积性函数与狄利克雷卷积

发布时间 2023-04-20 13:10:23作者: |Roland|

积性函数

定义

\(f\left(n\right)\)为数论函数,若:
\(\left(1\right)f\left(1\right)=1\)
\(\left(2\right)\)\(\left(a,b\right)=1,f\left(ab\right)=f\left(a\right)f\left(b\right)\)
则称\(f\)为积性函数。

重要性质

积性函数的和函数也是积性函数。
用数学语言表示:
\(f\)为积性函数,则\(F\left(n\right)=\sum\limits_{d\mid n}^{n}f\left(d\right)\)也是积性函数。
证明:
\(\ \ \ \ F\left(a\right)F\left(b\right)\)
\(=\sum\limits_{d_{1}\mid a}^{a}f\left(d_{1}\right)\sum\limits_{d_{2}\mid b}^{b}f\left(d_{2}\right)\)
\(=\sum\limits_{d_{1}\mid a}^{a}\sum\limits_{d_{2}\mid b}^{b}f\left(d_{1}\right)f\left(d_{2}\right)\)
\(\because \left(a,b\right)=1\)
\(\therefore \left(d_{1},d_{2}\right)=1\)
\(\therefore f\left(d_{1}\right)f\left(d_{2}\right)=f\left(d_{1}d_{2}\right)\)
\(\therefore F\left(a\right)F\left(b\right)=\sum\limits_{d_{1}\mid a}^{a}\sum\limits_{d_{2}\mid b}^{b}f\left(d_{1}d_{2}\right)\)
\(=\sum\limits_{d\mid ab}^{ab}f\left(d\right)\)
\(=\sum\limits_{d\mid n}^{n}f\left(d\right)=F\left(n\right)\)
\(\therefore F\left(n\right)\)为积性函数

常见积性函数

单位函数\(id\left(n\right)=n\)
幂函数\(I_{k}\left(n\right)=n^{k}\)
元函数\(\varepsilon\left(n\right)=\left\{\begin{array}{l} 1\cdots\left[n=1\right]\\0\cdots\left[otherwise\right]\end{array}\right.\)
因数和函数\(\sigma\left(n\right)=\sum\limits_{d\mid n}^{n}d\)
因数个数函数\(d\left(n\right)=\sum\limits_{d\mid n}^{n}1\)
\(n=\prod\limits_{i=1}^{k}p_{i}^{c_{i}}\)\(p_{i}\)为质数)
欧拉函数\(\varphi\left(n\right)=n\prod\limits_{i=1}^{k}\left(1-\frac{1}{p_{i}}\right)\)
莫比乌斯函数\(\mu\left(n\right)=\left\{\begin{array}{l}1\cdots\left [n=1\right]\\ 0\cdots \left[\exists i\in\left[1,k\right],c_{i}>0\right]\\ \left(-1\right)^{k}\cdots\left[\forall i\in\left[1,k\right],c_{i}=1\right]\end{array}\right.\)

狄利克雷卷积

定义

定义一种新运算\(*\)\(h=f*g\)的定义是\(h\left(n\right)=\sum\limits_{d \mid n}^{n}f\left(d\right)g\left(\frac{n}{d}\right)\)

性质

\((1)\)交换律\(f*g=g*f\)
\((2)\)结合律\(\left(f*g\right)*h=f*\left(g*h\right)\)
\((3)\)分配率\(f*\left(g+h\right)=f*g+f*h\)
\((4)\)积性函数的卷积仍是积性函数
简单不证