Would the following conjectures imply $\lim\inf_{n\to\infty}p_{n+k}-p_{n}=O(k\log k)$?

Assume Goldbach's conjecture. Then for every $$n\ge 2$$ there exists at least one non-negative integer $$r\le n-2$$ such that both $$n+r$$ and $$n-r$$ are primes. Let's write $$r_{0}(n):=\inf\{r\le n-2, (n-r,n+r)\in\mathbb{P}^{2}\}$$ and $$k_{0}(n):=\pi(n+r_{0}(n))-\pi(n-r_{0}(n))$$.

I call $$r$$ a primality radius of $$n$$, $$r_{0}(n)$$ the fundamental primality radius of $$n$$ and $$k_{0}(n)$$ the order of centrality of $$n$$. I say that $$n$$ is a $$k$$-central number if and only if $$k_{0}(n)=k$$.

Now, 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):

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

Could one deduce from this conjecture that $$\dfrac{\pi(x+\max_{n\le x}r_{0}(n))}{k}\sim\dfrac{\pi(x)}{k}$$?

If so, one would have $$\pi_{C,k}(x)\le \dfrac{\pi(x)}{k}(1+o(1))$$.

Hence $$\displaystyle{\mathcal{N}_{k}(x):=\sum_{l=0}^{k}\pi_{C,l}(x)\le\pi(x)(1+H_{k})(1+o(1))}$$, where $$H_{k}$$ is the $$k$$-th harmonic number.

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

Now, from the prime number theorem, $$\mathcal{N}_{k}(x)\sim x$$ for $$k$$ large enough and less than $$x$$. So, is it possible to prove rigorously that the conjunction of Goldbach's conjecture and NFPR conjecture would entail that $$\lim\inf_{n\to\infty} p_{n+k}-p_{n}=O(k\log k)$$ which, as stated in http://arxiv.org/pdf/1306.0948.pdf, follows from Hardy-Littlewood's prime k-tuples conjecture?

EDIT November 22nd 2013: I guess a better way to give an estimation of $$\pi_{C,k}(x)$$, and thus of $$\mathcal{N}_{k}(x)$$, would be to establish rigorously that $$\pi_{C,k}(x)\asymp \frac{\pi(x)}{k}$$ under NFPR conjecture (since the PNT shows that under this conjecture, $$\pi(x+r_{0}(x))\sim\pi(x)$$). Is such an asymptotics correct?

EDIT January 23rd 2014: Let's define $$\alpha(x,k)$$ 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 $$\forall(x,k)\vert \alpha(x,k)\vert. I call this statement "$$\alpha$$ conjecture".

A direct consequence of $$\alpha$$ conjecture is that one would have $$\pi_{C,k}(x)=\frac{\pi(x)}{k}-O(1)$$, which is even stronger than $$\pi_{C,k}(x)=\frac{\pi(x)}{k}(1+o(1))$$ and could give further evidence for the desired conclusion: indeed one would have $$\mathcal{N}_{k}(x)=\pi(x)(1+H_{k})-O(k)$$ and therefore $$\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})=k(1+H_{k})-O(k)\sim k\log k$$.

EDIT January 23rd 2014 (bis): I think I have a proof of the natural strenghtening of $$\alpha$$ conjecture (hence called "strong $$\alpha$$ conjecture") that says that $$\forall(x,k)\vert\alpha(x,k)\vert\leqslant 1$$.

Suppose indeed that $$n\leqslant x$$ is a $$k$$-central number with $$k>0$$. 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<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 $$\lim\inf_{n\to+\infty} p_{n+k}-p_{n}=O(\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n}))$$ and hopefully $$\lim\inf_{n\to+\infty} p_{n+k}-p_{n}\sim \mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})$$. Any help would be greatly appreciated.

Edit June 1st 2014: One has $$p_{n+k}-p_{n}=\mathcal{N}_{n+k}(p_{n+k})-\mathcal{N}_{n+k}(p_{n})$$ and thus $$\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})=p_{n+k}-p_{n}-(n+k)(H_{n+k}-H_{n})+n(H_{n+k}-H_n)+O(n+k)=p_{n+k}-p_{n}-k(H_{n+k}-H_{k})+O(n+k).$$

Hence $$\dfrac{p_{n+k}-p_{n}}{\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})}=1+O(\dfrac{n+k}{k\log k})+O(\dfrac{\log n}{\log k})$$. As Maynard proved that $$\lim\inf_{n\to\infty}p_{n+k}-p_{n}$$ only depends on $$k$$, one should obtain $$\lim\inf_{n\to\infty}\dfrac{p_{n+k}-p_{n}}{\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})}$$ substracting the divergent part of the error term above, hence $$\lim\inf_{n\to\infty}\dfrac{p_{n+k}-p_{n}}{\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})}=1+O(\dfrac{1}{\log k})$$ and thus $$\lim\inf_{n\to\infty}p_{n+k}-p_{n}\sim k\log k.$$

• I am sorry, may I ask You a question? I posted new topic: I checked the result as follows is true for $x=1, 2, \cdots, 9.5\times 10^8$ $g_n=P_{n+1} - P_{n} \leq n$ ** My question: ** Is the result well-known? I don’t known why some one voted down? Jul 6, 2018 at 14:43
• Should a positive answer be given to mathoverflow.net/questions/351649/…, a proof of NFPR conjecture would follow immediately. Feb 1, 2020 at 11:58