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0 时, 有Ec(x) x1/c exp ? D(log x)3/5 (log log x)?1/5 . 证见Ivi? c 文[2] 中的定理 14.2. 下面证明本文的定理. 令F(s) := ∞ n=1 n is square?full q (e) k (n) ns = ∞ n=1 q (e) k (n)f2(n) ns ( s > 1), 其中 f2(n) = 1, n 是square-full 数;

0, 否则. 因为 q (e) k (n) 是可乘函数, 由欧拉乘积可以得到 F(s) = ∞ n=1 n is square?full q (e) k (n) ns = ∞ n=1 q (e) k (n)f2(n) ns = p

1 + q (e) k (p)f2(p) ps + q (e) k (p2 )f2(p2 ) p2s + q (e) k (p3 )f2(p3 ) p3s q (e) k (p2k ?1 )f2(p2k ?1 ) p(2k?1)s = p

1 +

1 p2s +

1 p3s +

1 p4s

1 p(2k?1)s = ζ(2s) p

1 +

1 p3s ?

1 p2ks ?

1 p(2k+1)s = ζ(2s)ζ(3s) p

1 ?

1 p6s ?

1 p2ks ?

1 p(2k+1)s +

1 p(2k+3)s +

1 p(2k+4)s = ζ(2s)ζ(3s) ζ(6s) p

1 ?

1 p2ks ?

1 p(2k+1)s +

1 p(2k+3)s +

1 p(2k+4)s ?

1 p(2k+6)s ?

1 p(2k+7)s + ・ ・ ・ = ζ(2s)ζ(3s) ζ(6s) G(s),

868 数学杂志Vol.

37 其中 G(s) = p

1 ?

1 p2ks ?

1 p(2k+1)s +

1 p(2k+3)s +

1 p(2k+4)s ?

1 p(2k+6)s ?

1 p(2k+7)s 显然, 对于 k ≥ 3, 当s>18+时, G(s) 绝对收敛. 根据卷积原理 [3] , 定义 H(s) := ζ(2s)ζ(3s)G(s) = ∞ n=1 h(n) ns = ∞ n=1 n=ml d(2, 3;

m)g(l) ns ( s > 1), 其中 h(n) = n=ml d(2, 3;

m)g(l). 那么可以得到 n≤x h(n) = ml≤x d(2, 3;

m)g(l) = l≤x g(l) m≤ x l d(2, 3;

m). 由引理

1 得到 n≤x d(2, 3;

n) = ζ(

3 2 )x

1 2 + ζ(

2 3 )x

1 3 + ?(2, 3;

x) = ζ(

3 2 )x

1 2 + ζ(

2 3 )x

1 3 + O(x

2 15 ). 很容易得到 n≤x |h(n)| l≤x |g(l)| m≤ x l d(2, 3;

m) x

1 2 , (2.1) 而且有 n≤x h(n) = l≤x g(l) ζ(

3 2 )( x l )

1 2 + ζ(

2 3 )( x l )

1 3 + O ( x l )

2 15 = ζ(

3 2 )x

1 2 l≤x g(l) l1/2 + ζ(

2 3 )x

1 3 l≤x g(l) l1/3 + O(x

2 15 l≤x |g(l)| l2/15 ) = ζ(

3 2 )x

1 2 ∞ l=1 g(l) l1/2 + ζ(

2 3 )x

1 3 ∞ l=1 g(l) l1/3 + O(x

1 2 l>x |g(l)| l1/2 ) +O(x

1 3 l>x |g(l)| l1/3 ) + O(x

2 15 l≤x |g(l)| l2/15 ). 又由 G(s) 在σ>18+时绝对收敛, 则可设 M(l) := t≤l |g(t)| x

1 8 , 根据 Abel 分部求和得到 x

1 2 l>x |g(l)| l1/2 = x

1 2 ∞ x l?

1 2 d M(l) = x

1 2 l?

1 2 M(l) ∞ x + x

1 2 ∞ x M(l)d l?

1 2 x

1 8 , No.

4 杨丽等: 关于 square-full 数上的函数 q (e) k (n) 的均值估计

869 x

1 3 l>x |g(l)| l1/3 = x

1 3 ∞ x l?

1 3 d M(l) = x

1 3 l?

1 3 M(l) ∞ x + x

1 3 ∞ x M(l)d l?

1 3 x

1 8 , x

2 15 l≤x |g(l)| l2/15 = x

2 15 x

1 l?

2 15 d M(l) = x

2 15 l?

2 15 M(l) x

1 + x

2 15 x

1 M(l)d l?

2 15 x

1 8 , 从而由 G(s) = ∞ n=1 g(n) ns , 有n≤x h(n) = ζ(

3 2 )G(

1 2 )x

1 2 + ζ(

2 3 )G(

1 3 )x

1 3 + O(x

1 8 ). (2.2) 由于 F(s) = H(s) ζ(6s) = ∞ n=1 h(m) ms ∞ m=1 ?(d) d6s , 所以当 n 是square-full 数时, 有q(e) k (n) = n=md6 h(m)?(d). (2.3) 再由 Perron 公式 [3] 可以得到 n≤x n is square?full q (e) k (n) = (2πi)?1 1+

1 log x +iT 1+

1 log x ?iT ζ(2s)ζ(3s) ζ(6s) G(s) xs x ds + O( x log2 x T ). 取T=x, 并将积分线平移至 σ = ?1

2 , 则在 s =

1 2 , s =

1 3 处的留数分别为 ζ(3

2 )G(1

2 ) ζ(3) x

1 2 , ζ(2

3 )G(1

3 ) ζ(2) x

1 3 . 再由 (2.1)、 (2.2)、 (2.3) 式和引理 2, 可以得到 n≤x n is square?full q (e) k (n) = ζ(3

2 )G(1

2 ) ζ(3) x

1 2 + ζ(2

3 )G(1

3 ) ζ(2) x

1 3 +O x

1 6 exp ? D(log x)

3 5 (log log x)?

1 5 , 其中 D > 0, 并且 G(s) 当s>18+时是绝对收敛的. 这样就得到了本文的定理.

870 数学杂志Vol.

37 参考文献[1] 贺艳峰, 孙春丽. 奇完全数的两个性质 [J]. 数学杂志, 2015, 35(1): 135C140. [2] Ivi? c A. The Riemann Zeta function[M]. New York: Wiley, 1985. [3] Subbarao M. V. On some arithmetic convolutions[J].The. Arith. Funct., 1972, 251: 247C271. [4] Titchmarsh E. C. The theory of the Riemann Zeta-function[M]. Oxford: Oxford University Press, 1986. [5] T? oth L. On certain arithmetic functions involving exponential divisors[J]. Ann. Univ. Sci. Budapest. Sect. Comp., 2004, 24: 285C294. [6] T? oth L. On certain arithmetic function involing exponential divisors II[J]. Ann. Univ. Sci. Budapest. Sect. Comp., 2007, 27: 155C156. THE MEAN VALUE OF FUNCTION q (e) k (n) OVER SQUARE-FULL NUMBER YANG Li1,2,3 , FU Chun3 (1.School of Management, Nanchang University, Nanchang 330031,China) (2.School of Mathematics, Nanchang University, Nanchang 330031,China) (3.Center for Central China Economic Development Research, Nanchang University, Nanchang 330031, China) Abstract: In this paper, we study the mean value of the characteristic function q (e) k (n) of k-free over square-full number. By using the property of Riemann-Zeta function and residue theorem, we obtain the asymptotic formula of the mean value, which is the generalization of q (e) k (n) over integers. Keywords: square-full number;