The main purpose of this paper is using the elementary method to study the properties of the Pseudo-Smarandache function Z(n), and solve two conjectures posed by Kenichiro Kashihara in reference

Yani, Zheng

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The main purpose of this paper is using the elementary method to study the properties of the Pseudo Smarandache function Z(n), and solve two conjectures posed by Kenichiro Kashihara in reference

Yani Zheng

NEUROSURGERY ENTHUSIASTIC WOMEN SOCIETY

On the Pseudo Smarandache function and its two conjectures

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Scientia MagnaVol. 3 (2007), No. 4, 74-76On the Pseudo Smarandache function and itstwo conjecturesYani Zheng† ‡†Department of Mathematics, Northwest University, Xi’an, Shaanxi, P.R.China‡ Department of Mathematics, Xianyang Normal College, Xianyang, Shaanxi, P.R.ChinaAbstract For any positive integer n, the famous Pseudo Smarandache function Z(n) isdefined as the smallest integer m such that n evenly dividesm∑k=1k. That is, Z(n) =min{m : n|m(m+ 1)2, m ∈ N}, where N denotes the set of all positive integers. The mainpurpose of this paper is using the elementary method to study the properties of the PseudoSmarandache function Z(n), and solve two conjectures posed by Kenichiro Kashihara in ref-erence [2].Keywords Pseudo Smarandache function, conjecture, unbounded.§1. Introduction and ResultsFor any positive integer n, the famous Pseudo Smarandache function Z(n) is defined asthe smallest positive integer m such that n evenly dividesm∑k=1k. That is,Z(n) = min{m : n|m(m+ 1)2, m ∈ N},where N denotes the set of all positive integers. For example, the first few values of Z(n) are:Z(1) = 1, Z(2) = 3, Z(3) = 2, Z(4) = 7, Z(5) = 4, Z(6) = 3, Z(7) = 6, Z(8) = 15,Z(9) = 8, Z(10) = 4, Z(11) = 10, Z(12) = 8, Z(13) = 12, Z(14) = 7, Z(15) = 5,Z(16) = 31, Z(17) = 16, Z(18) = 8, Z(19) = 18, Z(20) = 15, · · · · · · .This function was introduced by David Gorski in reference [3], where he studied the ele-mentary properties of Z(n), and obtained a series interesting results. For example, he provedthat if p ≥ 2 is a prime, then Z(p) = p − 1; If n = 2k, then Z(n) = 2k+1 − 1. The othercontents related to the Pseudo Smarandache function can also be found in references [2], [4]and [5]. Especially in reference [2], Kenichiro Kashihara posed two problems as follows: Arethe following values bounded or unbounded?A) |Z(n+ 1)− Z(n)|,B)Z(n+ 1)Z(n).Vol. 3 On the Pseudo Smarandache function and its two conjectures 75About these two problems, it seems that none had studied it, at least we have not seenrelated papers before. In this paper, we use the elementary method to study these two problems,and prove that they are unbounded. That is, we shall prove the following conclusion:Theorem. For any positive number M large enough, there are infinitely positive integersn, such thatZ(n+ 1)Z(n)> M and |Z(n+ 1)− Z(n)| > M.From this theorem, we know that |Z(n + 1) − Z(n)| and Z(n+ 1)Z(n)are unbounded. Thissolved two problems posed by Kenichiro Kashihara in reference [2].§2. Proof of the theoremIn order to complete the proof of the theorem, we need the following important conclusion:Lemma. Let k and h are any positive integers with (h, k) = 1, then there are infinitelymany primes in the arithmetic progression nk + h, where n = 0, 1, 2, 3, · · · .Proof. This is the famous Dirichlet’s Theorem, see reference [6].Now we use this Lemma to complete the proof of our Theorem. In fact for any positivenumber M , we take positive integer m such that 2m > M . Note that(22m+1, 2m + 1)= 1,so from Dirichlet’s Theorem we can easily deduce that there are infinitely many primes in thearithmetic progression:22m+1k + 2m + 1, where k = 0, 1, 2, · · · .Therefore, there must exist a positive integer k0 such that 22m+1k0 + 2m + 1 = P be aprime. For this prime P , from the definition and properties of Z(n) we can deduce thatZ(P ) = P − 1 = 22m+1k0 + 2m,Z(P − 1) = Z(22m+1k0 + 2m) = Z(2m(2m+1k0 + 1)).Since2m+1k0∑i=1i =2m+1k0(2m+1k0 + 1)2and 2m(2m+1k0 + 2m) evenly divides2m+1k0∑i=1i, so we haveZ(P − 1) ≤ 2m+1k0.ThusZ(P )Z(P − 1) >22m+1k0 + 2m2m+1k0> 2m > M.SoZ(P )Z(P − 1) is unbounded.76 Yani Zheng No. 4Similarly, we have|Z(P )− Z(P − 1)| ≥ |Z(P )| − |Z(P − 1)|≥ 22m+1k0 + 2m − 2m+1k0= 2m+1k0(2m − 1) + 2m > 2m > M.So |Z(P )− Z(P − 1)| is also unbounded.Since there are infinitely positive integers m, such that 2m > M , so there are infinitelypositive integers n, such that |Z(n+ 1)− Z(n)| and Z(n+ 1)Z(n)are unbounded.This completes the proof of the theorem.References[1] F. Smarandache, Only Problems, Not Solutions, Chicago, Xiquan Publishing House,1993.[2] Kenichiro Kashihara, Comments and topics on Smarandache notions and problems,Erhus University Press, USA, 1996.[3] David Gorski, The pseudo Smarandache function, Smarandache Notions Journal, 13(2002), 140-149.[4] Liu Yanni, On the Smarandache Pseudo Number Sequence, Chinese Quarterly Journalof Mathematics, 21(2006), No. 4, 581-584.[5] A. W. Vyawahare and K.M.Purohit, Near Pseudo Smarandache function, SmarandacheNotions Journal, 14(2006), 42-59.[6] Tom M. Apostol, Introduction to analytical number theory, Spring-Verlag, New York,1976.[7] Zhang Wenpeng, The elementary number theory, Shaanxi Normal University Press,Xi’an, 2007.[8] Pan Chengdong and Pan Chengbiao, The elementary proof of the prime theorem, Shang-hai Science and Technology Press, Shanghai, 1988.

On the Pseudo Smarandache function and its two conjectures

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