Ofcourse all these abs[(-2)^n-1] numbers where n >=3 n odd and n an integer will be composite and divisible by +/-3.(-2-1). So divide out the common 3 factor and one does get primes. I found this out using a program written in Mathematica 3.0.1. Regards, Allan Menezes P.S. I do not know whether the resulting composite numbers are all square free.
Hi, I was wondering if we could be looser in the definition of mersenne numbers allowing it also to be defined as (-2)^n-1 for n >=2 and n a natural number and n odd. For example using +/-2 and n=3 yields 2^3-1=7 a mersenne prime. abs[(-2)^3-1]=9=3^2 which would make the alternative definition of mersenne numbers (looser definition ) not square free! What does this suggest to you? Regards, Allan Menezes
