Consider the following *somos-like* sequence
$$x\_n=\frac{x\_{n-1}^2+x\_{n-2}^2}{x\_{n-3}}.$$
It's known that $x\_n$ is a Laurent polynomial in $x\_0, x\_1$ and $x\_2$. I got interested in the denominators of the sequence $x\_n$. Some initial observations indicate particular structures regarding the exponents of the denominators, so I am tempted to ask:



> 
> **Question:** Is this true?
>  $$x\_n=P\_n(x\_0,x\_1,x\_2)x\_0^{1-F\_{n-3}}x\_1^{1-F\_{n-4}}x\_2^{1-F\_{n-5}},$$
>  where $P\_n$ is some polynomial and $F\_n$ is the Fibonacci sequence.
> 
> 
> 



Denominators of certain Laurent polynomials
Yes, this is true. These are the *denominator vectors* or *$d$-vectors* of the cluster algebra associated to the Markov quiver. The Markov quiver has vertices $\{1,2,3\}$ to two arrows $i \to i+1$ for each $i$ taken modulo $3$ (i.e. a directed $3$-cycle with all double arrows). This quiver has the property that whenever to mutate it the effect is the reversal of all arrows. The sequence in the question is obtained by cluster algebra theory by mutating at $1,2,3,1,2,3,1 \dots$


We can consider the $d$-vector recursion in (2.6) of [arXiv:1210.6299](https://arxiv.org/abs/1210.6299) to give a positive answer to the question. The recursion tells us to compute
$$2(F\_{k+2} - 1) - (F\_k - 1)$$
which equals $F\_{k+3} - 1$ as desired.


