20121012, 04:58  #12  
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3·29·83 Posts 
Quote:


20121012, 10:09  #14 
Banned
"Luigi"
Aug 2002
Team Italia
2·29·83 Posts 

20121012, 10:21  #15 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
22322_{8} Posts 
No, WK only takes reservations for the whole ensemble cast (as in a sieve followed by pfgw gxo), not GFN separately.

20121012, 12:40  #16  
"Jerry"
Nov 2011
Vancouver, WA
1,123 Posts 
Quote:
_______________________ S.B. EDIT: uploading to the mmffgfn/ folder... Ok, everyone, you are all set. For Windows, get the mmffgfnX0.26win32win64.zip and tests_and_cudart.zip files. Use separate folders for each base. Use sample worktodo.txt files from the tests_and_cudart.zip file. For Linux, you will be better off building your own binary (source is posted, too), but you can try the posted binaries (they were built in OpenSuSE 12.1, so they may not work for you). The rest is the same: Use separate folders for each base. Use the tests from tests_and_cudart.zip file. Good luck! Last fiddled with by Batalov on 20121013 at 08:29 

20121013, 21:16  #17 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·3·1,571 Posts 
The Russian expression 'biting elbows' will be painfully learnt by those who will come late to the party.

20121014, 08:26  #18 
Jul 2003
1142_{8} Posts 
hi,
i have found a factor and now i need an explanation on how to get the cannonical form using factordb 
20121014, 09:04  #19 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·3·1,571 Posts 
1. Put it in factordb > hopefully it is prime/PRP
2. add 1 to the end of the factordb input line > observe 2^N (and you may know N from the output logs anyway) 3. Add "(" around the factor "1)/2^N" (use the value of N) > you will learn the k value. 4. Run pfgw gxo q"k*2^N+1" You can also post the "has" string in the results thread  we will walk you through this. This only seems hard the first time, and then is like riding a bike. Last fiddled with by Batalov on 20121014 at 09:16 
20121023, 01:28  #20 
Apr 2010
Over the rainbow
2^{2}·641 Posts 
a factor  not composite  would be really nice.

20121023, 01:36  #21 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
24D2_{16} Posts 
There are some left. I am sure of it!
Try the probability of success calculator in gp: Code:
# gp k1=2e12; k2=10e12; n=102 p = log(k2/k1) / (log(k1*k2)/2 + n*log(2)1) %1 = 0.01628 # so you'd need ~60 of similar ranges to strike gold # When I started, I'be been luckier: k1=1e11; k2=2e12; n=102 p = log(k2/k1) / (log(k1*k2)/2 + n*log(2)1) %6 = 0.031035 
20121023, 01:40  #22 
"Jerry"
Nov 2011
Vancouver, WA
1,123 Posts 

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