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?

$\begingroup$ So, I found the second equality again on Wikiproofs. I think it would suffice if I could prove that the first sum is greater than zero. Does anybody think that's possible without going too deeply into the actual number of prime factors of a random divisor of $n$? $\endgroup$ – Alex Botros Aug 30 '11 at 3:21

2$\begingroup$ In my opinion this would be more appropriate at math.stackexchange. $\endgroup$ – Gjergji Zaimi Aug 30 '11 at 3:22

$\begingroup$ When $d$ is squarefree, then $2^{\omega(d)}=\tau(d)$, the number of divisors. If I am correct, you are computing $\sum_{dn'} \mu(d)\tau(d)/d=\prod_{pn'} (1+(1)\cdot 2/p)$, where $n'$ is the squarefree kernel of $n$. $\endgroup$ – Junkie Aug 30 '11 at 3:59

$\begingroup$ FANTASTIC!!!! That's true! thank you $\endgroup$ – Alex Botros Aug 30 '11 at 4:05

$\begingroup$ The question has been posted at math.stackexchange.com/questions/60648/… $\endgroup$ – Gjergji Zaimi Aug 30 '11 at 4:05
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).