0
$\begingroup$

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.

$\endgroup$

1 Answer 1

3
$\begingroup$

$N_{I_n}(k)\leq k$ 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 $m\leq k$.

$\endgroup$
7
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.