Math Paper

ON THE PRIMITIVE NUMBERS OF POWER P AND ITS MEAN VALUE PROPERTIES

Ding Liping Department of Mathematics, Northwest University Xi’an, Shaanxi, P.R.China

Abstract. Let p be a prime, n be any fixed positive integer. Sp (n) denotes the smallest positive integer such that Sp (n)! is divisible by pn . In this paper, we study the mean value properties of Sp (n) for p, and give a sharp asymptotic formula for it.

1. Introduction Let p be a prime, n be any fixed positive integer, Sp (n) denotes the smallest positive integer such that Sp (n)! is divisible by pn . For example, S3 (1) = 3, S3 (2) = 6, S3 (3) = 9, S3 (4) = 9, S3 (5) = 12, · · · . In problem 49 of book [1], Professor F. Smarandache asks us to study the properties of the sequence Sp (n). About this problem, some asymptotic properties of this sequence have been studied by Zhang Wenpeng and Liu Duansen [2], they proved that Sp (n) = (p − 1)n + O



 p log n . log p

The problem is interesting because it can help us to calculate the Smarandache function. In this paper, we use the elementary methods to study the mean value properties of Sp (n) for p, and give a sharp asymptotic formula for it. That is, we shall prove the following: Theorem. Let x ≥ 2, for any fixed positive integer n, we have the asymptotic formula   k−1 X nam x2 X nx2 nx2 , + +O Sp (n) = 2 log x m=1 logm+1 x logk+1 x p≤x

where am (m = 1, 2, · · · , k − 1) are computable constants. Key words and phrases. Primitive numbers; Mean value; Asymptotic formula.. *This work is supported by the N.S.F.(10271093) and P.N.S.F of P.R.China Typeset by AMS-TEX

1

2

DING LIPING

2. Some Lemmas To complete the proof of the theorem, we need the following: Lemma. Let p be a prime, n be any fixed positive integer. Then we have the estimate (p − 1)n ≤ Sp (n) ≤ np. Proof. Let Sp (n) = k = a1 pα1 + a2 pα2 + · · · + as pαs with αs > αs−1 > · · · > α1 ≥ 0 under the base p. Then from the definition of Sp (n) we know that p | Sp (n)! and the Sp (n) denotes the smallest integer satisfy the condition. However, let (np)! = 1 · 2 · 3 · · · p · (p + 1) · · · 2p · (2p + 1) · · · np = upl . where l ≥ n,p†u. So combining this and p | Sp (n)! we can easily obtain (1)

Sp (n) ≤ np.

On the other hand, form the definition of Sp (n) we know that p | Sp (n)! and pn † (Sp (n) − 1)!, so that α1 ≥ 1, note that the factorization of Sp (n)! into prime powr is Y k! = q αq (k) . q≤k

where

Q

denotes the product over all prime , and

q≤k

αq (k) =

 ∞  X k i=1

qi

because pn | Sp (n)!, so we have

n ≤ αp (k) =

 ∞  X k i=1

pi

≤

∞ X k k = . pi p−1 i=1

or (2)

(p − 1)n ≤ k

Combining (1) and (2) we immediately get the estimate (p − 1)n ≤ Sp (n) ≤ np. This completes the proof of the lemma.

ON THE PRIMITIVE NUMBERS OF POWER P AND ITS MEAN VALUE PROPERTIES3

3. Proof of the Theorem In this section, we complete the proof of Theorem. Based on the result of lemma (p − 1)n ≤ Sp (n) ≤ np we can easily get X

(p − 1)n ≤

p≤x

Let a(n) =

X

Sp (n) ≤

p≤x



1, 0,

X

np.

p≤x

if n be a prime; otherwise.

Then from [3] we know that for any positive integer k, X

n≤x

  k−1 X m! x x . a(n) = π(x) = (1 + m )+O k+1 logx log x log x m=1

By Abel’s identity we have X X p= a(m)m p≤x

m≤x

= π(x)x −

Z

x

π(t)dt 2

  Z x k−1 k−1 X m! t x2 X m! x2 x2 = + − (1 + )dt + O logx logx m=1 logm x logm t logk+1 x 2 logt m=1   k−1 X x2 x2 a m x2 +O = + , 2 log x m=1 log( m + 1)x logk+1 x where am (m = 1, 2, · · · , k − 1) are computable constants. From above we have X

(p − 1) =

p≤x

X

p≤x

  k−1 X x2 x2 a m x2 p − π(x) = +O . + 2 log x m=1 log( m + 1)x logk+1 x

Therefore X

p≤x

Sp (n) =

X

p≤x

  k−1 X x2 a m x2 x2 + k= +O . 2 log x m=1 log( m + 1)x logk+1 x

This completes the proof of the theorem. References 1. F. Smarandache, Only problems, not Solutions, Xiquan Publ. House, Chicago, 1993,pp. 42. 2. Zhang Wenpeng and Liu Duansen, On the primitive numbers of power P and its asymptotic property, Smarandache Notions Journal 13(2002), pp. 173-175.. 3. M.Ram Murty, Problems in Analytic Number Theory, Springer-Verlag New York, 2001, pp. 36.