常见求和公式及其推导 挑几个稍微经典一点的求和公式随便写点东西

要出去比赛了，看了看huchi的博客，挑了几个式子证一证。
时间比较紧，没时间写得很仔细了，随便写写，日后有时间写完整一点。

1.\(\sum_{a=1}^n\sum_{b=1}^mgcd(a,b)=\sum_{d=1}^{min(n,m)}\phi(d)\lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor\)

不妨设\(n\leq m\),枚举\(d=gcd(a,b)\),反演得到左边=\(\sum_{d=1}^n\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}d\mu(i)\lfloor\frac{n}{id}\rfloor\lfloor\frac{m}{id}\rfloor\),变形，用\(t\)替换\(i*d\),得\(\sum_{t=1}^n\lfloor\frac{n}{t}\rfloor\lfloor\frac{m}{t}\rfloor\sum_{d|t}d\mu(\frac{t}{d})\),后面那个式子是典型卷积，所以原公式成立。

2.\(\sum_{a=1}^n\sum_{b=1}^mlcm(a,b)=\frac{1}{4}\sum_{d=1}^{min(n,m)}d\lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor(\lfloor\frac{n}{d}\rfloor+1)(\lfloor\frac{m}{d}\rfloor+1)\sum_{i|d}i\mu(i)\)

反演：设\(F(t)=\sum_{i=1}^n\sum_{j=1}^mij[t|gcd(i,j)],f(t)=\sum_{i=1}^n\sum_{j=1}^mij[t=gcd(i,j)]\),则有\(F(t)=\sum_{i=1}^n\sum_{j=1}^mij[t|i][t|j]=\frac{\lfloor\frac{n}{t}\rfloor\lfloor\frac{m}{t}\rfloor(\lfloor\frac{n}{t}\rfloor+1)(\lfloor\frac{m}{t}\rfloor+1)}{4}\).因此答案=\(\sum_{d=1}^nd\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\mu(i)i^2\frac{\lfloor\frac{n}{id}\rfloor\lfloor\frac{m}{id}\rfloor(\lfloor\frac{n}{id}\rfloor+1)(\lfloor\frac{m}{id}\rfloor+1)}{4}\).\(t=id\)来替换\(d\),套路求解。记\(G(d)=\sum_{i|d}i\mu(i)\),显然是积性函数，对质数\(p\)有\(G(p^k)=1-p\),随便推一推就能用线性筛预处理，前面再一分块，就能\(O(\sqrt n)\)求解。

3.\(\sum_{a=1}^n\sum_{b=1}^nlcm(a,b)=\sum_{i=1}^n(-i+2\sum_{j=1}^ilcm(i,j))\).

预处理,O(1)输出。

4.\(\sum_{i=1}^ngcd(i,n)=\sum_{d|n}\frac{n}{d}\phi(d)\).

枚举\(d\),老套路了。

5.\(\sum_{i=1}^n\sigma(i)=\sum_{i=1}^ni\lfloor\frac{n}{i}\rfloor\).

根据\(\sigma\)的定义轻松得出。左边=\(\sum_{i=1}^n\sum_{j|i}j=\sum_{j=1}^nj\sum_{i=1}^{\lfloor\frac{n}{j}\rfloor}1\).