f(0) = 1
g(0) = 1
f(x+1) = ((f(x))^2 + (g(x))^2)
g(x+1) = f(x)g(x)

Is it possible to find a closed form for these two functions?

  • 2
    $\begingroup$ Written in this way it's not really two recursive sequences, but a recursive pair of sequences. $\endgroup$
    – YCor
    Sep 28 at 9:39

1 Answer 1


$f(n)$ is sequence A073833 and $g(n)$ is sequence A073834 in the OEIS, you can find all information there.

  • $\begingroup$ Just in case the two links become dead in the future: $f(n)$ and $g(n)$ are the numerator and denominator respectively of the sequence $b(n) = b(n-1) + \frac{1}{b(n-1)}$ for $n > 1$ and $b(1) = 1$. $\endgroup$
    – mhdadk
    Sep 28 at 17:14
  • $\begingroup$ @mhdadk I'd never imagine I'd say this, but as of 2023, the likelihood of those links becoming dead are less than this site becoming dead itself. Even after meta.mathoverflow.net/q/5737/70594 ... $\endgroup$
    – Glorfindel
    Sep 28 at 17:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.