20100326, 17:56  #1 
May 2009
Russia, Moscow
2·3^{3}·47 Posts 
F22 factored!
Yesterday Buckle have found first factor of F22: 64658705994591851009055774868504577 !
Congratulations! 
20100326, 18:11  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2×13×359 Posts 
...cofactor status?

20100326, 18:16  #3  
Oct 2004
Austria
2482_{10} Posts 
Quote:
This year seems to be a very good one for Fermat factors!! Last fiddled with by Andi47 on 20100326 at 18:17 

20100326, 18:23  #4 
"Tapio Rajala"
Feb 2010
Finland
3^{2}·5·7 Posts 
Congratulations to David Bessell!!
I am not that surprised that you were to one to find the factor with the massive amount of work you have done! 
20100326, 19:23  #5 
Banned
"Luigi"
Aug 2002
Team Italia
3·1,601 Posts 

20100326, 19:40  #6 
"Mark"
Apr 2003
Between here and the
2^{3}×11×71 Posts 

20100326, 20:01  #7 
∂^{2}ω=0
Sep 2002
República de California
2^{5}·3·11^{2} Posts 
Very nice ... note that it would have needed a p1 with stage 1 bound just over 10^9 (specifically, >= 1045429261) and stage 2 bound >= 52795084261 to find this factor.

20100326, 20:45  #8 
Oct 2009
Oulu, Finland
11110_{2} Posts 

20100326, 21:14  #9 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2·41·71 Posts 
I think I may have thought of a way to speed up the search for factors of fermat numbers. Currently when people search for fermat factors they test if numbers of the form k*2^n+1 are factors of the fermat number. They only try possible factors that are either proven prime or have at least been trial factored themselves in order to maximize throughput.
Now for my idea: Since the 2^n can easily be taken care of specially in P1, we know that if P1 has been run on the fermat number then the k of any factor must have a factor(or maybe more than one) that evades the bounds of the P1. Surely if we eliminated candidates that would have been already found with P1 then that would eliminate a lot of the candidate factors from testing. Is anything like this done already or am I missing something? If this idea works then maybe it could also be applied to the search for factors of mersenne numbers(factors of the form 2kp+1) 
20100326, 21:28  #10 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
4267_{10} Posts 
Interesting idea. I don't know how useful it'd be in application, (especially for Mersenne numbers, where I know that you only do a small amount of TF after P1, and individual candidates are quickly checked) but I see no reason why it wouldn't be correct.
To put it another way: When P1 factoring has proven that any potential k in k*2^n+1 or 2kp+1 must be smoother than certain bounds, you can skip checking k's that are smoother than those bounds. Last fiddled with by MiniGeek on 20100326 at 21:33 
20100327, 00:21  #11 
Nov 2003
2^{2}×5×373 Posts 

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