A week ago, George asked about probability in P-1 factoring.  I haven't
seen a single response, though I would guess that he has received many
private emails.  Could this please be included on the list?  I am interested
in this, as (I'm guessing) many others are.

> Extra credit:  Can someone tell me the probability that P-1 factoring
> will find an n-bit factor?

Shouldn't the probability of it finding a n-bit factor be immaterial,
as the fact that it finds a factor is all that matters.  The whole
n-bit concept should only matter if we speak of trial factoring.

> This is but one piece of the puzzle in
> determining the best trial factoring limits and P-1 bounds.  To compute
> this probability the k in the 2kp+1 factor must be "smooth", that is
> all factors must be less than bound #1 except for one factor that
> must be less than bound #2.  I'm sure I have the info around here somewhere,
> but everyone might be interested in the math behind deriving trial factoring
> limits and P-1 bounds.

The k must be smooth to B1, yes, but a more rigorous requirement is there
also.  What if k=2^70?  Smooth to any B1, but this number would not be found
by P-1 factoring.  I'm guessing that numbers that are smooth with high
exponents is low, so it should be about the same.

Here's what I was able to come up with while doing repetitive jobs at work:

A number p^x (p prime) is said to be "enough for" n if p^(x+1) does not
divide n.
Let f(p,a,b)="the sum from v=a to v=b of 1/p^v"
(note for those of you who don't know, f(p,0,b)=(1-(1/p)^n)/(1-1/p))

The probability that p^x is not enough for n is
p(p,x,n)=1-f(p,1,x)+f(p,x+1,[log_p(n)])

Thus the probability that value of Q=2^x_1*3^x_2...B1^x_(pi(B1)) will
be enough for n, for all p_i<B1 is
1-(prod(p(p_i,x_i,n), i=1, i=B1))
(I think this should be about 1-exp(-gamma)/log(B1)).

This should make the probability that P-1 will find a d bit factor
~(1-exp(-gamma)/log(B1))*.81*(1/d-1/(d+1))

-Lucas


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