Timeline for Is there an odd-order group whose order is the sum of the orders of the proper normal subgroups?
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| Jun 17, 2012 at 20:43 | comment | added | François Brunault | @Arnie : You should probably contact Gerhard Paseman directly, since only the author of the answer is noticed of the comments. | |
| Jun 17, 2012 at 19:46 | comment | added | Jose Arnaldo Bebita | @Gerhard - +1, but may I ask, what specific paper in the OPN literature are you referring to which could explain why the $m$ here should be a square? | |
| Feb 21, 2011 at 9:56 | history | edited | François Brunault | CC BY-SA 2.5 |
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| Feb 16, 2011 at 21:48 | comment | added | Gerhard Paseman | Francois (forgive my limited typeset) : my apologies for assuming N is abundant. However looking at the OPN literature may given foundation to my feeling that the ratio (abundancy index? deficiency index?) should contribute to an explanation for m being a square. Gerhard "Ask Me About System Design" Paseman, 2011.02.16 | |
| Feb 10, 2011 at 17:33 | comment | added | Max Horn | So the example by Tom De Medts almost fits in with this answer, except that gcd(N,pq) there is 3, not 1... | |
| Feb 10, 2011 at 17:21 | comment | added | François Brunault | @Gerhard : thanks for mentioning this result on odd perfect numbers which I didn't know. It would be indeed interesting to prove something similar here. Note that in my setting $N=m(1+q+pq)$ has to be deficient (nevertheless, the ratio $\sigma(N)/N$ should be close to 2, in my example it is $1.748\ldots$). | |
| Feb 10, 2011 at 16:47 | comment | added | Gerhard Paseman | Your conditions imply m(1+q+pq) is abundant and relatively prime to pq. If (1+q+pq) doesn't contribute much to abundancy, then m should; if m is small then 3 and preferably 9 should be a factor, and odd perfect numbers are (a prime times a)square, so likely such a small m should be a multiple of a square if not square itself. Gerhard "Ask Me About System Design" Paseman, 2011.02.10 | |
| Feb 10, 2011 at 15:51 | comment | added | Tom Leinster | And is there any interesting reason why p and q are Mersenne primes? | |
| Feb 10, 2011 at 15:48 | vote | accept | Tom Leinster | ||
| Feb 10, 2011 at 15:48 | comment | added | Tom Leinster | Bravo ! | |
| Feb 10, 2011 at 13:18 | comment | added | François Brunault | Is there any interesting reason why $m$ is a perfect square (here $m=627^2$) ? | |
| Feb 10, 2011 at 12:58 | history | answered | François Brunault | CC BY-SA 2.5 |