ON THE PRIMITIVE NUMBERS OF POWER P AND ITS ASYMPTOTIC PROPERTY ∗ Yi Yuan Research Center for Basic Science, Xi’an Jiaotong University Xi’an, Shaanxi, P.R.China yuanyi@mail.xjtu.edu.cn
Abstract
Let p be a prime, n be any positive integer, Sp (n) denotes the smallest integer m ∈ N + , where pn |m!. In this paper, we study the mean value properties of Sp (n), and give an interesting asymptotic formula for it.
Keywords:
Smarandache function; Primitive numbers; Asymptotic formula
§1. Introduction and results Let p be a prime, n be any positive integer, Sp (n) denotes the smallest integer such that Sp (n)! is divisible by pn . For example, S3 (1) = 3, S3 (2) = 6, S3 (3) = 9, S3 (4) = 9, · · · · · ·. In problem 49 of book [1], Professor F. Smarandache ask us to study the properties of the sequence {Sp (n)}. About this problem, Professor Zhang and Liu in [2] have studied it and obtained an interesting asymptotic formula. That is, for any fixed prime p and any positive integer n, ¶ µ p · ln n . Sp (n) = (p − 1)n + O ln p In this paper, we will use the elementary method to study the asymptotic properties of Sp (n) in the following form: 1X |Sp (n + 1) − Sp (n)| , p n≤x where x be a positive real number, and give an interesting asymptotic formula for it. In fact, we shall prove the following result: Theorem. For any real number x ≥ 2, let p be a prime and n be any positive integer. Then we have the asymptotic formula µ
1X 1 |Sp (n + 1) − Sp (n)| = x 1 − p n≤x p ∗ This
¶
µ
¶
ln x +O . ln p
work is supported by the Zaizhi Doctorate Foundation of Xi’an Jiaotong University.
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SCIENTIA MAGNA VOL.1, NO.1
§2. Proof of the Theorem In this section, we shall complete the proof of the theorem. First we need following one simple Lemma. That is, Lemma. Let p be a prime and n be any positive integer, then we have ½
p, if pn k m!; 0, otherwise,
|Sp (n + 1) − Sp (n)| =
where Sp (n) = m, pn k m! denotes that pn |m! and pn+1 †m!. Proof. Now we will discuss it in two cases. (i) Let Sp (n) = m, if pn k m!, then we have pn |m! and pn+1 †m!. From the definition of Sp (n) we have pn+1 †(m+1)!, pn+1 †(m+2)!, · · ·, pn+1 †(m+ p − 1)! and pn+1 |(m + p)!, so Sp (n + 1) = m + p, then we get |Sp (n + 1) − Sp (n)| = p. (ii)
(1)
Let Sp (n) = m, if pn |m! and pn+1 |m!, then we have Sp (n + 1) = m,
so |Sp (n + 1) − Sp (n)| = 0.
(2)
Combining (1) and (2), we can easily get ½
|Sp (n + 1) − Sp (n)| =
p, if pn k m!; 0, otherwise.
This completes the proof of Lemma. Now we use above Lemma to complete the proof of Theorem. For any real number x ≥ 2, by the definition of Sp (n) and Lemma we have X 1X 1 X |Sp (n + 1) − Sp (n)| = p= 1, p n≤x p n≤x n≤x pn km!
(3)
pn km!
where Sp (n) = m. Note that if pn k m!, then we have (see reference [3], Theorem 1.7.2) n =
¸ ∞ · X m i=1
= m·
pi X i≤logp
=
·
X
=
i≤logp m
m pi
¸
³ ´ 1 + O log m p pi m ¶
µ
m ln m . +O p−1 ln p
(4)
From (4), we can deduce that µ
¶
p ln n m = (p − 1)n + O . ln p
(5)
On the primitive numbers of power p and its asymptotic property 1
177
So that µ
1 ≤ m ≤ (p − 1) · x + O
¶
p ln x , ln p
if
1 ≤ n ≤ x.
Note that for any fixed positive integer n, if there has one m such that pn k m!, then pn k (m + 1)!, pn k (m + 2)!, · · ·, pn k (m + p − 1)!. Hence there have p times of m such that n = ³
O
p ln x ln p
∞ h i P m
i=1
´
pi
in the interval 1 ≤ m ≤ (p − 1) · x +
. Then from this and (3), we have
1X |Sp (n + 1) − Sp (n)| = p n≤x
X 1 X p= 1 p n≤x n≤x pn km!
µ
pn km!
µ
p ln x 1 (p − 1) · x + O = p ln p µ ¶ µ ¶ 1 ln x = x· 1− +O . p ln p
¶¶
This completes the proof of Theorem.
References [1] F.Smarandache, Only Problems, Not Solutions, Chicago, Xiquan Publ. House, 1993. [2] Zhang Wenpeng and Liu Duansen, primitive numbers of power p and its asymptotic property, Smaramche Notes Journal, 13 (2002) 173-175. [3] Pan Chengdong and Pan Chengbiao, The Elementary number Theory, Beijing University, Press Beijing, 2003. [4] Tom M.Apostol, Introduction to Analytic Number Theory, New York, Springer-Verlag, 1976.