Let $f(1) = f(2) = 1$ and for $n>2:$
$$f(n) = \frac{(f(n-1)+1)^2 (f(n-1)+2)^2}{(f(n-2)+1)(f(n-2)+2)^3}$$
So we get
- $f(3) = \frac{2}{3} = 0.6666...$
- $f(4) = 0.36579789..$ etc
Now this sequence does not go to infinity. And it does not converge to a constant either.
It oscillates between $0$ and some upper bound that is below $2 \pi$.
But when looking at its graph a strange thing seems to happen.
It seems to converge to a periodic function.
If we look at where $f(n)$ is close to zero, we seem to get a periodic pattern. The average distance between such low values is between $7$ and $9$.
So it seems we get a period $T$ strict between $7$ and $7.25$.
For instance $f(32) = 0.06..$ , $f(39) = 0.03..$ and $f(46) = 0.02$.
So it feels like our $f(n)$ are the integer imputs of a function $f(r)$ that gets closer and closer to a function $g(r)$ with $g(r) = g(r+T)$ where $r$ is a positive real $>3$ and $T$ is around $7$.
Is that true ?
And what is the value of $T$ ? Does $T$ have a closed form ?
See also the related question with a similar recursion and identical initial values.
Limit related to $f(n) = \frac{f(n-1)^2 (f(n-1)+1)^2}{f(n-2)(f(n-2)+1)^3}$
