20210425, 16:23  #1 
Einyen
Dec 2003
Denmark
3^{2}×349 Posts 
P+1 test is also P1 ?
The new P+1 test in Prime95 is also finding P1 factors. 4 of the 8 factors found so far are P1 smooth within B1/B2 and NOT P+1 smooth:
https://mersenneforum.org/showpost.p...&postcount=199 Will P+1 test in GMPECM also find P1 smooth factors? I never read anything about this before, I thought it was a special extra feature George added to his P+1 test, that it was taking more time to include P1 as well. If yes will P1 smooth factors only be found by half the seeds x0 same as P+1 smooth factors? (In a P+1 test). https://mersenneforum.org/showpost.p...&postcount=230 Last fiddled with by ATH on 20210425 at 16:24 
20210425, 16:26  #2 
Romulan Interpreter
Jun 2011
Thailand
2^{3}×1,193 Posts 
Nope. that's only coincidental (or accidental ). They find different types of factors. But the classes intersect.
Last fiddled with by LaurV on 20210425 at 16:26 
20210425, 17:05  #3  
Jun 2003
29×173 Posts 
Yes.
Quote:
Suppose a number N has two smooth prime factors f and g (and other not smooth prime factors, which we don't care about). f is + side smooth, g is  side smooth (i.e. f+1 and g1 are smooth to given b1/b2 bounds) Then there exist seeds such that: 1) they will find both f & g when doing a P+1 run to b1/b2 bounds 2) they will find f but not g ... 3) they will find g but not f ... 4) they will find neither f nor g ... Whether a given seed will find a given factor depends on both the relationship of the seed to the factor and the smoothness side of the factor. As long as they "align", the factor will be found. Probability of alignment is 50%, regardless of whether the smooth side is +1 or 1. Last fiddled with by axn on 20210425 at 17:06 

20210425, 17:39  #4 
Einyen
Dec 2003
Denmark
3^{2}×349 Posts 
Thank you.
How does this work when the test is designed for P+1 to be smooth? Why is this not in the GMPECM Readme under P+1 ?? Granted, I have not done much P+1 factoring, but I have never heard about this before, where has this information been hidden? I guess I should have read actual P+1 theoretical papers, but an important fact like this should be listed in P+1 software. 
20210425, 18:20  #5 
Jun 2003
1399_{16} Posts 
That's just the way the math works out.
A simple intro: https://en.wikipedia.org/wiki/Willia...2B_1_algorithm The thing that needs to be smooth is p(Dp). If (Dp) is 1, then p+1. If (Dp) is 1, then p1. Note: using P+1 to find P1 factors is inefficient because a) P+1 is slower, and b) you only have a 50% chance of success, where as pure P1 has 100% chance of success. Therefore, if you've already run P1, then P+1 with same bounds is not going to find any additional "p1" factors. 
20210425, 19:48  #6  
"Robert Gerbicz"
Oct 2005
Hungary
7×211 Posts 
Quote:
would you choose these starting values: D=3; D=77; and then D=51975 after two failures? You should really avoid this, the problem with this is that (51975/q)=1 if you know (3/q)=1 and (77/q)=1, and this is independent from the q value, what you don't(!) know. 51975=3^3*5^2*7*11, so (51975/q)=(3/q)^3*(5/q)^2*(7/q)*(11/q)=(3/q)*1*(77/q)=1 where we haven't used the value of q. So in general you have to avoid those D for that D*d_i1*...*d_ik is a square number for some earlier d_i1,..,d_ik seeds. Last fiddled with by R. Gerbicz on 20210425 at 19:50 

20210501, 15:18  #7 
"Pavel Atnashev"
Mar 2020
2^{2}×11 Posts 
There are two special seeds.
2/7 selects plus or minus side that is divisible by 6. 6/5 selects the side divisible by 4. These two seeds increase the probability of smoothness. For arbitrary numbers P+1 method with any of the two seeds finds more factors than P1 method. 
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