狄利克雷卷积是函数与函数之间的运算
运算定义为：$(f*g)(x) = \sum_{d|x}{f(d) * g(\frac{x}{d})}$
该运算有交换律：$f*g=g*f$
有结合律：$f*(g*h)=(f*g)*h$
有分配率：$f*(g+h)=f*g+f*h$
在这种新定义的运算中有许多特殊的元素
$e$ 满足 $e(n) = [n==1]$ 即当只有$n=1$时$e(n)=1$，否则为$0$
(对命题$P$，引入符号$[P]$，如果命题成立则值为$1$，否则为$0$.)
$d$ 满足$d(n) = \sum_{i<=n}\ \ [i|n]$
$\sigma$ 满足$\sigma(n) = \sum_{i|n}{i}$
$\varphi$ 满足$\varphi(n) = \sum_{i<=n}\ \ [gcd(i,n) == 1]$
$\mu$ 满足$\mu(n) = \left\{\begin{matrix} 1 & n == 1 \\ (-1)^k & a1 = a2 = …… = ak\\ 0 & otherwise \end{matrix}\right.$ 其中 $n = p1^{a1}*p2^{a2}*……*pk^{ak}$
$I$ 满足$I(n) = 1$
$id$ 满足$id(n) = n$
以上函数均为积性函数，(也易证)
函数简单运算
1.$e$为元，有$e*f=f$
解释：$e*f = \sum_{d|n}e(d)f(\frac{n}{d})$，当且仅当$d = 1$时$e(d) = 1$，此时$e(d)f(\frac{n}{d}) = e(1)f(n) = f(n)$，当$d!=1$时$e(d)=0$，所以$e*f = f$
2.$\mu*I = e$
解释：$\mu * I = \sum_{d|n}{\mu(n)I(\frac{n}{d})} = \sum_{d|n}{\mu(n)}$，又由$\mu(n)$为积性函数，当$n!=1$时
$\sum_{d|n}{\mu(n)} = (1+\mu(p1)+\mu(p1^2)+……+\mu(p1^{a1}))(……)(1+\mu(pk)+……+\mu(pk^{ak}))$，又由定义$\mu(p^i)=0\ \ (i>=2)$，$\mu(p)=1\ \ (p为质数)$
$\therefore \sum_{d|n}{\mu(n)} = (1+\mu(p1))(……)(1+\mu(pk)) = (1-1)(……)(1-1) = 0$
只有当$n == 1$时$\mu(n) = 1$，所以$\mu*I = e$
3.$\varphi*i = id$
解释：$\varphi*I = \sum_{d|n}\varphi(d)I(\frac{n}{d}) = \sum_{d|n}\varphi(d)$
$=(1+\varphi(p1)+\varphi(p1^2)+……+\varphi(p1^{a1}))(……)(1+\varphi(pk)+……+\varphi(pk^{ak}))$
而 $1+\varphi(p)+\varphi(p^2)+……+\varphi(p^a) = 1+p-1+p^2-p+p^3-p^2+……+p^a-p^{a-1} = p^a$
$\therefore 原式=p1^{a1}p2^{a2}……pk^{ak} = n = id(n)$
4.$I*I = d$
解释：$I*I = \sum_{d|n}I(d)I(\frac{n}{d}) = \sum_{d|n}1 = d(n)$
5.$\mu*id = \varphi$
解释：$\mu*id = \sum_{d|n}{\mu(d)id(\frac{n}{d})} = n\sum_{d|n}{\frac{\mu(d)}{d}}$
$= n(1+\frac{\mu(p1)}{p1}+……+\frac{\mu(p1^{a1})}{p1^{a1}})(……)(1+\frac{\mu(pk)}{pk}+……+\frac{pk^{ak}}{pk^{ak}})$
又由定义$\mu(p^i)=0\ \ (i>=2)$，$\mu(p)=1\ \ (p为质数)$
$\therefore 原式=n(1-\frac{1}{p1})(1-\frac{1}{p2})(……)(1-\frac{1}{pk}) = \varphi(n)$
6.$I*id = \sigma$
解释：$I*id = \sum_{d|n}{id(d)I(\frac{n}{d})} = \sum_{d|n}{d} = \sigma(n)$
函数乘法一览表
$\begin{matrix}
* & e(n) & d(n) & \sigma(n) & \varphi(n) & \mu(n) & I(n) & id(n) \\
e(n) & e(n) & d(n) & \sigma(n) & \varphi(n) & \mu(n) & I(n) & id(n) \\
d(n) & d(n) & - & - & \sigma(n) & - & - & - \\
\sigma(n) & \sigma(n) & - & - & - & - & - & - \\
\varphi(n) & \varphi(n) & \sigma(n) & - & - & - & id(n) & - \\
\mu(n) & \mu(n) & - & - & - & - & e(n) & \varphi(n) \\
I(n) & I(n) & - & - & id(n) & e(n) & d(n) & \sigma(n) \\
id(n) & id(n) & - & - & - & \varphi(n) & \sigma(n) & nd(n) \\
\end{matrix}$
莫比乌斯反演
若$g(n) = \sum_{d|n}f(d)$，则$f(n) = \sum_{d|n}{g(d)\mu(\frac{n}{d})}$
解释：由上$g=f*I$，则$g*\mu=f*I*\mu$，又$\mu*I=e，e*f=f$，所以$f=g*\mu=\sum_{d|n}{g(d)\mu(\frac{n}{d})}$
若$g(n) = \sum_{n|d}^{d<=MAX}\ \ {f(d)}$，则$f(n) = \sum_{n|d}^{d<=MAX}\ \ {g(d)\mu(\frac{d}{n})}$
解释：令$\frac{d}{n} = k$，则$\sum_{n|d}^{d<=MAX}\ \ {g(d)\mu(\frac{d}{n})} = \sum_{k}^{nk<=MAX}\ \ {g(nk)\mu(k)}$
$=\sum_{k}^{nk<=MAX}\ \ {\mu(k)\sum_{nk|t}^{t<=MAX}\ \ {f(t)}}$
$=\sum_{t}^{t<=MAX}\ \ {f(t)\sum_{nk|t}{\mu(k)}}$
又$\sum_{nk|t}{\mu(k)} = \sum_{k|\frac{t}{n}}{\mu(k)} = \mu(\frac{t}{n})I(\frac{t}{n}) = e(\frac{t}{n})$
$\therefore \sum_{t}^{t<=MAX}\ \ {f(t)\sum_{nk|t}{\mu(k)}} = \sum_{t}^{t<=MAX}\ \ {f(t)e{(\frac{t}{n})}}$
当且仅当$t = n$时$e != 0$，对求和有贡献
$\therefore \sum_{t}^{t<=MAX}\ \ {f(t)e{(\frac{t}{n})}}= f(n)$
