How to find: $g(n)= \sum_{d|n}\frac{\phi(d)}d=?$

Question

How to find: $$f(n)=\sum_{d\mid n} d \phi(d)=? $$,

$$ g(n)= \sum_{d\mid n}\frac{\phi(d)}d=?$$

Answer 1

$f(n)$ is a multiplicative function (that is, $f(uv) = f(u) f(v)$ for $\gcd(u,v) = 1$), so it is enough to compute it for the powers of a prime $p$. We have $$ \begin{align} f(p^{m}) &= \sum_{k=0}^{m} p^{k} \varphi(p^{k}) \\&= 1 + \sum_{k=1}^{m} p^{k} (p-1) p^{k-1} \\&= 1 + (p-1) p \sum_{k=1}^{m} p^{2k-2} \\&= 1 + p \frac{p^{2m} - 1}{p+1}. \end{align} $$

$g(n)$ is also multiplicative, and here a simpler calculation yields $$ g(p^{m}) = 1 + m \frac{p-1}{p}. $$