ON THE SMARANDACHE FUNCTION AND SQUARE COMPLEMENTS

ON THE SMARANDACHE FUNCTION AND SQUARE COMPLEMENTS ∗ Zhang Wenpeng Research Center for Basic Science, Xi’an Jiaotong University, Xi’an, Shaanxi, P.R.China wpzhang@nwu.edu.cn

Xu Zhefeng Department of Mathematics, Northwest University, Xi’an, P.R.China zfxu@nwu.edu.cn

Abstract

The main purpose of this paper is using the elementary method to study the mean value properties of the Smarandache function, and give an interesting asymptotic formula.

Keywords:

Smarandache function; Square complements; Asymtotic formula.

§1. Introduction Let n be an positive integer, if a(n) is the smallest integer such that na(n) is a perfect square number, then we call a(n) as the square complements of n.

The famous Smarandache function S(n) is defined as following: S(n) = min{m : m ∈ N, n|m!}.

In problem 27 of [1], Professor F. Smarandache let us to study the properties of the square complements. It seems no one know the relation between this sequence and the Smaradache function before. In this paper, we shall study the mean value properties of the Smarandache function acting on the square complements, and give an interesting asymptotic formula for it. That is, we shall prove the following conclusion: ∗ This

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

2

SCIENTIA MAGNA VOL.1, NO.1 Theorem. For any real number x ≥ 3, we have the asymptotic formula X

π 2 x2 S(a(n)) = +O 12 ln x n≤x

Ã

!

x2 . ln2 x

§2. Proof of the theorem To complete the proof of the theorem, we need some simple Lemmas. For convenience, we denote the greatest prime divisor of n √by p(n).

Lemma 1. If n is a square free number or p(n) > n, then S(n) = p(n).

Proof. (i) n is a square free number. Let n = p1 p2 · · · pr p(n), then pi |p(n)!,

i = 1, 2, · · · , r.

So n|p(n)!, but√p(n) † (p(n) − 1)!, so n † (p(n) − 1)!, that is, S(n) = p(n); (ii) p(n) > n. Let n = pα1 1 pα2 2 · · · pαr r p(n), so we have √ pα1 1 pα2 2 · · · pαr r < n then

pαi i |p(n)!, i = 1, 2, · · · , r.

So n|p(n)!, but p(n) † (p(n) − 1)!, so 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

Ã

!

x2 . ln2 x

Proof. Let π(x) denotes the number of the primes up to x. Noting that µ

¶

x x +O , ln x ln2 x from the Abel’s identity [2], we have Z x X √ √ p = π(x)x − π( x) x − √ π(t)dt π(x) =

√

x

x≤p≤x

=

1 x2 x2 − +O ln x 2 ln x

=

x2 +O 2 ln x

Ã

à !

x2 ln2 x

!

x2 . ln2 x

This proves Lemma 2.

Now we prove the theorem. First we have X

S(a(n)) =

n≤x

X

m2 n≤x

=

X

S(n)|µ(n)| X

√ m≤ x n≤ mx2

S(n)|µ(n)|.

(1)

On the Smarandache function and square complements 1

3

To the inner sum, using the above lemmas we get X

=

S(n)|µ(n)|

x m2

n≤

X

³

np≤ x2 √m p≥ np

=

m2

X

√

|µ(n)|

x m2

X n≤ln2 x

=

3

´

p|µ(n)| + O x 2

√mx2

n≤

=

´

x

np≤

=

³

X

p≥

3

p|µ(n)| + O x 2

³

X

√

3

´

p + O x2

x ≤p≤ x 2 m2 nm

|µ(n)|x2 x + 2n2 m4 ln nm 2

ζ(2)x2 +O 2ζ(4)m4 ln x

Ã

X ln2 x<n≤

√

!

x m2

|µ(n)|x2 x +O 2n2 m4 ln nm 2

Ã

x2 m4 ln2 x

x2 . m4 ln2 x

!

(2)

Combining (1) and (2), we have  X

S(a(n)) =

n≤x

m≤ x

= Noting that ζ(2) =

π2 6 ,



ζ(2)x2 X 1 x2 X 1   + O 2ζ(4) ln x √ m4 √ m4 ln2 x ζ(2)x2 +O 2 ln x

Ã

m≤ x

!

x2 . ln2 x

so we have

X

π 2 x2 S(a(n)) = +O 12 ln x n≤x

Ã

!

x2 . ln2 x

This completes the proof of Theorem.

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

[2] Tom M. Apostol, Introduction to Analytic Number Theory, SpringerVerlag, New York, 1976, pp. 77.

[3] Pan Chengdong and Pan Chengbiao, Element of the Analytic Number Theory, Science Press, Beijing, 1991.