Mersenne-Like
 

Let rho(n) = number of prime factors of n counting multiplicities.

Then rho(mn) = rho(m)+rho(n) [trivial]
and rho(2^n-1) >= rho(n) [easy].

Besides the Mersenne series primes, for which rho(2^p-1) = rho(p) = 1,
what other n satisfy rho(2^n-1) = rho(n) ?

So far, other than Mersenne exponents, only prime powers:
1, 4, 8, 9, 16, 27, 32, 49, ...

Doesn't work for higher terms. Fails for 11^2 and 13^2.

Though it appears that empirically this happens for Mersenne prime exponents and a [I]subset[/I] of the powers of primes. Brute forcing in Maple, and so far nothing but Mersenne prime exponents and the powers of primes you listed, up to 256.

[QUOTE=Kevin;160753]Doesn't work for higher terms. Fails for 11^2 and 13^2.[/QUOTE]

Sorry, I didn't mean that it worked for all prime powers, just that I haven't discovered any term that worked other than a prime power. I already found that it failed for those two, having tested up to 200 or 500 or something like that.

Sine 2^ab-1 is divisible by 2^a-1 and 2^b-1,
and since 2^p-1 has at least 2 factors if prime p is not Mersenne,
only n=p^m for p=Mersenne prime exponent (not just any prime)
need be considered. Also, only the smallest exponents m need
be considered, because if m fails so does m+1.

Thus in addition to the solutions given, the squares of the Mersenne
prime exponents (e.g. 19^2, 31^2, etc,) should be checked, and only
if one satisfies the criterion does then the cube need to be checked, etc.
This should speed up the search.

Isn't all the Gimps traffic fascinating?

A world-wide interaction of people and computers
all working to answer some very interesting mathematical questions.

So answer: are there an infinite number of Mersenne primes?

For the answer to that question, see the YJ-Conjecture in Conjectures R Us
and the Wagstaff Conjecture in Math.

The answer is: yes, there are, and it either is or will be proven soon.