# 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?

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Cross posted at MO... –  draks ... Jul 15 '12 at 18:36

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,

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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).