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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.

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    $\begingroup$ Not quite. In an OEIS sequence A064098 comment of mine from 2013, I explicitly state the Laurent property with denominators $a_1^{e_n}a_2^{e_{n-1}}a_3^{e_{n-2}}$ where $e_n:=F_n-1$. The initial values in OEIS are $a_1,a_2,a_3$ but otherwise the sequences are the same. Thus the question equation should be $x_n=P_n(x_0,x_1,x_2)x_0^{1-F_{n}}x_1^{1-F_{n-1}}x_2^{1-F_{n-2}}$. $\endgroup$ – Somos Sep 1 '18 at 19:40
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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 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.

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Suppose that $$\, x_n = c\,x_{n-1}x_{n-2} - x_{n-3} \tag{1}$$ for all $\, n \in \mathbb{Z}\,$ where $\, x_0, x_1, x_2 \,$ are variables and $\, c := (x_0^2 + x_1^2 + x_2^2) / (x_0 x_1 x_2). \,$ Then $$ x_n x_{n-3} = x_{n-1}^2 + x_{n-2}^2 \tag{2}$$ for all $\, n \in \mathbb{Z}.\,$ Define $\, P_n := x_n\, x_0^{e_n} x_1^{e_{n-1}} x_2^{e_{n-2}} \,$ for all $\, n \in \mathbb{Z} \,$ where $\, e_n := -1 + |F_n|. \,$ Notice that $\, x_{2-n}, P_{2-n} \,$ are the same as $\, x_n, P_n \,$ except that $\, x_0 \,$ and $\, x_2 \,$ are swapped. Now we get $$ P_n = (x_0^2 + x_1^2 + x_2^2)\, P_{n-1}P_{n-2} - \big(x_0^{2F_{n-2}} x_1^{2F_{n-3}} x_2^{2F_{n-4}}\big)\, P_{n-3} \tag{3}$$ for all $\, n>3 \,$ from equation $(1)$. Now $\, P_0 = P_1 = P_2 = 1, P_3 = x_1^2 + x_2^2 \,$ and the exponents in equation $(3)$ are non-negative, thus $\, P_n \,$ is a polynomial in $\, x_0,x_1,x_2 \,$ for all $\, n \in \mathbb{Z} \,$ by induction.

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