题目
神仙题\(emmm\)
做法
求\(\sum_{i=1}^n\sum_{j=1}^m\varphi(i,j)\)
首先有一个很神奇的性质：
\(\varphi(ij)=\dfrac{\varphi(i)\varphi(j)gcd(i,j)}{\varphi(gcd(i,j))}\)
证明：
\(\varphi(i)\varphi(j)=i\prod_{p|i}\dfrac{p-1}{p}j\prod_{p|j}\dfrac{p-1}{p}\)
\(~~~~~~~~~~~~~~=ij\prod_{p|ij}\dfrac{p-1}{p}\prod_{p|gcd(i,j)}\dfrac{p-1}{p}\)
\(\therefore \varphi(i)\varphi(j)gcd(i,j)=\varphi(ij)\varphi(gcd(i,j)) \Longrightarrow \varphi(ij)=\dfrac{\varphi(i)\varphi(j)gcd(i,j)}{\varphi(gcd(i,j))}\)
\(Ans=\sum_\limits{i=1}^n\sum\limits_{j=1}^m\frac{\varphi(i)\varphi(j)gcd(i,j)}{\varphi(gcd(i,j))}\)
\(~~~~~~~=\sum_\limits{d=1}^{min(n,m)}\sum_\limits{i=1}^n\sum_\limits{j=1}^m \dfrac{\varphi(i)\varphi(j)d}{\varphi(d)}[gcd(i,j)=d]\)
\(~~~~~~~=\sum_\limits{d=1}^{min(n,m)}\frac{d}{\varphi(d)}\sum_\limits{i=1}^n\sum_\limits{j=1}^m \varphi(i)\varphi(j)[gcd(i,j)=d]\)
\(~~~~~~~=\sum_\limits{d=1}^{min(n,m)}\frac{d}{\varphi(d)}\sum_\limits{i=1}^{\left\lfloor\frac{n}{d}\right\rfloor}\sum_\limits{j=1}^{\left\lfloor\frac{m}{d}\right\rfloor} \varphi(i)\varphi(j)[gcd(i,j)=1]\)
\(~~~~~~~=\sum_\limits{d=1}^{min(n,m)}\frac{d}{\varphi(d)}\sum_\limits{i=1}^{\left\lfloor\frac{n}{d}\right\rfloor}\sum_\limits{j=1}^{\left\lfloor\frac{m}{d}\right\rfloor} \varphi(i)\varphi(j)\sum\limits_{w|i,w|j}\mu(w)\)
\(~~~~~~~=\sum_\limits{d=1}^{min(n,m)}\frac{d}{\varphi(d)}\sum\limits_{w|d}\mu(\frac{d}{w})\sum_\limits{i=1}^{\left\lfloor\frac{n}{d}\right\rfloor}\sum_\limits{j=1}^{\left\lfloor\frac{m}{d}\right\rfloor} \varphi(i)\varphi(j)\)