Hi,
gmp-ecm version 5 can read ECM residues written by Prime95 to perform
stage 2 on them. Stage 2 is asymptotically faster in gmp-ecm, but
Prime95 has a much faster multiplication for large numbers, so this
feature is mostly interesting for relatively small numbers (i.e. of a
few
thousand bits) t
Alex Kruppa wrote:
>
> Hi,
>
> 2^n+1 numbers (compared to 2^n+1 numbers).
^
Ack! Typos, I _hate_ them! Must be 2^n-1 of course.
Ciao,
Alex.
_
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Hi,
Ernst Mayer once mentioned to me that Prime95 needs twice the FFT size for 2^n+1
numbers (compared to 2^n+1 numbers). Does that mean that George is using the identity
2^(2n)-1 = (2^n+1)(2^n-1) ? I was wondering why ECM on 2^n+1 numbers took much longer
than on 2^n-1 of the same size..
That
Hello all,
I have set up a web page with the current factoring status for Mersenne numbers with
prime exponents >= 33219281 and <3600, the ten megadigit mersenne prime candidates.
The URL is
http://www.informatik.tu-muenchen.de/~kruppa/M33M/index.html
The status information currently cons
Lucas Wiman wrote:
> P-1 factoring is based upon Fermat's little theorem which states that
>
> [...]
> What if we use 2 as a base? Well, for mersenne numbers, this might be to
The problem is that 2 is not a primitive root (mod Mp). From a computers
point of view, the mod Mp operation is a rota
Hi,
from stuff I saw on the list earlier I put together this estimate:
The chance for Mp being prime if it has no factor below 2^q is
2q/p (if q is reasonably big, at i.e. 2^q much bigger than 2p+1).
If we test the Mersennes in the 10M-digits range for factors
to, say, 2^80, the chance for, say
> 33219281 _is_ prime, the status of 2^33219281 is (so far as I know)
> not known at this time ... unless someone found a factor bigger than
> my 2^40 search limit.
I tried up to 46695341939693537 ~= 2^55, but no factor.
Ciao,
Alex.
___
Hi all,
I have a different question concerning P-1 and ECM.
Some time ago I asked which power to put small primes
into when multiplying them into E ( factor = gcd(a^E-1,N) ).
Paul Leyland, I believe, replied that the power for prime p should
be trunc( ln(B1) / ln(p) ) ( log(B1) with base p ),
wh
Alex Kruppa wrote:
> P.S. I have no idea what the factor 55 (mod 120) is doing there. I'll add
> something to my program to print out offending factors when I have time.
Now I know how the 55 (mod 120) got there, it's:
M( 310169 )C: 3486114749130405725455
and the problem is t
Hi all,
I wrote a little program to show the distribution of factors among
congruence classes, especially (mod 8) and (mod 120) (all others were
as uninteresting as you would expect). What surprised me was that the
distribution is not smooth at all, some congruence classes contain a
lot more fact
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