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A high-school student named Erna Fekete made a conjecture to me via email three years ago, which I could not answer. I've since lost touch with her. I repeat her interesting conjecture here, in case anyone can provide updated information on it.

Here is how she phrased it. Let $b(0) = 1$ and $b(n)= \tan( b(n-1) )$. In other words, $b(n)$ is the repeated application of $\tan(\;)$ to 1: $$\tan(1) = 1.56, \; \tan(\tan(1)) = 74.7, \; \tan^3(1) = -0.9, \; \ldots $$

Let $a(n) = \lfloor b(n) \rfloor$. Her conjecture is:

Every integer eventually appears in the $a(n)$ sequence.

This sequence is not unknown; it is A000319 in Sloane's integer sequences. Essentially hers is a question about the orbit of 1 under repeated $\tan(\;)$-applications. Her and my investigations at the time led us to believe it was an open problem.

  • 11
    How about a more daring conjecture, that the b(n) themselves are dense in R? At least it does away with the nasty floor function. Is it obviously false?2010-10-14
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    @Alon: Good point! Not only is it not obviously false---I think the orbit _is_ dense!2010-10-14
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    Is it obvious that the sequence is everywhere defined? (I think it must be true but I don't see how to prove it.)2010-10-14
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    @Qiaochu: Not obvious to me either!2010-10-15
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    @Qiaochu, @Joseph O'Rourke: Suppose we consider the set $X$ of all $x \in \mathbb{R}$ such that some $\tan^{n}(x)$ is not defined. Because $X$ is the union of repeated (set-valued) applications of $\tan^{-1}$ on $\{(2k+1)\pi/2\}$, it seems to me that $X$ is countable but dense in $\mathbb{R}$. Is it plausible to conjecture that $X$ contains no rational number?2010-10-20
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    Generalizing to an arbitrary starting point x, rather than just starting at 1, the (possibly terminating) sequence [a(0),a(1),a(2),...] looks a lot like a kind of continued fraction representation. In fact, it gives a one-to-one map between real x and possibly terminating sequences of integers. So there are uncountably many starting points where every integer appears, and uncountably many where only a finite set of integers occur. There's uncountably many x for which only 0 and 1 appear. However, I think, for almost every x, the sequence a(n) will tend to some fixed distribution.2010-10-21
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    So, this question sounds like a much more difficult version of problems involving regular continued fractions. It is not even known if the continued fraction of $\pi$ contains every positive integer (according to Sloane: http://akpublic.research.att.com/~njas/sequences/A032523). It seems likely that this question is much harder than that.2010-10-21

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