So where does this leave us? Does the same kind of (non)-argument suggest that there are only finitely many Fermat primes? (Maybe Fermat knew them all, since no new ones have been discovered since.*) What should we expect about Mr. In Between, the number of n!+1 primes? Sorry, I didn't understand the Kleber non-argument. Presumably at some point it must have used the fact that the Mersennes grow at most exponentially. Where? I don't think we should scoff too much at this kind of (non)reasoning. Polya had a wonderful paper in the monthly in 1959 called "Heuristic Reasoning in the Theory of Numbers" in which suggested that not only are their infinitely many prime twins but the should have a well defined assymptotic density. * weird; an indepenent clause that begins and ends with "since", first as a conjunction, then as an adverb. At 08:08 AM 8/10/2004 -0600, you wrote:
Doesn't work for Fermat primes, 'cos they grow super-exponentially, and the series is like sum 1/2^n which is finite. R.
On Tue, 10 Aug 2004, Michael Kleber wrote:
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
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