| I thought about this for a second, and it seems that you are right: if the primes were distributed randomly, the twim prime conjecture would be true (see below). Are there any analytic number theorists in the audience to explain what the quote in the parent post is supposed to mean and/or fix my mistakes? The prime number theorem [1] states that the probability that a natural number N is prime is approximately 1/log(N). So, the probability that N & N+2 are both prime is about 1/log(N)^2. If they were distributed randomly, there would be about integral(1/log(x)^2 dx) from x=2 to x=N twin primes between 1 and N. This integral converges to infinity as N grows large[2] (wish I remembered enough caclulus to do it by hand). So, there is some N for which you'd expect to have 10^100 twin primes less than N, etc. So it seems that if you decided by a coin toss whether a number should be prime (where you'd make it prime with probability 1/log(N)), there would be infinitely many twin primes. Wikipedia seems to confirm it's a reasonable estimate [3]. UPDATE: By the way, the same argument implies that if primeness was decided by a coin toss, there would be infinitely many N such that N and N+1 are prime. There is, however, only one such N (namely 2). So, primeness is not decided by a coin toss, and the above is not a proof of the twin primes conjecture :). [1]: http://en.wikipedia.org/wiki/Prime_number_theorem [2]: http://www.wolframalpha.com/input/?i=Integrate[1%2FLog[x]^2%...] [3]: http://en.wikipedia.org/wiki/Twin_prime#First_Hardy.E2.80.93... |
I a cousin comment, Retric proves that the integral of 1/log(x)^2 diverges. The point is that log(n)^2 < n for x > 2 (you can prove this by calculus: the only maximum of the function log(x)^2/x is at x=e). So, 1/log(x)^2 > 1/x, and the integral of 1/x diverges.
The most fun way to prove the last statement is that the integral of 1/x from 1 to infinity is greater than 1/2 + 1/3 + 1/4 + 1/5 + ... > 1/2 + 1/4 + 1/4 + 1/8 + ... = 1/2 + 2 * 1/4 + 4 * 1/8 + ... = 1/2 + 1/2 + 1/2 + ....