On Saturday 28 June 2003 18:47, you wrote: > Will the 64-bit residue be the SAME when a given exponent > was originally Lucas-Lehmer tested with a 384K FFT, but > the double-check is performed using a 448K FFT ?
Hopefully - in fact the whole 2^p-1 bit residue R(p) should be the same! R(2)=4 R(n+1) = R(n)^2 -2 modulo 2^p-1 I don't see how the FFT run length should affect this ... in fact the FFT is only used at all because it's the most efficient method known of doing the multiplication of very large numbers. If the residues calculated with 384K FFT & 448K FFT are different, then: - most likely, at least one of the runs has been affected by a "glitch" of some sort; - or, the 384K FFT run length is too small & some data value was rounded to the wrong integer during the run - I do not think that this is very likely unless serious abuse has been made of the SoftCrossoverAdjust parameter; - or, there is a systematic error in the program code for at least one of the run lengths. Since short runs (400 or 1000 iterations) have been crosschecked with various FFT run lengths, again I do not think this is very likely. Regards Brian Beesley _________________________________________________________________________ Unsubscribe & list info -- http://www.ndatech.com/mersenne/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers