I would like to know if some widely believed conjecture, be it GRH, Hardy-Littlewood conjecture, or any other would imply the following statement for some $l>1$:

$l$-th power radioprimal growth conjecture ($l$-PRG conjecture for short)

Let $r_{0}(n):=\inf\{r>0,(n-r,n+r)\in\mathbb{P}^{2}\}$, $r_{i+1}(n):=\inf\{r>r_{i}(n),(n-r,n+r)\in\mathbb{P}^{2}\}$ and $k_{i}(n):=\pi(n+r_{i}(n))-\pi(n-r_{i}(n))$. There exists $C_{l}>1$ such that there are infinitely many $n$ such that $i\leq C_{l}$ implies $k_{i-1}(n)=i^{l}$.

I would also like to know if $C_{l}$ can be taken arbitrarily large, and if the hypothetical existence of a maximal $l$ such that the $l$-PRG conjecture holds implies $r_{0}(n)<K(\varepsilon)\log^{l+\varepsilon} n$ for all sufficiently large $n$ and all $\varepsilon>0$.


1 Answer 1


Assuming the strengthened version of Hardy-Littlewood conjecture I discuss here (which follows from Dickson's conjecture), the following much stronger result holds: let $a_0,\dots,a_m$ be any sequence of natural numbers with $a_0\geq 1$ and $a_{i+1}\geq a_i+2$ for $i<m$. Then there are infinitely many integers $n$ such that $k_i(n)=a_i$ for $i\leq m$.

Let $N$ be some fixed natural number to be chosen later. Consider the set $T=\{N,2N,\dots,(a_m-m)N\}\cup\{-(a_i-i)N\mid 0\leq i\leq m\}$, and $S$ the set of other numbers in the interval $[-(a_m-m)N,(a_m-m)N]$. If $N$ is divisible by all small enough primes, then $T$ is an admissible tuple. Let $n$ be any integer given by the strengthened Hardy-Littlewood conjecture above for these $T,S$. Then clearly its first $m+1$ primality radii are $(a_i-i)N$ for $0\leq i\leq m$, and you can count there are $a_i$ primes between $n-(a_i-i)N$ and $n+(a_i-i)N$, so $k_i(n)=a_i$.

  • $\begingroup$ Many thanks Woj. I expected a link to the staircase numbers, though I could not be sure the question could be answered that easily. I'll study your answer more deeply a bit later in the evening. $\endgroup$ Dec 26, 2021 at 18:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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