A Strange Recursion – Mattia Giuri's bizarre blog

This is a really nice recursion I wanted to share.

$\begin{cases} a_0 = 1 \\ a_{n+1} = 2a_n +\sqrt{3a_n^2 - 2} \end{cases}$

We have to find the general formula for the nth term of the succession.

We have:

$(a_{n+1} - 2a_n)^2 = 3a_n^2-2$

Also:

$(a_{n} - 2a_{n-1})^2 = 3a_{n-1}^2-2$

By subtracting the two expressions and simplifying we get:

$a_{n+1}^2-a_{n-1}^2 -4a_n(a_{n+1}-a_{n-1}) = 0$

Now, since $a_n > a_{n-1}$ we can further simplify to get the final form:

$a_{n+1} = 4a_n - a_{n-1}$

By solving the recursion’s equation we can find the nth term of the sequence.

$t^2-4t+1 = 0$

Its solutions are $2+\sqrt{3}, 2-\sqrt{3}$ .

Now by solving the following system we can find constants x and y such that $a_n = x(2+\sqrt{3})^n + y(2-\sqrt{3})^n$ .

$\begin{cases} x + y = 1 \\ (2+\sqrt{3})x + (2-\sqrt{3})y = 3 \end{cases}$

We find

$\begin{cases} x = \frac{1+\sqrt{3}}{2\sqrt{3}} \\ y = \frac{\sqrt{3}-1}{2\sqrt{3}} \end{cases}$

So the nth term equals

$a_n = \frac{1}{2\sqrt{3}}[(1+\sqrt{3})(2+\sqrt{3})^n + (\sqrt{3}-1)(2-\sqrt{3})^n]$

$a_n = \frac{1}{\sqrt{3}}[(\frac{1+\sqrt{3}}{2})^{2n+1} - (\frac{1-\sqrt{3}}{2})^{2n+1}]$

And the problem is solved.