# Recursive Question in AMC 2009 (12A)

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?

• I look at that equation, and see tangents. – Lord Shark the Unknown Feb 6 at 6:30

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