 
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, PseudoSmarandache function, composite
function, difference.
For any positive integer n, let S(n), Z(n) denote the Smarandache function
and the PseudoSmarandache 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) = p1.
We see from (1) and (2) that S(Z(n)) > Z(S(n)) for any odd prime p.
It is a wellknown 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(p1) < p1
and
(4) Z(S(n)) = Z(S(p)) = Z(p) = p1.
By (3) and (4), we get S(Z(n)) < Z(S(n)) for any p. Thus, the theorem
is proved.
Reference:
[1] C. Ashbacher, Problems, Smarandache Notions Journal, 9(1998),
144151.
