k ≤n≤x

X

h(n)

0≤n<x−M k

 X

 X

h(n) + O 

t=1 n≤g(t) M −1 X

X

h(fk (n)) +

 X

t=1 n≤g(t)

h(n)

0≤n≤(M +1)k −M k

h(n) + O 

 X n≤g(M )

h(n) ,

On the additive k -th power part residue function k−1 X

121

!

1 k i where g(t) = t and M = [x k ]. i i=1 This completes the proof of Lemma 2.

§3. Proof of the theorems In this section, we shall use the elementary method to complete the proof of 1 the theorems. First we prove Theorem 1. Let h(n) = n and M = [x k ], note 1 that x k − M = O (1), then from Lemma 2 and the Euler summation formula (see [4]) we obtain

X

M −1 X

fk (n) =

 X

n+O

t=1 n≤g(t)

n≤x

1 2

= =

M −1 X

 X

n

n≤g(M ) 2

k 2 t2k−2 + O(x2− k )

t=1 k2

2(2k − 1)

1

2

x2− k + O(x2− k ).

This completes the proof of Theorem 1. Now we prove Theorem 2. From Lemma 1 and Lemma 2 we have

X

d(fk (n))

n≤x

=

M −1 X

 X

d(n) + O 

t=1 n≤g(t)

=

M −1 h X

 X

d(n)

n≤g(M )

1

ktk−1 ln ktk−1 + (2γ − 1)ktk−1 + O (ktk−1 ) 2

i

t=1 1

+O x1− k ln x =

M −1 X

k(k − 1)tk−1 ln t + (2γ + ln k − 1)ktk−1 + O(t

t=1 1

+O x1− k ln x = k(k − 1)

M −1 X

tk−1 ln t + (2γ + ln k − 1)k

t=1

1− k1

+O x

M −1 X t=1

ln x .

tk−1

k−1 2

)

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RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

Then from the Euler summation formula, we can easily get 1 1 1 x ln x + 2γ + ln k − 2 + x + O x1− k ln x . d(fk (n)) = 1 − k k n≤x X

This completes the proof of the theorems.

References [1] F.Smarandache. Only Problems, not Solutions. Xiquan Publishing House, Chicago, 1993. [2] Xu Zhefeng. On the Additive k-power Complements. Research on Smarandache Problems in Number Theory, Hexis 13-16, 2004. [3] G.L.Dirichlet. Sur I’usage des Series infinies dans la Theorie des nombres. Crelle’s Journal, 1938 [4] Tom M. Apostol. Introduction to Analytic Number theory. New York: Springer-Verlag, 1976. [5] Pan Chengdong and Pan Chengbiao. Foundation of Analytic Number Theory. Beijing: Science Press, 1997, P98.

AN ARITHMETICAL FUNCTION AND THE K -TH POWER COMPLEMENTS Huang Wei Department of Engineering Course, Baoji Vocational Technology College, Baoji, Shaanxi, P.R.China

Abstract

The main purpose of this paper is using the analytic methods to study the asymptotic properties of an arithmetical function acting on the k-th power complements, and give an interesting asymptotic formula for it.

Keywords:

Arithmetical function; Asymptotic formula; k-th power complements.

§1. Introduction and results For any positive integer n, let D(n) denotes the number of the solutions of the equation n = n1 n2 with (n1 , n2 ) = 1. That is, D(n) =

X

1.

d|n (d, n )=1 d

It is clear that D(n) is a multiplicative function, and has many interesting arithmetical properties. Now we define another arithmetical function Ak (n). For any fixed positive integer k, let Ak (n) denotes the k-th power complements. That is, Ak (n) denotes the smallest positive integer such that nAk (n) is a perfect k-th power. For example, A2 (1) = 1, A2 (2) = 2, A2 (3) = 3, A2 (4) = 1, A2 (5) = 5, A2 (6) = 6, A2 (7) = 7, A2 (8) = 2, ¡ ¡ ¡ ¡ ¡ ¡. In reference [1], professor F.Smarandache asked us to study the properties of the sequences {Ak (n)}. About this problem, some people had studied it before, and obtained some interesting results, see references [4] and [5]. In this paper, we use the analytic methods to study the mean value properties of the arithmetical function D(n) acting on the set {Ak (n)}, and obtain a sharper asymptotic formula. That is, we shall prove the following: Theorem . For any real number x ≼ 1, we have the asymptotic formula 2 6Îś(k)x ln x Y 1− k D(Ak (n)) = 2 Ď€ p + pk−1 p n≤x X

1

+ C(k)x + O(x 2 + ),

where C(k) is a computable constant,Y Îś(k) is the Riemann zeta-function, denotes any fixed positive number, and denotes the product over all primes. p

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RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

From this theorem we may immediately deduce the following: Corollary. For any real number x ≼ 1, we have the asymptotic formula X

Y Ď€2 2 x ln x 1− 4 15 p + p3 p

D(Ak (n)) =

n≤x

1

+ C(4)x + O(x 2 + ).

§2. A Lemma Before the proof of the theorem, a Lemma will be useful. Lemma . For any positive integer number n ≼ 1, we have the identity D(n) = 2v(n) , where v(n) denotes the number of all distinct prime divisors of n, i.e.

v(n) =

0, if n = 1; k, if n = pι1 1 pι2 2 pι3 3 ¡ ¡ ¡ pιk k .

Proof. Let n = pι1 1 pι2 2 pι3 3 ¡ ¡ ¡ pιk k denotes the factorization of n into prime powers. Note that D(n) is a multiplicative function and D(pι ) = 2, so from the definition of v(n), we have X

D(n) =

1 = Ck0 + Ck1 + ¡ ¡ ¡ + Ckk = 2v(n) .

d|n )=1 (d, n d

This proves the lemma.

§3. Proof of the theorem In this section, we shall complete the proof of the theorem. Let Dirichlet’s series f (s) =

∞ X D(Ak (n)) n=1

ns

.

For any real number s > 1, it is clear that f (s) is absolutely convergent. So from the lemma and the Euler’s product formula [2] we have !

f (s) =

Y p

D(Ak (p)) D(Ak (p2 )) 1+ + + ¡¡¡ ps p2s

!

=

Y p

=

Y p

D(pk−1 ) D(pk−2 ) D(1) 1+ + + ¡ ¡ ¡ + ks + ¡ ¡ ¡ s 2s p p p k−1 )

2v(p 1+ ps

k−1 )

2v(p + p2s

!

2v(1) + ¡ ¡ ¡ + ks + ¡ ¡ ¡ p

An arithmetical function and the k -th power complements Y

=

p

125

2 2 1 2 1 + s + 2s + ¡ ¡ ¡ + ks + (k+1)s + ¡ ¡ ¡ p p p p

Y

= Îś(s)Îś(ks)

p

1 2 1 + s − ks p p

Îś 2 (s)Îś(ks) Y 2 1 − ks , Îś(2s) p + p(k−1)s p

=

where Μ(s) is the Riemann zeta-function. Therefore by Perron’s formula [3] with s0 = 0, b = 2, T = 23 , we have 1 D(Ak (n)) = 2πi n≤x

2+iT

Z

X

2−iT

where R(s) =

Y

1 Îś 2 (s)Îś(ks) xs R(s) ds + O(x 2 +Îľ ), Îś(2s) s

2 . ks p + p(k−1)s

1−

p

To estimate the main term 1 2Ď€i

Z

Îś 2 (s)Îś(ks) xs R(s) ds, Îś(2s) s

2+iT

2−iT

we move the integral line from s = 2 Âą iT to s =

1 2

Âą iT , then the function

Îś 2 (s)Îś(ks) xs R(s) Îś(2s) s have a second order pole point at s = 1 with residue Y 2 Îś(k) 1− k x ln x Îś(2) p + pk−1 p

+ C(k)x,

where C(k) is a computable constant. So we have 1 2Ď€i =

Z

2+iT

Z

+ 2−iT

6Μ(k)x ln x π2

1 +iT 2

Z

+

2+iT

Y

1−

p

2+iT 1 −iT 2

2 pk + pk−1

Z

+

2−iT

!

1 −iT 2

Îś(s)Îś(ks)R(s)

xs ds s

+ C(k)x.

We can easily get the estimate 1 2Ď€i

Z

1 +iT 2

2+iT

Z

+

2+iT 1 −iT 2

Z

+

2−iT 1 −iT 2

!

Îś(s)Îś(ks)R(s)

1 xs ds x 2 +Îľ . s

Now note that Îś(2) = Ď€ 2 /6, we may immediately obtain the asymptotic formula X Y 1 6 2 D(Ak (n)) = 2 Îś(k)x ln x 1− k + C(k)x + O(x 2 + ), k−1 Ď€ p +p p n≤x

126

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

where C(k) is a computable constant. This completes the proof of the theorem.

References [1] F. Smarandache. Only Problems, Not solutions. Chicago: Xiquan Publishing House, 1993. [2] Apostol T M. Introduction to Analytic Number Theory. New York: Springer- Verlag, 1976. [3] Pan Chengdong and Pan Chengbiao. Foundation of Analytic Number Theory. Beijing: Science press, 1997. [4] Zhu Weiyi. On the k-power complement and k-power free number sequence. Smarandache Notions Journal, 2004, 14: 66-69. [5] Liu Hongyan and Lou Yuanbing. A Note on the 29-th Smarandache’s problem. Smarandache Notions Journal, 2004, 14: 156-158.

ON THE TRIANGLE NUMBER PART RESIDUE OF A POSITIVE INTEGER Ji Yongqiang Danfeng Teacher’s School, Shangluo, Shaanxi, P.R.China

Abstract

For any positive integer n, let a(n) denotes the triangle number part residue of n. That is, a(n) = n − k(k+1) , where k is the greatest positive integer such that 2 k(k+1) ≤ n. In this paper, we using the elementary methods to study the mean 2 value properties of the sequences {a(n)}, and give two interesting asymptotic formulae for it.

Keywords:

The triangle number part residue; Mean value; Asymptotic formula.

§1. Introduction and results For any positive integer n, let a(n) denotes the triangle number part residue of n. That is, a(n) = n − k(k+1) , where k is the greatest positive integer such 2 k(k+1) that 2 ≤ n. For example, a(1) = 0, a(2) = 1, a(3) = 0, a(4) = 1, a(5) = 2, a(6) = 0, · · · · · ·. In reference [1], American-Romanian number theorist Professor F. Smarandache asked us to study the arithmetical properties of this sequences. About this problem, it seems that none had studied it, at least we have not seen any related papers before. In this paper, we using the elementary methods to study the mean value properties of this sequences, and give two interesting asymptotic formulae for it. That is, we shall prove the following: Theorem 1. For any real number x ≥ 3, we have the asymptotic formula √ X 2 3 x 2 + O(x). a(n) = 3 n≤x Theorem 2. formula

For any real number x ≥ 3, we also have the asymptotic

2 1 ln 2 − 3 d(a(n)) = x ln x + 2γ + x + O(x 3 ), 2 2 n≤x

X

where d(n) is the Dirichlet divisor function (provide d(0) = 0), and γ is the Euler constant.

128

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

Note: It is clear that if there exists a mean value formula for any arithmetical functionP f (n), then using our methods we can also obtain an asymptotic formula for n≤x f (a(n)).

§2. Proof of the theorems In this section, we shall complete the proof of the theorems. First we prove Theorem 1. For any real number x ≥ 3, let M be a fixed positive integer such that M (M + 1) (M + 1)(M + 2) ≤x< . 2 2 Then from the definition of a(n), we have X

a(n) =

n≤x

M X k=1

=

X (k+1)(k+2) k(k+1) ≤n< 2 2

= = =

M X k X

a(n)

(M +1)(M +2) x<n< 2

M X k=1 i≤

X

a(n) −

X

i+O 

k(k+1) (k+1)(k+2) −1− 2 2

i + O M2

s≤

 X (M +1)(M +2) M (M +1) − 2 2

s 

k=1 i=0 M X

1 k(k + 1) + O M 2 2 k=1 1 3 M + O M2 . 6

(1)

On the other hand, note that the estimates M M2 3 ≤x− < M + 1. 2 2 2

(2)

Combing (1) and (2), we have √ X

a(n) =

n≤x

2 3 x 2 + O(x). 3

This completes the proof of Theorem 1. Now we prove Theorem 2. Using the similar method of proving Theorem 1, we have X n≤x

d(a(n)) =

M X k X k=1 i=0

d(i) +

X M (M +1) <n≤x 2

d(a(n)).

129

On the triangle number part residue of a positive integer

Note that the asymptotic formula (see reference [2]) X

1

d(n) = x ln x + (2γ − 1)x + O x 3 ,

n≤x

where γ is the Euler constant. We have X

d(a(n))

n≤x

=

M X

1

k ln k + (2γ − 1)k + O k 3

+O 

k=1

=

X i≤x−

d(i) 

M (M +1) 2

4 1 2 1 1 M ln M − (M 2 − 1) + (2γ − 1)M 2 + O(M 3 ). 2 4 2

(3)

Now combining (2) and (3) we may immediately get 2 ln 2 − 3 1 x + O(x 3 ). d(a(n)) = x ln x + 2γ + 2 2 n≤x

X

This completes the proof of Theorem 2.

References [1] F.Smarandache. Only Problems, not Solutions. Chicago: Xiquan Publ. House, 1993. [2] Pan Chengdong and Pan Chengbiao. Foundation of Analytic Number Theory. Beijing: Science Press, 1999. [3] Tom M.Apostol. Introduction to Analytic Number Theory. SpringerVerlag: New York, 1976.

AN ARITHMETICAL FUNCTION AND THE K -FULL NUMBER SEQUENCES Zhao Jiantang Department of Mathematics, Xianyang Teacher’s College, Xianyang, Shaanxi, P.R.China

Abstract

The main purpose of this paper is using the elementary methods to study the mean value properties of an arithmetical function acting on the k-full number sequences, and give an interesting asymptotic formula for it.

Keywords:

k-full number sequence; Asymptotic formula; Mean value.

§1. Introduction For any prime p, let ep (n) denotes the largest exponent ( of power p ) which divides n. In problem 68 of reference [1], Professor F.Smarandache asked us to study the properties of this arithmetical function. About this problem, many scholars showed great interest in it and obtained some interesting results. For example, Lv Chuan [2] used the elementary methods to studied the asymptotic P properties of the mean value n≤x em p (n), and gave the following asymptotic formula: X p−1 Ax + O(logm+1 x), em p (n) = p n≤x where A is a computable constant. Professor Zhang [3] studied the mean value of ep (Sp (n)), where Sp (n) denotes the smallest integer m such that pn |m!, and obtained X n≤x

ep (Sp (n)) =

p+1 x + O(ln3 x). (p − 1)2

In this paper, we shall use the elementary methods to study the mean value properties of ep (n) acting on the k-full number sequences, and give an interesting asymptotic formula for it. For convenience, first we give the definition of the k-full number. In fact, a number n is called a k-full number if p|n ⇐⇒ pk |n for any prime divisor p of n. Let A denotes the set of all the k-full numbers. It seems that no one had studied the relations between the arithmetical function ep (n) and the k-full numbers. In this paper, we shall prove the following:

132

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

Theorem. Let p be a prime, then for any real number x ≥ 1, we have the asymptotic formula X

1

1

ep (n) = C(p, k)ap (k)x k + x 2k +ε ,

n≤x n∈A

where 6k C(p, k) = 2 π

!

1

1− p−p

k−1 k

Y

+1

1+

q

!

1 1

(q + 1)(q k − 1)

,

a p (k) is a computable positive constant, ε > 0 is any fixed real number, and Y denotes the product over all prime q. q

§2. Proof of the theorem In this section, we shall complete the proof of Theorem. Let a(n) denotes the character function of k-full number. That is,

a(n) =

1, if n is a k-full number; 0, otherwise .

It is clear that X

1=

n≤x n∈A (n,p)=1

Let Dirichlet series f (s) =

X

a(n).

n≤x (n,p)=1

∞ X

a(n) . ns

n=1 (n,p)=1

It is clear that this series is convergent if Re s > 1. From the Euler product formula [4] and the definition of a(n), we have f (s) !

=

Y q6=p

=

Y q6=p

=

a(q k ) a(q k+1 ) 1 + ks + (k+1)s + · · · q q 1 1 1 + ks · q 1 − q1s

!

Y ζ(ks) 1 1 1 − ks 1 + ,(1) ζ(2ks) (q ks + 1)(q s − 1) p − p(k−1)s + 1 q

where ζ(s) is the Riemann zeta-function.

An arithmetical function and the k -full number sequences

133

Obviously, we have

∞

X a(n))

1

≤ Ďƒ

n Ďƒâˆ’ n=1

|a(n)| ≤ 1,

1 k

,

where Ďƒ is the real part of s. Therefore, by the Perron’s formula [4] we have X

a(n) =

n≤x (n,p)=1

1 2Ď€i

b+iT

b−iT

1 Îś(ks) xs R(s) ds + O(x 2k +Îľ ), Îś(2s) s

1

Y

pks − p(k−1)s + 1

q

where R(s) = 1 −

Z

Now moving the integral line from b to a =

1 . 1 + ks (q + 1)(q s − 1)

1 2k ,

we have

!

xs xs 1 ds = Res f (s) , . s s k a−iT a+iT b+iT b−iT (2) 1 Note that lim Îś(ks) s − k = 1, we may immediately get

1 2Ď€i

Z

b+iT

Z

a+iT

Z

a−iT

b−iT

Z

f (s)

+

+

+

s→ k1

1

kx k 1 xs 1 , = R . s k Îś(2) k

Res f (s)

(3)

Combining (1), (2), (3) and the following estimates

1 Z a−iT 1 xs

f (s) ds x 2k +

2Ď€i a+iT s

and

1 Z a+iT Z b−iT 1 xs

f (s) ds x 2k + +

2Ď€i b+iT s

a−iT

we can easily get 1

X

1

a(n) = C(p, k)x k + O x 2k + ,

n≤x (n,p)=1

where 6k C(p, k) = 2 π

!

1

1− p−p

Y

k−1 k

+1

q

1+

!

1 1

(q + 1)(q k − 1)

Based on the definition of ep (n) and the above estimate, we have X n≤x n∈A

ep (n) =

X k≤ι≤logp x

Îą

X n≤ pxι (n,p)=1

a(n)

.

134

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

X

=

Îą

k≤ι≤logp x

x pÎą

1

k

C(p, k) + O

x pÎą

1 + 2k





= C(p, k)x

1 k

X k≤ι≤logp



X 1 Îą x 2k + Îą + O k xp k≤ι≤log

∞ X

X n n − k n=1 p Îą>log

1 = C(p, k)x k 

p



!!

p

ι ι k xp



1 X Îą Îą  + O x 2k + − Îą Îą k k p p x Îą<k 

∞ 1 X Îą + logp x X Îą n 1   + O x 2k + = C(p, k)x − n − Îą Îą 1 k k pk p k [logp x] Îą=1 n=1 p Îą<k p 1 k

1

∞ X

1

1

= C(p, k)x k ap (k) + O(x− k log x) + O x 2k + 1

1

= C(p, k)ap (k)x k + x 2k + , where 6k C(p, k) = 2 π

!

1

1− p−p

Y

k−1 k

ap (k) =

+1

∞ X n n=1

p

n k

1+

q

−

!

1 1

(q + 1)(q k − 1)

,

X Îą Îą

Îą<k

pk

is a computable positive constant. This completes the proof of Theorem.

References [1] F.Smarandache. Only problems, not Solutions. Xiquan Publ.House, Chicago, 1993. [2] Lv Chuan. A number theoretic function and its mean value, Research on Smarandache problems in number theory (collected papers). Hexis, 2004, 33-35. [3] Zhang Wenpeng. An arithmetic function and the primitive number of power p, Research on Smarandache problems in number theory (collected papers). Hexis, 2004, 1-4. [4] T.M.Apostol. Introduction to Analytic Number Theory. Springer-Verlag, 1976.

ON THE HYBRID MEAN VALUE OF SOME SPECIAL SEQUENCES∗ Li Yansheng and Gao Li College of Mathematics and Computer Science, Yan’an University, Yan’an, Shaanxi, P.R.China

Abstract

The main purpose of this paper is to define a new arithmetic function by the m-th power complement numbers and the k-th power free numbers, and use the analytic methods to obtain some interesting asymptotic formulae for them.

Keywords:

m-th power complement numbers; k-th power free numbers; Asymptotic formula; Arithmetic function.

§1. Introduction and main results Let m be a fixed positive integer with m ≥ 2. For any positive integer n, we define the m-th power complement numbers am (n) of n as the smallest positive integer such that nam (n) is a perfect m-th power. We also call a positive integer n as k-th power free number if it can not be divided by any pm , where p be a prime.XWe denotes the k-th power free number by ck (n). It µ(d), where µ(d) is the M¨ obius function. is clear that ck (n) = n dk |n

In reference [1], professor F.Smarandache asked us to study the properties of the m-th power complement numbers and the k-th power free number sequences. About these sequences, some people had studied them, and obtained many interesting results, see references [2], [3], [4] and [5]. In this paper, we introduce a new sequences f (n) = am (n)ck (n), then we use the analytic method to study the mean value properties of this sequences, and obtain some sharp asymptotic formulae. That is, we shall prove the following: Theorem 1. For any real numberx ≥ 1, we have the asymptotic formula

X n≤x

∗ This

am (n)ck (n) =

1 6xm+1 R(m + 1) + O(xm+ 2 +ε ), (m + 1)π 2

work is supported by the P.N.S.F(00JK301)

136

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

where ε denotes any fixed positive number, and !

R(m + 1) =

p(k−2)(m+1) − 1 1+ , (p + 1)(p(k−1)(m+1) − p(k−2)(m+1) )

Y p

if m ≥ k;   Y  R(m + 1) = 1 +  p 

2 (pm −1

− 1) +

− 1)

(p + 1)(pm(m+1) − k−1 X

+

(pm(m+1)

i−1 X

pjm+m+1

p(jm+1)(m+1) j=1 pm2 −1 )

pim+m+1  pj(m+1)  j=im+1  

p+1

,  

if m < k. Theorem 2. For any real numberx ≥ 1, we have the asymptotic formula X

φ(am (n)ck (n)) =

n≤x

1 6xm+1 R∗ (m + 1) + O(xm+ 2 +ε ), 2 (m + 1)π

where φ(n) is the Euler function, ε denotes any fixed positive number, and !

∗

R (m + 1) =

p(p(k−2)(m+1) − 1) − (p(k−1)(m+1) − 1) 1+ , p(p + 1)(p(k−1)(m+1) − p(k−2)(m+1) )

Y p

If m ≥ k; ∗

R (m+1) =

Y p

2 −1

(pm

!

− 1) + (pm(m+1) − 1)H(m + 1) G(m + 1) + , p+1 (p + 1)(pm(m+1) − pm2 −1 )

if m < k; H(m + 1) =

i−1 X j=1

and G(m + 1) =

pjm+m+2 − pjm+m+1 1 − p p(jm+1)(m+1)

!

k−1 X

pim+m+1 − pim+m . pj(m+1) j=im+1

Taking k = 2 in Theorem 1 and Theorem 2, we may immediately deduce the following two Corollaries:

On the hybrid mean value of some special sequences1

137

Corollary 1. For any real number x ≥ 1 and integer m ≥ 2, we have the asymptotic formula

X

am (n)c2 (n) =

n≤x

1 6 xm+1 + O(xm+ 2 +ε ). 2 (m + 1)π

Corollary 2. For any real numberx ≥ 1 and integer m ≥ 2, we have the asymptotic formula

X

φ(am (n)c2 (n)) =

n≤x

where

R∗ (m

+ 1) =

Y p

1 6xm+1 R∗ (m + 1) + O(xm+ 2 +ε ), 2 (m + 1)π

1 1− . p(p + 1)

§2. Proof of the theorem In this section, we shall complete the proof of the theorems. First we prove Theorem 1. Let

f (s) =

∞ X am (n)ck (n) n=1

ns

,

It is clear that am (n) and ck (n) are multiplicative functions of n, so am (n)ck (n) is also a multiplicative function of n. If the real part of s is large enough, then the Dirichlet’s series f (s) is absolutely convergent. So for m ≥ k, from the Eurler’s product formula [6] we have

f (s) am (pk−1 )ck (pk−1 )) am (p)ck (p) am (p2 )ck (p2 )) = 1+ + + · · · + ps p2s p(k−1)s p Y 1 1 1 1 = 1 + s−m + · · · + (2s−2) (1 + s + · · · + (k−3)s ) p p p p p Y

!

ζ(s − m) Y p(k−2)s − 1 = 1 + (s−m) ; ζ(2(s − m)) p (p + 1)(p(k−1)s − p(k−2)s )

!

138

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

if m ≤ k (im ≤ k < (i + 1)m, i > 1), then f (s) Y am (p)ck (p) am (pm )ck (pm ) = 1+ + · · · + ps pms p m+1 )c (pm+1 ) am (p2m )ck (p2m ) k + ··· + + am (p p(m+1)s p2ms ! im+1 im+1 am (p )ck (p ) am (pk−1 )ck (pk−1 ) +··· + + ··· + p(im+1)s p(k−1)s Y 1 1 1 1 = 1 + s−m + 2s−m (1 + s + · · · + (m−2)s ) p p p p p 1 1 1 + (m+1)s−2m (1 + s + · · · + (m−1)s ) p p p P p(i+1)m 1 + · · · + p((i−1)m+1)s−im + k−1 js j=im+1 p 

(p(m−1)s − 1) + (pms − 1)

ζ(s − m) Y   = 1 + ζ(2(s − m)) p  

(ps−m + 1)(pms −

k−1 X

+

i−1 X pjm+s

p(jm+1)s j=1 p(m−1)s )

pim+s  pjs  j=im+1  ps−m +1

.  

Obviously, we have the inequality

2

|am (n)ck (n)| ≤ n ,

∞

X a (n)c (n)

1 m

k ,

< σ

n σ − m −1 n=1

where σ > m + 1 is the real part of s. So by Perron formula [7] we have: X a(n) n≤x

ns0

=

1 2πi

Z

b+iT

b−iT

xs f (s + s0 ) ds + O s

xb B(b + σ0 ) T

log x +O x H(2x)min 1, T x −σ0 . +O x H(N )min 1, T ||x||

1−σ0

!

On the hybrid mean value of some special sequences2

139

1 when N is the nearest integer to x, N = x− , kxk = |x−N |. Taking s0 = 0, 2 3 1 b = m + 2, T = x 2 , H(x) = x2 , B(σ) = , we have: σ−m−1 X

am (n)ck (n) =

n≤x

1 2iπ

Z

m+2+iT

m+2−iT

1 ζ(s − m) xs R(s) ds + O(xm+ 2 +ε ), ζ(2(s − m)) s

where  ! (k−2)s − 1 Y  p    , 1 + s−m   (p + 1)(p(k−1)s − p(k−2)s )  p     i−1  X  pjm+s  (m−1)s − 1) + (pms − 1)  (p     p(jm+1)s  Y  j=1    1 +    (ps−m + 1)(pms − p(m−1)s ) p  R(s) =       k−1  X pim+s        pjs   j=im+1    + ,   ps−m +1        

m ≥ k;

m < k.

To estimate the main term Z m+2+iT 1 1 ζ(s − m) xs R(s) ds + O(xm+ 2 +ε ), 2iπ m+2−iT ζ(2(s − m)) s 1 we move the integral line from s = m + 2 ± iT to s = ± iT . This time, the 2 function ζ(s − m)xs f (s) = R(s) ζ(2(s − m))s have a simple pole point at s = m + 1 with residue

xm+1 R(m + 1). (m + 1)ζ(2)

So we have 1 2iπ

Z

m+2+iT

Z

+

Z

+

m+2+iT

m+2−iT

=

m+ 21 +iT

m+ 21 −iT

m+ 21 +iT

Z

+

m+2−iT

m+ 21 −iT

!

ζ(s − m)xs R(s)ds ζ(2(s − m))s

xm+1 R(m + 1). (m + 1)ζ(2)

Taking T = x3/2 , we can easy get the estimate

Z m+ 1 +iT Z m+2−iT !

1

2 ζ(s − m)xs

+ R(s)ds

1

2πi

ζ(2(s − m+2+iT m+ 2 −iT

m))s Z m+2

1 x2

xm+2

ζ(σ − m + iT ) R(s) dσ = xm+ 2

T

T m+ 1 ζ(2(σ − m + iT )) 2

140

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

and

1 Z m+ 12 −iT ζ(s − m)xs

R(s)ds

2πi m+ 1 +iT ζ(2(s − m))s

2

Z T

m+ 12

1

ζ(1/2 + it) x

dt xm+ 2 +ε .

ζ(1 + 2it) t 0

Note that ζ(2) =

X

π2 , from the above estimates we have 6

am (n)ck (n) =

n≤x

1 6xm+1 R(m + 1) + O(xm+ 2 +ε ). (m + 1)π 2

This completes the proof of Theorem 1. Now we prove Theorem 2. Let

f1 (s) =

∞ X φ(am (n)ck (n))

ns

n=1

,

Then from Euler product formula [6] and the definition of φ(n), we also have if m ≥ k, then

f1 (s) ϕ(pm−(k−1) · pk−1 ) ϕ(pm−1 · p) ϕ(pm−2 · p2 ) = 1+ + + · · · + ps p2s p(k−1)s p ! Y pm − pm−1 pm − pm−1 pm − pm−1 = 1+ + + ··· + ps p2s p(k−1)s p Y 1 1 1 1 = 1 + s−m + 2s−m (1 + s + · · · + (k−3)s ) p p p p p Y

1 − ps−m+1 (1 +

1 ps

+ ··· +

1

) p(k−2)s

!

ζ(s − m) Y p(p(k−2)s − 1) − (p(k−1)s − 1) = 1+ ; ζ(2(s − m)) p p(ps−m + 1)(p(k−1)s − p(k−2)s )

!

On the hybrid mean value of some special sequences3

141

if m < k (im ≤ k < (i + 1)m, i > 1), then pm − pm−1 pm − pm−1 + · · · + ps pms p 2m 2m−1 2m −p2m−1 p −p + p p(m+1)s + ··· + p2ms p(i+1)m −p(i+1)m−1 p(i+1)m −p(i+1)m−1 +··· + + · · · + (im+1)s (k−1)s p p Y 1 1 1 1 = 1 + s−m + 2s−m 1 + s + · · · + (m−2)s p p p p p

f1 (s) =

Y

1+

1 1 1 + p1s + · · · + p(m−1)s − ps−m+1 1 1 1 + · · · + (m+1)s−2m 1 + s + · · · + (m−1)s p p p 1 1 1 − p(m+1)s−2m+1 1 + ps + · · · + p(m−1)s 1 1 1 + · · · + ((i−1)m+1)s−im 1 + s + · · · + (m−1)s p p p 1 1 1 − ((i−1)m+1)s−im+1 1 + s + · · · + (m−1)s p p p Pk−1 p(i+1)m −p(i+1)m−1 + j=im+1 pjs

!

(p(m−1)s − 1) + (pms − 1)H(s) G(s) ζ(s − m) Y 1+ + s−m = , (m−1)s s−m ms ζ(2(s − m)) p p +1 (p + 1)(p − p ) where H(s) =

i−1 jm+s+1 X p − pjm+s

(

j=1

and G(s) =

p(jm+1)s

1 − ) p

k−1 X

pim+s − pim+s−1 . pjs j=im+1

By Perron formula [7] and the method of proving Theorem 1, we also have 1 φ(am (n)ck (n)) = 2iπ n≤x X

Z

m+2+iT

m+2−iT

1 ζ(s − m) xs R∗ (s) ds+O(xm+ 2 +ε ), ζ(2(s − m)) s

where !

p(p(k−2)s − 1) − (p(k−1)s − 1) , p(ps−m + 1)(p(k−1)s − p(k−2)s ) (p(m−1)s − 1) + (pms − 1)H(s) (ps−m + 1)(pms − p(m−1)s ) G(s) , ps−m + 1

 Y    1+    p    Y R∗ (s) = 1+    p       +

This completes the proof of Theorem 2.

m ≥ k;

m < k.

142

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

References [1] F. Smarandache. Only problems, Not solutions. Chicago: Xiquan Publ. House, 1993, P27. [2] Zhang Tianping. On the m-th power free number sequence. Smarandache Notions Journal, Vol.14(2004), 62-65. [3] Zhu Weiyi. On the k-th power complement and k-th power free number sequences. Smarandache Notions Journal, Vol.14(2004), 66-69 [4] Zhang Tianping. On the cubic residues numbers and k-th power complement numbers. Smarandache Notions Journal, Vol.14(2004), 147-152. [5] Yao Weili. On the mean value of the m-th power free numbers. Smarandache Notions Journal, Vol.14(2004), 271-273. [6] Tom M. Apstol. Introduction to Analytic Number Theory. New York: Springer-Verlag, 1976. [7] Pan Chengdong and Pan Chengbiao. Foundation of Analytic Number Theory. Beijing: Science Press, 1997, P98.

ON THE MEAN VALUE OF A NEW ARITHMETICAL FUNCTION Ma Junqing Department of Mathematics, Xianyang Teacher’s College Xianyang, Shaanxi, P.R.China

Abstract

The main purpose of this paper is using the analytic method to study the mean value properties of δm (f (n)), and obtain an interesting asymptotic formula for it.

Keywords:

Arithmetic function; k-power free numbers; Mean value.

§1. Introduction and main results For any fixed positive integer q and any positive integer n, we define the arithmetical function f (n) = f (q, n) = (q, n), where (q, n) denote the greatest common divisor of q and n. Obviously it is a multiplicative arithmetical function. Another famous multiplicative function is defined as δt (n) = max{d : d | n, (d, t) = 1}, where t is any fixed positive integer. In reference [1], Professor J.Herzog and T.Maxsein studied the mean value of the error term Et (n) =

X n≤x

where σ(t) =

X

δt (n) −

tx2 , 2σ(t)

d, and proved that

d|t

X n≤x

Et (n) =

tx2 + O x lnω(t) x , 4σ(t)

where ω(t) denotes the number of all different prime divisors of n. In this paper, we want to study the mean value properties of δm (f (n)), and obtain an interesting asymptotic formula for it. First we need to introduce a special number: k-power free number. A positive integer n is called a k-power free number if it can not be divided by any pk , where p is a prime number.

144

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II

In reference [2], Professor F. Smarandache asked us to study the properties of the k-power free number sequence. About this problem, many scholars had studied it before (see reference [3] and [4]). Furthermore, there exists an interesting identity δt (n) = δα(t) (n), where α(t) is a square free number. In this paper, we shall use the analytic method to prove the following conclusion. Theorem. Let A denotes the set of all k-power free numbers (k > 1). Then for any fixed positive integer m and any real number x ≥ 1, we have the asymptotic formula X

δm (f (n)) =

n≤x n∈A

1 R(1) x + O x 2 +ε , ζ(k)

where ζ(k) is the Riemann-zeta function, ε denotes any fixed positive number, and R(s) 

=

Y  p|q p†m

1−

1

1 pks − p1s

Y

1+

pβ kq

1 ps−1

+ ··· +

1 pβ(s−1)

pβ − pβ−(k−β−1)s + p(β+1)s − pβs

!

p†m β<k

×

Y pβ kq p†m β≥k

p p2 pk−1 1 + s + 2s + · · · + (k−1)s p p p

!

,

where pβ kq denotes pβ |q and pβ+1 †q. From this theorem we may immediately deduce the following three Corollaries: Corollary 1 Let q be a square free number, then for any fixed positive integer m and any real number x ≥ 1, we have x Y δm (f (n)) = ζ(k) p|q n≤x X

n∈A

pk − 1 pk − pk−1

!

1 − p2−k 2+ p−1

!

1

+ O x 2 +ε .

p†m

Corollary 2 Let m be any fixed positive integer with (q, m) = 1. If q be a k-power free number, then we have X

δm (f (n))

n≤x n∈A

=

x Y pk − 1 k ζ(k) p|q p − pk−1

! Y pβ kq

1 − pβ+1−k β+1+ p−1

!

1

+ O x 2 +ε .

145

On the mean value of a new arithmetical function

Corollary 3 For any fixed positive integer q and any real number x ≥ 1, we have 1 X x δq (f (n)) = + O x 2 +ε . ζ(k) n≤x n∈A

§2. Proof of the theorem In this section, we shall complete the proof of Theorem. For convenience, we define a new arithmetical function a(n) as follows:

a(n) =

n, if n=1, ornisak−f reenumber, 0, otherwise.

It is clear that X

n=

n≤x n∈A

X

a(n).

n≤x

Let F (s) =

∞ X δm (f (a(n))) n=1

ns

.

From the Euler product formula [5] and the definition of δm (f (a(n))) we have F (s) =

δm (f (p)) δm (f (p2 )) δm (f (pk−1 )) 1+ + + · · · + ps p2s p(k−1)s

Y p

=

δm ((q, pk−1 )) δm ((q, p)) δm ((q, p2 )) + + · · · + 1+ ps p2s p(k−1)s

Y p

=

Y

1+

p†q

×

1 1 + · · · + (k−1)s s p p

Y p|q p†m

=

!

Y p|(m,q)

1+

1 1 + · · · + (k−1)s s p p

δm ((m, pk−1 )) δm ((m, p)) 1+ + · · · + ps p(k−1)s 

!

!

1 ζ(s) Y  1 − pks  Y p p2 pk−1 1 + + + · · · + ζ(ks) p|q ps p2s p(k−1)s 1 − p1s pβ kq p†m

×

Y pβ kq p†m β<k

!

p†m β≥k

p pβ pβ pβ 1 + s + · · · + βs + (β+1)s + · · · + (k−1)s p p p p

!

146

RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY II 

1 p ζ(s) Y  1 − pks  Y p2 pk−1 1 + + + · · · + ζ(ks) p|q ps p2s p(k−1)s 1 − p1s pβ kq

=

p†m

Y

×

!

p†m β≥k

1

1+

ps−1

pβ kq p†m β<k

+ ··· +

1 pβ(s−1)

pβ − pβ−(k−β−1)s + p(β+1)s − pβs

!

.

Obviously, we have inequality

∞

X δ (f (a(n)))

1 m

,

< σ

n σ−1 n=1

|δm (f (a(n)))| ≤ n,

where σ > 1 is the real part of s. So by Perron formula ? we have X a(n) n≤x

1 2iπ

=

ns0

Z

b+iT

b−iT

xs f (s + s0 ) ds + O s

xb B(b + σ0 ) T

!

log x +O x H(2x) min(1, ) T x ) , +O x−σ0 H(N ) min(1, ||x||

1−σ0

where N is the nearest integer to x, kxk = |x − N |. Taking s0 = 0, b = 2, 3 1 T = x 2 , H(x) = x2 , B(σ) = σ−1 , we have 1 δm (f (n)) = 2iπ n≤x X

Z

2+iT

2−iT

1 ζ(s) xs R(s) ds + O(x 2 +ε ), ζ(ks) s

n∈A

where 

R(s) =

Y  p|q p†m

×

1−

1

1 pks − p1s

Y

p p2 pk−1 1 + s + 2s + · · · + (k−1)s p p p

Y

1+

pβ kq

pβ kq p†m β≥k

1 ps−1

+ ··· +

1 pβ(s−1)

pβ − pβ−(k−β−1)s + p(β+1)s − pβs

p†m β<k

To estimate the main term 1 2iπ

Z

2+iT

2−iT

!

ζ(s) xs R(s) ds, ζ(ks) s

!

.

147

On the mean value of a new arithmetical function

we move the integral line from s = 2 ± iT to s = function ζ(s)xs f (s) = R(s) ζ(ks)s has a simple pole point at s = 1 with residue 1 2iπ

Z

2+iT

Z

+ 2−iT

1 +iT 2

Z

+

2+iT

1 −iT 2 1 +iT 2

Z

+

R(1) ζ(k) x.

2−iT 1 −iT 2

!

1 2

± iT . This time, the

So we have

ζ(s)xs R(1)x ds = . ζ(ks)s ζ(k)

We can easily get the estimate

Z 1 +iT Z 2−iT !

1 2 ζ(s)xs

+ ds

1

2πi ζ(ks)s

2+iT −iT 2

Z 2

2

1 x2

ζ(σ − 1 + iT ) x

= x2

dσ 1 ζ(k(σ − 1 + iT )) T

T 2

and

Z T

1 Z 12 −iT ζ(s)xs

ds

2πi 1 +iT ζ(ks)s

0 2

ζ(1/2 + it) x 12

1

dt x 2 +ε .

ζ(k/2 + kit) t + 1

Combining the above estimates we have X n≤x n∈A

δm (f (n)) =

1 R(1) x + O x 2 +ε . ζ(k)

This completes the proof of Theorem.

References [1] J.Herzog and T.Maxsein. On the behavior of a certain error-term, Arch. Math., vol. 50, 1988, 145-155. [2] F. Smarandache. Only problems, not Solutions, Xiquan Publ. House, Chicago, 1993, 24-25. [3] Zhang Tianping. On the k-power free number sequence, Smarandache Notions Journal, vol. 14, 2004, 62-65. [4] Zhu Weiyi. On the k-power complement and k-power free number sequence, Smarandache Notions Journal, vol. 14, 2004, 66-69. [5] Tom M. Apostol, Introduction to Analytic Number Theory, SpringerVerlag, New York, 1976. [6] Pan Chengdong and Pan Chengbiao. Foundation of Analytic Number Theory, Science Press, Beijing, 1997.

This book contains 34 papers, most of which were written by participants to the First Northwest Number Theory Conference held in Shangluo Teacher‘s College, China, in March, 2005. In this Conference, several professors gave a talk on Smarandache Problems and many participants lectured on them both extensively and intensively. All these papers are original and have been refereed. The themes of these papers range from the mean value or hybrid mean value of Smarandache type functions, the mean value of some famous number theoretic functions acting on the Smarandache sequences, to the convergence property of some infinite series involving the Smarandache type sequences. (The Editors)

List of the Contributors Zhang Wenpeng Yi Yuan Xue Xifeng Ren Ganglian Li Jie Liu Yanni Li Zhanhu Wang Ting Ren Dongmei Lu Yaming Li Junzhuang Li Chao Zhao Jian Wang Nianliang Lou Yuanbing Yang Qianli Ren Zhibin Zhao Xiaopeng Feng Zhiyu Chen Guohui Fu Ruiqin Zhang Shengsheng Ji Yongqiang Zhao Jiantang Ma Junqing

Xu Zhefeng Ding Liping Ma Jinping Yang Hai Liu Duansen Yang Cundian Gao Nan Yang Mingshun Wang yongxing Du Jianli Huang Wei Li Yansheng

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