狄利克雷卷积

积性函数\(f\)和\(g\)，
狄利克雷卷积的形式:

\[(f\ast g)(n) = \sum\limits_{d|n}f(d)g(\frac{n}{d}) \]

或者

\[(f\ast g)(n) = \sum\limits_{i\times j=n}f(i)g(j) \]

\(\\\)
\(\\\)

它满足

• 结合律 \((f \ast g) \ast h = f \ast ( g \ast h )\)

证明

\[(f\ast g) \ast h = \sum\limits_{n|m}\sum\limits_{d|n}f(d)g(\frac{n}{d})h(\frac{m}{n}) \]

等同于

\[(f\ast g) \ast h = \sum\limits_{i\times j \times k=n}f(i)g(j)h(k) \]

这个式子满足轮换对称，\(i,j,k\)怎么换都一样

\(\\\)
\(\\\)

\([a]\)表示 \(a\) 成立为 \(1\) ,否则为 \(0\)

有一些基本的积性函数:

• \[\epsilon(n) = [n = 1] \]

• \[1(n) = 1 \]

• \[id(n) = n \]

• \[d(n) = \sum\limits_{d|n}1 \]

• \[\sigma(n) = \sum\limits_{d|n}d \]

• \[d_w(n) =\sum\limits_{d|n}1^w \]

• \[\sigma_w(n) = \sum\limits_{d|n}d^w \]

• \[g(n) = \sum\limits_{i=1}^{n}gcd(n,i) \]

常用的卷积:

• \[f \ast \epsilon = f \]

证明:

\[\sum\limits_{d|n}f(d)\epsilon(\frac{n}{d}) \]

当且仅当\(d=n\)时，\(\epsilon(1) = 1\)

\[=f(n) \]

\(\\\)
\(\\\)

• \[\mu \ast 1 = \epsilon \]

证明:

\[=\sum\limits_{d|n}\mu(d)1(\frac{n}{d}) \]

\[=\sum\limits_{d|n}\mu(d) \]

\[=[n=1] \]

特别的，\(n=1\)时,\(\mu(1)=1\),任意积性函数的\(1\)都是\(1\)

当\(n \ne 1\)时，

令\(d = p_{1}^{x_1}p_2^{x_2}...p_k^{x_k}\)

根据\(\mu\)的定义，

\(\mu(i) = \left\{ \begin{aligned} 1\quad \quad \quad ,n=1 \\ (-1)^k \quad ,\prod_{i=1}^{k}x_i=1 \\ 0 \quad,\prod_{i=1}^{k}x_i\ne1\\ \end{aligned} \right. \)

当\(i > 1\)时，只有中间的一项有贡献

\[\sum\limits_{i=0}^{k}(_ {i}^{k})(-1)^i \]

意义是从\(k\)个中选有\(i\)个质因子的数的个数,\(\mu\)如果有奇数个质因子，则为\(-1\)，\(i = 0\)意味着不选，也就是\(\mu(1)\)的贡献。

\[=\sum\limits_{i=0}^{k}(_ {i}^{k})(-1)^i1^{k-i} \]

\[=(1-1)^k \]

\[=0 \]

\(\\\)
\(\\\)

• \[\varphi\ast 1 = id \]

证明:

\[\sum\limits_{d|n}\varphi(d)1(\frac{n}{d}) \]

\[=\sum\limits_{d|n}\varphi(d) \]

根据\(\varphi(i)\)的定义，

\(\varphi(i)=\) \(1\)~\(i\) 与 \(i\) 互质的个数

考虑构造有\(n\)个元素的集合\(A=\left\{\frac{i}{n}|i\in Z,i\in [1,n] \right\}\)

有\(\frac{x}{n}\in A\),把它化解到最简\(\frac{a}{b}\),显然每个 \(\frac{a}{b}\) 仅会对应一个\(\frac{x}{n}\)

显然\(b|n\),因为同时除以了公因数。

由于是最简形式，对于每个分母是\(b\)的分数, 显然\(a\)都与\(b\)互质，
所以\(a\)的个数为\(\varphi(b)\)

所以

\[\sum\limits_{d|n}\varphi(d) = |A| = n \]

\(\\\)
\(\\\)

• \[\varphi = id \ast \mu \]

证明:

同乘\(1\)

\[\varphi \ast 1= id \ast \mu \ast 1 \]

化简得

\[id = id \ast \epsilon \]

得

\[id = id \]

\(\\\)
\(\\\)

• \[1 \ast 1 = d \]

证明:

\[=\sum\limits_{d|n}1(d)1(\frac{n}{d}) \]

\[=\sum\limits_{d|n}1 \]

\(\\\)
\(\\\)

• \[1 = d \ast \mu \]

证明:

\[1 = 1 \ast 1 \ast \mu \]

\[1 = 1 \ast \epsilon \]

\[1 = 1 \]

\(\\\)
\(\\\)

• \[id \ast 1 = \sigma \]

证明:

\[= \sum\limits_{d|n}id(d)1(\frac{n}{d}) \]

\[= \sum\limits_{d|n}d \]

\(\\\)
\(\\\)

• \[id = \sigma \ast \mu \]

证明:

\[id = id \ast 1 \ast \mu \]

\[id = id \ast \epsilon \]

\[id = id \]

\(\\\)
\(\\\)

• \[id_w \ast 1 = \sigma_{w} \]

证明:

\[= \sum\limits_{d|n}id_w(d)1(\frac{n}{d}) \]

\[= \sum\limits_{d|n}d^w \]

\(\\\)
\(\\\)

• \[id_w = \sigma_w \ast \mu \]

证明:

\[id_w = id_w \ast 1 \ast \mu \]

\[id_w = id_w \ast \epsilon \]

\[id_w = id_w \]

\(\\\)
\(\\\)

• \[\sigma = \varphi \ast d \]

证明:

\[1\ast id =id \ast \mu \ast d \]

\[1\ast 1 \ast id = id \ast \mu \ast 1 \ast d \]

\[d \ast id = id \ast \epsilon \ast d \]

\[d \ast id = id \ast d \]

\(\\\)
\(\\\)

• \[g = \varphi \ast id \]