整数$n>1$有标准分解形式: $n=p^{a_{1}}_{1}p^{a_{2}}_{2}\cdots p^{a_{r}}_{r}$.若满足$n=p^{a_{1}}_{1}p^{a_{2}}_{2}\cdots p^{a_{r}}_{r}$, 其中$a_{1}\geq k, a_{2}\geq k, \cdots, a_{r}\geq k$, 则称正整数$n$为$k$-full数, 当$k=2$时也称为square-full数.令$f_{k}(n)$是$k$-full数的特征函数, 即

$ f_{k}(n)=\left\{ \begin{array}{ll} 1,&\hbox{$n$ 是$k$-full数;} \\ 0,&\hbox{否则.} \end{array} \right. $

正整数$n=p^{a_{1}}_{1}p^{a_{2}}_{2}\cdots p^{a_{r}}_{r}$称为指数$k$-full数, 若所有的指数$a_{1}, a_{2}, \cdots, a_{r}$都是$k$-full数.正整数$n=p^{a_{1}}_{1}p^{a_{2}}_{2}\cdots p^{a_{r}}_{r}$称为指数$k$-free数, 若所有的指数$a_{1}, a_{2}, \cdots, a_{r}$都是$k$-free数.

令$q_{k}^{(e)}(n)$是指数$k$-free数的特征函数.由Tóth^[5-6]可以知道, 函数$q_{k}^{(e)}(n)$是可乘的且对每一素数幂$p^{a}$有

$ q_{k}^{(e)}(p)=q_{k}^{(e)}(p^{2})=q_{k}^{(e)}(p^{3})=\cdots=q_{k}^{(e)}(p^{2^{k}-1})=1, q_{k}^{(e)}(p^{2^{k}})=0. $

Tóth^[5-6]对该函数进行了研究并给出了其在整数上的均值估计

$ \sum\limits_{n\leq x}q_{k}^{(e)}(n)=D_kx+O(x^{1/2^k\delta(x)}), $

这里

$ D_k=\prod\limits_p\le(1+\sum\limits_{2^k}^\infty\frac{q_k(a)-q_k(a-1)}{p^a}), $

$q_k^{(e)}(n)$是$k$-free数的特征函数.贺艳峰和孙春丽^[1]也做了相关研究.

本文主要给出square-full数上$q_{k}^{(e)}(n) \ \ (k\geq3)$的均值估计, 得到

$ \sum\limits_{n\leq x \atop n \ {\rm is \ square-full} } q_{k}^{(e)}(n) \;=\; \sum\limits_{n\leq x} q_{k}^{(e)}(n)f_{2}(n) $

的渐近公式.即得到下面的定理.

定理  当$D > 0$时, 对$k\geq3$, 有

$ \sum\limits_{n\leq x \atop n \ {\rm is \ square-full}} q_{k}^{(e)}(n) \;=\; \frac{\zeta(\frac{3}{2})G(\frac{1}{2})}{\zeta(3)} x^{\frac{1}{2}} + \frac{\zeta(\frac{2}{3})G(\frac{1}{3})}{\zeta(2)} x^{\frac{1}{3}} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+ O\Big( x^{\frac{1}{6}}\exp\big(-D(\log x)^{\frac{3}{5}}(\log\log x)^{-\frac{1}{5}}\big)\Big), $

其中$G(s)=\prod_{p}\big(1-\frac{1}{p^{2^{k}s}}-\frac{1}{p^{(2^{k}+1) s}}+\frac{1}{p^{(2^{k}+3) s}}+\frac{1}{p^{(2^{k}+4) s}}-\frac{1}{p^{(2^{k}+6) s}}-\frac{1}{p^{(2^{k}+7) s}}+\cdots\big)$在$\Re s > \frac{1}{8}+\epsilon$时绝对收敛.

注  本文中, $\epsilon$表示一个任意小的的正常数, 在不同的式中不必相同.

为了证明定理, 需要以下的一些引理.

引理1  设$a, b $是整数, 且$1 \leq a < b$, 定义

$ d(a, b;k)=\sum\limits_{n_{1}^{a}n_{2}^{b}=k}1, \ \ D(a, b;x)=\sum\limits_{1\leq k\leq x}d(a, b;k), $

有

$ D(a, b;x)=\sum\limits_{1\leq k\leq x}d(a, b;k)=\zeta(\frac{b}{a})x^{\frac{1}{a}}+\zeta(\frac{a}{b})x^{\frac{1}{b}}+\triangle(a, bx), $

其中

$ \triangle(a, bx)=-\sum\limits_{n^{a+b\leq x}}\{{\psi((\frac{x}{n^{b}})^{\frac{1}{a}})}+{\psi((\frac{x}{n^{a}})^{\frac{1}{b}})}\}+O(1), $

而且

$ \triangle(a, bx) \ll \left\{ \begin{array}{ll} x^{\frac{2}{3a+3b}},&\hbox{ $b\ < \ 2a$;} \\ x^{\frac{2}{9a}}\log x,&\hbox{ $b=2a$;} \\ x^{\frac{2}{5a+2b}},&\hbox{ $b\ > \ 2a$.} \end{array} \right. $

证  本引理的证明见Ivić文[2]中的第14.3节和定理14.4.

引理2  假设$f(n)$是算术函数, 满足

$ \begin{eqnarray*} &&\sum\limits_{n\leq x}f(n)=\sum\limits_{j=1}^{l}x^{a_{j}}P_{j}(\log x) + O(x^{a}), \\ &&\sum\limits_{n\leq x}|f(n)|=O(x^{a_{1}}\log^{r} x), \end{eqnarray*} $

其中

$ a_{1} \geq a_{2} \geq \cdots \geq a_{l} > 1/c > a \geq 0, r \geq 0, P_{1}(t), \cdots, P_{l}(t) $

是关于$t$的次数不超过$r$的多项式, 并且$c \geq 1$, $b \geq 1$是固定的整数.如果$h(n) = \sum\limits_{d^{c}|n}\mu(d)f(n/d^{c}), $那么

$ \sum\limits_{n\leq x}h(n)=\sum\limits_{j=1}^{l}x^{a_{j}}R_{j}(\log x) + E_{c}(x), $

其中$R_{1}(t), \cdots, R_{l}(t)$是关于$t$的次数不超过$r$的多项式, 并且当$D > 0$时, 有

$ E_{c}(x) \ll x^{1/c}\exp\Big(\big(-D(\log x)^{3/5}(\log\log x)^{-1/5}\big)\Big). $

证  见Ivić文[2]中的定理14.2.

下面证明本文的定理.令

$ F(s):= \sum\limits_{n=1 \atop n \ {\rm is \ square-full} }^{\infty} \frac{q_{k}^{(e)}(n)}{n^{s}} = \sum\limits_{n=1}^{\infty}\frac{q_{k}^{(e)}(n)f_{2}(n)}{n^{s}}~~(\Re s > 1), $

其中

$ f_{2}(n)=\left\{ \begin{array}{ll} 1, \hbox{$n$ 是 square-full 数;} \\ 0, \hbox{否则.} \end{array} \right. $

因为$q_{k}^{(e)}(n)$是可乘函数, 由欧拉乘积可以得到

$ \nonumber {F(s)} =\sum\limits_{\substack{n=1 \\ n~{\rm is \ square-full} }}^{\infty} \frac{q_{k}^{(e)}(n)}{n^{s}}= \sum\limits_{n=1}^{\infty}\frac{q_{k}^{(e)}(n)f_{2}(n)}{n^{s}}\\ \;\;\;\;\;\;\;\;=\;\prod\limits_{p}\bigg(1+\frac{q_{k}^{(e)}(p)f_{2}(p)}{p^{s}}+\frac{q_{k}^{(e)}(p^{2})f_{2}(p^{2})}{p^{2s}}+\frac{q_{k}^{(e)}(p^{3})f_{2}(p^{3})}{p^{3s}}\nonumber\\ \;\;\;\;\;\;\;\;\;\;\;\;+\cdots+\frac{q_{k}^{(e)}(p^{2^{k}-1})f_{2}(p^{2^{k}-1})}{p^{(2^{k}-1)s}}\bigg)\nonumber\\ \nonumber \;\;\;\;\;\;\;\;=\;\prod\limits_p\big(1+\frac{1}{p^{2s}}+\frac{1}{p^{3s}}+\frac{1}{p^{4s}}+\cdots+\frac{1}{p^{(2^{k}-1)s}}\bigg)\\ \nonumber \;\;\;\;\;\;\;\;=\;\zeta(2s)\prod\limits_p\bigg(1+\frac{1}{p^{3s}}-\frac{1}{p^{2^{k}s}}-\frac{1}{p^{(2^{k}+1)s}}\bigg)\\ \nonumber \;\;\;\;\;\;\;\;=\;\zeta(2s)\zeta(3s)\prod\limits_p\bigg(1-\frac{1}{p^{6s}}-\frac{1}{p^{2^{k}s}}-\frac{1}{p^{(2^{k}+1)s}}+\frac{1}{p^{(2^{k}+3)s}}+\frac{1}{p^{(2^{k}+4)s}}\bigg)\\ \nonumber \;\;\;\;\;\;\;\;=\;\frac{\zeta(2s)\zeta(3s)}{\zeta(6s)}\prod\limits_{p}\bigg(1-\frac{1}{p^{2^{k}s}}-\frac{1}{p^{(2^{k}+1)s}}+\frac{1}{p^{(2^{k}+3)s}}\nonumber\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+\frac{1}{p^{(2^{k}+4)s}}-\frac{1}{p^{(2^{k}+6)s}}-\frac{1}{p^{(2^{k}+7)s}}+\cdots\bigg)\\ \nonumber \;\;\;\;\;\;\;\;=\;\frac{\zeta(2s)\zeta(3s)}{\zeta(6s)}G(s), $

其中

$ G(s)=\prod\limits_{p}\big(1-\frac{1}{p^{2^{k}s}}-\frac{1}{p^{(2^{k}+1)s}}+\frac{1}{p^{(2^{k}+3)s}}+\frac{1}{p^{(2^{k}+4)s}}-\frac{1}{p^{(2^{k}+6)s}}-\frac{1}{p^{(2^{k}+7)s}}+\cdots\big). $

显然, 对于$k\geq3$, 当$\Re s>\frac{1}{8}+\epsilon$时, $G(s)$绝对收敛.

根据卷积原理^[3], 定义

$ H(s) := \zeta(2s)\zeta(3s)G(s)= \sum\limits_{n=1}^{\infty}\frac{h(n)}{n^{s}}= \sum\limits_{n=1}^{\infty}\frac{\sum\limits_{n=ml}d(2, 3;m)g(l)}{n^{s}}~~ (\Re s > 1), $

其中$h(n) = \sum\limits_{n=ml}d(2, 3; m)g(l)$.那么可以得到

$ \sum\limits_{n\leq x}h(n) = \sum\limits_{ml\leq x}d(2, 3;m)g(l) = \sum\limits_{l\leq x}g(l)\sum\limits_{m\leq \frac{x}{l}}d(2, 3;m). $

由引理1得到

$ \begin{align*} \sum\limits_{n\leq x}d(2, 3;n)=\zeta(\frac{3}{2})x^{\frac{1}{2}} + \zeta(\frac{2}{3})x^{\frac{1}{3}} + \Delta(2, 3;x)\nonumber\\ =\zeta(\frac{3}{2})x^{\frac{1}{2}} + \zeta(\frac{2}{3})x^{\frac{1}{3}} + O(x^{\frac{2}{15}}).\nonumber \end{align*} $

很容易得到

$ \sum\limits_{n\leq x}|h(n)|\ll \sum\limits_{l\leq x}|g(l)|\sum\limits_{m\leq \frac{x}{l}}d(2, 3;m)\ll x^{\frac{1}{2}}, $

(2.1)

而且有

$ \sum\limits_{n\leq x}h(n) \;=\;\sum\limits_{l\leq x}g(l)\Big[\zeta(\frac{3}{2})(\frac{x}{l})^{\frac{1}{2}} + \zeta(\frac{2}{3})(\frac{x}{l})^{\frac{1}{3}} + O\big((\frac{x}{l})^{\frac{2}{15}}\big)\Big]\\ \;\;\;\;\;\;\;\;=\;\zeta(\frac{3}{2})x^{\frac{1}{2}}\sum\limits_{l\leq x}\frac{g(l)}{l^{1/2}} + \zeta(\frac{2}{3})x^{\frac{1}{3}}\sum\limits_{l\leq x}\frac{g(l)}{l^{1/3}} + O(x^{\frac{2}{15}}\sum\limits_{l\leq x}\frac{|g(l)|}{l^{2/15}})\\ \;\;\;\;\;\;\;\;=\;\zeta(\frac{3}{2})x^{\frac{1}{2}}\sum\limits_{l=1}^{\infty}\frac{g(l)}{l^{1/2}} + \zeta(\frac{2}{3})x^{\frac{1}{3}}\sum\limits_{l=1}^{\infty}\frac{g(l)}{l^{1/3}} + O(x^{\frac{1}{2}}\sum\limits_{l>x}\frac{|g(l)|}{l^{1/2}}) \\ \;\;\;\;\;\;\;\;\;\;\;\;+ O( x^{\frac{1}{3}}\sum\limits_{l>x}\frac{|g(l)|}{l^{1/3}}) + O(x^{\frac{2}{15}}\sum\limits_{l\leq x}\frac{|g(l)|}{l^{2/15}}). $

又由$G(s)$在$\sigma>\frac{1}{8}+\epsilon$时绝对收敛, 则可设

$ M(l) := \sum\limits_{t\leq l}|g(t)| \ll x^{\frac{1}{8}}, $

根据Abel分部求和得到

$ \begin{aligned} x^{\frac{1}{2}}\sum\limits_{l>x}\frac{|g(l)|}{l^{1/2}} &= x^{\frac{1}{2}}\int_{x}^{\infty}l^{-\frac{1}{2}}d\big(M(l)\big)\\ &= x^{\frac{1}{2}}\big(l^{-\frac{1}{2}}M(l)\big)\Big|_{x}^{\infty} + x^{\frac{1}{2}}\int_{x}^{\infty}M(l)d\big(l^{-\frac{1}{2}}\big)\\ &\ll x^{\frac{1}{8}}, \end{aligned} $

$ \begin{aligned} x^{\frac{1}{3}}\sum\limits_{l>x}\frac{|g(l)|}{l^{1/3}} &= x^{\frac{1}{3}}\int_{x}^{\infty}l^{-\frac{1}{3}}d\big(M(l)\big)\\ &= x^{\frac{1}{3}}\big(l^{-\frac{1}{3}}M(l)\big)\Big|_{x}^{\infty} + x^{\frac{1}{3}}\int_{x}^{\infty}M(l)d\big(l^{-\frac{1}{3}}\big)\\ &\ll x^{\frac{1}{8}}, \end{aligned} $

$ \begin{aligned} x^{\frac{2}{15}}\sum\limits_{l\leq x}\frac{|g(l)|}{l^{2/15}} &= x^{\frac{2}{15}}\int_{1}^{x}l^{-\frac{2}{15}}d\big(M(l)\big)\\ &= x^{\frac{2}{15}}\big(l^{-\frac{2}{15}}M(l)\big)\Big|_{1}^{x} + x^{\frac{2}{15}}\int_{1}^{x}M(l)d\big(l^{-\frac{2}{15}}\big)\\ &\ll x^{\frac{1}{8}}, \end{aligned} $

从而由$G(s)= \sum\limits_{n=1}^{\infty}\frac{g(n)}{n^{s}}$, 有

$ \sum\limits_{n\leq x}h(n)=\zeta(\frac{3}{2})G(\frac{1}{2})x^{\frac{1}{2}} + \zeta(\frac{2}{3})G(\frac{1}{3})x^{\frac{1}{3}} + O(x^{\frac{1}{8}}). $

(2.2)

由于

$ F(s)=\frac{H(s)}{\zeta(6s)}=\sum\limits_{n=1}^{\infty}\frac{h(m)}{m^{s}}\sum\limits_{m=1}^{\infty}\frac{\mu(d)}{d^{6s}}, $

所以当$n$是square-full数时, 有

$ q_{k}^{(e)}(n)=\sum\limits_{n=md^{6}}h(m)\mu(d). $

(2.3)

再由Perron公式^[3]可以得到

$ \sum\limits_{n\leq x \atop n \ {\rm is \ square-full}} q_{k}^{(e)}(n) = (2\pi i)^{-1}\int_{1+\frac{1}{\log x}-iT}^{1+\frac{1}{\log x}+iT}\frac{\zeta(2s)\zeta(3s)}{\zeta(6s)}G(s)\frac{x^{s}}{x}ds +O(\frac{x\log^{2}x}{T}). $

取$T=x$, 并将积分线平移至$\sigma=-\frac{1}{2}$, 则在$s=\frac{1}{2}, \ s=\frac{1}{3}$处的留数分别为

$ \frac{\zeta(\frac{3}{2})G(\frac{1}{2})}{\zeta(3)} x^{\frac{1}{2}}, \ \ \frac{\zeta(\frac{2}{3})G(\frac{1}{3})}{\zeta(2)} x^{\frac{1}{3}}. $

再由(2.1)、(2.2)、(2.3) 式和引理2, 可以得到

$ \sum\limits_{n\leq x \atop n \ {\rm is \ square-full}} q_{k}^{(e)}(n) \;\;=\;\; \frac{\zeta(\frac{3}{2})G(\frac{1}{2})}{\zeta(3)} x^{\frac{1}{2}} + \frac{\zeta(\frac{2}{3})G(\frac{1}{3})}{\zeta(2)} x^{\frac{1}{3}} \\ \;\;\;\;\;\;\;\;\;\;\;\;+O\Big( x^{\frac{1}{6}}\exp\big(-D(\log x)^{\frac{3}{5}}(\log\log x)^{-\frac{1}{5}}\big)\Big), $

其中$D > 0$, 并且$G(s)$当$\Re s > \frac{1}{8}+\epsilon$时是绝对收敛的.

这样就得到了本文的定理.