k-central numbers and bounded gaps between primes

As in the previous articles, let’s assume Goldbach’s conjecture so as to define properly the notion of k-central number. The number of k-central numbers less than x \pi_{C,k}(x) should verify the following relation:

\pi_{C,k}(x)=\vert\{n\le x, k_{0}(n)=k\}\vert

and thus \pi_{C,k}(x)\le\dfrac{\pi(x+\max_{n\le x}r_{0}(n))}{k}(1+o(1)).

I now formulate the following conjecture:

Negligible fundamental primality radius conjecture (NFPR conjecture for short):

\forall\varepsilon>0,\forall x>2, \max_{n\le x}r_{0}(n)=O_{\varepsilon}(x^{\varepsilon})

One can deduce from this conjecture and the prime number theorem that \dfrac{\pi(x+\max_{n\le x}r_{0}(n))}{k}\sim\dfrac{\pi(x)}{k}.

Thus \pi_{C,k}(x)\le \dfrac{\pi(x)}{k}(1+o(1)).

Hence, defining \mathcal{N}_{k}(x) as the quantity equal to:

\displaystyle{\sum_{l=0}^{k}\pi_{C,l}(x)},  one gets:


where H_{k} is the k-th harmonic number.

So that one should have:

\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})\le k(1+H_{k})(1+o(1))


\mathcal{N}_{k}(x)=O(k\log k).

Now, from the prime number theorem, one has:

\mathcal{N}_{k}(x)\sim x for k large enough and less than x

So, it should be possible to prove rigorously that the conjunction of Goldbach’s conjecture and NFPR conjecture would entail that:

\displaystyle{\liminf_{n\to\infty} p_{n+k}-p_{n}=O(k\log k)}

This last relation, as stated in http://arxiv.org/pdf/1306.0948.pdf, follows from Hardy-Littlewood’s prime k-tuples conjecture.

Let’s now define the quantity \alpha(x,k) as follows:


It seems that there exists C>0 (and possibly not much bigger than 1) such that:

\forall(x,k)\vert \alpha(x,k)\vert0

Then there exists a unique m such that n=\frac{p_{m}+p_{m+k}}{2}.

One has obviously p_{m}<n<p_{m+k}, hence:

m\leqslant\pi(n)\leqslant m+k

and thus m\geqslant \pi(n)-k

Moreover m\leqslant\pi(x), so that the total number of k-central numbers below x verifies:

\pi_{C,k}(x)=\delta\vert\{m', \frac{p_{m}+p_{m+k}}{2}\leqslant x\}\vert+h_{k}(x)

where \delta is the probability for n'=\frac{p_{m'}+p_{m'+k}}{2} to be k-central,

and 0\leqslant \vert h_{k}(x)\vert\leqslant 1

There are k possibilities for the value of k_{0}(n'), namely k_{0}(n')=1, 2, \cdots, k

Since n' is k-central if and only if k_{0}(n')=k, one gets:


Thus \pi_{C,k}(x)\geqslant\frac{\pi(x)-k}{k}

Since \pi_{C,k}(x)=\frac{\pi(x)}{k}-\alpha(x,k) one finally gets:

\vert\alpha(x,k)\vert\leqslant max(\vert h_{k}(x)\vert,1)

hence \vert\alpha(x,k)\vert\leqslant 1.

Obviously the next step consists in showing that:

\displaystyle{\liminf_{n\to+\infty} p_{n+k}-p_{n}=O(\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n}))}

and hopefully:

\displaystyle{\liminf_{n\to+\infty} p_{n+k}-p_{n}\sim \mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})}

One has:


and thus:




Thus \dfrac{p_{n+k}-p_{n}}{\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})}=1+O(\dfrac{n+k}{k\log k})+O(\dfrac{\log n}{\log k})

As Maynard proved that the quantity


only depends on k, one should obtain:


by substracting the divergent part of the error term above.

So that we finally get:

\displaystyle{\liminf_{n\to\infty}\dfrac{p_{n+k}-p_{n}}{\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})}=1+O(\dfrac{1}{\log k})}

and thus:

\displaystyle{\liminf_{n\to\infty}p_{n+k}-p_{n}\sim k\log k}.