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In the paper of Green and Tao "Restriction Theory of the Selberg Sieve, with applications," their theorem 6.1 states: Let $N$ be a large integer. Then the number of Chen primes in the interval $(N/2,N)$ is at least $c_1N/\ln^2N$, for some absolute constant $c_1>0$.

My question is, what the heck is $c_1$? Is it Brun's constant, or is that just wishful thinking?

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    $\begingroup$ For anyone who wants to see for herself, the paper is at emis.de/journals/JTNB/2006-1/article09.pdf [new paragraph] It looks to me like $c_1$ is exactly what it says it is; some absolute constant. I don't see any reason to expect it to be related to any other particular constant. Also note that "their" Theorem 6.1 is a quotation from some notes of Iwaniec. [new paragraph] Unless you have something to add, I'm leaning toward a vote to close. $\endgroup$ Dec 1, 2010 at 8:26
  • $\begingroup$ Sounds good. Just wanted to know if anyone knew of any work done that set some bounds on this constant, or approximated it in any way. I know their paper doesn't, but that's simply because their proof of 6.1 does not require any. $\endgroup$ Dec 1, 2010 at 23:26

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Although I don't have the reference convenient, I believe that the last chapter of Halberstam and Richert's book Sieve Methods states (and proves) Chen's theorem with an explicit value of c_1.

As I recall, it is roughly 3/11 times the "expected" constant from probabilistic arguments.

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  • $\begingroup$ Okay, that paper is good, but I'm not smart enough to figure out how the lower bound of |p:p≤N,N−p=P2| Where p is prime, and N a large enough even integer gives you the lower bound of the number of Chen primes in (N/2,N]. Any thoughts? $\endgroup$ Dec 11, 2010 at 1:46
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Okay, that paper is good, but I'm not smart enough to figure out how the lower bound of

$\vert p: p\leq N , N-p=P_2\vert$

Where $p$ is prime, and N a large enough even integer gives you the lower bound of the number of Chen primes in $(N/2,N]$. Any thoughts?

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  • $\begingroup$ Regarding "I'm not smart enough" -- Chen's paper was a brilliant and unexpected achievement; if you want to understand the proof there is probably no shortcut to mastering the entire contents of H+R. $\endgroup$ Dec 6, 2010 at 20:26

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