Talk:Brun's theorem

Proposed merger

The following discussion is closed. Please do not modify it. Subsequent comments should be made in a new section. A summary of the conclusions reached follows.

Agreement for merge. -- P 1 9 9 • TALK 18:01, 25 March 2011 (UTC)[reply]

There's a significant overlap between Brun's theorem and Brun's constant. Richard Pinch (talk) 20:46, 4 August 2008 (UTC)[reply]

Merge After you have read Brun's theorem, Brun's constant boils down to a sentence or two: Brun's constant is hard to estimate, and some people have calculated a lower bound of NNNNNNNNNNN. --Uncia (talk) 21:24, 4 August 2008 (UTC)[reply]

merge seems an easy call. Jason Quinn (talk) 20:13, 13 February 2011 (UTC)[reply]

I think it is ok to merge, but please make a redirect from constant to this article. —Preceding unsigned comment added by 24.245.20.61 (talk) 04:02, 28 January 2010 (UTC)[reply]

The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

Proof

Can someone add a basic description of the proof, or a link to the proof? It's the first thing I was looking for, but I can't find it anywhere that isn't behind a paywall. — Preceding unsigned comment added by 24.165.127.173 (talk) 00:37, 6 July 2011 (UTC)[reply]

pi_2(x)

This function is not defined in the article. 137.226.198.75 (talk) 09:29, 16 August 2011 (UTC)[reply]

Yes, it is.

Let $\pi _{2}(x)$ denote the number of primes p ≤ x for which p + 2 is also prime ....

— Arthur Rubin (talk) 02:02, 17 August 2011 (UTC)[reply]

Rationality

Actually, if the sum of the reciprocals is rational, couldn't there still be an infinite number of twin primes, as there are an infinite number of powers of two, even though the sum of their reciprocals is rational, namely 1 or two, depending on if you count 2⁰. But if the sum is irrational, then certainly there must be an infinite number of twin primes, as you cannot add together a finite number of rational numbers and arrive at an irrational number (Q is closed under addition). Captain Gamma (talk) 00:39, 13 October 2012 (UTC)[reply]

Good point. I have asked the respected editor who added it. PrimeHunter (talk) 01:32, 13 October 2012 (UTC)[reply]

Ok, yes, I agree. If there are infinitely many, then there is no reason to expect their sum to converge to an irrational number. —David Eppstein (talk) 03:23, 13 October 2012 (UTC)[reply]

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