There are a couple of "fine points" to be mentioned here. There are odd numbers K such that 2^N+K is composite for all large N. I think the smallest such K is about 75000. The proof is elementary: Split N = 24 M + R, 0<=R<24, and check that each R leads to a small factor. [I may be wrong about 24, it could be 12 or 16.] 2^N+1 is special because N cannot have odd divisors > 1, and must itself be a power of 2. 2^N+3 and its siblings will need a correction factor relative to typical numbers of the same size: Never divisible by 2 or 3, 1/4 chance of 5-factor, 2/6 of 7-factor, 1/10 of 11-factor, 1/12 of 13-f, never 17, etc. We assume that the product of the correction factors is convergent, making only a modest change to the expected number of primes. 2^N+9 is either the sum of two squares or A^2+2B^2; this excludes divisors 8K+7. 2^N-9 must have N odd to be prime, excluding divisors 8K+-3. 2^P-1 has additional complications: No divisor < 2P+1; divisors must be 2kP+1 and also 8K+-1. On the other hand, divisors of these forms seem to have matching increases in likelihood. I don't know of any adjustments for N!+1 except divisor>N, but would be happy to learn. Rich -----Original Message----- From: David Gale To: math-fun@mailman.xmission.com Sent: 8/10/2004 5:08 AM Subject: [math-fun] Infinitely many primes, maybe In the excellent article on Mersenne primes http://www.utm.edu/research/primes/mersenne/ <http://www.utm.edu/research/primes/mersenne/> I found the following, to me, cryptic passage "Are there infinitely many Mersenne primes? The answer is probably[ital] yes (because the harmonic series diverges). Could someone please explain the "because"? Does the same argument work for Fermat primes? for primes of the form 10^n + 3? for (a^ n)+b, gcd[a,b]=1?