A good estimate (in some sense) is $$\frac{\alpha_{2r}C_2}{\ln^2N}$$ where $$C_2 = \prod_{p\ge 3} \frac{p(p-2)}{(p-1)^2} \approx 0.66016.$$ and $$\alpha_{2r}=\prod_{p \mid 2r}\frac{p-1}{p-2}$$ is the product is over all *odd* primes $p$ which divide $2r.$

Notice that $\alpha_2=\alpha_4=\alpha_8=\dots=1$ and $\alpha_6=\alpha_{12}=\alpha_{18}=\alpha_{24}=\dots=2$

**More careful statement**: Let $\pi_{2r}(N)$ be the number of primes $p$ up to $N$ with $p+2r$ also prime and $\alpha_{2r}=\lim_{N \to \infty}\frac{\pi_{2r}(N)}{\pi_2(N)}.$ The twin prime conjecture *is* indeed open: there is no proof that $\lim_{N \to \infty} \pi_2(N)=\infty.$ However there is every reason to expect that

- $\pi_2(N) \sim C_2\frac{N}{(\ln{N})^2}.$
- $\alpha_{2r}=\prod_{p \mid 2r}\frac{p-1}{p-2}$ where the product is over all the
*odd* primes which divide $r.$

The great article Heuristic Reasoning in the Theory of Numbers (read it!) discuss the heuristics and some computational (circa 1959) evidence behind this.