Let $g(x) = x^6 - 30 x $
Let $h(x) = x^6 $
Let $f(x) = x^2 - 2 $
Let $r$ be a reduced fraction $0 < \frac{p}{q} < 2 $ with integers $p,q > 1$
Let $f_{n+1}(x) = f(f_n(x)) = f_n(f(x)) , f_0(x) = x$.
Now consider for n going to infinity the following ' averages ' :
$$ a(r,v) = \lim n^{-1} \sum_{i=1}^n f_n(r)^{2v+1} $$
For $v > -1 $ an integer.
$$ b(r) = \lim n^{-1} \sum_{i=1}^n g(f_n(r)) $$
$$ c(r) = \lim n^{-1} \sum_{i=1}^n h(f_n(r)) $$
Now Independant of our choices of $r,v$ we get
$$ a(r,v) = 0 $$ (property a)
However it appears that for most $v$ the choice $r = 13/20 $ converges as one of the fastest !?
That is a bit mysterious to me.
From property a , it is easy to see that $b(r) = c(r)$. However ... It seems that usually $b(r) $ converges Faster than $c(r)$.
That is the reason d'être of b. Another mystery.
The third mystery is that again for $b,c$ that value $r = 13/20 $ is one of the fastest.
The Fourth mystery is that apparently for all $r$ ;
$$ b(r) = c(r) = 20 $$
I have been told this related to the logistic map, but i do not know how.
Since i can only ask one question at a time the main question is this
Is it true that
$$ b(13/20) = 20 $$
And if so , how to prove it ?
Update.
Gerry Myerson's hint that
$$\alpha=(x\pm\sqrt{x^2-4})/2$$
$$ f_n(x)=\alpha^{2^n}+\alpha^{-2^n} $$
Together with $(c + 1/c)^6 = * + 20$
Proves where they number $20$ came from.
The average of even Powers of iterations of $f$ are the central binomial coefficients. Nice.
So that solved Mystery 4.
Below is however another mystery :
Mystery 5 :
It appears that for $t = 13/20 $
$$ \lim \sum^n \frac{ \frac{1}{t} + \frac{1}{f(t)} + ... \frac{1}{f_n(t)} }{n} = \frac{-1}{2} $$