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The best result concerning bounded gaps between primes, whose existence was first proved by the seminal work of Yitang Zhang two years ago, are to my knowledge all of the form $\liminf_{n \rightarrow \infty} \left(p_{n+1} - p_n\right) \leq C$, where the record is currently $C = 246$, according to this source: http://michaelnielsen.org/polymath1/index.php?title=Bounded_gaps_between_primes . These results imply that for at least one value of $k$, $k\leq 123$, that there are infinitely many primes separated by exactly $2k$.

Suppose we can prove that for some positive integer $k$, that there are infinitely many consecutive primes $p_n, p_{n+1}$ such that $p_{n+1} - p_n = 2k$. Can we prove that there are infinitely many consecutive primes separated by exactly $2k+2, 2k+4, \cdots$ etc.? That is, does the existence of infinitely many twin primes for example imply that all even gaps must occur infinitely often? Certainly, whatever technique was used to prove the existence of infinitely many two primes should allow us to just as easily prove it for any even gap, but my question is a bit more specific: namely, can one directly prove the existence of infinitely many primes separated by $2k$ knowing only that there are infinitely many primes separated by $\max\{2, 2k-2\}$?

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The current technology works as follows. For any admissible finite set $\mathcal{H}$ of size at least $50$, one of the differences that occur within $\mathcal{H}$ also occurs as a difference of two (not necessarily consecutive) primes infinitely often. One can draw many interesting conclusions from this principle, but what you are looking for seems to be beyond this technology.

In particular, the set of numbers that occur as a difference of two (not necessarily consecutive) primes infinitely often has lower asymptotic density exceeding $\frac{1}{354}$. This is the current record in the sense that, to my knowledge, this lower bound has not been improved to $\frac{1}{353}$. The actual fraction here is not that important, but if we could prove what you are looking for, then we knew that the density is precisely $\frac{1}{2}$, because in that case all but finitely many even numbers would belong to the set of differences in question. Even assuming the generalized Elliott-Halberstam conjecture, we have no proof so far that the above lower asymptotic density is $\frac{1}{2}$.

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    $\begingroup$ Under general EH we can replace 50 with 3 and $1/354$ with $1/6$, if I remember correctly. The obstruction would be to disprove that all gaps that occur infinitely often are multiples of $6$. $\endgroup$ – Will Sawin Jun 13 '15 at 18:10
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    $\begingroup$ In fact under this method the opposite is true - if we know that certain short gaps go not appear, we can prove that more long gaps appear. $\endgroup$ – Will Sawin Jun 13 '15 at 18:12
  • $\begingroup$ @WillSawin: Regarding the density, you are right. For $k$ in place of $50$ we have the lower bound $\geq\frac{1}{k-1}\prod_{p\leq k}\left(1-\frac{1}{p}\right)$ for the lower asymptotic density. See arxiv.org/abs/1410.8198 $\endgroup$ – GH from MO Jun 13 '15 at 18:27

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