Cramer’s conjecture implies twin prime conjecture

Let’s define the quantity G(x):=\sup\{p_{n+1}-p_{n}\mid \frac{p_{n}+p_{n+1}}{2}\le x\}, and assume Cramer’s conjecture. Then K_{m}:=\sup\{\frac{G(x)}{\log^2 x}\mid x>2\}<\infty. Let’s denote by U(n) the event n is 1-central, and by H(n) the event (n-1,n+1) is a couple of twin primes. Then from de Bayes’ theorem, one gets P(H(n))=\frac{P(H(n)\cap U(n))}{P(U(n)/H(n))}=P(H(n)\cap U(n)), since if n is the half sum of a couple of twin primes, then necessarily n is 1-central. But P(H(n)\cap U(n)) is equal to the ratio of the number of quantities 2r_{0}(n) such that n is the half sum of twin primes to the number G(n) as defined above, hence P(H(n))\ge\frac{1}{K_{m}\log^{2} n}. As P(H(n))=\frac{1}{2}\frac{\pi_{2}(n)}{n}, one finally gets \pi_{2}(x)\ge\frac{2x}{K_{m}\log^{2} x}. Hence the assumption of Cramer’s conjecture implies that there are infinitely many twin primes.

 

 

 

 

 

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3 réflexions sur “Cramer’s conjecture implies twin prime conjecture

  1. Could you clarify what you mean by « the number of quantities 2r_{0}(n) such that n is the half sum of twin primes »? Also, how do you deduce P(H(n))\ge\frac{1}{K_{m}\log^{2} n}?

  2. A major issue in here is treating H(n) and U(n) like random events. If you _really_ mean that H(n) is the event that (n-1,n+1) is a twin prime pair, then P(H(n)) is always 0 or 1 – this is a deterministic event and there is no probability involved here. If you want to make it valid, then you have to consider H(n) to be the event « a random number below n is the middle of two twin primes ». Even if this were the case, isn’t 2r_0(n) always 2 for n lying between twin primes, so that r from your previous comment must be equal to 1? That being said, I have no clue why you claim that P(H(n) cap U(n)) = r/G(n).

    Indeed, I am convinced that whatever you are trying to prove is wrong. You aren’t using any intrinistic properties of primes here, so the argument should work more generally. But if you consider « paraprimes » to be precisely the numbers 1 mod 4, then K_m = 0 < infinity, so the same argument should show there are infinitely many "paraprimes". But there are none at all!

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