TITLE: Is there an explicit example of such a real number with the following property?
QUESTION [16 upvotes]: In Diophantine approximation, for a given positive real number $\alpha$ let $[a_0, a_1, \cdots]$ denote its continued fraction expansion and let $p_n/q_n = [a_1, \cdots, a_n]$. Then it is known that $q_n$ grows at least exponentially. In 1935, Paul Levy proved that in fact for almost all real $\alpha$, we have $\displaystyle \lim_{n \rightarrow \infty} q_n^{1/n} = \exp(\pi^2/12 \log 2)$. The result was proved using Ergodic theory. Now my question is, does there exist a single known example of a real number $\beta$ such that $\beta = [b_0, b_1, \cdots]$, $p_n/q_n = [b_0, \cdots, b_n]$, and $\displaystyle \lim_{n \rightarrow \infty} q_n^{1/n} = \exp(\pi^2/12 \log 2)$?
For example, one might expect $\beta = \exp(\pi^2/12 \log 2)$ to do the trick...

REPLY [5 votes]: Here is a sort-of-but-not-really explicit answer, following up on Helge's idea. Set $h = e^{\pi^2/(12 \log 2)}$. Recursive define the $b_i$ and $q_i$ as follows: $q_{-2} =1$, $q_{-1}=0$ and $q_i = b_i q_{i-1} + q_{i-2}$. If $q_{i}^{1/i} > h$, then $b_{i+1} = 2$, otherwise, $b_{i+1}=4$. This is clearly a well defined recursion; you can decide whether or not you think the result of this process is explicit.
Lets first study, in general, the situation where every $b_i$ is either $2$ or $4$. Write $r_i = q_i/q_{i-1}$. Then $r_i = b_i + r_{i-1}^{-1}$ so $b_i < r_i < b_i + 1$. Note that $3 < h \approx 3.28 < 4$.
So, our algorithm has the property that, if $q_i^{1/i} < h$, then $r_{i+1}>h$ and vice versa.
Set $s_n = \log q_n = \sum_{i=0}^{n-1} \log r_i$. So, if $s_n < (\log h) n$, then $s_{n+1} = s_n + \log r_i \in [s_n+\log 4, s_n + \log 5]$ and we see that $s_{n+1} - (\log h) (n+1) \in $[(s_n - (\log h) n) + \log 4 - \log h, (s_n - (\log h) n) + \log 5 - \log h]$. Writing $t_n = s_n - (\log h) n$, we see that

If $t_n <0$, then $t_{n+1} \in [t_n + (\log 4 - \log h), t_n + (\log 5 - \log h)]$.

Similarly,

If $t_n > 0$, then $t_{n+1} \in [t_n + (\log 2 - \log h), t_n + (\log 3 - \log h)]$

In particular, once $t_n$ gets into the interval $[\log 2 - \log h, \log 5 - \log h]$, it stays there. I leave it to you to check that it gets there.
So $t_n$ is bounded, $\log q_n = (\log h) n + O(1)$ and $\lim_{n \to \infty} q_n^{1/n} = h$.