[总结] 一些推式子的中间过程

\(\sum_{i=1}^n\sum_{j=1}^m\sigma_k(ij)\)

\[\begin{aligned} &\sum_{i=1}^n\sum_{j=1}^m\sigma_k(ij)\\ =&\sum_{i=1}^n\sum_{j=1}^m\sum_{x\mid i}\sum_{y\mid j}[\gcd(x,y)=1](\frac{i}{x}\cdot y)^k\\ =&\sum_{i=1}^n\sum_{j=1}^m\sum_{x\mid i}\sum_{y\mid j}\sum_{d\mid \gcd(x,y)}\mu(d)(\frac{i}{x}\cdot y)^k\\ =&\sum_{i=1}^n\sum_{j=1}^m\sum_{d\mid \gcd(i,j)}\mu(d)\sum_{x\mid\frac{i}{d}}\sum_{y\mid \frac{j}{d}}(\frac{i}{dx}\cdot dy)^k\\ =&\sum_{i=1}^n\sum_{j=1}^m\sum_{d\mid \gcd(i,j)}\mu(d)\sum_{x\mid\frac{i}{d}}(\frac{i}{x})^k\sum_{y\mid \frac{j}{d}}y^k\\ =&\sum_{i=1}^n\sum_{j=1}^m\sum_{d\mid \gcd(i,j)}\mu(d)d^k\sum_{x\mid\frac{i}{d}}(\frac{\frac{i}{d}}{x})^k\sum_{y\mid \frac{j}{d}}y^k\\ =&\sum_{i=1}^n\sum_{j=1}^m\sum_{d\mid \gcd(i,j)}\mu(d)d^k\cdot\sigma_k(\frac{i}{d})\cdot\sigma_k(\frac{j}{d})\\ =&\sum_{d=1}^n\mu(d)d^k\sum_{i=1}^\frac{n}{d}\sum_{j=1}^\frac{m}{d}\sigma_k(i)\cdot\sigma_k(j)\\ =&\sum_{d=1}^n\mu(d)d^k\sum_{i=1}^\frac{n}{d}\sigma_k(i)\sum_{j=1}^\frac{m}{d}\sigma_k(j)\\ \end{aligned} \]