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 would mean that I can do ECM on, for example, P773 and M773 at the same time by
doing ECM on M1546, and it will take just as long as ECM on only P773, right? When I
find a factor, I'll just have see which number it divides.
Ciao,
Alex.
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