The Cohen-Lenstra statistics describe how often a prime divides the class number of quadratic number field $\mathbb{Q}[\sqrt{d}]$

$$ \mathbb{P}\big[h(d) \not\equiv 0\; (\mod p) \big] = \prod_{k \geq 2}\left( 1 - \frac{1}{p^k}\right) $$

As can be seen in Section 6.3 of this Number Theory course by Andrew Granville.

I would like to know if there is a similar conjecture for the continued fraction convergents of $1 + \sqrt{2}$. Notice that

$$ \sqrt{2}+1 = 2 + \frac{1}{\sqrt{2}+1} = [2,2,2,\dots] $$

Therefore, the convergents of the continued fraction can be generated by recursion

$$ \frac{1}{1}, \frac{2}{1}, \frac{5}{3}, \frac{12}{7}, \frac{29}{17}, \dots$$

In particular, the numberators satisfy the recursion $a_{n+1} = 2 a_n + a_{n-1}$, and it seems likely the divisibility properties of this number should be "random".

The distribution of the primes dividing $a_n$ satisfy pattern like Cohen-Lenstra?