On the Difference S(Z(n))

On the Difference S(Z(n)) - Z(S(n))

Maohua Le
Department of Mathematics
Zhanjiang Normal College
Zhanjiang, Guangdong
P.R. China

Abstract: In this paper, we prove that there exist infinitely many positive integers n satisfying
S(Z(n))> Z(S(n)) or S(Z(n)) < Z(S(n)).

Key words: Smarandache function, Pseudo-Smarandache function, composite function, difference.

For any positive integer n, let S(n), Z(n) denote the Smarandache function and the Pseudo-Smarandache function of n respectively. In this paper, we prove the following results:

Theorem 1: There exist infinitely many n satisfying S(Z(n)) > Z(S(n)).

Theorem 2: There exist infinitely many n satisfying S(Z(n)) < Z(S(n)).

The above mentioned results solve Problem 21 of [1].

Proof of Theorem 1.
Let p be an odd prime. If n = ½p(p+1), then we have

(1) S(Z(n)) = S(Z(½p(p+1))) = S(p) = p

and

(2) Z(S(n)) = Z(S(½p(p+1))) = Z(p) = p-1.

We see from (1) and (2) that S(Z(n)) > Z(S(n)) for any odd prime p. It is a well-known fact
that there exist infinitely many odd primes p. Thus, the theorem is proved.

Proof of Theorem 2.
If n = p, where p is an odd prime, then we have

(3) S(Z(n)) = S(Z(p)) = S(p-1) < p-1

and

(4) Z(S(n)) = Z(S(p)) = Z(p) = p-1.

By (3) and (4), we get S(Z(n)) < Z(S(n)) for any p. Thus, the theorem is proved.