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Old 2004-09-23, 18:29   #1
philmoore
 
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Default Odd Perfect numbers - a factoring challenge

For those of you interested in factoring by ECM/P-1, 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. Gmp-ecm would probably be the preferred tool here.
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Old 2004-09-23, 19:17   #2
akruppa
 
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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 P-1 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 gmp-ecm, please help factor these numbers!

Alex
Attached Files
File Type: zip Hare_composites.zip (1.4 KB, 393 views)
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Old 2004-09-23, 19:32   #3
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I'm running gmp-ecm on the C301 at the 30-digit 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 35-digit 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
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Old 2004-09-23, 19:45   #4
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OK then, I'm doing 500@25e4 on the C789, this will take a few hours.

Dave
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Old 2004-09-23, 21:27   #5
xilman
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Quote:
Originally Posted by dleclair
I'm running gmp-ecm on the C301 at the 30-digit 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 35-digit 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
I loaded the 2 smaller numbers into my home ECMNET soon after the NMBRTHRY moderator let loose the posting. Not having any information about the work done so far I assumed that no work had been done.

Pari-gp 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
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Old 2004-09-23, 21:33   #6
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Quote:
Originally Posted by dleclair
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.
I recently ran Dario Alpern's applet to 350, which gets 25 curves at the 35 digit level but jumps there straight from the 25 digit level. I'm not running any more at the present.

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(127108)
sigma(16192)
sigma(280178)

or if his approach doesn't need these.

Last fiddled with by wblipp on 2004-09-23 at 21:37
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Old 2004-09-23, 21:51   #7
dave_dm
 
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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
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Old 2004-09-23, 22:36   #8
akruppa
 
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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
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Old 2004-09-23, 23:25   #9
dave_dm
 
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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
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Old 2004-09-24, 00:06   #10
akruppa
 
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Yes, perfect sense! Good point.

According to Pari, both phi_17(phi_19(x)) and phi_23(phi_13(x)) are irreducible.

Alex
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Old 2004-09-24, 00:49   #11
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For the C301, I've finished 1100 curves at B1=1M with gmp-ecm 5.0.3 so it is unlikely that there are any 35-digit or smaller factors.

A few minutes ago my machines started 2900 curves at 3M (the 40-digit level). That should be done late tomorrow.

If no factors are found I plan to stop when the 40-digit level is complete.

-Don Leclair
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