Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
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
2  
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 1

up vote 4 down vote accepted

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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.