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...

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    $\begingroup$ I'm guessing Zev means, why might one expect $\beta$ to do the trick. Since one expects almost all real numbers to do the trick, I guess one might expect $\beta$ to do it, but I wouldn't expect it to be easier to prove for $\beta$ than for other numbers. $\endgroup$ Feb 15, 2011 at 3:32
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    $\begingroup$ For what it's worth, the Levy references are Sur les lois probabilites dont dependent les quotients complets et incomplets d'une fraction continue, Bull Soc Math France 57 (1929) 178-194 and Sur la developpement en fraction continue d'un nombre choisi au hasard, Compositio Math 3 (1936) 286-303. Both papers are also in v. 6 of his Oeuvres, 266-282 and 285-302. $\endgroup$ Feb 15, 2011 at 3:41
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    $\begingroup$ From $q_n = a_n q_{n-1} + q_{n-2}$ and that the constant is $>2$. One should be able to define $a_n$ by some recursive formula. Of course, I am unsure if this is "explicit" enough. $\endgroup$
    – Helge
    Feb 15, 2011 at 5:50
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    $\begingroup$ It's interesting to note that $\exp(\pi^2/(12\log 2))\approx 10$. This means that to compute $N$ terms of the continued fraction expansion, you need approximately $N$ decimal digits... $\endgroup$ Feb 15, 2011 at 18:03
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    $\begingroup$ Uhm, $\exp(\pi^2/(12 log 2)) \approx 3.2758229187$. See mathworld.wolfram.com/Khinchin-LevyConstant.html $\endgroup$ Feb 15, 2011 at 19:50

2 Answers 2


Yes. There is a classical construction in number theory due to Champernowne of a number that has the right frequency of each block in its decimal expansion. The number is just 0.12345678910111213141516171819202122$\ldots$. You have to do a certain amount of work to check this property. Such a number is called a normal number base 10.

Adler, Keane and Smorodinsky in 1981 constructed a "continued fraction normal number" analogous to the Champernowne number for the continued fraction transformation in a reasonably explicit way - they gave an essentially explicit description of the $b_n$'s that appear. This continued fraction normality is (much) stronger than the condition that you are asking for: it implies that the denominators grow at the correct rate and also any other average quantity belonging to a very wide class defined on the basis of the underlying dynamical system takes the same value for this number as it does for a set of Lebesgue measure 1 in [0,1].

Incidentally, experimentally $\pi$ has the correct denominator growth, whereas $e$ has an anomalous denominator growth rate (provably).

  • $\begingroup$ Typo: The last specified digit of your Champernowne expansion should be 2, not 1. $\endgroup$ Feb 15, 2011 at 10:04
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    $\begingroup$ Is there any obvious way to see that a normal number of a particular base satisfies the question? $\endgroup$ Feb 15, 2011 at 13:14
  • $\begingroup$ Then why is this an answer to the question? The question was give a normal which ideally is continued fraction normal or, at minimum, has the same value of this growth statistic. $\endgroup$ Feb 15, 2011 at 16:43
  • $\begingroup$ My mistake, question remains open. $\endgroup$ Feb 15, 2011 at 16:53
  • $\begingroup$ Ah, I see. The AKS paper does answer the question, and they make an analogy between the number they construct and the Champernowne number. However, their number is NOT the Champernowne number and the notion of base b normality doesn't actually show up in their proof. Link to the paper: ams.org/mathscinet-getitem?mr=602450 $\endgroup$ Feb 15, 2011 at 18:10

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


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 $\lim_{n \to \infty} q_n^{1/n} = h$.

  • $\begingroup$ Of course, this has nothing to do with the particular value of $h$: the same construction works for any $h>2$ (and, if you are clever, you can get down to $h>(1+\sqrt{5})/2 \approx 1.618$.) $\endgroup$ Feb 15, 2011 at 14:13
  • $\begingroup$ This looks more complicated then what I had in mind. I was going to define $b_n = \lfloor \frac{h^n - q_{n-2}}{q_{n-1}}\rfloor$, where $h = \exp(...)$. That way $q_{n} < h^n$ and since $h > 2$, we have $b_n \geq 1$. I wonder what values $b_n$ takes in this construction, i.e. if also $2$ and $4$ suffice... $\endgroup$
    – Helge
    Feb 15, 2011 at 18:46

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