20170116, 11:03  #1 
Dec 2011
After milion nines:)
2757_{8} Posts 
Prime95 P1/Pfactor using to find factors on CRUS bases
When Masser suggested that very light P1 factoring would further significantly reduce the likelihood of an unfactored, large exponent term with algebraic factors, I decide to try on this for me, new area of interest.
I know that Pfactor and Pminus1 is not the same, but it looks like they work similar on Prime95. So I try on my npg sequence that have 2M1 digits and above ( bases 155, 212) in order to find same factors. I run it all night, and in the morning I have found few new factors. So now when I have test data, I try to optimize it, so I run different setting , so find what settings find all factors I have ( in my test data) and to be faster then it was first time. At the end I found that Pfactor=4,212,n,1,45,8 was winning combination. Since Prime95 write in log what is B1 and B2 choose for optimal: I try also to play with Pminus1 option in Prime95. Got pretty much same result. And last night I try to go much more deeper ( B2 is set to 2000000) and in the morning I was surprised that I found only one new factor ( yes I know there is no need to get same B1, B2 setting and process it few times since it will be (almost) always find same factors. I also play with base2 candidates, and in that case ( and since I sieving that sequence with GPU) I found no new factors, but find it is much faster then on any other bases. So I have question: 1. why much more deeper B2 ( from 700000  2000000) gives such small number of new candidate?  Yes, it must be some reason why Masser suggested very light P1 factoring ( in my case, and my experiment suggest values for very light P1 will be B1=20000 , B2=400000). Some of you maybe thing it is useless to use P1/Pfactor to find candidates, and maybe sr x sieve will find it faster, but it is experiment, and every candidate less is good thing So if you gave any idea, suggestion be free to write it , thanks for reading! Last fiddled with by pepi37 on 20170116 at 11:04 
20170116, 15:05  #2 
Dec 2011
After milion nines:)
7^{2}×31 Posts 
Additional info
From CUDA PPS sieve I know for two factors 451228441149281  28561*2^4020784+1 478743809616001  83521*2^4011252+1 Prime95 successfully detect one of them , but not the other, regardless,am I using Pfactor or Pminus1 ( and regardless of low boundary value) Pminus1 P1 found a factor in stage #2, B1=20000, B2=700000, E=6. 83521*2^4011252+1 has a factor: 478743809616001 (P1, B1=20000, B2=700000, E=6) 28561*2^4020784+1 completed P1, B1=20000, B2=700000, E=6, We4: 6815670D Pfactor [Mon Jan 16 16:02:46 2017] 83521*2^4011252+1 completed P1, B1=55000, B2=660000, E=6, We4: 67E373DB 28561*2^4020784+1 completed P1, B1=55000, B2=660000, E=6, We4: 68157375 It look like I need to learn more :) 
20170116, 18:24  #3 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{3}×1,229 Posts 
P1 is effective for the inputs that have a known partial factorization of all future factors. A trivial example is Mersenne primes, as well as Wagstaff's, GFNs, GUs, GMs, EMs, All of them have factors P, with P1 known to have 2025 bits already factored. You get this 2025 bit boost for free and on top of it you get an additional factor of P1, and as a result you get a factor P which is above what you can get with TF/sieving.
For most CRUS candidates, you only know that P1 has a factor 2 (which is no new information, and hence no boost), and a P1 curve will be as effective as 1 curve of ECM. Occasionally, yes, you will get a factor in line with random probability, 
20170116, 18:44  #4 
Dec 2011
After milion nines:)
7^{2}·31 Posts 
So sequence like 4*155^n+1 or 4*20^n+1 pr 4*50^n+1 etc etc will gain some boost?

20170116, 18:47  #5 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9832_{10} Posts 
Yes, a 1 bit boost. They are "GFNs" of form x^2+1, so all their factors are of form 4k+1.

20170116, 18:54  #6 
Dec 2011
After milion nines:)
7^{2}×31 Posts 
So does I do something wrong if I got only 1 bit boost, and you say above, 2025 bit boost.Seems like I dont use Prime95 properly.

20170117, 05:42  #7 
Romulan Interpreter
"name field"
Jun 2011
Thailand
3^{5}×41 Posts 

20170117, 07:01  #8 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{3}·1,229 Posts 
Your "M" does have a factor M, for which M1 is indeed = 2kp.
Please by all means continue correcting my missing commas and other typos. You are indispensable in this role. 
20170117, 10:46  #9 
Dec 2011
After milion nines:)
1519_{10} Posts 
Just do not begin fight , everything else is allowed

20170120, 14:31  #10 
Dec 2011
After milion nines:)
7^{2}×31 Posts 
After few days of playing using different B1 and B2 values: I came to conclusion that removal rate is about constant 2.2% , so for small numbers it is very slower, much slower then LLR, but on huge numbers ( like Batalov say) it has some sense!
And of course it is good to learn something new. Thanks masser for point me to P1. Last fiddled with by pepi37 on 20170120 at 14:32 
20170120, 16:04  #11  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{3}·1,229 Posts 
Quote:
I was just being polite. On these, it most likely doesn't. I explained on which candidates "it has some sense". 

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