most recent 30 from http://mathoverflow.net 2013-05-26T07:03:51Z http://mathoverflow.net/feeds/question/74035 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74035/sum-of-mobius-function-and-omega-function Alex Botros 2011-08-30T03:06:33Z 2011-08-30T04:48:51Z

<p>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?</p>

http://mathoverflow.net/questions/74035/sum-of-mobius-function-and-omega-function/74039#74039 Greg Martin 2011-08-30T04:48:51Z 2011-08-30T04:48:51Z

<p>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).</p>