与酉最小公倍数有关的和函数

The Sum Function Related to the Least Unitary Common Multiple

摘要: 最大公因数和最小公倍数问题是数论的经典问题之一，那么对任意的正整数n，若$d|n$ 且$\left(d,\frac{n}{d}\right)=1$ ，我们就称d是n的酉除数(unitary divisor)，用${\left[m,n\right]}^{\ast }$ 表示两个数m, n的酉最小公倍数，本文主要考虑当${\left[m,n\right]}^{\ast }\le x$ 时，满足该条件的点$\left(m,n\right)$ 的和函数的渐近公式。

Abstract: The greatest common divisor and the least common multiple are one of the classical problems in number theory. For any positive integer n, if $d|n$ and $\left(d,\frac{n}{d}\right)=1$ , we call d is the unitary divisor of n. In this paper, using ${\left[m,n\right]}^{\ast }$ to represent the unitary least common multiple of two numbers m, n, and this article mainly considers the asymptotic formula of the sum function of points that satisfy this condition when ${\left[m,n\right]}^{\ast }\le x$ .

1. 引言

最小公倍数问题是数论中的经典问题之一。整数环Z上最小公倍数的和函数定义为：

$L\left(n\right)={\displaystyle \sum _{i=1}^{n}lcm\left(i,n\right)}$

那么$L\left(n\right)$ 的和函数的渐近公式为

${\displaystyle \sum _{n\le x}L\left(n\right)}=\frac{\zeta \left(3\right)}{8\zeta \left(2\right)}{x}^4+O\left(x^3{\left(\log x\right)}^{2/3}{\left(\log \log x\right)}^{4/3}\right)$

参见([1], Th.6.3)。

lcm和函数的其他推广结果可参考[2] [3]。

如果对任意的正整数n，若$d|n$ 且$\left(d,\frac{n}{d}\right)=1$ ，那么称d是n的酉除数(unitary divisor)，其基本性质可见[4]。

那么，同样也可以考虑${\displaystyle {\sum }_{\left[m,n\right]\le x}1}$ 相关的计数问题。H. Jager [5]在1962年得到该和式的渐近公式。Suryanarayana D, Sitaramachandrarao R [6], WG Nowak, M Schmeier [7]分别在黎曼假设的前提下将该渐近公式的余项改进至$x^{\frac{2-\alpha }{5-4\alpha }}\exp \left(A\frac{\log x}{\log \log x}\right)$ ，$x^{\frac{53}{116}+ϵ}$ 。

基于以上结果的启发，本文研究当两个数的酉最小公倍数小于等于x时满足该条件的数对$\left(m,n\right)$ 有多少个，即对${\displaystyle {\sum }_{{\left[m,n\right]}^{\ast }\le x}1}$ 进行计算，得到如下定理。

定理：${\displaystyle {\sum }_{{\left[m,n\right]}^{\ast }\le x}1}=A_{1}x{\left(\log x\right)}^2+A_{2}x\log x+A_{3}x+O\left(\sqrt{x}{\left(\log x\right)}^2\right)$ ，其中$A_1,A_2,A_3$ 为常数。

2. 基本引理

引理2.1：${\displaystyle \sum _{k\le n}\frac{1}{k}}=\log n+\gamma +O\left(\frac{1}{n}\right)$

证明：参见([8]，定理0.8)

引理2.2：${\displaystyle \sum _{d|n}\mu \left(d\right)}=\{\begin{array}{l}1,若n=1\\ 0,若n>1\end{array}$

证明：参见([8]，定理2.8)

引理2.3：(a) ${\displaystyle \sum _{n\le x}\frac{\mu \left(n\right)}{n^2}}=\frac{6}{{\pi }^2}+O\left(\frac{1}{x}\right)$

(b) ${\displaystyle \sum _{k\le x}\frac{1}{n^s}}=\frac{x^{1-s}}{1-s}+\zeta \left(s\right)+O\left(x^{-s}\right)$ ，$s>0$ 且$s\ne 1$

证明：参见([9]，P61，定理3.2)

引理2.4：${\displaystyle \sum _{n\le x}\frac{\log n}{n}}=\frac{1}{2}{\log }^{2}x+A+O\left(\frac{\log x}{x}\right)$ ，A为常数

证明：由Euler求和公式即可得出

$\begin{array}{c}{\displaystyle \sum _{n\le x}\frac{\log n}{n}}={\displaystyle \underset{1}{\overset{x}{\int }}\frac{\log t}{t}\text{d}t}+{\displaystyle \underset{1}{\overset{x}{\int }}\left(t-\left[t\right]\right)}\frac{1-\log t}{t^2}\text{d}t+\frac{\log x}{x}\left(\left[x\right]-x\right)\\ ={\displaystyle \underset{1}{\overset{x}{\int }}\frac{\log t}{t}\text{d}t}+{\displaystyle \underset{1}{\overset{\infty }{\int }}\left(t-\left[t\right]\right)}\frac{1-\log t}{t^2}\text{d}t-{\displaystyle \underset{x}{\overset{\infty }{\int }}\left(t-\left[t\right]\right)}\frac{1-\log t}{t^2}\text{d}t+\frac{\log x}{x}\left(\left[x\right]-x\right)\\ =\frac{1}{2}{\log }^{2}x+{\displaystyle \underset{1}{\overset{\infty }{\int }}\left(t-\left[t\right]\right)}\frac{1-\log t}{t^2}\text{d}t-{\displaystyle \underset{x}{\overset{\infty }{\int }}\left(t-\left[t\right]\right)}\frac{1-\log t}{t^2}\text{d}t+O\left(\frac{\log x}{x}\right)\end{array}$

由于${\displaystyle \underset{x}{\overset{\infty }{\int }}\left(t-\left[t\right]\right)\frac{1-\log t}{t^2}\text{d}t}\le {\displaystyle \underset{x}{\overset{\infty }{\int }}\frac{1-\log t}{t^2}\text{d}t}\le {\displaystyle \underset{x}{\overset{\infty }{\int }}\frac{1}{t^2}\text{d}t}=\frac{1}{x}$

故${\displaystyle \underset{1}{\overset{x}{\int }}\left(t-\left[t\right]\right)}\frac{1-\log t}{t^2}\text{d}t$ 收敛，记为常数A

$原式=\frac{1}{2}{\log }^{2}x+A+O\left(\frac{\log x}{x}\right)$

引理2.5：${\left[m,n\right]}^{\ast }=\frac{mn}{{\left(m,n\right)}^{\ast }}$

证明：见文献[4]

引理2.6：${\displaystyle {\sum }_{n\le x}{\tau }_3\left(x\right)}=x{P}_3\left(\log x\right)+O\left(x^{\frac{43}{96}+\epsilon }\right)$ ，$P_3\left(\log x\right)$ 为$\log x$ 的一个二次多项式

证明：见文献[10]

3. 定理证明

证明：首先由引理2.5将所求和式转化为下式

${\displaystyle {\sum }_{{\left[m,n\right]}^{\ast }\le x}1}={\displaystyle {\sum }_{\frac{mn}{{\left(m,n\right)}^{\ast }}\le x}1}={\displaystyle {\sum }_{\begin{array}{c}mn\le kx\\ {\left(m,n\right)}^{\ast }=k\end{array}}1}$

这里令${\left(m,n\right)}^{\ast }=k$ 。下面由引理2.2，我们将上式转化以下和式

$\begin{array}{l}={\displaystyle \sum _{\begin{array}{c}{m}_1{n}_1\le \frac{x}{k}\\ \left(m_1,n_1\right)=k\end{array}}1}{\displaystyle \sum _{d|\left(m_1,n_1\right)}\mu \left(d\right)}\\ ={\displaystyle \sum _{d\le \sqrt{x}}\mu \left(d\right)}{\displaystyle \sum _{\begin{array}{c}d|m_1,d|n_1\\ m_1{n}_1\le \frac{x}{k}\end{array}}1}\\ ={\displaystyle \sum _{d\le \sqrt{x}}\mu \left(d\right)}{\displaystyle \sum _{d^2{m}_2{n}_2\le x/k}1}\\ ={\displaystyle \sum _{d\le \sqrt{x}}\mu \left(d\right)}{\displaystyle \sum _{l\le x/d^2}{\tau }_3\left(l\right)}\left(l=m_2{n}_{2}k\right)\\ ={\displaystyle \sum _{d\le \sqrt{x}}\mu \left(d\right)}\left\{\frac{x}{d^2}{C}_1{\left(\log \frac{x}{d^2}\right)}^2+C_2\frac{x}{d^2}\log \frac{x}{d^2}+C_3\frac{x}{d^2}+O\left({\left(\frac{x}{d^2}\right)}^{43/96+\epsilon }\right)\right\}\\ =S_1+S_2+S_3+S_4\end{array}$

其中

$\begin{array}{l}{S}_1={\displaystyle {\sum }_{d\le \sqrt{x}}\mu \left(d\right)\frac{x}{d^2}{C}_1{\left(\log \frac{x}{d^2}\right)}^2}\\ S_2={\displaystyle {\sum }_{d\le \sqrt{x}}\mu \left(d\right)\frac{x}{d^2}{C}_2\log \frac{x}{d^2}}\\ S_3={\displaystyle {\sum }_{d\le \sqrt{x}}\mu \left(d\right)\frac{x}{d^2}{C}_3}\\ S_4={\displaystyle {\sum }_{d\le \sqrt{x}}\mu \left(d\right)O\left({\left(\frac{x}{d^2}\right)}^{43/96+\epsilon }\right)}\end{array}$

下面对$S_1,S_2,S_3,S_4$ 分别进行计算，从而得到所需结果

A、对$S_4$ 的计算：

$\left|{\displaystyle {\sum }_{d\le \sqrt{x}}\mu \left(d\right)O\left({\left(\frac{x}{d^2}\right)}^{\frac{43}{96}+ϵ}\right)}\right|\ll x^{\frac{1}{2}+ϵ}$ (1)

B、对$S_3$ 的计算：

${\displaystyle \sum _{d\le \sqrt{x}}\frac{\mu \left(d\right)}{d^2}{c}_{3}x}=\left(\frac{6}{{\pi }^2}+O\left(\frac{1}{\sqrt{x}}\right)\right)c_{3}x=c_{4}x+O\left(\sqrt{x}\right)$ (2)

C、对$S_2$ 的计算：

$\begin{array}{c}{S}_2={\displaystyle \sum _{d\le \sqrt{x}}\mu \left(d\right)}\frac{x}{d^2}{c}_2\left(\log \frac{x}{d^2}\right)=c_{2}x{\displaystyle \sum _{d\le \sqrt{x}}\frac{\mu \left(d\right)}{d^2}}\left(\log x-\log d^2\right)\\ =S_{21}+S_{22}\end{array}$

$S_{21}=c_{2}x\log x{\displaystyle \sum _{d\le \sqrt{x}}\frac{\mu \left(d\right)}{d^2}}=c_{5}x\log x+O\left(\sqrt{x}\log x\right)$

$\begin{array}{c}{S}_{22}=c_{2}x{\displaystyle \sum _{d\le \sqrt{x}}\frac{\mu \left(d\right)}{d^2}}\log d^2=c_{2}x\left({\displaystyle \sum _{d=1}^{\infty }\frac{\mu \left(d\right)}{d^2}\log d^2}-{\displaystyle \sum _{d>\sqrt{x}}\frac{\mu \left(d\right)}{d^2}}\log d^2\right)\\ =c_{2}x\left(A+O\left(\frac{\log x}{\sqrt{x}}\right)\right)=c_{6}x+O\left(\sqrt{x}\log x\right)\end{array}$

所以

$S_2=c_{5}x\log x+c_{6}x+O\left(\sqrt{x}\log x\right)$ (3)

D、对$S_1$ 的计算：

$S_1={\displaystyle \sum _{d\le \sqrt{x}}\mu \left(d\right)}\frac{x}{d^2}{c}_1{\left(\log \frac{x}{d^2}\right)}^2=c_{1}x{\displaystyle \sum _{d\le \sqrt{x}}\frac{\mu \left(d\right)}{d^2}\left({\left(\log x\right)}^2-2\log x\log d^2+4\log d^2\right)}=S_{11}+S_{12}+S_{13}$

$\begin{array}{l}{S}_{11}=c_{1}x{\left(\log x\right)}^2{\displaystyle \sum _{d\le \sqrt{x}}\frac{\mu \left(d\right)}{d^2}}=c_{7}x{\left(\log x\right)}^2+O\left(\sqrt{x}{\left(\log x\right)}^2\right)\\ S_{12}=c_{1}x\log x{\displaystyle \sum _{d\le \sqrt{x}}\frac{\mu \left(d\right)}{d^2}}\log d=c_{8}x\log x+O\left(\sqrt{x}\log x\right)\\ S_{13}=c_{1}x{\displaystyle \sum _{d\le \sqrt{x}}\frac{\mu \left(d\right)}{d^2}}{\left(\log d\right)}^2=c_{9}x+O\left(\sqrt{x}{\left(\log x\right)}^2\right)\end{array}$

综合上式有：

$S_1=c_{7}x{\left(\log x\right)}^2+c_{8}x\log x+c_{9}x+O\left(\sqrt{x}{\left(\log x\right)}^2\right)$ (4)

综合(1)~(4)式定理得证。

符号说明

${[m,n]}^{\ast }$	两个数的最小酉公倍数
$\zeta \left(s\right)$	黎曼zeta函数
$\mu \left(n\right)$	莫比乌斯函数
$\log \left(x\right)$	以自然常数e为底的对数
${\tau }_3\left(n\right)$	三维除数函数
$\gamma$	欧拉常数
$f=O\left(g\right)$	$\left|f\left(x\right)\right|\le Cg\left(x\right)$
${\left(m,n\right)}^{\ast }$	两个数的最大酉公因数