François Perruchaud wrote:
An old book of mine gives without proof an example of Fibonacci Sequence
that countains no primes, but where U(1) and U(2) are co-prime.
The sequence was found by R. L. Graham.
Reference :
"A Fibonacci-like sequence of composite numbers",
R.L. Graham, Math. Mag. 37,
why is f(0) in an ll test = 4
The value of f(0) must be such that f(0)-2 is a quadratic residue mod Mp
and f(0)+2 is a quadratic nonresidue mod Mp. 4 is the smallest value
which works for all p; 10 is the next, followed by 52. You could use
f(0) = 3 provided that 5 is a quadratic nonresidue mod
The behavior of the recursion x[n+1] = x[n]^2 - 2 can be precisely
analyzed. (In fact, it is because of this that the LL-test works at
all.) For a fixed prime p, the periodicity depends on the factorization
of either p-1 (if x[1]^2-4 is a quadratic residue mod p) or p+1 (if
x[1]^2-4 is a
Peter-Lawrence.Montgomery wrote:
Problem A3 in Richard Guy's `Unsolved Problems in Number Theory'
includes this question, by D.H. Lehmer:
Let Mp = 2^p - 1 be a Mersenne prime, where p 2.
Denote S[1] = 4 and S[k+1] = S[k]^2 - 2 for k = 1.
Then S[p-2] == +-
From: "Olivier Langlois" [EMAIL PROTECTED]
Date: Wed, 6 Jan 1999 22:31:24 -0500
Subject: Mersenne: Questions on Crandall algorithm
Hi,
I'm trying to understand Crandall's algorithm but I'm not sure to get the
idea behind it. So let me explain you what I've understand and I would
The usual concept of an integer FFT is to choose a prime q such that q-1
is divisible by a reasonably large power of 2, say N = 2^n, find a
primitive (integer) N-th root of 1 mod q, say w, then use w and
arithmetic mod q to calculate the FFT. If it also happens that there is
an integer N-th root
