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 $\sup\{\Re(s)\mid\zeta(s)=0\}\leqslant\frac{1}{r}$