# Lower bound for the 'de Polignac constant'

Let's introduce the 'de Polignac constant' $K_{Pol} : =\sum_{i>0}2^{-k_{i}}$ , where $k_{i}$ is the $i$ -th positive integer such that $2k_{i}$ is a Polignac number, i.e a number that is the difference of two consecutive primes in infinitely many ways. De Polignac's conjecture is equivalent to $K_{Pol}=1$ .

Do we know a non trivial lower bound for $K_{Pol}$?

Edit : a proof that $K_{Pol}>1/2$ would entail the truth of the twin prime conjecture.

• It should be "Polignac constant", without "de", for the same reason you don't say "de Fermat's theorem". – YCor May 24 '17 at 14:51
• This would be a good issue to raise on an English language Q&A website ! In French, even though everyone says "dernier théorème de Fermat", some people insist on the fact that one should say "conjecture de de Polignac" and not "conjecture de Polignac", hence my phrasing. – Sylvain Julien May 24 '17 at 15:22
• It already exists: en.wikipedia.org/wiki/French_name#Particles (another useful keyword: onomastic). The people you mention ignore the standard rules for particles. "conjecture de de Polignac" is doubly false: first because the particle "de" should be dropped in this case, and in case rules impose it after French preposition "de", the particle takes a capital: thus we write "La cohomologie de De Rham fut introduite par de Rham". Of course this latter subtlety concerns only French writing! – YCor May 24 '17 at 16:11

We can give much better estimates on $K_{Pol}$ than ones mentioned in Charles answer. All the results used are already proven in the same paper by Polymath, but instead of the results cited in the introduction, we can use stronger results encapsulated in Theorem 3.2, the relevant parts of which I will quote for completeness.

Definition Denote by $DHL[k,j]$ the statement "For any admissible $k$-tuple $\mathcal H=(h_1,\dots,h_k)$ there exist infinitely many translates $n+\mathcal H=(n+h_1,\dots,n+h_k)$ of $\mathcal H$ which contain at least $j$ primes".

Theorem 3.2 The following hold:

• $DHL[50,2]$, unconditionally.
• $DHL[3,2]$, under the assumption of GEH conjecture.

Also, this result isn't listed in the theorem, but on page 46 it is mentioned that EH implies $DHL[6,2]$, as proven by Goldston, Pintz and Yıldırım. I won't be doing full calculations in all cases, but I will prove an estimate using the result under GEH.

Claim Under GEH, either $2$ is a Polignac number, or, for every $n\geq 1$, either $6n$ or $6n-2$ is a Polignac number.

Proof The triple $(0,2,6n)$ is rather clearly admissible, so under GEH infinitely many translates of this triple contain two primes. If $2$ is not Polignac, then for each $n$, such pairs of primes must have difference $6n-2$ or $6n$. One of these must appear infinitely often.

Therefore either $K_{Pol}$ is at least $2^{-1}$, or at least $\sum_{n=1}^{\infty}2^{-3n}=\frac{1}{7}$. In either case, $K_{Pol}\geq\frac{1}{7}$. I don't think you can improve this bound, since every admissible triple will have a difference divisible by $3$, so it appears to be consistent that Polignac numbers are precisely the ones divisible by $6$.

Thanks to the Polymath project we know that infinitely often $p_{n+1}-p_n \le 246$, which means that some gap {2, 4, 6, ..., 244, 246} must occur infinitely often. In the worst case, only 246 would occur infinitely often, contributing 2^-246 to the sum. From Theorem 1.4 we get more precisely:

$$K_{Pol} \ge 2^{-246} + 2^{-398130} + 2^{-24797814} + 2^{-1431556072} + 2^{-80550202480} + \sum_{m=6}^\infty 2^{-Cm\exp((4-28/157)m)}$$ for "an absolute (and effective) constant $C$".

For practical purposes the first term gives a sufficient approximation: $$K_{Pol} \ge 8.8434366\times10^{-75}.$$

On the Elliott-Halberstam conjecture the bound improves to $$K_{Pol} \ge^? 2^{-12} + 2^{-270} + 2^{-52116} + 2^{-474266} + 2^{-4137854} + \sum_{m=6}^\infty 2^{-Cme^{2m}} \approx 0.000244140625$$

and on a generalized EH it is further improved to $$K_{Pol} \ge^? 2^{-6} + 2^{-252} + 2^{-52116} + 2^{-474266} + 2^{-4137854} + \sum_{m=6}^\infty 2^{-Cme^{2m}} \approx 0.015625.$$

 D. H. J. Polymath, Variants of the Selberg sieve, and bounded intervals containing many primes

• Thank you but I don't understand why it's not $2^{-123}$ instead of $2^{-246}$ since I take half the prime gap as the exponent of $2$ . – Sylvain Julien Mar 13 '17 at 17:30
• @SylvainJulien Ah, I didn't see that, you can halve all the exponents. – Charles Mar 13 '17 at 17:49