20121013, 15:11  #1 
Aug 2010
Kansas
547 Posts 
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, P1 would find factors as long as the k is bsmooth? Is the limitation of this the ~1e15 k value, the size of 2^2^n, or something else that I'm missing? Thanks, Johann.

20121018, 03:52  #2 
"Matthew Anderson"
Dec 2010
Oregon, USA
2^{4}×3^{2}×5 Posts 
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 
20121018, 04:23  #3  
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
16065_{8} Posts 
Quote:


20121018, 05:26  #4  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·53·89 Posts 
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:
In fact, P1 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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