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?
$\begingroup$
$\endgroup$
1
-
2$\begingroup$ Written in this way it's not really two recursive sequences, but a recursive pair of sequences. $\endgroup$– YCorCommented Sep 28, 2023 at 9:39
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
$f(n)$ is sequence A073833 and $g(n)$ is sequence A073834 in the OEIS, you can find all information there.
answered Sep 28, 2023 at 7:49
-
$\begingroup$ 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 $b(1) = 1$. $\endgroup$– mhdadkCommented Sep 28, 2023 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 ... $\endgroup$ Commented Sep 28, 2023 at 17:57