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 $\lim\inf\_{n\to\infty}p\_{n+k}-p\_{n}=O(k\log k)$?](https://mathoverflow.net/questions/132973/would-the-following-conjectures-imply-lim-inf-n-to-inftyp-nk-p-n-ok-lo) 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.



Upper bound for the number of $k$-central numbers in a prime gap
$N\_{I\_n}(k)\leq k$ 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 $m\leq k$.


