一些反演求和过程？

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问题$$\sum _{i,j<=n}{i}^a{j}^a(i,j{)}^b$$记$$\begin{gathered} f(d)=\sum _{(i,j)=d}{i}^a{j}^a \\ F(d)=\sum _{d|i,d|j}{i}^a{j}^a \end{gathered}$$可得$$F(d)=\sum _{k}f(kd)$$$$f(d)=\sum _{k}F(kd)\mu (k)$$带入得到$$\sum _{d=1}^n{d}^{b}f(d)$$$$=\sum _{d=1}^n{d}^b\sum _{k}F(kd)\mu (k)$$$$=\sum _{ij<=n}F(ij)i^b\mu (j)$$$$=\sum _{T=1}^{n}F(T)\sum _{d|T}{d}^b\mu (\frac{T}{d})$$因为$$F(z)=z^{2a}\sum _{iz<=n,jz<=n}{i}^a{j}^a$$$$=z^{2a}\ast (\sum _{i=1}^{\frac{n}{z}}{i}^a{)}^2$$
可以得到$$G(x)=\sum _{[\frac{n}{z}]=x}F(z)=(\sum _{i=1}^{\frac{n}{z}}{i}^a{)}^2\ast (\sum _z{z}^{2a})$$
预处理伯努利数，可以$O(\sqrt{n}a)$计算前缀和
让后考虑计算$\mu \ast i{d}^b$
注意到$I\ast \mu \ast i{d}^b=i{d}^b$，考虑直接杜教筛即可

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$$\sum _{i=1}^n\sum _{j=1}^n(ij{)}^{(i,j)}$$
$$=\sum _{d=1}^n\sum _{(i,j)=d}(ij{)}^d$$
$$=\sum _{d=1}^n{d}^{2d}\sum _{id<=n}\sum _{jd<=n}(ij{)}^d[(i,j)=1]$$
$$=\sum _{d=1}^n{d}^{2d}\sum _{id<=n}\sum _{jd<=n}(ij{)}^d\sum _{t|i,t|j}\mu (t)$$
$$=\sum _{d=1,t=1}^n{d}^{2d}\mu (t)\sum _{i,j<=\frac{n}{dt}}(it{)}^d(jt{)}^d$$
$$=\sum _{d=1,t=1}^n{d}^{2d}{t}^{2d}\mu (t)\sum _{i,j<=\frac{n}{dt}}(ij{)}^d$$
记$F(x,d)={\sum }_{i,j<=x}(ij{)}^d$
$$=\sum _{dt<=n}{d}^{2d}{t}^{2d}\mu (t)F(n/dt,d)$$
发现有效的$dt$只有$O(n\log n)$，问题变为如何计算$F(x,d)$
$$F(x,d)=(\sum _{i=1}^x{i}^d{)}^2$$发现需要$\min (O(d),O(x))$的复杂度来计算
于是先枚举$t$，再卡常一下就可以了