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 mobius 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 first sum?



Sum of Mobius function and omega function
Whenever $f(n)$ is a multiplicative function, so is $g(n) = \sum\_{d\mid n} f(d)$. Therefore to evaluate your function, you only need to know its values on prime powers. Since
$$
\sum\_{d\mid p^k} \frac{2^{\omega(d)}}d \mu(d) = \sum\_{j=0}^k \frac{2^{\omega(p^j)}}{p^j} \mu(p^j) = 1 - \frac2p,
$$
it follows that
$$
\sum\_{d\mid n} \frac{2^{\omega(d)}}d \mu(d) = \prod\_{p\mid n} \bigg( 1 - \frac2p \bigg),
$$
as Junkie commented. In particular, it equals zero if $n$ is even (which you can see in hindsight by pairing each odd divisor $d$ with its double $2d$ and realizing that the corresponding summands cancel out, while summands corresponding to multiples of 4 vanish individually).


