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## Lower bounds (or less) for the period of \sqrt(D) and related sequences.

This is a continuation of

http://mathoverflow.net/questions/23134/ which is a continuation of

http://mathoverflow.net/questions/22811/

Yesterday, I came across a nice article with many problems on continued fractions of Michel Mendes France (from 1993):

http://retro.seals.ch/digbib/fr/view?rid=ensmat-001:1993:39::498&id=&id2=&id3=

In problem 6 he asks: Let $x$ be quadratic irrational and $\pi(x)$ the length of its period. Is $sup_n(\pi(x^n))=\infty$?

Is this problem still open? I saw many articles about upper bounds for $\pi(\sqrt{D})$ but none on lower bounds (expect the Golubeva one that is mentioned in the first mentioned question, but due to lack of background, I fail to understand the results there).

As Will Jagy asked: Do we even know whether $[\pi(\sqrt{n})):n\in \mathbb N]$ is bounded or not?

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*Edit: The following argument has an error in the second paragraph. I claimed that the set $\{ D^n\sqrt{D} (\mod 1)\}$ has uncountably many accumulation points, but as Sam Nead points out in the comments, this doesn't follow from $\sqrt{D}$ being irrational. On the other hand, it is true that the set $\{ \sqrt{n}(\mod 1)\}$ is uniformly distributed, and therefore that $\sup \pi(\sqrt{n}) =\infty$. This is because $\sqrt{n+1}-\sqrt{n} \to 0$ as $n\to \infty$. See a paper of Elkies and McMullen for related information on the distribution of this set. The issue is here that the set of accumulation points of $\{ x^n \mod 1\}$ for $x$ irrational may be a countable set. This question is actually discussed in Mendes' paper referred to in the question (I should have looked at it more closely before!). As pointed out in that paper, this can occur for Pisot numbers. Also, the statement is proved by Grisel for many quadratic numbers (as mentioned in a note to Mendes paper). * See also another paper of Grisel.

I think I can show that $sup_n(\pi(x^n))=\infty$ when $x=\sqrt{D}$.

Suppose that $x$ is an irrational, non-quadratic number in the interval $(0,1)$, and $n_i$ is a sequence of integers such that $\underset{i\to\infty}{\lim} \sqrt{n_i} (\mod 1) = x$. Then the claim is that $\pi(\sqrt{n_i})$ is unbounded. Basically, since $x$ has non-repeating continued fraction expansion, one can show that the continued fraction expansions of $\sqrt{n_i}$ have unbounded period. One can see this geometrically, by noting that there is an element of $PGL_2(\mathbb{Z})$ fixing $\pm \sqrt{n_i}$, and fixing the hyperbolic geodesic in $\mathbb{H}^2$ joining these two points. If $a^2-n_i c^2=\pm 1$ is the minimal solution to the Pell equation, then the matrix is: $$\begin{bmatrix} a & cn_i \\ c & a\end{bmatrix}$$ Moreover, the pattern of horoballs which the geodesic penetrates in the Farey packing tells you the periodic continued fraction. When we take $\sqrt{n_i} (\mod 1)$, these geodesics shifted by $-\left[ \sqrt{n_i} \right]$ approach the vertical geodesic through $x$, whose Farey pattern gives the continued fraction expansion for $x$. Therefore $\pi(\sqrt{n_i})$ is unbounded.

Now we have to understand the continued fractions of $\sqrt{D^{2n+1}} = D^n \sqrt{D}$, where $D$ is squarefree. The accumulation points of $D^n \sqrt{D} (\mod 1)$ correspond to the truncations of the base $D$ representation of $\sqrt{D}$. Since $\sqrt{D}$ is irrational, the limit points of these numbers in $[0,1]$ will be uncountable, and therefore contains a non-quadratic irrational $x$ which is a limit of $\sqrt{D^{2n_i+1}} (\mod 1)$, as $n_i\to \infty$. So we see that $\pi(\sqrt{D^{2n_i+1}}) \to \infty$.

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I'm not sure I understand the statement "Since $\sqrt{D}$ is irrational, the limit points of these numbers in $[0,1]$ will be uncountable". <p>For example, consider the number $\alpha = \sum 1/2^{2^n}$ and then the set $2^n \cdot \alpha\,(\mod 1)$. It looks to me like the closure of this set is countable, with accumulation points only at $1/2^m$. Am I missing something? <p>If not, then perhaps your argument can be saved using the fact that $\sqrt{D}$ is a much nicer number than $\alpha$. In any case, I'm off to plot the points $2^n\sqrt{2}\,(\mod 1)$ on my computer. Fun! – Sam Nead Jul 23 2010 at 20:30
I vaguely remember another reason to worry here: the powers of $(1 + \sqrt{5})/2$, taken modulo $1$, only accumulate on zero and one? – Sam Nead Jul 23 2010 at 20:34
I used Sage to spit out the first 10000 bits of $\sqrt{2}$ and then ran a few tests. Looks pretty random to me! I also found various papers (eg Good and Grover) that ran statistical tests on the binary expansion, thus experimentally verifying its normality. – Sam Nead Jul 23 2010 at 22:46
Probably a better way to think about it is the Farey tessellation. This perspective is discussed here: math.cornell.edu/~hatcher/TN/TNpage.html – Ian Agol Jul 30 2010 at 19:47
@ Menny: $((1+\sqrt{5})/2)^n + ((1-\sqrt{5})/2)^n \in \mathbb{Z}$. – Ian Agol Aug 11 2010 at 2:10

I just came across an interesting Ph.D. thesis by Roger Patterson which is very relevant to this topic: http://arxiv.org/pdf/math/0703519v1

There are some $D$ which have unusually short continued fraction periods:

If $D = n^2 + 1$, and the CF of $\sqrt{D}$ is $[n,\overline{2n}]$

If $D = n^2 + 4$ and $n$ odd then the CF of $\sqrt{D}$ is $[n,\overline{\frac{n-1}{2},1,1,\frac{n-1}{2},2n}]$

There are others given in the thesis that I refer to. He treats various families with unusually short periods and discusses a guess of Kaplansky that these are the only ones with unusually short periods (more precise definitions are given in the paper).

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