The general term of such a recursion formula

Question

The general term formula of such a recursion formula is what? I wanted to know how to prove it has the general term formula and what it is? $$ \frac{1}{a_n^2}+a_n^2=a_{n+1}^2$$ $$a_0=1$$

Answer 1

The numerators and denominators of $a_n^2$ are sequences A073833 and A073834 in the OEIS. There doesn't seem to be a (known) closed form for your sequence. The OEIS gives a few references that you could try looking at:

H. L. Montgomery, Ten Lectures on the Interface Between Analytic Number Theory and Harmonic Analysis, Amer. Math. Soc., 1996, p. 187.

D. J. Newman, A Problem Seminar, Springer; see Problem #60.

J. H. Silverman, The arithmetic of dynamical systems, Springer, 2007, see p. 113 Table 3.1

Edit: Neither Newman nor Silverman gives an explicit formula, but Newman does give very good bounds.

Proposition: Let $x_0 = 1$ and $x_{n+1} = x_n + 1/x_n$. Then $$ \sqrt{2n + 2} \lt x_n \lt \sqrt{2n + \frac52 + \frac12 \ln(n-1)}$$ for all $n \ge 2$.

Proof. Let $y_n = x_n^2 - 2n$. A short calculation shows that $$y_{n+1} = y_n + \frac{1}{y_n + 2n}$$ for all $n \ge 0$. It follows that $$y_n \lt y_{n+1} < y_n + \frac{1}{2n}$$ for all $n \ge 1$. Thus $$2 \lt y_n \lt 2 + \frac{1}{2} \sum_{k=1}^{n-1} \frac{1}{k} < 2 + \frac{1}{2}(\ln(n-1) + 1) = \frac52 + \frac12 \ln(n-1)$$ for all $n \ge 2$. Rewriting this in terms of $x_n$ yields the desired inequality. $\square$

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A final remark: I have found one paper where the recurrence $x_{n+1} = x_n + 1/x_n$ is mentioned explicitly. I suppose you could try emailing the author, but I wouldn't get my hopes up if I were you.