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 $l$?

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.

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 $g_{n}\leq\frac{l(n)(l(n+1)}{2}$.

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 $n$.

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 $l$?