Recall that for a real number $\alpha$, it has a continued fraction expansion usually written as

$$\displaystyle \alpha = [a_0; a_1, a_2, \cdots].$$

Moreover, $\alpha$ is rational if and only if its continued fraction expansion is finite, and is a quadratic irrational if and only if the continued fraction expansion is eventually periodic. The numbers $a_0; a_1, a_2, \cdots$ are called partial quotients of $\alpha$.

It is known that with respect to Lebesgue measure, almost all real numbers $\alpha$ have the property that the partial quotients of $\alpha$ follow a specific distribution: the frequency that $k$ appears in the sequence $\{a_0; a_1, a_2, \cdots\}$ is $\frac{1}{\log 2} \log \left(\frac{(k+1)^2}{k(k+2)} \right)$.

Consider the interval $I_X = (X, 2X]$ for some (large) positive number $X$. For each integer $n \in I_X$, consider the continued fraction expansion of $\sqrt{n}$, say $\sqrt{n} = [a_0; a_1, a_2, \cdots]$. We know that this sequence is eventually periodic, and in fact $\sqrt{n} = [a_0; \overline{a_1, a_2, \cdots, a_2, a_1, 2a_0}]$ where the bar denotes the periodic part and the string $a_1, a_2, \cdots, a_2, a_1$ denotes a palindrome.

My question is: as $n$ runs over $I_X$, how are the partial quotients of $\sqrt{n}$ distributed? In other words, for each positive integer $k$ what is the frequency of appearance of $k$ as partial quotients of $\sqrt{n}$ as $n$ runs over $I_X$?

Note that the frequency is 0 if $k \gg X^{1/2}$, since it is known that the partial quotients of $\sqrt{n}$ are at most $2 \sqrt{n}$ in size.


1 Answer 1


It is not an answer, but probably it is the only known information about Gauss-Kuz'min statistics for quadratic irrationals.

For a reduced quadratic irrational $\omega$ (which has a purely periodic representation in the form of a continued fraction) denote by $\rho(\omega)$ the length of $\omega$ which is the length of the corresponding closed geodesic on modular surface $\mathbb{H}/PSL_2(\mathbb Z)$, $\mathbb{H}=\{(x,y):\ y>0\}$ (the projection of the geodesic joining $\omega$ and $\omega^*$, where $\omega^{*}$ is the number conjugate to $\omega$). The answer is positive in the case when you average Gauss-Kuz'min statistics over all $\omega$ such that $\rho(\omega)\le X$. More precisely, let $x, y \in [0, 1]$ be real numbers and $$ r(x,y;N)=\sum_{\substack{\omega\in\mathcal{R}\\ \varepsilon_0(\omega)\leqslant N}} [\omega\leqslant x,\ -1/\omega^{*}\leqslant y]. $$ Here $\mathcal{R}$ is the set of reduced quadratic irrationals, $\varepsilon_0(\omega)=\frac{1}{2}(x_0+\sqrt{\Delta}y_0)$ is the fundamental solution of Pell’s equation $$X^2-\Delta Y^2=4,$$ $\Delta=B^2-4AC$, where $AX^2+BX+C$ is the minimal polynomial of $\omega$; moreover, $[A]$ stands for $1$ if the statement $A$ is true and for $0$ otherwise. The fundamental solution $\varepsilon_0(\omega)$ is clesely connected to the length: $\rho(\omega)=2\log\varepsilon_0(\omega)$. Then (see Theorem 3 from Spin chains and Arnold's problem on the Gauss-Kuz'min statistics for quadratic irrationals) $$ r(x,y;N)=\frac{\log(1+xy)}{2\zeta(2)}N^2+{O}(N^{3/2}\log^4N). $$

  • $\begingroup$ I believe there are two differences between your answer and what I am asking about. First, your height function is essentially the size of the fundamental unit of a quadratic order, which is typically much much larger than the discriminant, whereas I am counting by discriminant. Second, you are dealing with all quadratic irrationals, whereas I am focused only on square roots of positive integers (or very roughly speaking, only quadratic forms in the principal class). $\endgroup$ Feb 13, 2020 at 15:23
  • $\begingroup$ @StanleyYaoXiao Yes, but you question looks like unsolved problem, That is why I gave the information which is known for me. I've added this remark to my answer. $\endgroup$ Feb 14, 2020 at 4:54

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