probability of m to be a primality radius of n

Question

Disclaimer : this is a crosspost from MSE, as the question got one upvote but no comment or answer whatsoever.

Assuming Goldbach's conjecture, let's denote by $r_{ 0}(n):=\inf\{r\geq 0,(n−r,n+r)\in\mathbb{P}^{2}\} $ . Then Cramer's model allows to write that the probability $ P((n−k,n+k)\in\mathbb{P}^{2})=\dfrac{c_{k}}{\log^{2}n} $ and $ r_{0}(n)=\inf\{m,\sum_{k=0}^{m}\dfrac{c_{k}}{log^{2}n}≥1\} $ .

I have two questions :

1) Would a proof that $ \forall k, c_{k}>0 $ entail that $ r_{0}(n)=O(\log^{2}n) $ ?

2) Building on the answer to Upper and lower bounds of sequences whose product of terms is asymptotically equal to their arithmetic mean, would an affirmative answer to 1) imply that the involved constant in $ O(\log^{2}n) $ actually equals $ 1 $?

Answer 1

I don't think it makes sense to focus on a particular $n.$ All that follows is speculative and depends on heuristics which are unproven yet widely believed to be true and well supported empirically.

One could discuss the distribution of $r_0(n)$ for large $n$ in some range and make strong predictions. But that would tell you nothing about possible outliers. It would be interesting to see how the distribution changes when restricted to those $n$ with a specified set of small odd prime divisors.

One could perhaps speculate on $$\limsup \frac{r_0(n)}{(\log{n})^2}.$$

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The definition of $c_k$ is somewhat problematic.

One could, for each $k>1$ look at $\pi_{2k}(x)=|\{2m \lt x \mid (2m-k,2m+k)\in \mathbb{P}\}|.$ Then one expects for each $k$ that there is a $c_k$ with $$\pi_{2k}(x)\sim \frac{2c_kx}{(\log x)^2}$$ in the sense that the ratio goes to $1.$ Here $c_k$ depends only on the set of prime divisors of $k.$ I think $$c_k=\prod\left(1-\frac1{(p-1)^2}\right)$$ where the product ranges over the odd primes which are relatively prime to $k.$

So indeed $c_{2k} \gt 0$ for all $k$ and, further, $c_{2k} \geq c_2 \approx 0.66$ with equality exactly when $k$ is a power of $2.$