# An inequality of the Smarandache function.

Abstract For any positive integer n, the famous Smarandache function S(n) is defined as the smallest positive integer m such that n|m!. That is, S(n) = min{m : m [element] N, n|m!}. In an unpublished paper, Dr. Kenichiro Kashihara asked us to solve the following inequalitiesS ([x.sup.n.sub.1]) + S ([x.sup.n.sub.2]) + ... + S ([x.sup.n.sub.n]) [greater than or equal to] nS ([x.sub.1]) * S ([x.sub.1]) ... S ([x.sub.n]).

In this paper, we using the elementary method to study this problem, and prove that for any integer n [greater than or equal to] 1, the inequality has infinite group positive integer solutions ([x.sub.1], [x.sub.2], ..., [x.sub.n]).

Keywords F.Smarandache function, inequalities, solution, necessary condition.

[section] 1. Introduction and Results

For any positive integer n, the famous F.Smarandache function S(n) is defined as the smallest positive integer m such that n | m!. That is, S(n) = min|m : n|m!, n [element] N}. For example, the first few values of S(n) are S(1) = 1, S(2) = 2, S(3) = 3, S(4) = 4, S(5) = 5, S(6) = 3, S(7) = 7, S(8) = 4, S(9) = 6, S(10) = 5, S(11) = 11, S(12) = 4, ... About the elementary properties of S(n), many authors had studied it, and obtained some interesting results, see reference [2], [3], [4] and [5]. For example, Wang Yongxing [3] studied the mean value properties of S(n), and obtained a sharper asymptotic formula about this function:

[summation over [n [less than or equal to]] S(n) = [[pi].sup.2] /12 [x.sup.2]/lnx + O ([x.sup.2]/[ln.sup.2]x)

Lu Yaming [4] studied the solutions of an equation involving the F.Smarandache function S(n), and proved that for any positive integer k [greater than or equal to] 2, the equation

S([m.sub.1] + [m.sub.2] + ... + [m.sub.k]) - S([m.sub.1]) + S([m.sub.2]) + ... + S([m.sub.k])

has infinite group positive integer solutions ([m.sub.1], [m.sub.2], ... + S([[m.sub.k]).

Jozsef Sandor [5] proved for any positive integer k [greater than or equal to] 2, there exist infinite group positive integers ([m.sub.1], [m.sub.2], ... + [m.sub.k]) satisfied the following inequality:

S([m.sub.1] + [m.sub.2] + ... + [m.sub.k]) > S([m.sup.1]) + S([m.sub.2]) + ... + S([M.sub.k])

Also, there exist infinite group positive integers ([m.sup.1], [m.sup.2], ... + [m.sub.k]) such that

S([m.sub.1] + [m.sub.2] + ... + [m.sub.k]) < S([m.sub.1]) + S([m.sub.2]) + ... + S([m.sub.k]).

In [6], Fu Jing proved more deeply conclusion, i.e., if the positive integer k and m satisfying one of the following conditions:

(a) k > 2 and m [greater than or equal to] 1 are all odd numbers.

(b) k [greater than or equal to] 5 is odd, m [greater than or equal to] 2 is even.

(c) Any even numbers k [greater than or equal to] 4 and any positive integer m; then the equation

m * S([m.sub.1] + [m.sub.2] + ... + [m.sub.k]) = S([m.su.b.1]) + S([m.sub.2]) + ... + S([m.sub.k])

has infinite group positive integer solutions ([m.sub.1], [m.sub.2], ... , [m.sub.k]).

On the other hand, Xu Zhefeng [7] studied the value distribution properties of S(n), and obtained a more interesting result. That is, he proved the following conclusion:

Let P(n) be the largest prime factor of n, then for any real numbers x [greater than or equal to] 1, we have the asymptotic formula:

[summation over (n [greater than or equal to] x) [(S(n) - P(n).sup.2] = 2[zeta] (3/2) [x.sup.3/2]/3 ln x + O (x.sup.3/2/[ln.sup.2]x)

where [zeta](s) is the Riemann zeta-function.

In an unpublished paper, Dr. Kenichiro Kashihara asked us to solve the following inequalities

S([x.sup.n.sub.1]) + S([x.sup.n.sub.2]) + .. + S ([x.sup.n.sub.n]) [greater than or equal to] nS ([x.sub.1]) * S([x.subb.1]) ... S([x.sub.n]). (1)

About this problem, it seems that none had studied it yet, at least we have not seen any related papers before. The main purpose of this paper is using the elementary methods to study this problem, and prove the following:

Theorem 1. For any fixed positive integer n [greater than or equal to] 1, the inequality (1) has infinite group positive integer solutions ([x.sub.1], [x.sub.2], ..., [x.sub.n]).

Theorem 2. For any fixed positive integer n [greater than or equal to] 3, if ([x.sub.1], [x.sub.2], ..., [x.sub.n]) satisfying the inequality (1), then at least n - 1 of [x.sub.1], [x.sub.2], ... , [x.sub.n] are 1.

It is clear that the condition n [greater than or equal to] 3 in Theorem 2 is necessary. In fact if n = 2, we can take [x.sub.1] = [x.sub.2] = 2, then we have the identity

S([x.sup.2.sub.1]) + S([x.sup.2.sub.2]) - S([2.sup.2]) + S([2.sup.2]) = 4 + 4 = 8 = 2S(2)S(2) = 2S([x.sub.1])S([x.sub.2]).

So if n = 2, then Theorem 2 is not correct.

[section] 2. Proof of the theorems

In this section, we shall prove our theorems directly. First we prove Theorem 1. If n = 1, then this time, the inequality (1) become S([x.sub.1]) [greater than or equal to] S([x.sub.1]), and it holds for all positive integers [x.sub.1]. So without lose of generality we can assume that n [greater than or equal to] 2. We taking [x.sub.1] = [x.sub.2] = ... [x.sub.n-1] = 1, [x.sub.n] = p > n, where p be a prime. Note that s(1) = 1, S(p) = p and S([p.sup.n]) = np, so we have

S([x.sup.n.sub.1]) + S([x.sup.n.sub.2) + ... + S([x.sup.n.sub.n]) = n-1 + np (2)

nS([x.sub.1]) * S ([x.sub.1]) ...... S([x.sub.n]) = nS(p) = np

From (2) and (3) we may immediately deduce that

S([x.sup.n.sub.1]) + S([x.sup.n.sub.2]) + ...... + S ([x.sup.n.sub.n]) [greater than or equal to] ns([x.sub.1])*S([x.sub.1]) ...... S([x.sub.n). (4)

Since there are infinite primes p > n, so all positive integer groups

([x.sub.1], [x.sub.2], ..., [x.sub.n]) - (1, 1, ..., p)

are the solutions of the inequality (1). Therefore, the inequality (1) has infinite group positive integer solutions ([x.sub.1], [x.sub.2], ..., [x.sub.n]). This proves Theorem 1.

Now we prove Theorem 2. Let n [greater than or equal to] 3, if ([x.sub.1], [x.sub.2], ..., [x.sub.n]) satisfying the inequality (1), then at least n - 1 of [x.sub.1], [x.sub.2], ..., [x.sub.n] are 1. In fact if there exist [x.sub.1] > 1, [x.sub.2] > 1, ..., [x.sub.k] > 1 with 2 [less than or equal to] k [less than or equal to] n such that the inequality

S ([x.sup.n.sub.1]) + S ([x.sup.n.sub.2]) + ...... + S ([x.sup.n.sub.n]) [greater than or equal to] ns ([x.sub.1]) * S([x.sub.1]) ...... S([x.sub.n]). (5)

Then from the definition and properties of the function S(n) we have S([x.sub.i]) > 1 and S ([x.sup.n.sub.i]) [less than or equal to] nS([x.sub.i]), i = 1, 2, ..., k. Note that [a.sub.1] + [a.sub.2] + ... + [a.sub.k] < [a.sub.1][a.sub.2] ... [a.sub.k] if [a.sub.i] > 1 and k [greater than or equal to] 3, i = 1, 2, ..., k; If k = 2, then [a.sub.1] + [a.sub.2] [greater than or equal to] [a.sub.1][a.sub.2], and the equality holds if and only if [a.sub.1] = [a.sub.2] = 2 ([a.sub.1] > 1, [a.sub.1] > 1). So this time, the inequality (5) become

n - k + S([x.sup.n.sub.1]) + S([x.sup.n.sub.2]) + ....... + S([x.sup.n.sub.k]

If k [greater than or equal to] 3, then from (6) and the properties of S(n) we have

n - k + n [S ([x.sub.1]) + S ([x.sub.2]) + ....... + S ([x.sub.k])] [greater than or equal to] nS([x.sub.1])S([x.sub.2]) ... S([x.sub.k])

or

n - k/n + S ([x.sub.1]) + S ([x.sub.2]) + ....... + S ([x.sub.k]) [greater than or equal to] S([x.sub.1])S([x.sub.2]) ... S([x.sub.k]). (7)

Note that 0 [greater than or equal to] n-k/n < 1, so the inequality (7) is not possible, because

S([x.sub.1])S([x.sub.2]) ... S([x.sub.k]) [greater than or equal to] S([x.sub.1]) + S([x.sub.2]) + ... + S([x.sub.k]) + 1.

If k = 2, then the inequality (6) become

n - 2 + S ([x.sup.n.sub.i]) + S ([x.sup.n.sub.2]) [greater than or equal to] nS([x.sub.1])S([x.sub.2]). (8)

Note that S([x.sup.n]) [less than or equal to] nS(x), S([x.sub.1]) + S([x.sub.2]) [less than or equal to] S([x.sub.1]) = S([x.sub.2]) and the equality holds if and only if [x.sub.1] = [x.sub.2] = 2, so if S([x.sub.1]) > 2 or S([x.sub.2]) > 2, then (8) is not possible. If S([x.sub.1]) = S([x.sub.2]) = 2, then [x.sub.1] = [x.sub.2] = 2. Therefore, the inequality (8) become

S([2.sup.n]) [greater than or equal to] 3n/2 + 1. (9)

Let S([2.sup.n]) = m, then m [greater than or equal to] 4, if n [greater than or equal to] 3. From the definition and properties of S(n) we have

[[infinity] summation over (i = 1)] [m - 1/[2.sup.i]] < n [less than or equal to] [[infinity] summation over [i = 1]] [m/[2.sup.]].

Thus,

n [greater than or equal to] 1 + [[infinity] summation over [i = 1]] > m - 1/2/2 + m - 1/4 = 3(m - 1)/4,

from (9) we have

m = S([2.sup.n]) [greater than or qual to] 3n/2 + 1 [greater than or equal to] 9/8(m - 1) + 1 = m + m-1/8 > m.

This inequality is not possible. So if n [greater than or equal to] 3 and ([x.sub.1], [x.sub.2], ..., [x.sub.n]) satisfying the inequality (1), then at least n - 1 of [x.sub.1], [x.sub.2], ..., [x.sub.n] are 1. This completes the proof of Theorem 2.

References

[1] F. Smarandache, Only Problems, Not Solutions, Chicago, Xiquan Publishing House, 1993.

[2] Farris Mark and Mitchell Patrick, Bounding the Smarandache function, Smarandache Nations Journal, 13(2002), 37-42.

[3] Wang Yongxing, On the Smarandache function, Research on Smarandache Problem In Number Theory (Edited by Zhang Wenpeng, Li Junzhuang and Liu Duansen), Hexis, 11(2005), 103-106.

[4] Lu Yaming, On the solutions of an equation involving the Smarandache function, Scientia Magna, 2(2006), No.1, 76-79.

[5] Jozsef Sandor, On certain inequalities involving the Smarandache function, Scientia Magna, 2(2006), No.3, 78-80.

[6] Fu Jing, An equation involving the Smarandache function. Scientia Magna, 2(2006), No.4, 83-86.

[7] Xu Zhefeng, On the value distribution properties of the Smarandache function, Acta Mathematica Sinica (in Chinese), 49(2006), No.5, 1009-1012.

[8] F. Smarandache, Sequences of numbers involving in unsolved problem, Hexis, 2006, 17-18.

[9] M.L.Perez, Florentin Smarandache, Definitions, solved and unsolved problems, conjectures and theorems in number theory and geometry, Xiquan Publishing House, 2000.

[10] Kenichiro Kashihara, Comments and topics on Smarandache notions and problems, Erhus University Press, USA, 1996.

[11] Zhang Wenpeng, The elementary number theory (in Chinese), Shaanxi Normal University Press, Xi'an, 2007.

[12] Tom M. Apostol, Introduction to Analytic Number Theory, New York, Springer-Verlag, 1976.

Weiyi Zhu

College of Mathematics, Physics and Information Engineer, Zhejiang Normal University

Aanghua, Zhejiang, P.R.China

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Author: | Zhu, Weiyi |
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Publication: | Scientia Magna |

Article Type: | Report |

Geographic Code: | 1USA |

Date: | Jan 1, 2008 |

Words: | 2088 |

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