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Old 2012-10-13, 15:11   #1
c10ck3r
 
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Default Stupid Question Re: Fermat Factor search

Okay, so all factors of a Fermat number are of the form k*2^n+1, right? So, theoretically, P-1 would find factors as long as the k is b-smooth? Is the limitation of this the ~1e15 k value, the size of 2^2^n, or something else that I'm missing? Thanks, Johann.
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Old 2012-10-18, 03:52   #2
MattcAnderson
 
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"Matthew Anderson"
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Hi,

Fermat numbers have the form 2^(2^m) + 1.
Factors of Fermat numbers have the form k*2^n + 1, where n >= m+2.

My reference is http://www.fermatsearch.org/

Using a trial division software downloaded from the above website, I have been searching n=36 and k>1.4e16

Luigi Morelli has a new software that runs about 50 times faster than this, and can find factors of Fermat and Double Mersenne numbers, but it requires a Graphics Processing Unit, and can be downloaded from -

http://www.doublemersennes.org/

Regards,
Matt
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Old 2012-10-18, 04:23   #3
Dubslow
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Quote:
Originally Posted by MattcAnderson View Post
Hi,

Fermat numbers have the form 2^(2^m) + 1.
Factors of Fermat numbers have the form k*2^n + 1, where n >= m+2.

My reference is http://www.fermatsearch.org/

Using a trial division software downloaded from the above website, I have been searching n=36 and k>1.4e16

Luigi Morelli has a new software that runs about 50 times faster than this, and can find factors of Fermat and Double Mersenne numbers, but it requires a Graphics Processing Unit, and can be downloaded from -

http://www.doublemersennes.org/

Regards,
Matt
Thank you Captain Obvious. (To quote Batalov.)
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Old 2012-10-18, 05:26   #4
Batalov
 
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Captain Obvious adds that if "Luigi has a new software", then in that sense everyone has it. The author who had it (as a baby) was one Woltman.

But anyway, that had nothing to do with the OP's question.
Quote:
Originally Posted by c10ck3r View Post
Okay, so all factors of a Fermat number are of the form k*2^n+1, right? So, theoretically, P-1 would find factors as long as the k is b-smooth? Is the limitation of this the ~1e15 k value, the size of 2^2^n, or something else that I'm missing? Thanks, Johann.
To that, the answer is "yes".

In fact, P-1 had been run with high limits already, so one would not expect to succeed where others* failed. Sometimes, though, - as the Russian proverb goes, even a stick shoots.

_______________
*to quote from "Almost Famous": " I'm about to boldly go where... ummm... many men have gone before"
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