# Another sum involving totient function or gcd

Motivated by this question, My question pertains to closed form of the sum $$\sum_{d|n}\frac{\phi(d)}{d}$$

There are some formulae and expressions for $$\sum_{n}\frac{\phi(n)}{n}$$, but what about when we replace $$n$$ by divisors of $$n$$ By seeing this question, I think the sum is multiplicative. Any hints? Thanks beforehand.

Here is not a closed form, but an alternative representation. Below, $$(x,y)={\rm gcd}(x,y)$$.

I claim that $$\sum_{k=1}^n (k,n)=n\sum_{d\mid n}\frac{\phi(d)}{d}.$$ To see this, observe that if $$d\mid n$$ is a divisor of $$n$$ $$(k,n)=d \iff \left(\frac{k}{d},\frac{n}{d}\right)=1.$$ In particular, the divisor, $$d$$, appears exactly $$\phi(n/d)$$ times in the summation. Thus, $$\sum_{k=1}^n (k,n)=\sum_{d\mid n}\phi(\frac{n}{d})d.$$ Now, letting $$d'=n/d$$, the sum can be rewritten as, $$\sum_{k=1}^n (k,n)=n\sum_{d'\mid n}\frac{\phi(d')}{d'}.$$ Hence, this object has the given operational meaning.

Now, for multiplicative property, take coprime $$(m,n)$$. Our goal is to show, $$\left(\sum_{d\mid mn}\frac{\phi(d)}{d}\right) = \left(\sum_{d\mid m}\frac{\phi(d)}{d}\right)\left(\sum_{d\mid n}\frac{\phi(d)}{d}\right).$$ Check that, the left hand side sum has $$\phi(mn)=\phi(m)\phi(n)$$ terms, so do the right hand side. Furthermore, each divisor $$d\mid mn$$ can be uniquely decomposed into $$d=d_1d_2$$ where $$d_1\mid m$$ and $$d_2\mid n$$, since $$(m,n)=1$$. Hence, we deduce the given function is indeed multiplicative.

Theorem: For any positive integer $$a$$, there exists a positive integer $$n$$, such that, $$\sum_{d\mid n}\frac{\phi(d)}{d}=a.$$
After a little observation, the sum is similar to evaluating $$\sum_{d|n}d\phi(d)$$. Since the function $$\sum_{d|n}\frac{\phi(d)}{d}$$ is (weakly) multiplicative, therefore, it suffices to evaluate the sum for prime powers. We have, $$\sum_{d|p^k}\frac{\phi(d)}{d}=1+\frac{p-1}{p}+\frac{p^2(p-1)}{p^3}+\ldots+\frac{p^k(p-1)}{p^{k+1}}$$ $$=1+k\frac{p-1}{p}$$ Hence the form of the sum for any general $$n=\prod_ip_i^{k_i}$$ is: $$\sum_{d|n}\frac{\phi(d)}{d}=\prod_{i}\left(1+k_i\frac{p_i-1}{p_i}\right)$$ where $$n$$ is the product of $$p_i$$ distinct primes with multiplicity $$k_i$$