The main purpose of this paper is using the analytic method to study the n-ary sieve sequence, and solved one conjecture about this sequence

Yuan, Yi

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THE 97-TH PROBLEM OF F.SMARANDACHE ∗Yi YuanResearch Center for Basic Science, Xi’an Jiaotong University Xi’an, Shaanxi, P.R.Chinayuanyi@mail.xjtu.edu.cnAbstract The main purpose of this paper is using the analytic method to study the n-arysieve sequence, and solved one conjecture about this sequence.Keywords: Quad 97-th problem of F.Smarandache; n-ary sieve sequence; Conjecture.§1. Introduction and resultsIn 1991, American-Romanian number theorist Florentin Smarandache in-troduced hundreds of interesting sequences and arithmetical functions, andpresented 105 unsolved arithmetical problems and conjectures about these se-quences and functions in book [1]. Already many researchers studied these se-quences and functions from this book, and obtained important results. Amongthese problems, the 97-th unsolved problem is:Let n be any positive integer with n ≥ 2, starting to count on the naturalnumbers set at any step from 1:— delete every n-th number;— delete from the remaining ones, every (n2)-th number;· · · · · ·;and so on: delete from the remaining ones, every (nk)-th number, k =1, 2, 3, · · · .For this special sequence, there are two conjectures:(1) there are an infinity of primes that belong to this sequence;(2) there are an infinity of number of this sequence which are not prime.In this paper, we shall use the analytic method to study the n-ary sieve se-quence, and solved conjecture (2). That is, we have the following conclusion:Theorem. For any positive integer n ≥ 2, the conjecture (2) of n-arysequence is true.∗This work is supported by the Zaizhi Doctorate Foundation of Xi’an Jiaotong University116 SCIENTIA MAGNA VOL.1, NO.1§2. Proof of TheoremIn this section, we shall complete the proof of Theorem. For any fixedreal number x ≥ 1 and positive integer k, let Ak(x) denotes the number ofremaining ones after deleting (nk)-th number from the interval [1, x]. In theinterval [1, x], for any n ∈ [1, x], first we delete n-th number from the interval[1, x], then the number of remaining ones isA1(x) = [x]−[xn],where [x] denotes the greatest integer which is not exceeding x, and x − 1 ≤[x] ≤ x+ 1.Note thatA1(x) = [x]−[xn]≤ x+ 1− xn= x(1− 1n)+ 1, (1)if we delete every (n2)-th number from the remaining ones, then the numberof remaining ones isA2(x) = [x]−[xn]−[[x]− [xn ]n2].From (1), we have the inequality[x]−[xn]−[[x]− [xn ]n2](2)≤[x(1− 1n)+ 1]−x(1− 1n)+ 1n2≤ x(1− 1n)+ 2−x(1− 1n)+ 1n2= x(1− 1n)(1− 1n2)+(2− 1n2)≤ x(1− 1n)(1− 1n2)+ 2.· · · · · ·, and so on: if we delete every (nk)-th number, from the remaining ones,we also have the inequalityAk(x) = x(1− 1n)(1− 1n2)· · ·(1− 1nk)+ k. (3)Similarly, we can also deduce thatx(1− 1n)− 1 = x− 1− xn≤ A1(x) = [x]−[xn], (4)The 97-th problem of F.Smarandache 1 117x(1− 1n)(1− 1n2)− 2 ≤ A2(x) = [x]−[xn]−[[x]− [xn]n2], (5)· · · · · ·, and so on:x(1− 1n)(1− 1n2)· · ·(1− 1nk)− k ≤ Ak(x). (6)Combining (5) and (6), we have the asymptotic formulaAk(x) = x(1− 1n)(1− 1n2)· · ·(1− 1nk)+O(k). (7)Note that k ¿ lnx, so we haveAk(x) = x(1− 1n)(1− 1n2)· · ·(1− 1nk)+O(lnx). (8)Let pi(x) denotes the number of the primes up to x, then we have (see refer-ence [2])pi(x) =xlnx+O(xln2 x). (9)Note that(1− 1n) (1− 1n2)· · ·(1− 1nk)is convergence if k → +∞, soAk(x)− pi(x)→ +∞, if x→ +∞.That is, there are an infinity of number of this sequence which are not prime.This completes the proof of Theorem.References[1] F. Smarandache, Only Problems, Not Solutions, Xiquan Publishing HouseChicago, 1993.[2] Tom M. Apostol, Introduction to Analytic Number Theory, New York,Springer-Verlag, 1976.

Introduction to Analytic Number Theory,

Only Problems, Not Solutions,

THE 97-TH PROBLEM OF F. SMARANDACHE

https://zenodo.org/record/8775/files/97thProblem.pdf

THE 97-TH PROBLEM OF F. SMARANDACHE

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