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

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

Aalok Thakkar
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1 Answers1

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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.

However, a lot is known about your particular sequence. See OEIS sequence A000058 and references there.

Robert Israel
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