# Closed form for the given recurrence relation

Let $a_0 = 2$. Let $a_k = (a_{k-1})(a_{k-1} - 1) + 1$.

Is there a closed form solution to evaluate $a_k$?

AFAIK, the only cases where a quadratic recurrence $a_k = f(a_{k-1})$ has a closed form general solution are $f(x) = x^2$ (which gives you $a_k = a_0^{2^k}$) and $f(x) = 2 x^2-1$ (which gives you the Chebyshev polynomials $a_k = T_{2^k}(a_0)$), and the conjugates of these. Yours is not one of those.