# Sum of Mobius function and omega 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 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?

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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 $n$? –  Alex Botros Aug 30 '11 at 3:21
In my opinion this would be more appropriate at math.stackexchange. –  Gjergji Zaimi Aug 30 '11 at 3:22
I'll give it a try. Thanks –  Alex Botros Aug 30 '11 at 3:38
When $d$ is squarefree, then $2^{\omega(d)}=\tau(d)$, the number of divisors. If I am correct, you are computing $\sum_{d|n'} \mu(d)\tau(d)/d=\prod_{p|n'} (1+(-1)\cdot 2/p)$, where $n'$ is the squarefree kernel of $n$. –  Junkie Aug 30 '11 at 3:59
FANTASTIC!!!! That's true! thank you –  Alex Botros Aug 30 '11 at 4:05

## 1 Answer

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).

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