Let $I_{n}:=]p_{n},p_{n+1}[$ be the open interval between the $n$-th and $(n+1)$-th prime. Under Goldbach's conjecture, denote by $r_{0}(m)$ the smallest positive integer $r$ such that both $m-r$ and $m+r$ are prime for any large enough composite integer $m$ and by $k_{0}(m)$ the quantity $\pi(m+r_{0}(m))-\pi(m-r_{0}(m))$. Any integer $m$ such that $k_{0}(m)=k$ will be called a $k$-central integer and $k$ its order of centrality.

There is exactly one $1$-central integer $m$ in $I_{n}$, namely $\frac{p_{n}+p_{n+1}}{2}$. Moreover, letting $l(n)$ be $\sup_{m\in I_{n}}\{k_{0}(m)\}$ one can expect that the number of $k$-central integers in $I_{n}$, denoted by $N_{I_{n}}(k)$, is upper bounded by some constant $C_{k}$.

Can one prove one has $N_{I_{n}}(k)\leq k$ for all $1\leq k\leq l(n)$?

In that case the prime gap $g_{n}:=p_{n+1}-p_{n}$ would fulfill $g_{n}\leq 1+\frac{l(n)(l(n)+1)}{2}$.

Edit: the following link Would the following conjectures imply $\lim\inf_{n\to\infty}p_{n+k}-p_{n}=O(k\log k)$? may help upper bound $l(n)$ in terms of $n$.

Edit: say $g_{n}$ is an $l$-grade prime gap if $l(n)=l$. Can one find an upper bound for every $l$-grade prime gap depending only on $l$?

I thus propose the following conjecture: Grade conjecture $\forall l>0$, there are infinitely $l$-grade prime gaps.

Note that for $l=1$, we recover the twin prime conjecture.


1 Answer 1


$N_{I_n}(k)\leq k$ indeed always holds. Let $a_1<\dots<a_m$ be $m$ elements in $I_n$ such that $k_0(a_i)=k$ for each $k$. Let $b_i=a_i-r_0(a_i),c_i=a_i+r_0(a_i)$ so that $b_i,c_i$ are primes and there are exactly $k-1$ primes strictly between $b_i$ and $c_i$ for each $i$.

I claim $b_1<\dots<b_m$. Indeed, if $b_i>b_j$ for some $i<j$, then from $a_i<a_j$ we deduce $c_i=2a_i-b_i<2a_j-b_j=c_j$. But then we find there are more primes between $b_j,c_j$ than between $b_i,c_i$, hence the claim.

Finally observe $b_m\leq p_n<p_{n+1}\leq c_1$. Since there are $k-1$ primes between $b_1$ and $c_1$, and $b_2,\dots,b_m$ all lie in this interval, we get $m\leq k$.

  • $\begingroup$ Assuming the prime tuple conjecture, the bound $N_{I_n}(k)\leq k$ will also be optimal for every $k$. I doubt an unconditional proof is possible here though. $\endgroup$
    – Wojowu
    Mar 2, 2021 at 12:57
  • $\begingroup$ That's merely astonishing. I didn't quite get your argument, but I'll get back to it later. Thanks a lot! $\endgroup$ Mar 2, 2021 at 14:24
  • $\begingroup$ Can we thus get somewhat closer to Cramer's conjecture? $\endgroup$ Mar 2, 2021 at 14:32
  • $\begingroup$ I have no idea what relation whatsoever to Cramer's conjecture this bears. If you are intending to use the bound in terms of $l(n)$ you propose, then I doubt it given there is no way to derive any nontrivial bounds from Goldbach conjecture. $\endgroup$
    – Wojowu
    Mar 2, 2021 at 14:55
  • 1
    $\begingroup$ Goldbach conjecture doesn't guarantee anything beyond $l(n)=O(n)$. True order of growth is probably smaller but we would need stronger assumptions to prove that. $\endgroup$
    – Wojowu
    Mar 2, 2021 at 15:10

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