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Old 2020-10-07, 12:49   #1
Nov 2014

3·13 Posts
Default P-1 factor probability distribution given n, B1 and B2 bounds

I want to calculate how likely a potential factors in a given bitrange will be missed by P-1, given B1, B2 and n. For ECM, I can get this information from here. It can't be the same as ECM for P-1, because P-1 will use the fact that the factor is of form 2kn+1.

Please read about the first, I will use the same notations.
\Psi(x,B1) should be the number of B1-smooth numbers below x, similiar for \Psi(x,B1,B2) .
Assume n is prime and large, too.

So I need to calculate p(n,B1,B2,k_{max}) = \frac{\left\{k\in\mathbb{N}|k<k_{max}\text{ and }2nk+1\in\mathbb{P}\text{ and }k\text{ is B1/B2-powersmooth}\right\}}{\left\{k\in\mathbb{N}|k<k_{max}\text{ and }2nk+1\in\mathbb{P}\right\}}.

The denominator should be (if Chebyshev bias is not considered) \pi(2nk_{max}+1) divided by \varphi(2n)=n-1 or around \frac{(2nk_{max}+1)}{(n-1)log(2nk_{max}+1)} or around \frac{(2k_{max}+1)}{log(2nk_{max}+1)}.

However, I can't come up with a method to count the elements in the nominator, not even if I consider stage1 only. By definition there are \Psi(k_{max},B1,B2) potential values for k where k will be power-smooth with B1/B2 bound, but I don't know how to count for how many of those 2kp+1 will be prime, too. Any ideas? And if I consider Brent-Suyama extension, then I am completly lost...

PS: Reading the source code of Prime95, it seems that it the P-1 factor probability for a mersenne number is the same as the P-1 factor probability for a random number with n bits less. However, I do not yet see why this is.

Last fiddled with by gLauss on 2020-10-07 at 12:57
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Old 2020-10-07, 20:53   #2
preda's Avatar
"Mihai Preda"
Apr 2015

53·11 Posts

Another P-1 calculator is here:

Oh, I think the two are related or similar, so no new insights.

Last fiddled with by preda on 2020-10-07 at 20:55
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