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](http://www.dms.umontreal.ca/~andrew/Courses/RationalPtsOnCurves.pdf) 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 [$1 + \sqrt{2}$](https://math.stackexchange.com/questions/14617/proving-the-continued-fraction-representation-of-sqrt2). 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?



Prime Divisors of the $x \mapsto 2x+1$ Recursion
The divisibility properties of the numbers defined by $a\_{n+1}=2a\_n+a\_{n-1}$ (and $a\_1=1$ and $a\_2=2$) are not random. One reason for this is that this sequence has the very non-random property of being eventually periodic mod $N$, for every integer $N$. In particular, since
$$
a\_n = \frac{(1+\sqrt{2})^n-(1-\sqrt{2})^n}{2\sqrt{2}},$$
we see that if $p$ is an odd prime then $a\_n\equiv 0\pmod{p}$ if and only if $n$ is a multiple of the order of
$$
\frac{1+\sqrt{2}}{1-\sqrt{2}} = -(1+\sqrt{2})^2
$$
in the multiplicative group of $\mathbf{F}\_p(\sqrt{2})$. (Note that the order of
this element does not depend on the choice of a square root of $2$ in $\mathbf{F}\_{p^2}^\*$.) Therefore if $p$ is an odd prime then
$$
\mathbb{P}\bigl[a\_n\equiv 0\pmod{p}\bigr] = \frac{1}{c\_p}
$$
where $c\_p$ is the order of $-(1+\sqrt{2})^2$ in $\mathbf{F}\_{p^2}^\*$.
For completeness, I note that the terms of the sequence $(a\_n)$ alternate in parity, so that half of them are even and half are odd.


The next question is to understand the distribution of the numbers $c\_p$, where $p$ is an odd prime. This connects with the number field analogue of Artin's primitive root conjecture. First consider primes $p$ with $p\equiv 1\pmod{8}$: for such primes, both $-1$ and $2$ are squares in $\mathbf{F}\_p^\*$, so that also $-(1+\sqrt{2})^2$ is a square and thus $c\_p\mid (p-1)/2$.
It is expected that there are infinitely many primes $p\equiv 1\pmod{8}$ for which $1+\sqrt{2}$ generates $\mathbf{F}\_p^\*$ (so that $c\_p= (p-1)/2$), and further there is a heuristically predicted (positive) value for the density of such primes $p$. Similar remarks apply when $p\equiv -1\pmod{8}$. Next if $p\equiv 3\pmod{8}$ then $2$ is a nonsquare in $\mathbf{F}\_p^\*$, but every unit in $\mathbf{Z}[\sqrt{2}]$ has norm $\pm 1$ and hence its image in $\mathbf{F}\_{p^2}^\*$ has order dividing $2(p+1)$, so that $c\_p\mid (p+1)$. Further, there is a heuristically predicted (positive) value for the density of the set of primes $p\equiv 3\pmod{8}$ for which $c\_p=(p+1)$. Again, similar remarks apply when $p\equiv -3\pmod{8}$. These heuristic predictions have been proved under the assumption of suitable Generalized Riemann Hypotheses. For references, see Pieter Moree's comprehensive survey on Artin's primitive root conjecture, which contains a wealth of further information.


