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I'll start by noting that I am not at all an expert on number theory. However I do use it in a proof and would like your assistance if possible.

Yitang Zhang breakthrough result established that there is a number $k<7\times10^7$ for which there are infinite pairs of primes $(p,p+k)$. This $k$ was later improved. My question is: Can I find a bound $X$ and a constant $k$ such that for a given number $n$, I may find n disjoint pairs of prime numbers $(p_1,p_2),(p_3,p_4),\dots,(p_{2n-1},p_{2n})$, where:

$$ p_1 < p_2 <\dots< p_{2n-1} < p_{2n} < X \\\text{and}\\ p_{2i} \leq p_{2i-1}+k, i=1,\dots,n $$

That is, a set of $n$ prime pairs with a gap less or equal to some constant k between the element of each pair and all primes are not greater than $X$.

I would like to find $X$ as a function of $n$ (hopefully bounded by a polynomial of $n$).

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marked as duplicate by Eric Naslund, Boris Bukh, Marco Golla, Lucia, Yoav Kallus Jul 28 '15 at 18:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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For any positive integer $m$, James Maynard and Terry Tao independently proved that there exists a positive number $h(m)$ (in Maynard's original paper, one can take $h(m) = m^3 e^{4m}$) such that there are infinitely many tuples of consecutive primes $p_n, \cdots, p_{n+m}$ such that $p_{n+m} - p_n = O(h(m))$, where the implied constant is absolute.

Applying this theorem to your question, we set $m = 2n$ and $k = \lceil O(h(2n)) \rceil$, with the same implied constant above. Then according to the Maynard-Tao theorem, there exist infinitely many tuples of primes $(q_1, \cdots, q_{2n})$ such that $q_{2n} - q_1 \leq k$, whence $q_{2j-1} \leq q_{2j} + k$ for $j = 1, \cdots, n$.

Looking at Maynard's original paper, it seems that $X$ is exponential in $n$ (namely, one can take $X = C n^3 e^{12n}$ for some positive number $C$).

Reference: http://annals.math.princeton.edu/2015/181-1/p07

http://arxiv.org/abs/1311.4600

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  • $\begingroup$ And maybe the OP would be glad to know that conjecturally, one can take $h(m)=m\log m$. $\endgroup$ – Sylvain JULIEN Jul 27 '15 at 16:43
  • $\begingroup$ Dear Stanley, Thank you for your answer. it is important for me to find a bound that is not exponential in n. I stumbled upon this answer which I think might be exactly what I need. However, not being a mathematician, it would help me a lot to get an approval for this bound. mathoverflow.net/questions/176875/… $\endgroup$ – Liron Yedidsion Jul 27 '15 at 18:49
  • $\begingroup$ Right now there is no polynomial guarantee that such a bound exists. The Hardy-Littlewood conjectures concerning prime k-tuples and associated calculations can tell you how many such are expected to appear (something like $Cx/(\log x)^{2n}$ for an effectively computable C ). If you are willing to take those conjectures into account, you may get what you want. Gerhard "And Hopefully What You Need" Paseman, 2015.07.27 $\endgroup$ – Gerhard Paseman Jul 27 '15 at 21:37
  • $\begingroup$ This is what I feared. It makes a proof that I am working on a bit weaker as it is based on a conjecture. However, I still don't understand how does that comply with the answer in this link mathoverflow.net/questions/176875/… by GH from MO. Can you clarify this for me please? What am I missing? $\endgroup$ – Liron Yedidsion Jul 28 '15 at 20:09
  • $\begingroup$ @LironYedidsion The answer given in your link is not what you're looking for I don't think. Zhang's proof can be strengthened (as can all such similar results) to obtain an asymptotic lower bound for the number of prime pairs that are say at most 70 million apart. This bound is of course at most $x$, since $x$ is the length of the interval. What you are looking for is a polynomial bound for the parameter $h(m)$ which guarantees that there are infinitely many $m$-tuples of primes which are $h(m)$ apart, which is not at all the same thing. $\endgroup$ – Stanley Yao Xiao Jul 28 '15 at 21:52

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