Let $n$ divide $c$. I would like to find a closed form for the expression
$$\sum_{k \mid c} \frac{\varphi(kn)}{\varphi(k)},$$
where $\varphi$ is the Euler phi function. Because $n$ divides $c$, it is not always the case that $\varphi(kn)=\varphi(k)\varphi(n)$, so we cannot simplify it that way. The method I attempted was to use the product formula
$$ \varphi(k) = k \prod_{p \mid k} \left(1-\frac{1}{p}\right). $$
The sum can be rewritten as
$$ \begin{align}
\sum_{k \mid c} \frac{\varphi(kn)}{\varphi(k)}
&= \sum_{k \mid c} \frac{kn \prod_{p \mid kn} \left(1-\frac{1}{p}\right)}{k \prod_{p \mid k} \left(1-\frac{1}{p}\right)} \\
& = n \sum_{k \mid c} \prod_{\substack{p \mid n \\ p \not\mid k}} \left(1-\frac{1}{p}\right).
\end{align} $$
I wasn't able to do much with this, though. How can either expression be simplified?
Best Answer
Given $c,n$ with $n\mid c,$ you can define the above sum as $f(c,n).$ Now, if $c,d$ are relatively prime, and $n\mid c,m\mid d$ we can show relatively easily that $f(cd,mn)=f(c,n)f(d,m).$
So we can reduce to the case $f(p^a,p^b)$ with $b\leq a$ and $p$ prime.
But $$f(p^a,p^b)=\sum_{i=0}^{a}\frac{\phi(p^{b+i})}{\phi(p^i)}=\phi(p^b)+\sum_{i=1}^a p^{b}=\phi(p^b)+ap^{b}$$
This gives $$f(p^a,p^b)=\begin{cases}a+1&b=0\\(a+1)p^b-p^{b-1}&b>0\end{cases}$$
So, if $c=p_1^{a_1}\cdots p_j^{a_j}$ and $n=p_1^{b_1}\cdots p_j^{b_j}$ then $$f(c,n)=\prod_{i=1}^{j} f\left(p_i^{a_i},p_i^{b_i}\right)=\tau(c)\prod_{p_i\mid n}\left(p_i^{b_i}-\frac{p_i^{b_i-1}}{a_i+1}\right)$$
Where $\tau(c)$ is the number of positive divisors of $c,$ which is $\tau(c)=(a_1+1)\cdots(a_j+1).$
This gives the inequality:
$$\tau(c)\phi(n)\leq f(n,c) \leq \tau(c) n$$