Divisor sum of totient function

Question

Is there any closed form expression for $\displaystyle\sum_{d|n} d\phi(d)$?

I have tried a lot but can only reduce it to $\displaystyle\sum_{k=1}^{n}\frac{n}{(k,n)}$ where $(k,n)$ is the greatest common divisor. But it cannot be simplified.

Answer 1

Your sum is (weakly) multiplicative: if $(m,n)=1$ then any divisor $d$ of $mn$ can be written uniquely as $d=d_1d_2$ where $d_1|n$ and $d_2|m$.

Therefore it suffices to compute your sum for prime powers. We have $$\sum_{d|p^k}d\varphi(d)=\sum_{i=0}^k p^i\varphi(p^i)$$

Writing this out we get (barring arithmetic error) $$1+p(p-1)+p^3(p-1)+\dots =1+(p-1)p\left[1+p^2+\dots +p^{2(k-1)}\right]$$ $$=1+(p-1)p\frac {p^{2k}-1}{p^2-1}=1+p\frac {p^{2k}-1}{p+1}=\frac {p^{2k+1}+1}{p+1}$$