Limit related to $f(n) = \frac{f(n-1)^2 (f(n-1)+1)^2}{f(n-2)(f(n-2)+1)^3}$

Let $f(1) = f(2) = 1$and for $n>2:$

$$f(n) = \frac{f(n-1)^2 (f(n-1)+1)^2}{f(n-2)(f(n-2)+1)^3}$$

So we get

$f(3) = 1/2,$$f(4) = 0.07031..f(5) = 0.00335..f(6)= etc$

and the sequence is going to $0$ in the limit.

Now we are getting to zero at an exponential rate

$$ \lim_{n \to \infty} \frac{f(n+1)}{f(n)} = C$$

Where $C$ is a constant

$$C = 0.003879..$$

What is the closed form for $C$ ??

C is probably algebraic !!

see the analogue simpler limits that give algebraic numbers :

$ a_n = \frac{a_{n-1}(a_{n-1} + 1)}{a_{n-2}}.$ and $ T = 3.73205080..$?