Timeline for Perfect numbers $n$ such that $2^k(n+1)$ is also perfect
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 11, 2013 at 2:47 | comment | added | Jose Arnaldo Bebita | Luis, I worked out the details (for this particular, remaining case) and came to the conclusion that $p$ has to be an odd prime. Please examine my answer below if it is correct. | |
| Aug 11, 2013 at 1:52 | comment | added | Jose Arnaldo Bebita | Luis, the remaining case is for $m$ even and $n$ odd, correct? | |
| May 10, 2011 at 17:51 | vote | accept | Luis H Gallardo | ||
| May 9, 2011 at 0:49 | comment | added | Luis H Gallardo | You are perhaps too optimistic gerhard; these congruences seems not to create contradictions. | |
| May 9, 2011 at 0:45 | comment | added | Gerhard Paseman | I dimly recall some congruence conditions for OPNs. You might find them and use them to finish off the problem. Gerhard "Ask Me About System Design" Paseman, 2011.05.08 | |
| May 9, 2011 at 0:38 | comment | added | Luis H Gallardo | You are right gerhard: Indeed we have $$ n =2^{p+1}-3 $$ with $p$ prime. | |
| May 9, 2011 at 0:37 | comment | added | Todd Trimble | Gerhard, I get the same result. | |
| May 9, 2011 at 0:30 | comment | added | Gerhard Paseman | Also, if I haven't deluded myself, n would be 3 less than a power of two. This would make n (not a multiple of 3) really big. I think this means no other known examples. Gerhard "Ask Me About System Design" Paseman, 2011.05.08 | |
| May 9, 2011 at 0:20 | comment | added | Gerhard Paseman | If n is odd and perfect, then n+1 is twice an odd number. I suspect this will not yield any more solutions. Gerhard "Ask Me About System Design" Paseman, 2011.05.08 | |
| May 8, 2011 at 23:58 | history | answered | Gerry Myerson | CC BY-SA 3.0 |