20060106, 21:20  #1 
Sep 2002
406_{8} Posts 
trial factoring and P1
When doing P1 factoring the b1 is actually b1*M
suppose we test M=67 to b1=1000 It actually look for factor of the form (2*....*M*b1*M+1) If it is (2*...*M+1) then the factor is found right away. Would it become faster to do P1 testing rather than trial at higher mersenne numbers? ex. 79299959 has been P1 tested to 8000000 so 2*...*79299959*8000000*79299959+1 76bits+ trial shows 79299959,72 or Have I got the whole thing wrong ps I never do stage 2 as it messes stage 1 save files. I keep my files as it continues to higher level from where it leftoff. 
20060108, 04:29  #2  
"Richard B. Woods"
Aug 2002
Wisconsin USA
7692_{10} Posts 
Quote:
My understanding is that, in the Prime95 implementation of the P1 algorithm, b1 is the upper limit on the prime factors of the "k" of potential factors 2kp+1 of 2^{p}1 that are to be found by the P1 method. That is, stage 1 P1 with b1 = 10000 performed on 2^{p}1 will find any factor 2kp+1 of 2^{p}1 in which the largest prime factor of k is less than (or equal to, if b1 were prime itself) 10000. Last fiddled with by cheesehead on 20060108 at 04:42 

20060108, 05:45  #3  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·641 Posts 
Quote:
My understanding is that, in the Prime95 implementation of the P1 algorithm, b1 is the upper limit on the powerofaprime factors of the "k" of potential factors 2kp+1 of 2^{p}1 that are to be found by the P1 method. That is, stage 1 P1 with b1 = 10000 performed on 2^{p}1 will find any factor 2kp+1 of 2^{p}1 in which the largest powerofaprime factor of k is less than (or equal to, if b1 were prime itself) 10000. Example: 59704785388637019242567 is a factor of 2^{6049993}  1. 59704785388637019242567 = 2 * 4934285493275531 * 6049993 + 1. Prime95's P1 stage 1 with b1 = 4000 would find this factor because the largest primepower factor of 4934285493275531 is less than 4000. 4934285493275531 = 61^{2} * 593 * 983 * 1153 × 1973. 61^{2} = 3721. In this example the factor 59704785388637019242567 could have been found in stage 1 with b1 as low as 3721. Also, Prime95's P1 stage 2 with b1 = 2000 and b2 = 4000 would find this factor because the largest primepower factor of 4934285493275531 is less than 4000 and all other primepower factors are less than 2000. (In fact, b1/b2 as low as b1 = 1973, b2 = 3721 would have worked.) 

20060108, 08:14  #4  
Jun 2003
12FD_{16} Posts 
Quote:


20060108, 16:16  #5  
Sep 2002
106_{16} Posts 
Quote:


20060109, 02:35  #6  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·641 Posts 
Quote:
jocelynl, I presume you've noted axn1's correction to my erroneous paragraph about stage 2. Last fiddled with by cheesehead on 20060109 at 02:37 

20060122, 18:48  #7  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×641 Posts 
BTW, let this be a lesson.
I failed to actually TRY running Pminus1=6049993,2000,4000,0,0 to confirm that stage 2 would find the 61^{2} = 3721 primepower factor of 4934285493275531 before I made my erroneous posting: Quote:


20060122, 19:47  #8  
Aug 2002
Buenos Aires, Argentina
17·79 Posts 
Quote:


20060201, 14:12  #9  
Sep 2002
100000110_{2} Posts 
Quote:
Last fiddled with by jocelynl on 20060201 at 14:13 

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