20040923, 18:29  #1 
"Phil"
Sep 2002
Tracktown, U.S.A.
19×59 Posts 
Odd Perfect numbers  a factoring challenge
For those of you interested in factoring by ECM/P1, I saw the following today on the nmbrthry listserve:
http://listserv.nodak.edu/scripts/wa...y&F=&S=&P=1064 Basically, the author says that he has proven that any odd perfect number must have at least 47 prime factors (including repetitions), and that improving this result depends upon finding factors of the three numbers listed in the posting. Gmpecm would probably be the preferred tool here. 
20040923, 19:17  #2 
"Nancy"
Aug 2002
Alexandria
100110100011_{2} Posts 
I was about to post this, too. I've emailed Kevin G. Hare, asking for info on previous factoring effort so that we can choose bounds accordingly, but haven't received a reply yet.
I've done P1 on the c301 with B1=100M, and on the c789 and c927 with B1=10M. I'm attaching files with the numbers to factor. Unfortunately I don't have a lot of cpu time available at present, but I'll start off with 100 curves at B1=1M on the c301. If you can run gmpecm, please help factor these numbers! Alex 
20040923, 19:32  #3 
Mar 2003
2^{4}×5 Posts 
I'm running gmpecm on the C301 at the 30digit level. I have about 15 (slowish) CPUs on it at moment and it will be done soon. If nothing is found I'll bump it up to the 35digit level and then 40 digits.
If anyone else is doing ECM work on the C301, let me know so we don't duplicate too much work at the same level. Don Leclair 
20040923, 19:45  #4 
May 2004
2^{4}×5 Posts 
OK then, I'm doing 500@25e4 on the C789, this will take a few hours.
Dave 
20040923, 21:27  #5  
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
2×5×11×107 Posts 
Quote:
Parigp is amazingly slow at calculating the largest number. I gave up after about 2 hours cpu and loaded only the 2 smaller ones. Kevin Hare has since mailed me the decimal representation of the largest. Since then a few hundred curves at B1=50K have completed, as have around 50 at B1=250K. Needless to say, no factors have yet been found. Paul 

20040923, 21:33  #6  
"William"
May 2003
Near Grandkid
5^{3}·19 Posts 
Quote:
These roadblocks suggest he may have found some other factors that are not in Richard Brent's data base. I wonder if he has found factors of sigma(127^{108}) sigma(16^{192}) sigma(2801^{78}) or if his approach doesn't need these. Last fiddled with by wblipp on 20040923 at 21:37 

20040923, 21:51  #7 
May 2004
2^{4}·5 Posts 
I did a quick check to ensure there are no algebraic factors (of polynomials in 11, 547, 3221 respectively). Indeed there aren't. Should this be obvious to me?
Dave 
20040923, 22:36  #8 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
I've done 100@1M for the c301. I'm doing 100@1M on the c927 now.
Note that sigma(11^18), sigma(547^18) and sigma(3221^12) are prime. sigma(p^n) is simply p^0+p^1+...+p^n, i.e. if n+1 is prime, the (n+1)st cyclotomic polynomial evaluated at p, which also explanins the absence of algebraic factors. Alex 
20040923, 23:25  #9 
May 2004
2^{4}×5 Posts 
I'm talking about (for instance) the poly in 11 being irreducible, not the poly in sigma(11^18).
For example, take sigma(sigma(2^4)^2). Here sigma(2^4) = 31 is prime, also 2+1 is prime and sigma(p^2) = p^2 + p + 1 which is irreducible in Z[p]. So in this case the composition of two irreducible cyclotomic polynomials has an algebraic factor: (x^2  x + 1)(x^6 + 3x^5 + 5x^4 + 6x^3 + 7x^2 + 6x + 3) This is really what I meant. Am I making sense? Dave 
20040924, 00:06  #10 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
Yes, perfect sense! Good point.
According to Pari, both phi_17(phi_19(x)) and phi_23(phi_13(x)) are irreducible. Alex 
20040924, 00:49  #11 
Mar 2003
2^{4}×5 Posts 
For the C301, I've finished 1100 curves at B1=1M with gmpecm 5.0.3 so it is unlikely that there are any 35digit or smaller factors.
A few minutes ago my machines started 2900 curves at 3M (the 40digit level). That should be done late tomorrow. If no factors are found I plan to stop when the 40digit level is complete. Don Leclair 
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