对于定义在\(\mathbb{N}\)上的函数\(F(n)\)和\(f(n)\)，若满足：
\(F(n) = \sum\limits_{d\mid n}f(d)\)
则有：
\(f(n) = \sum\limits_{d\mid n}\mu(d)F(\frac{n}{d})\)
其中\(\mu(d)\)为莫比乌斯函数：
令 \(x = p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}}\cdot\cdot\cdotp_{k-1}^{a_{k-1}}p_{k}^{a_{k}},p_{1},p_{2},p_{3} \cdot \cdot \cdot p_{k-1},p_{k}\in \mathbb{P}\)
则满足：
\(\mu(x) = \left\{\begin{matrix}&1 &,x = 1\\ &(-1)^{k} &,\forall i\in [1,k],a_{i}=1\\ &0 &,\exists i\in [1,k],a_{i}>1\end{matrix}\right.\)
证明如下：
\(\sum\limits_{d\mid n}\mu(d)F(\frac{n}{d}) = \sum\limits_{d\mid n}\mu(d)\sum\limits_{d^{'}\mid\frac{n}{d}}f(d^{'}) = \sum\limits_{d\mid n}\sum\limits_{d^{'}\mid\frac{n}{d}}\mu(d)f(d^{'}) = \sum\limits_{d^{'}\mid n}\sum\limits_{d\mid\frac{n}{d^{'}}}\mu(d)f(d^{'}) = \sum\limits_{d^{'}\mid n}f(d^{'})\sum\limits_{d\mid\frac{n}{d^{'}}}\mu(d) = f(n)\)
原文：https://www.cnblogs.com/dummyummy/p/9210515.html