Interesting recursive sequence: $a_{n+1}=\frac{a_{n}+a_{n-1}}{1-a_n a_{n-1}}$

Question

I found the following problem interesting (and rather satisfying to solve), so I thought that I would share it with the MSE community.

Let $a_n$ be a sequence with $a_0=a_1=\sqrt 3$ and $$a_{n+1}=\frac{a_{n}+a_{n-1}}{1-a_n a_{n-1}}$$ Find an explicit formula for $a_n$ in terms of $F_n$, the fibonacci numbers, with $F_0=F_1=1$.

I've already solved this, and I'm posting it as a "recreational" math problem, so have fun!

Answer 1

\begin{align} a_0=a_1&=\sqrt 3\\ a_{n+1}&=\frac{a_{n}+a_{n-1}}{1-a_n a_{n-1}}\\ a_n &= \tan(x) \qquad\text{The range of $\tan(x)$ is all real numbers, so we can make this substitution}\\ a_{n-1} &= \tan(y)\\ a_{n+1} &= \tan(x+y)\\ a_0=a_1=\sqrt3&=\tan(\pi/3)\\ a_2 &= \tan(\pi/3 + \pi/3) = \tan (2\pi/3)\\ a_3 &= \tan ((1+2) \pi/3)\\ a_n &= \tan(F_2 \cdot \pi/3) \end{align}

Answer 2

Let $b_n=\tan^{-1}a_n$. Then the recurrence becomes $$\tan b_{n+1}=\frac{\tan b_n+\tan b_{n-1}}{1-\tan b_n\tan b_{n-1}} =\tan(b_n+b_{n-1})$$ and so $$b_{n+1}=b_n+b_{n-1}+\langle\hbox{a multiple of $\pi$}\rangle\ .$$ With the initial conditions this gives $$b_n=F_n\frac\pi3+\langle\hbox{a multiple of $\pi$}\rangle$$ and so $$a_n=\tan\Bigl(F_n\frac\pi3\Bigr)\ .$$

Answer 3

I have a minor dispute with the previously posted solutions. Although the using the tangent transformation converts the recurrence to the Fibonacci form, the result is not the Fibonacci sequence itself, as a consequence of the initial conditions. In fact, if we have the equation $f_n=f_{n-1}+f_{n-2}$ with initial conditions $f_0=f_1$ we can show that

$$f_n=f_0(F_n+F_{n-1})=f_0F_{n+1}$$

where $F_n$ is the Fibonacci sequence. Thus, in the present case I find that

$$a_n=\tan\left(F_{n+1}\frac{\pi}{3} \right)$$

I have verified numerically that

$$a_n=\frac{a_{n-1}+a_{n-2}}{1-a_{n-1} a_{n-2}}=\tan\left(F_{n+1}\frac{\pi}{3} \right)$$

for $a_0=a_1=\sqrt{3}$.