MEAN VALUE OF A NEW ARITHMETIC FUNCTION
Liu Yanni Department of Mathematics, Northwest University Xi’an, Shaanxi, P.R.China
Gao Peng School of Economics and Management,Northwest University Xi’an, Shaanxi, P.R.China
Abstract
The main purpose of this paper is using elementary method to study a new arithmetic function, and give an interesting asymptotic formula for it.
Keywords:
Arithmetic function; Mean value; Asymptotic formula
§1. Introduction For any positive integer n, we have n = uk v, where v is a k-power free number. Let bk (n) be the k-power free part of n. Let p be any fixed prime, n be any positive integer, ep (n) denotes the largest exponent of power p. That is, ep (n) = m,if pm |n and pm+1 † n. In [1], Professor F.Smarandache asked us to study the properties of these two arithmetic functions. It seems that no one knows the relationship between these two arithmetic functions before. The main purpose of this paper is to study the mean value properties of ep (bk (n)), and obtain an interesting mean value formula for it. That is, we shall prove the following conclusion: Theorem. Let p be a prime, k be any fixed positive integer. Then for any real number x ≥ 1, we have the asymptotic formula X n≤x
Ã
ep (bk (n)) =
!
³ 1 ´ pk − p k−1 − x + O x 2 +² , (pk − 1)(p − 1) pk − 1
where ² denotes any fixed positive number. Taking k = 2 in the theorem, we may immediately obtain the following
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SCIENTIA MAGNA VOL.1, NO.1
Corollary. For any real number x ≥ 1, we have the asymptotic formula X
ep (bk (n)) =
n≤x
³ 1 ´ 1 x + O x 2 +² . p+1
§2. Proof of the theorem In this section, we shall use analytic method to complete the proof of the theorem. In fact we know that ep (n) is not a multiplicative function, but we can use the properties of the Riemann zeta-function to obtain a generating function. For any complex s, if Re(s) > 1, we define the Dirichlet series f (s) =
∞ X ep (bk (n)) n=1
ns
.
Let positive integer n = pα n1 , where (n1 , p) = 1, then from the definition of ep (n) and bk (n), we have: ep (bk (n)) = ep (bk (pα n1 )) = ep (bk (pα )). From the above formula and the Euler product formula (See Theorem 11.6 of [3]) we can get f (s) = =
∞ X ep (bk (n)) n=1 ∞ X α=0
ns ∞ X n1 =1 (n1 ,p)=1
ep (bk (pα )) pαs ns1
∞ 1 X ep (bk (pα ) = ζ(s)(1 − s ) . p α=1 pαs
Let A
=
∞ X ep (bk (pα ) α=1
=
pαs
2 k−1 1 1 2 k−1 + 2s + · · · + (k−1)s + (k+1)s + (k+2)s + · · · + (2k−1)s s p p p p p p 2 k−1 1 + · · · + (uk+1)s + (uk+2)s + · · · + (uk+k−1)s p p p
=
∞ X
X r 1 k−1 uks p prs u=0 r=1
=
1 1 1 − p(k−1)s k−1 − ks . 1 1 s p −1 p 1 − pks 1 − ps
1
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Mean value of a new arithmetic function
So we have f (s) =
∞ X ep (bk (n))
ns
n=1
Ã
=
!
pks − ps k−1 − ks ζ(s). ks s (p − 1)(p − 1) p − 1
Because the Riemann zeta-function ζ(s) have a simple pole point at s = 1 with s the residue 1, we know f (s) x´s also have a simple pole point at s = 1 with the ³ residue
pk −p (pk −1)(p−1)
−
k−1 pk −1
x. By Perron formula (See [2]), taking s0 = 0,
b = 23 , T > 1, then we have X
ep (bk (n)) =
n≤x
1 2πi
we move the integral line to Re s = X
Ã
ep (bk (n)) =
n≤x
Ã
=
3 +iT 2 3 −iT 2
1 2
xs f (s) ds + O s
Ã
3
x2 T
!
,
+ ², then taking T = x, we can get !
pk − p k−1 − k x k (p − 1)(p − 1) p − 1 1 + 2πi
Z
1 +²+iT 2 1 +²−iT 2
f (s)
³ 1 ´ xs ds + O x 2 +² s !
pk − p k−1 − k x k (p − 1)(p − 1) p − 1
¯ ¯ 1 +² ! ³ 1 ´ ¯ 1 ¯ x2 ¯ ¯ f ( + ² + it) dt + O x 2 +² +O ¯ 2 ¯ 1 + |t| −T Ã ! ³ 1 ´ pk − p k−1 +² ÃZ
=
Z
T
(pk − 1)(p − 1)
−
pk − 1
x + O x2
.
This completes the proof of Theorem.
References [1] F.Smarandache, Only Problems,Not Solutions, Chicago, Xiquan Publishing House, 1993. [2] Pan Chengdong and Pan Chengbiao, Foundation of Analytic Number Theory, Beijing, Science Press, 1991. [3] Tom M. Apstol, Introduction to Analytic Number Theory, New York, Springer-Verlag, 1976.