A deformation

\[ \sum_{i=1}^nf*g(i)=\sum_{i=1}^nf(i)\sum_{j=1}^{\lfloor\frac{n}{i}\rfloor}g(j)\\ \begin{align} \sum_{i=1}^n\phi(i)\sum_{j=1}^{\lfloor\frac{n}{i}\rfloor}\phi(j)&=\sum_{i=1}^n\mu*\phi(i)\sum_{j=1}^{\lfloor\frac{n}{i}\rfloor}id(j)\\ &=\sum_{i=1}^n\mu*\phi(i)\frac{\lfloor\frac{n}{i}\rfloor(\lfloor\frac{n}{i}\rfloor+1)}{2}\\ &=\frac{\sum_{i=1}^n\lfloor\frac{n}{i}\rfloor^2\mu*\phi(i)+\sum_{i=1}^n\lfloor\frac{n}{i}\rfloor\mu*\phi(i)}{2}\\ &=\frac{\sum_{i=1}^n\lfloor\frac{n}{i}\rfloor^2\mu*\phi(i)+\sum_{i=1}^n\phi(i)}{2}\\ \end{align}\\ \sum_{i=1}^n\lfloor\frac{n}{i}\rfloor^2\mu*\phi(i)=2\times\sum_{i=1}^n\phi(i)\sum_{j=1}^{\lfloor\frac{n}{i}\rfloor}\phi(j)-\sum_{i=1}^n\phi(i) \]