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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/

Links to various references: http://michaelnielsen.org/polymath1/index.php?title=Bounded_gaps_between_primes

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

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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
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Perhaps that's a conversation to be had on meta? –  HJRW Jun 3 '13 at 13:12
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I have an impression that this topic is more suitable for Polymath project: polymathprojects.org –  Nurdin Takenov Jun 3 '13 at 13:19
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@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
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1 Answer 1

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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Maybe we should try to have a reasonably good chronology of improvements in this or another answer? –  François G. Dorais Jun 4 '13 at 22:15
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There is a good chronology at michaelnielsen.org/polymath1/… –  Scott Morrison Jun 5 '13 at 4:52
    
Here is a nice brief description by Gergely Harcos of the skeleton of the argument. –  Andres Caicedo Oct 11 '13 at 16:12
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