# Tightening Zhang's bound [closed]

Inspired by a blogpost by Scott Morrison and ongoing discussion there I decided to create this community wiki to track progress on the original bound of Yitan Zhang.

The original bound was $70,000,000$. The accepted answer should contain latest known improvement.

As of 4.6. 2013 there is a polymath project devoted to improving this bound. The proposal can be found at http://polymathprojects.org/2013/06/04/polymath-proposal-bounded-gaps-between-primes/

A good place to start is to read notes by Terence Tao and his blog post on the topic.

## closed as off-topic by user9072, Willie Wong, Todd Trimble♦Oct 29 '14 at 13:08

• This question does not appear to be about research level mathematics within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

• I voted the question down because I think that the question/answer format of MathOverflow is not right medium for this kind of collaboration. – Boris Bukh Jun 3 '13 at 13:08
• Perhaps that's a conversation to be had on meta? – HJRW Jun 3 '13 at 13:12
• I have an impression that this topic is more suitable for Polymath project: polymathprojects.org – Nurdin Takenov Jun 3 '13 at 13:19
• – François G. Dorais Jun 3 '13 at 17:20
• @v08ltu, it looks like a polymath project is starting up, and this is exactly the sort of analysis needed. Could I suggest you post your comments either on a blog (if you have access to one), or at the polymath proposal blog polymathprojects.org/2013/06/04/…, or at one of the other blogs posts which is sure to appear soon? – Scott Morrison Jun 4 '13 at 4:43

## 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 $4,802,222$.