A sum involving the Möbius function I am trying to find some work done on the following: $$\sum_{d \vert n}\frac{2^{\omega(d)}}{d}\mu(d)$$ where $\omega(d)$ is the number of distinct prime factors of $d$ and $\mu$ is the Möbius function. I saw something about $$\sum_{d \vert n}\frac{\mu(d)}{d}=\phi(n)/n$$ (where $\phi$ is the Euler phi function) on planetmath, but I'm not entirely certain how to use it. Does anyone know of any work done on the first sum?

Let $n=p_1^{a_1}\cdots p_k^{a_k}$. We can prove that
$$\sum_{d|n}\frac{m^{\omega(d)}}{d}\mu(d)=\prod_{i=1}^k\left(1-\frac{m}{p_i}\right)$$
(notice that for $m=1$ this is $\varphi(n)/n$). The proof is easy, just check the corresponding coefficients of $m^{r}$ on both sides.
Indeed, there is the well-known identity
$$\prod_{i=1}^k (1+tx_i)=\sum_{r=1}^k t^r e_r(x_1,\dots,x_k)$$
where the $e_r$'s are the elementary symmetric polynomials. If we let $S(n)$ denote the squarefree part of $n$, then by putting $x_i=\frac{1}{p_i}$ we get
$$\sum_{d|S(n)}\frac{t^{\omega(d)}}{d}=\sum_{r=1}^k t^r e_r(\frac{1}{p_1},\dots,\frac{1}{p_k})=\prod_{i=1}^k\left(1+\frac{t}{p_i}\right)$$
Now to get your identity just use $t=-m$, and the fact that $\mu(d)=(-1)^{\omega(d)}$ when $d$ is squarefree. For $t=-2$ we get
$$\left(1-\frac{2}{p_1}\right)\cdots\left(1-\frac{2}{p_k}\right)$$
which is always non-negative (it is $=0$ only when $n$ is even).
