# 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 $\forall(x,k)\vert \alpha(x,k)\vert<C$. 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 $p_{m}<n<p_{m+k}$ 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 $0\leqslant \vert h_{k}(x)\vert<1$. 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? – Đào Thanh Oai Jul 6 '18 at 14:43