Quote:
Originally Posted by ppo
if S is the generic term of the LL sequence and M is the Mersenne number to be tested, when S is bigger than 2^(n1)

First, I'll assume you mean that the Mersenne number being tested is M = 2^n1 (so n is the exponent of 2 for that number), and that by "S" you mean one of the S(i) terms
mod M.
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when S is bigger than 2^(n1) it is possible to replace it by MS, so reducing the size of the numer to be squared.

But this reduction in size is trivial. When S(i) is just below 2^n1, it's only about twice as big as 2^(n1)  i.e., one bit longer. Squaring a 12,016,057bit number is no slower than squaring a 12,016,056bit number. (The speed jumps come at FFT size boundaries.)
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Can this result in speedingup the test, or it is something already considered?

1) no (well, it could if you could use a smaller FFT by reducing one bit, but that would happen too rarely to be worth the time needed to program the change), and 2) yes.