# 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: $\mathcal{N}_{k}(x)\le\pi(x)(1+H_{k})(1+o(1))$

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

So that one should have: $\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})\le k(1+H_{k})(1+o(1))$

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

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: $\alpha(x,k):=\frac{\pi(x)}{k}-\pi_{C,k}(x)$

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}, 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: $\delta=\frac{1}{k}$

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: $p_{n+k}-p_{n}=\mathcal{N}_{n+k}(p_{n+k})-\mathcal{N}_{n+k}(p_{n})$

and thus: $\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})=p_{n+k}-p_{n}-(n+k)(H_{n+k}-H_{n})+n(H_{n+k}-H_n)+O(n+k)$

hence: $\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})=p_{n+k}-p_{n}-k(H_{n+k}-H_{k})+O(n+k)$

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 $\displaystyle{\liminf_{n\to\infty}p_{n+k}-p_{n}}$

only depends on $k$, one should obtain: $\displaystyle{\liminf_{n\to\infty}\dfrac{p_{n+k}-p_{n}}{\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})}}$

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