Let $x$ and $y$ be variables. Consider the following recurrence:
\begin{equation}
u_{n}:=
\begin{cases}
\displaystyle{\frac{1+u_{n-1}^b}{u_{n-2}}} & if\ n\ \text{is even},\\
&\\
\displaystyle{\frac{1+u_{n-1}^c}{u_{n-2}}} & if\ n\ \text{is odd}.
\end{cases}
\end{equation}
with $b,c\in\mathbb{N}^{+}$ (non zero naturals numbers), $u_{0}=x$ and $u_{1}=y$. Suppose $bc\geq 3$. Consider the homogeneous linear recurrence relations of order 4 given by the formulas:
$$\alpha_{n}=(bc-2)\cdot\alpha_{n-2}-\alpha_{n-4},\ \ \forall n\geq 5$$
with initial conditions:
$$(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4})=(0,1,c,bc-1);$$
and
$$ \beta_{n}=(bc-2)\cdot\beta_{n-2}-\beta_{n-4},\ \ \forall n\geq 5 $$
with initial conditions:
$$ (\beta_{1},\beta_{2},\beta_{3},\beta_{4})=(0,1,b,bc-1).$$
I found $\alpha_n$ and $\beta_n$ experimentally. Notice that both recurrences $\alpha_n$ and $\beta_n$ reduce to the formula of Definition 1
from https://arxiv.org/pdf/1106.0952.pdf when $b=c$ (they call $r$ the common value when $b=c$).
Using Maple I verified several cases where for some polynomials $P_{n,b,c}, Q_{n,b,c}\in\mathbb{R}[x,y]$ and for $n\geq 4$, we have the reduced expressions:
$$ u_n=\frac{\displaystyle{P_{n,b,c}(x,y)}}{x^{\beta_{n-1}}y^{\alpha_{n-2}}},\quad n\geq 4$$
and also
$$ u_{3-n}=\frac{\displaystyle{Q_{n,b,c}(x,y)}}{y^{\beta_{n-1}}x^{\alpha_{n-2}}},\quad n\geq 4$$
I was trying to tackle the following questions:
Can we show that if the recurrence $u_n$ is periodic (that is, $u_{n+p}=u_{n}$ for all $n\in \mathbb{Z}$ and some fixed positve integer $p$) then $bc\leq 3$ without using Cluster Algebras and the Fomin-Zelevinsky results ? It is clear that $bc\leq 3$ implies that the recurrence for $u_n$ is periodic since it is well know that: $$bc=1 \Rightarrow p=5,\ \textrm{that is}\ \ u_{n+5}=u_{n},\ \textrm{for all}\ n\in\mathbb{Z}$$ $$bc=2 \Rightarrow p=6,\ \textrm{that is}\ \ u_{n+6}=u_{n},\ \textrm{for all}\ n\in\mathbb{Z}$$ $$bc=3 \Rightarrow p=8,\ \textrm{that is}\ \ u_{n+8}=u_{n},\ \textrm{for all}\ n\in\mathbb{Z}$$ for any non zero initial conditions $u_0=x$ and $u_1=y$ I proved this directly from definition. The converse result is a well known result of Fomin-Zelevinsky but I was wondering if we can find a method to prove it without Cluster Algebras.
Can we find formulas for $P_{n,b,c}$ and $Q_{n,b,c}$ like: $$P_{n,b,c}(x,y)=\sum_{\gamma\in\mathcal{F}_n} x^{b\vert\gamma\vert_{1}}y^{c(\alpha_{n-1}-\vert\gamma\vert_{2})}$$ $$Q_{n,b,c}(x,y)=\sum_{\gamma\in\mathcal{F}_n} y^{b\vert\gamma\vert_{1}}x^{c(\alpha_{n-1}-\vert\gamma\vert_{2})}$$ where the sum is taken over a family $\mathcal{F}_n$ of subpaths of Dyck pahts ?
How to show that $x^{\beta_{n-1}}y^{\alpha_{n-2}}$ is the denominator of the reduced fraction $u_n$. The same for $u_{3-n}$.
Can we find a solution for $P_{n,b,c}$ and $Q_{n,b,c}$ using the notion of Greedy elements (see https://arxiv.org/pdf/1208.2391.pdf) ? Can we express these polynomials as a sum over a set of compatible pairs in $\mathcal{D}^{\beta_{n-1}\times\alpha_{n-2}}$ ?
From that fact that both recurrences $\alpha_n$ and $\beta_n$ have the same characteristic polynomial: $$C_{b,c}(x)=x^4+(2-bc)x^2+1$$ whose factorization ca be written as: $$ x^4+(2-bc)x^2+1=(x^2-\sqrt{bc}\,x+1)(x^2+\sqrt{bc}\,x+1), $$ can we write a nice formula for a closed solution of them ?From that fact that both recurrences $\alpha_n$ and $\beta_n$ have the same characteristic polynomial: $$C_{b,c}(x)=x^4+(2-bc)x^2+1$$ whose factorization ca be written as: $$ x^4+(2-bc)x^2+1=(x^2-\sqrt{bc}\,x+1)(x^2+\sqrt{bc}\,x+1), $$ can we show somehow that $u_{n}$ is periodic for some $p$ if $bc\leq 3$, that is equivalent to the fact that $(x^2\pm\sqrt{bc}\,x+1)$ have complex roots, because the discriminant of $(x^2\pm\sqrt{bc}\,x+1)$ is $bc-4$. As $b$ and $c$ are naturals $bc<4$ iff $bc\leq 3$.
This is motivated by the work of Lee and Schiffler see https://arxiv.org/pdf/1106.0952.pdf and https://arxiv.org/pdf/1208.2391.pdf