I am not convinced that the definition of $c_k$ makes sense.

Consider a simpler problem:

Since there are infinitely many primes we can define $g(n)=\min\{k \ge 0 \mid n+k \in \mathbb{P}\}.$

• Would it make sense to say $P(n+k \in \mathbb{P})=\frac{d_k}{\log n}?$
• At any rate, would you entertain the notion that $g(n)=O(\log n)?$

Actually it is known that is false. In fact, Cramer's conjecture is that $g(n)=O((\log n)^2).$ The heuristic supports the stronger claim $\limsup \frac{g(n)}{(\log n)^2}=1.$ Of course only prime $n$ need be considered.

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 $$\limsup \frac{r_0(n)}{(\log{n})^2}.$$

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 $2.$