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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.

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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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I had made the same conjecture as Fekete, apparently around the same time -- mid-2007. In 2008 I verified that the first twenty million terms do not include 319. (I actually pushed the verification further, but I can't find the more recent records at the moment.)

Because $\tan(x) - x = x^3/3 + O(x^5)$, the function spends a lot of its time in a small neighborhood around $0$. It escapes when it nears $\pi/2$ and quickly returns for many iterations.

A mostly-unexplained phenomenon presumably related to the above: there are long spans of small numbers followed by short, 'productive' spans with large numbers. $\tan^k(1)$ is "below 20 or so" (according to a 2008 email I sent) for $360110\le k\le1392490$ but in the next 2000 numbers there are five which are above 20.

More theory is needed!

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    @Joseph: No. I was in the process of my 25 million digit effort when Tony Noe added his comment and I never bothered to send an update.2010-10-22
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This isn't a proof, but's too long for a comment, and may just be a restatement of the problem.

For contradiction, let $k$ be any integer such that $b(n) = k$ never holds. This means $k \leq a(n) < k+1$ never holds.

Since $a(n)$ can't be between $k$ and $k+1$, $\arctan a(n)$ can't be either.

Thus, there's an interval between $-\pi/2$ and $\pi/2$ that a(n) may not touch. Let's call it $[c,d)$.

Since tan is periodic, $a(n)$ must also avoid $m\pi+[c,d)$.

Since $\pi$ is irrational, $m\pi+[c,d)$ must contain an infinite number of integers (pretty sure this is true, but I could be wrong).

Therefore, there are an infinite number of intervals (approaching $\pi/2$) that $a(n)$ must avoid. Further, $a(n)$ must avoid the arctans of these intervals, and the arctans of those intervals, etc. The repeated arctan intervals approach 0.

Of course, $a(n)$ also has to avoid those intervals plus any multiple of $\pi$.

This non-proof actually applies to any interval $a(n)$ misses, so, if true, shows that $a(n)$ is dense in $\mathbb{R}$. Hope that helps.