Find the general term of the sequence $a_n = a_{n-1}^2 +1$

Question

I need to find the general rule of the sequence $1,2,5,26,677,...$ I know that the recursive definition is $a_n = a_{n-1}^2 +1$.

Could someone please explain HOW to get to the general form. Not WHAT it is. Thanks.

Answer 1

I doubt that a closed form is possible.

For the simpler sequence

$$b_n=b_{n-1}^2$$

the general term is

$$b_n=(b_0)^{2^n}$$

which gives you a first asymptotic estimate.

As for large $a_n$ the term $1$ is neglectible, you can use this approximation of your sequence

$$a_n=(a_k)^{2^{n-k}}$$ for some $k<n$.

Now if you consider the expression

$$\sqrt[2^n]{a_n},$$ it tends to a constant, about $c=1.2259024$, and $a_n\approx c^{2^n}$.

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After a little more research, it turns out that a closed formula is

$$\color{green}{a_n=\left\lfloor c^{2^n}\right\rfloor}$$ where

$$c=\lim_{n\to\infty}\sqrt[2^n]a_n,$$ because the approximation is so good that the error is always smaller than one unit, by excess.