Just a quick note to thank Peter for his wonderful illustration of the
"repeating LL test remainders" question, with what I hope was a somewhat
surprising result!
> That is, S(32341) == S(1) mod M23.
> This is far more than 23 LL steps before the sequence repeats.
> EXERCISE: Repeat this analysis modulo M11 = 23 * 89.
> Find the period modulo M29 = 233 * 1103 * 2089, after
getting
> the individual periods for each prime factor.
*smiles* I will not pre-empt the result for anyone who wants to calculate
these. The important observation is to look at each factor in turn just as
Peter suggests. One can almost work in reverse as well. What we see in a
(Mersenne number) Lucas-Lehmer test really is a very small sample of the
sequence (2+sqrt(3))^n. If a cycle were detectable from this small sample,
then all the factors of the number under test must satisfy some very
stringent, perhaps intractable conditions. As Brian Beesley points out, I
was guilty of a lot of guesswork, and didn't go into details of the
calculation of the period of the L-sequence. Hopefully Peter's excellent
examples show that the chances of the L-sequence period being spotted in a
much briefer S-sequence are very small indeed.
Chris
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