Timeline for Integers n such that sigma(n)=omega(n)n and omega(n) divides n
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 14, 2013 at 20:31 | answer | added | kiskis | timeline score: 2 | |
| May 14, 2013 at 19:05 | answer | added | Gerhard Paseman | timeline score: 4 | |
| May 14, 2013 at 17:58 | vote | accept | Sylvain JULIEN | ||
| May 14, 2013 at 17:56 | comment | added | Gerhard Paseman | Even if you relax the first equation to n being multiply perfect, I suspect there will still be sharp limits for divisibility, e.g. omega might need to be even and possibly abundant itself in order for it to be a factor of n and significantly greater than 2. Gerhard "Feels That's How It Is" Paseman, 2013.05.14 | |
| May 14, 2013 at 17:54 | answer | added | Dietrich Burde | timeline score: 5 | |
| May 14, 2013 at 17:48 | comment | added | Gerhard Paseman | No. Consider how the product p/(p-1) grows where the product is taken over the first omega many primes. This puts a sharp upper bound on omega and should imply n is even for the first equation to hold. If omega is greater than 2, an exhaustive search should finish it off. Gerhard "Ask Me About Pi Inverse" Paseman, 2013.05.14 | |
| May 14, 2013 at 16:54 | history | asked | Sylvain JULIEN | CC BY-SA 3.0 |