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The first two terms of a sequence are $a_1 = 1$ and $a_2 = \frac {1}{\sqrt3}$. For $n\ge1$, $$a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}.$$

What is $|a_{2009}|$?

The simplest solution for this question was to just work out the sequence and find that it repeats with a period of 24. However, I don't think many people would work out to many terms, just to see if there is a repeating cycle

Does anyone know if there is any way to know if a recursive sequence will be cyclic just by looking at the equation?

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    $\begingroup$ I look at that equation, and see tangents. $\endgroup$ – Lord Shark the Unknown Feb 6 at 6:30
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If we make the substitution $a_n = \tan \theta_n$, where we restrict $0 \le \theta_n < \pi$, then we have $\theta_1 = \dfrac{\pi}{4}$, $\theta_2 = \dfrac{\pi}{6}$, and $$\tan \theta_{n+2} = \dfrac{\tan \theta_n + \tan \theta_{n+1}}{1-\tan\theta_n\tan\theta_{n+1}} = \tan(\theta_n+\theta_{n+1}),$$ i.e. $$\theta_{n+2} \equiv \theta_n + \theta_{n+1} \pmod{\pi}.$$ It shouldn't take too much more work to show that $\theta_n$ is periodic. Regardless, it is probably easier to compute the first several terms of $\theta_n$ and see a pattern than it is to compute the first several terms of $a_n$ directly and see the pattern.

EDIT: If we substitute $b_n = \dfrac{12}{\pi}\theta_n$, then we have $b_1 = 3$, $b_2 = 2$, and $$b_{n+2} \equiv b_n + b_{n+1} \pmod{12}.$$ Now it is really easy to see that this must be periodic.

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  • $\begingroup$ So simple ... when you see such an elegant solution ! $\to +1$ $\endgroup$ – Claude Leibovici Feb 6 at 7:11

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