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Given an integer $k\geq 2$, and $k+1$ invertible initial values $s_0,s_1,\ldots,s_k$ in some commutative ring $\mathcal A$ we set $$s_{n+1}=\frac{\sum_{j=1}^ks_{n+1-j}^2+q \sum_{j=1}^{k-1}s_{n+1-j}s_{n-j}} {s_{n-k}}$$ as long as $s_0,s_1,\ldots,s_{n-k}$ are invertible in $\mathcal A(q)$.

We get seemingly $s_n\in \mathcal A[q]$ with coefficients in $\mathcal A$ up to denominators which are products of $s_0,\ldots,s_k$. (It is of course also possible to choose $s_0,\ldots,s_k$ in $ \mathcal A[q]$ or in $\mathcal A(q)$. Denominators of $s_n$ involve seemingly also only factors occuring in $s_0,\ldots,s_k$.)

In particular, if $s_0,\ldots,s_k$ are all in $\{\pm 1\}$, we get seemingly integral polynomials $s_n$ in $\mathbb Z[q]$.

Has anybody a clue why this could be true?

Complements: For $s_0=s_1=\ldots=s_k=1$ the evaluation at $q=0$ yields integral sequences related to solutions of so-called Hurwitz equations (for $k=2,3,4$ we get respectively sequences A64098 (which is also connected to Markov numbers, Frieze-patterns, cluster algebras and so on), A72878 and A72879 of the OEIS).

For $k=2$ (and $s_0=s_1=s_2=1$), the evaluations at $q=1$ and $q=2$ give A165903 and A72882 of the OEIS, other evaluations seem to be absent from the OEIS.

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    $\begingroup$ Well, you already basically noted this, but it seems like it could be related to the Laurent phenomenon in cluster algebra theory, with close connections to things like the Somos sequences, etc. $\endgroup$ Commented Dec 26, 2023 at 20:15
  • $\begingroup$ Resembles Somos polynomials: grail.eecs.csuohio.edu/~somos/somospol.html $\endgroup$ Commented Dec 27, 2023 at 5:41

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Recurrences like these often times have conserved quantities. For your particular case the quantity $$A_n=\frac{s_{n+1}+qs_n+qs_{n-k+1}+s_{n-k}}{s_ns_{n-1}\cdots s_{n-k+1}}$$ remains constant for any $n\geq k$.

Proof: We can calculate the following induction step $$s_{n+2}=\frac{s_{n+1}^2+s_n^2+\cdots +s_{n+2-k}^2+qs_{n+1}s_n+\cdots qs_{n+3-k}s_{n+2-k}}{s_{n+1-k}}$$ $$=\frac{s_{n+1}s_{n-k}+s_{n+1}^2-s_{n-k+1}^2+qs_{n+1}s_n-qs_{n-k+1}s_{n-k+2}}{s_{n+1-k}}$$ $$=\frac{s_{n+1}s_{n-k}+s_{n+1}(A_ns_n\cdots s_{n-k+1}-qs_n-qs_{n-k+1}-s_{n-k})-s_{n-k+1}^2+qs_{n+1}s_n-qs_{n-k+1}s_{n-k+2}}{s_{n+1-k}}$$ $$=A_ns_{n+1}\cdots s_{n-k+2}-qs_{n+1}-qs_{n-k+2}-s_{n-k+1}$$ which ultimately gives us $A_{n+1}=A_n$.


This means that your sequence can also be generated with the following recurrence relation $$s_{n+1}==A_ks_{n}\cdots s_{n-k+1}-qs_{n}-qs_{n-k+1}-s_{n-k}$$ and since $A_k$ is a fixed Laurent monomial in the initial $s_i$'s, your claim follows.

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