20201106, 00:12  #12  
Feb 2019
China
59 Posts 
Quote:


20201106, 00:28  #13  
Feb 2019
China
59 Posts 
Quote:
so we can test a primality on p=k*2^(n+2)+1 before we calculate Fn Mod p, do you have any algorithm which can be more faster to get factor? 

20201106, 02:23  #14 
"Curtis"
Feb 2005
Riverside, CA
3·1,579 Posts 
Yes. And the code is already written for you. mmff, like you've been told.

20201106, 02:50  #15 
Feb 2019
China
59 Posts 

20201106, 03:44  #16 
"Viliam Furík"
Jul 2018
Martin, Slovakia
110111110_{2} Posts 

20201106, 03:48  #17 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
3·3,163 Posts 

20201106, 03:57  #18 
"Gary"
Aug 2015
Texas
2^{6} Posts 
Going back to your original question, the answer depends on what you meant by "get the F118 factor":
 If you just want to rediscover the factor of F118 for fun, you can do that using any of the "Sieve+trial division" programs listed at FermatSearch/Download by giving it a search range of N = 120, K = 1527888802614951. This will take just a fraction of a second to run.  If you want to find a new factor of F118 that is of similar size as 1527888802614951*2^120+1, that is no longer possible because for N = 120129 all K values up to 3.2 * 10^15 have already been searched. You can find what has already been searched for any range of N at FermatSearch/STATS/Merge Tables.  If you want to try to find a new factor of F118 (or ay other Fermat number), go to the FermatSearch Merge Tables and, for your range of N, find the highest K searched so far. Then select a range of K just above that. For example, for N = 120 you could select 3.2 * 10^15 to 3.3 *10^15. Reserve the range via FermatSearch, select an appropriate program via the Download page, and then start your search. And then hope you are very, very lucky 
20201106, 04:45  #19  
Feb 2019
China
59 Posts 
Quote:


20201106, 05:15  #20  
"Ben"
Feb 2007
2^{2}×23×37 Posts 
Quote:
Trial division is the process that actually determines the factor. It uses modular exponentiation to check if the Fermat number is 0 modulo the potential factor. The process is similar to that for Mersenne's, which is well described here: https://www.mersenne.org/various/mat...rial_factoring The sieve portion is a preprocessing step that eliminates numbers that can't be factors, leaving fewer to check for divisibility. 

20201106, 05:24  #21  
Feb 2019
China
59 Posts 
Quote:
Quote:
so we can use primality to sieve any potential p=k*2^(n+2)+1, is there any more efficient method to sieve potential p=k*2^(n+2)+1? 

20201106, 05:32  #22  
"Gary"
Aug 2015
Texas
2^{6} Posts 
Quote:
Hope that helps a bit. 

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