lcm/gcd

$A(n)=\sum_{i=1}^n\frac{lcm(i,n)}{gcd(i,n)}$,$F(n)=\sum_{i=1}^nA(i)$,求$F(n)$%1000000007

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$$\begin{align} A(n)=\sum_{i=1}^n\frac{lcm(i,n)}{gcd(i,n)}&=\sum_{i=1}^n\frac{n*i}{(i,n)^2}\\ &=\sum_{i=1}^n\sum_{d|n}[(i,n)=d]\frac{n*i}{d^2}\\ &=\sum_{d|n}\sum_{i=1}^{[\frac nd]}[(i,\frac nd)=1]\frac{n*i}d\\ &=\sum_{d|n}d\sum_{i=1}^d[(i,d)=1]*i\\ &=\frac 12(1+\sum_{d|n}d^2\varphi(d)) \end{align}$$

实即求$\sum_{i=1}^n\sum_{d|i}d^2\varphi(d)=\sum_{i=1}^n\sum_{d=1}^{[\frac ni]}d^2\varphi(d)$.
令$\phi'(n)=\sum_{i=1}^ni^2\varphi(i)$.

由于：

$$\sum_{d|n}d^2\varphi(d)*(\frac nd)^2=n^2\sum_{d|n}\varphi(d)=n^3$$
所以：
$$\begin{align} \sum_{i=1}^ni^3=(\frac{n(n+1)}2)^2&=\sum_{i=1}^n\sum_{d|i}d^2\varphi(d)*(\frac id)^2\\ &=\sum_{i=1}^ni^2\sum_{d=1}^{[ \frac ni]}d^2\varphi(d)\\ &=\sum_{i=1}^ni^2\phi'([\frac ni]) \end{align}$$

所以得出：$\phi'(n)=(\frac{n(n+1)}2)^2-\sum_{i=2}^ni^2\phi'([\frac ni])$,可以先预处理前$n^{2/3}$,原问题可以再$O(n^{2/3}log(n))$解决。