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2020-09-11, 13:22   #11
Dr Sardonicus

Feb 2017
Nowhere

416010 Posts

Quote:
 Originally Posted by Uncwilly Is there enough data yet to suggest that Mersenne non-primes 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, long-known data suggest "no contest."

I'll assume, for the sake of discussion, that you mean Mersenne numbers Mp 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 2n - 1 for odd n < 1200 I have on file, a quick scan finds that Mp 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 M1061 as a C320 so the number of factors was only known to be 2 or more at the time.

It seems that 21061 - 1 was subsequently found to be the product of two prime factors, so Mp 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 2p - 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.

Last fiddled with by Dr Sardonicus on 2020-09-11 at 13:30 Reason: Omit unnecessary words!