I have a simple question regarding a particular form of a sum and I was hoping someone could provide some insights or guidance.
I was wondering if there was any other way to express the following sum by using mobius inversion or some other form of transformation:
$$\sum_{d|n} \frac{{n/d}}{\varphi\left(n/d\right)}$$
thanks in advance.
Playing around.
$\begin{array}\\ s(n) &=\sum_{d|n} \frac{{n/d}}{\varphi(n/d)}\\ &=\sum_{d|n} \frac{d}{\varphi(d)}\\ s(p) &=\sum_{d|p} \frac{d}{\varphi(d)}\\ &=\frac{{1}}{\varphi\left(1\right)}+\frac{p}{\varphi(p)}\\ &=1+\frac{p}{p-1}\\ &=\frac{2p-1}{p-1}\\ s(p^m) &=\sum_{d|p^m} \frac{d}{\varphi(d)}\\ &=\sum_{k=0}^m\frac{p^k}{\varphi(p^k)}\\ &=1+\sum_{k=1}^m\frac{p^k}{p^{k-1}(p-1)}\\ &=1+\sum_{k=1}^m\frac{p}{(p-1)}\\ &=1+\frac{mp}{(p-1)}\\ &=\frac{(m+1)p-1}{(p-1)}\\ &=\frac{(m+1)(p-1)+m+1-1}{(p-1)}\\ &=m+1+\frac{m}{(p-1)}\\ \end{array} $
Since $d$ and $\varphi(d)$ are multiplicative, so is $\frac{d}{\varphi(d)}$ and so, if $n=\prod p_i^{m_i}$ then $s(n)=\prod s(p_i^{m_i})$.
So, if $n=\prod p_i^{m_i}$,
$\begin{array}\\ s(n) &=s(\prod p_i^{m_i})\\ &=\prod (m_i+1+\frac{m_i}{(p-1)}))\\ \end{array} $