Supremum of $ a_n = a_{n-1}^3 - a_{n-2} $ Let $a_1=0$ and let $ - \ln(2) < a_2 < \ln(2) $ 
Define
$$ a_n = a_{n-1}^3 - a_{n-2} $$
Then 
$$ \sup_{n>2} a_n = a_2 $$
And
$$ \inf_{n>2} a_n = - a_2 $$
How to prove that ?
 A: For $b:=a_2$, the GCD of the polynomials $a_{10}=a_{10}(b)$ and $a_9(b)+b$ is 
$$b \left(b^2+1\right) \left(b^{78}-b^{76}+b^{74}-b^{72}-8 b^{70}+8 b^{68}-8 b^{66}+8 b^{64}+28 b^{62}-28
   b^{60}+28 b^{58}-28 b^{56}-59 b^{54}+59 b^{52}-59 b^{50}+59 b^{48}+85 b^{46}-85 b^{44}+85 b^{42}-85
   b^{40}-86 b^{38}+86 b^{36}-86 b^{34}+86 b^{32}+61 b^{30}-61 b^{28}+61 b^{26}-60 b^{24}-30 b^{22}+30
   b^{20}-30 b^{18}+27 b^{16}+9 b^{14}-9 b^{12}+9 b^{10}-6 b^8-3 b^6+3 b^4-3 b^2+1\right). 
$$
The only root of this GCD in $(0,1)$ is $b_*\approx0.637295\in(0,\ln2)$. 
So, for $a_2=b_*$ we have $a_{10}=0=a_1$ and $-a_9=a_{11}=a_2$, so that the sequence $(a_n)$ is of period $9$; we also have $\max_n a_n=a_2$ and $\min_n a_n=-a_2$. 
A: I encountered a similar problem with the iteration $u_n=\sqrt{u_{n-1}^2+1}-u_{n-2}$, where there is fundamentally a $9$-cycle in the shape of a maple leaf
(replace the +1 by 0 to see this). I asked an expert 20 years ago who told me that using the
KAM (Kolmogorov-Arnold-Moser) theorem, one could prove that the "curve" drawn
above by R. Israel is not a curve at all, but a very thin strip of width 0.07
or something. Don't ask me for the proof, I have no idea.
A: The question is really about the iteration behaviour of the maps $(x,y) \mapsto (y, y^3-x)$ with various starting points.  We have a fixed point $(0,0)$ and a $6$-cycle $$(1,0),(0, -1), (-1, -1), (-1, 0), (0, 1), (1, 1)$$
It appears that for something like $-0.797 < x < 0.797$, $(0,x)$ is on an invariant curve.  For example, here are $10000$ iterates starting at $(0,0.796)$:

For starting values with slightly larger $x$, the process seems to become unstable.
