----- Original Message -----
From: "Alexander Kruppa" <[EMAIL PROTECTED]>
To: "Daran" <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: Saturday, December 15, 2001 2:05 AM
Subject: Re: [Fwd: Mersenne: P-1 Stage 2]

> > However, there's another generalisation that occurs to me that would be
> > quite cheap to implement.  The purpose of stage two is to catch those
> > factors which are 'just out of reach' of stage 1.  However another way a
> > factor could be 'just out of reach' would be if the power of exactly one
of
> > the primes < B1 was exactly 1 more than stage 1 would find.  Stage 2
would
> > find these factors *as well* if the algorithm were run for primes in
[2,B2]
> > instead of [B1,B2]

Actually that really ought to be [2,~B2/B1] and then [B1,B2].  Another power
on top of a prime in [~B2/B1,B1] will take it above B2.  Since B2 is
typically only an order of magnitude greater than B1 this is not going to
make a lot of difference.

> > Has this been investigated?
>
> Good point. The implementation I described finds a factor f iff the
> factorization of ord(a) mod f has only primes and prime powers <= B1 and
> at most one prime <= B2.
> It would be more reasonable if it could have primes and prime powers <=
> B1 and at most one prime _or prime power_ <= B2. That prime power has,
> after all, a probability >=1/B2 of dividing a random integer.

This is a generalisation of the above.

> However, that is awful to implement. In stage 1, you exponentiate each
> prime p by floor(log_p(B1)). In stage 2, you'd have to include small
> primes p <= sqrt(B2), exponentiated by floor(log_p(B2)) -
> floor(log_p(B1)).

Except that for primes above ~B2/B1, this will be zero, so we could ignore
them.

[...]

> If you want to make really sure that all powers of primes <B2 are
> included, you can modify stage 1 to do so, i.e. exponentiate each prime
> p<B1 by floor(log_p(B2)). This won't take a lot of extra time and is
> *much* easier to implement.

I agree.  The question is, which algorithm gives us the greatest factors to
work ratio?

> Alex

Daran


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