On Aug 10, 2004, at 7:08 AM, David Gale wrote:
"Are there infinitely many Mersenne primes? The answer is probably[ital] yes (because the harmonic series diverges). Could someone please explain the "because"?
I think this is the kind of sloppy heuristic I'm guilty of posting here on many occasions. The actual number of Mersenne primes is the sum, over all primes p, of the function "1 if 2^p-1 is prime, 0 otherwise". But if we replace this by the function "the probability that a number around 2^p-1 is prime", then we can hope that we get about the same sum -- assuming the "2^p-1"-ness of the numbers is in some sense independent of their chances of being prime in the first place. The latter sum is basically \sum 1/p, which diverges "because" the harmonic series does, by the usual argument. (And so we can weaken the condition in the previous paragraph: we expect the number of Mersenne primes to be infinite as long as the "2^p-1"-ness of a number does not decrease its chance of being prime compared to other numbers of the same magnitude.)
Does the same argument work for Fermat primes? for primes of the form 10^n + 3? for (a^ n)+b, gcd[a,b]=1?
Sure -- except for the part about it being an argument in the first place! --Michael Kleber