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?

Note that since $\omega$ is additive and $\mu$ is multiplicative, the function $\sum_{d \, | \, n} \frac{ 2^{\omega(d)} \mu(d) }{d}$ is multiplicative, so it suffices to compute it at prime powers.
At prime powers, one easily computes $1 - \frac 2p$, since 
$$
\sum_{d \, | \, p^{\alpha}} \frac{ 2^{\omega(d)} \mu(d) }d = \sum_{i=0}^{\alpha} \frac{ 2^{\omega(p^i)} \mu(p^i) }{p^i} = 1 - \frac 2p
$$
so that
$$
\sum_{d \, | \, n} \frac{ 2^{\omega(d)} \mu(d) }{d} = \prod_{p \, | \, n} \left( 1 - \frac 2p \right).
$$
Hope that helps,
