Timeline for Sum of Mobius function and omega function
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 30, 2011 at 4:51 | vote | accept | Alex Botros | ||
| Aug 30, 2011 at 4:48 | answer | added | Greg Martin | timeline score: 5 | |
| Aug 30, 2011 at 4:13 | comment | added | Alex Botros | So, what would then happen if $n$ were even? | |
| Aug 30, 2011 at 4:05 | comment | added | Gjergji Zaimi | The question has been posted at math.stackexchange.com/questions/60648/… | |
| Aug 30, 2011 at 4:05 | comment | added | Alex Botros | FANTASTIC!!!! That's true! thank you | |
| Aug 30, 2011 at 3:59 | comment | added | Junkie | 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 $n$. | |
| Aug 30, 2011 at 3:22 | comment | added | Gjergji Zaimi | In my opinion this would be more appropriate at math.stackexchange. | |
| Aug 30, 2011 at 3:21 | comment | added | Alex Botros | 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 $n$? | |
| Aug 30, 2011 at 3:06 | history | asked | Alex Botros | CC BY-SA 3.0 |