Quote:
Originally Posted by Uncwilly
Is there enough data yet to suggest that Mersenne nonprimes have a higher likelihood to have 3 only factors vs 2 only?

Exactly three versus
exactly two prime factors, not sure. Exactly two versus more than two, longknown data suggest "no contest."
I'll assume, for the sake of discussion, that you mean Mersenne numbers M
_{p} with
prime exponent p. For composite exponents n, the algebraic factorization alone is very likely to provide more than two proper factors, which, in turn, are very likely to provide more than two prime factors.
From an old factorization table of 2
^{n}  1 for odd n < 1200 I have on file, a quick scan finds that M
_{p} was known to be prime for 13 odd prime p, known to have exactly two prime factors for 38 prime p, and known to have 3 or more prime factors for 143 odd prime p. The table lists M
_{1061} as a C320 so the number of factors was only known to be 2 or more at the time.
It seems that 2
^{1061}  1 was subsequently found to be the product of two prime factors, so M
_{p} is known to be prime for 13 odd prime p < 1200, known to have exactly two prime factors for 39 odd primes p < 1200, and known to have 3 or more prime factors for 143 odd prime p < 1200. That accounts for all 195 odd primes p < 1200.
So I would say that the data suggest that more than two factors is much more likely than exactly two.
There are only about 400 primes p for which the
exact number of factors of 2
^{p}  1 is known or "probably" known. I'm too lazy to check for how many p the number of factors is known to be
at least 3.