メビウス関数の和

$ \sum_{d|n}\mu (d)=\left\{ \begin{array} {l}1 &(n=1) \\ 0 & (n>1)\end{array} \right.

$ \sum_{k=0}^n (-1)^k \binom{n}{k} = 0

$ (-1 + 1) ^ n = \sum_{k=0}^n (-1)^k 1^{n-k} \binom{n}{k} = \sum_{k=0}^n (-1)^k \binom{n}{k}

$ (-1 + 1) ^ n = 0^n = 0

$ \sum_{d|n} \mu(n/d) g(d) = \sum_{d|n} \mu(n/d) \sum_{l|d} f(l) = \sum_{d|n} \sum_{l|d} \mu(n/d) f(l)

$ \sum_{d|n} \sum_{l|d} \mu(n/d) f(l) = \sum_{l|n} f(l) \sum_{(n/d)|(n/l)} \mu(n/d)

eq1より