## Original approach

A set of integers $H$ is called *admissible* if it avoids at least one residue class modulo $p$ for each prime $p$. In other words
$$
\forall p \in \mathcal{P} :\text{cardinality of} \, \lbrace x \bmod p \, | \, x \in H \rbrace \leq p-1.
$$

Let $Q(k_0)$ denote the assertion that for any admissible set $H$ of cardinality $k_0$ there are infinitely many translates $n+H$ that contain at least two primes. The bound on the gap is then $\mathrm{diam}\, H$.

Zhang deduces his bound from the following result:

**T1:** $Q(3,500,000)$ is true

In Zhang's paper the length $k_0$ is determined by the following inequality (1) that has to hold for some natural number $l_0$

$$
(1+4\varpi) (1-\kappa_2) > \left(1 + \frac{1}{2l_0+1}\right) \left(1 + \frac{2l_0+1}{k_0}\right) (1+\kappa_1),
$$
where

$$
\kappa_1 = \delta_1 \left( 1 + \delta_2^2 + k_0 \log\Bigl(1+\frac{1}{4\varpi} \Bigr) \right) \binom{k_0+2l_0}{k_0}
$$

$$
\kappa_2 = \delta_1 (1+4\varpi) \left(1 +\delta_2^2 + k_0 \log\Bigl(1+\frac{1}{4\varpi} \Bigr) \right) \binom{k_0+2l_0+1}{k_0-1}
$$

$$
\varpi = 1/1168
$$

and

$$
\delta_1 = (1+1/4\varpi)^{-k_0}
$$

$$
\delta_2 = \sum_{j=0}^{1/4\varpi} \frac{k_0\log(1+\frac{1}{4\varpi}))^j}{j!}.
$$

The admissible set that Zhang uses is $H = \{ p_{k_0+1}, \ldots, p_{2k_0}\}.$

## Current record

Terence Tao & Scott Morrison: 4,802,222

Terence Tao established another inequality on $k_0$ that manages to remove most of inefficiency of Zhang estimate.
$$
1+4\varpi > \left(1 + \frac{1}{2l_0+1}\right) \left(1 + \frac{2l_0+1}{k_0}\right) (1+\kappa)
$$
where
$$
\kappa := \sum_{1 \leq n < 2 + \frac{1}{2\varpi}} \Bigl(1 - \frac{2n \varpi}{1 + 4\varpi}\Bigr)^{k_0/2 + l_0} \prod_{j=1}^{n} \left(1 + 3k_0 \log\Bigl(1+\frac{1}{j}\Bigr)\right).
$$
Moreover $l_0$ is allowed to be real number. Scott Morrison then found that for $l_0 = 291.7$ one gets $k_0 = 341,640$ which is the best possible $k_0$ for given $\varpi = 1/1168$.

Paper by Richards suggest to take as admissible set $H_m = \{ \pm 1, \pm p_{m+1}, \ldots, \pm p_{m+k_0/2+1} \}$ for $m$ large enough. This leads to bound
$$
2p_{m+\lceil k_0/2 \rceil + 1} \quad \text{for } k_0 \text{ even}
$$
and
$$
p_{m+\lfloor{k_0/2}\rfloor-1} + p_{m+\lfloor{(k_0+1)/2}\rfloor-1} \text{ for } k_0 \text{ odd.}
$$
For given $k_0=341,640$ program written by Scott Morrison found that $m=5553$ gives the smallest bound of $4,802,222$.

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