Assuming Goldbach's conjecture, denote as usual by $r_{0}(n)$ for any large enough positive integer $n$ the smallest positive integer $r$ such that both $n-r$ and $n+r$ are prime.

Let's define the notion of "staircase number" as any such integer $n$ such that the elements of the sequence $r_{0}(n), r_{0}(n)^2,\cdots,r_{0}(n)^{r_{0}(n)}$ are the first integers $r$ such that $n-r$ and $n+r$ are simultaneously prime. Say $n$ is an $r$-staircase number if $n$ is a staircase number and $r_{0}(n)=r$.

I now formulate the following conjectures:

Weak staircase conjecture: There are infinitely many staircase numbers.

Middle strength staircase conjecture: There exists some $r>0$ such that there are infinitely many $r$-staircase numbers.

Strong staircase conjecture: For all positive integer $r$, there are infinitely many $r$-staircase numbers.

Note that $15$ is a $2$-staircase number, $70$ is a $3$-staircase number, and any half sum of two twin primes is obviously a $1$-staircase number.

My question is: does the Elliott-Halberstam conjecture or some generalization thereof together with proven results following Zhang's 2013 breakthrough imply at least one of these conjectures?

Edit: it would suffice to prove that the number of $r$-staircase numbers below $x$ is asymptotically $S_{r}(x)\sim\frac{x}{\log^{O_{r}(1)}x}$ to entail the strong staircase conjecture. Numerically, I found that for $r\in\{2,3\}$, this number is close to $\tilde{S}_{r}(x):=\frac{x}{\log^{2+r^{(r-1)\log (1+\gamma)}}x}$, where $\gamma$ is the Euler-Mascheroni constant.

Edit February 16th, 2021: defining antistaircase numbers by permuting the bases and exponents in the definition of a staircase number, hence if the first $r_{0}(n)$ primality radii of $n$ are $1^{r_{0}(n)},\cdots r_{0}(n)^{r_{0}(n)}$, then necessarily $r_{0}(n)=1$. So if we could prove that the weak staircase conjecture is equivalent to the existence of infinitely many antistaircase numbers, then it is equivalent to the twin prime conjecture.

Edit March 20th, 2021: I have a heuristics suggesting there should be infinitely many $r$-staircase numbers for $r\in\{2,3\}$ and no $r$-staircase number for $r>3$. Namely for $r$ large enough there should be at least around $\frac{r^r}{\log^{2}(r^{r})}=\frac{r^{r}}{(r\log r)^2}=\frac{r^{r-2}}{\log^{2}r}=r^{r-2-\varepsilon}$ primality radii of an integer greater or equal to $r^{r}$ up to $r^{r}$. But in the case of an $r$-staircase number, there are exactly $r$ primality radii up to $r^{r}$. This implies $r-2-\varepsilon\approx 1$ hence $r=3$. If $r=2$ one has $\log^{2}r\approx 1/2$, hence we get $r\approx 2r^{r-2}\approx r^{r-1}\approx r$, which is coherent.

Numerical evidence is consistent with that.

Edit April 28th 2021: as the first $r$ primality radii of an $r$-staircase number make an arithmetic progression, one may consider the sum of their reciprocals and hope to show there are infinitely many $r$-staircase numbers iff this sum is greater than some absolute constant. Perhaps one can establish a link with RH proving there are infinitely many $r$-staircase numbers iff $\sum_{n>0}\frac{1}{r^n}\gt\sup\{\Re(s)\mid\zeta(s)=0\}$. Were this firmly established, this would prove both the twin prime conjecture and that there are finitely many $r$-staircase numbers whenever $r>2$. Note also that the existence of infinitely many $2$-staircase numbers is predicted by Hardy-Littlewood $k$-tuple conjecture, as those numbers correspond to the constellation $(0,2,6,8)$: so that this conjecture may imply a quasi-Riemann hypothesis, i.e. that there exists $\varepsilon>0$ such that $\zeta(s)=0\Longrightarrow\Re(s)\lt 1-\varepsilon$.

I thus dare formulate the following conjecture: Riemann staircase conjecture

There are infinitely many $r$-staircase numbers iff $\sup\{\Re(s)\mid\zeta(s)=0\}\leqslant\frac{1}{r}$


1 Answer 1


Strong staircase conjecture follows from a the following version of the $k$-tuple conjecture: for any admissible tuple $T$ and a finite set $S$ disjoint from it, there are infinitely many integers $n$ such that $n+T$ contains only primes, and $n+S$ contains only composite numbers. This version follows from the Dickson's conjecture, by an argument similar to the one presented in this answer to an old question of mine. This conjecture also follows from the full first Hardy-Littlewood conjecture (the version predicting the frequency of prime tuples of given length)

For any $r\in\mathbb N$ let $T=\{\pm r,\pm r^2,\dots,\pm r^r\}$ and let $S$ be the set of all remaining elements of the interval $[-r^r,r^r]$. Any $n$ such that $n+T$ consists of primes and $n+S$ has no primes is an $r$-staircase number. Therefore we only have to verify that $T$ is an admissible tuple. But this is obvious - for any prime $p$, either all elements of $T$ are divisible by it or none are, so $T$ can't cover all residue classes modulo $p$.


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