Explicit formula for $a_0=1$, $a_{n+1}=a_n^2+1$

Question

Is there an explicit formula in elementary function for the sequence $a_n$ where $a_0=1$, $a_{n+1}=a_n^2+1$? How does one prove or disprove such claims?

Edit: I think my question may be formulated in the following way: Is there an elementary function $f$ which satisfies $f(0)=1$ and $f(n+1)=f(n)^2+1$?

A003095 The page does not give an explicit formula but there are some links to articles and a list of topics where this sequence arises. — Jean-Claude Arbaut, Dec 07 '18 at 14:16
This may also be relevant: https://math.stackexchange.com/questions/1918972/finding-an-explicit-formula-for-a-n-defined-recursively-by-a-n1-a-n26 — David K, Dec 07 '18 at 14:19
See also https://math.stackexchange.com/questions/2525139/closed-form-for-the-given-recurrence-relation — Robert Israel, Dec 07 '18 at 14:34
@RobertIsrael Thanks. But how do one prove that such a sequence does not have an closed form expression? — Jiu, Dec 07 '18 at 14:37
@Jiu: by trying all possible expressions and showing that they don't match ? — , Dec 07 '18 at 14:38
@Jiu Since "closed form" is not precisely defined, one can't. — Robert Israel, Dec 07 '18 at 14:41

score 4 · Accepted Answer · answered Dec 07 '18 at 14:58

4

According to A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fib. Quart., 11 (1973), 429-437, $$ a(n+1) = \text{round}(b^{2^n})$$ where $b$ is a certain real number, approximately $2.25851845058946539883779624006373187243427469718511465966\ldots$

answered Dec 07 '18 at 14:58

Robert Israel

470,583

With a simple calculator, $2.2585184505894653988377962400637$, estimated from $a_7$. – Dec 07 '18 at 15:17
Based on formula from your link , $b = \exp\left[\sum\limits_{k=0}^\infty \frac{1}{2^k}\log\left(1 + \frac{1}{a_k^2}\right)\right]$ – achille hui Dec 07 '18 at 15:26

score 1 · Answer 2 · answered Dec 07 '18 at 14:43

As the sequence is very quickly growing, past a certain term, let $a_m$, the $+1$ becomes really tiny and the iterates are well approximated by

$$x_n=(x^m)^{2^{n-m}}.$$

An interesting question is if it holds that

$$x_n\sim c^{2^n}$$ where $c$ would be

$$\lim_{n\to\infty}\sqrt[2^n]x_n.$$

Explicit formula for $a_0=1$, $a_{n+1}=a_n^2+1$

2 Answers2