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 $n$).



Counting function for prime pair with bounded gaps between them
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>


