莫比乌斯反演式子变形

1:\(\sum_{d|gcd}f(d)=\sum_{d=1}f(d)[d|gcd(i,j)]\)
(这个显然吧。。。)

2:\(\sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{d=1}f(d)[d|gcd(i,j)]=\sum_{d=1}f(d)\sum_{i=1}^{n}\sum_{j=1}^{m}[d|gcd(i,j)]\)

3:\(\sum_{i=1}^{n}\sum_{j=1}^{m}[d|gcd(i,j)]=\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{j=1}^{\lfloor \frac{m}{d} \rfloor}[1|gcd(i,j)]=\lfloor \frac{n}{d} \rfloor\lfloor \frac{m}{d} \rfloor\)

4：\(\sum_{i=1}^{n}\sum_{j=1}^{m}ij[gcd(i,j)=d]=d^2*\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{j=1}^{\lfloor \frac{m}{d} \rfloor}ij[gcd(i,j)=1]\)

5:\(\sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)=1]=\sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{d|gcd(i,j)}\mu(d)\)
（\(\mu\)的性质）

6:\(\sum_{k|i}\mu(\frac{i}{k})\lfloor \frac{n}{i} \rfloor\lfloor \frac{m}{i} \rfloor=\sum_{i=1}^{\lfloor \frac{n}{k} \rfloor}\mu(i)\lfloor \frac{n}{k*i} \rfloor\lfloor \frac{m}{k*i} \rfloor\)