# Identifying a sequence of polynomials

Studying a specific quantum cluster algebra, I have come across the following sequence of polynomials :

$$X_1$$ $$q^{-1/2}(X_1X_2-1)$$ $$q^{-1/2}(X_1X_2X_3-X_3-X_1)$$ $$q^{-1}(X_1X_2X_3X_4-X_3X_4-X_1X_2-X_1X_4+1)$$ $$q^{-1}(X_1X_2X_3X_4X_5 - X_1X_4X_5 - X_1X_2X_3 - X_1X_2X_5 - X_3X_4X_5 + X_1 + X_3 + X_5)$$

Either finding a simple expression, or just other places where these polynomials occur would be interesting. They differ only from a $q$ factor from their commutative analogues, which can be caracterized as determinants of the following matrix:

$$\begin{pmatrix} x_n & 1 & 0 & \cdots & 0 \\\\ 1 & x_{n-1} & 1 & \cdots & 0\\\\ 0 & 1 & x_{n-2} & \cdots & 0\\\\ \vdots & \vdots & \vdots & \ddots & \vdots \\\\ 0 & 0 & 0 & \cdots & x_1 \end{pmatrix}$$

I read about a quantum analog of determinants, but it involves quantum matrices and this is not a quantum matrix since it would need to satisfy $X_1 = qX_1$.

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I'm not sure if I understand: are you saying that the sequence of the factors in front of the polynomials, starting from $i=0$, is $1,1,q^{-1/2},q^{-1/2},q^{-1},q^{-1}, q^{-3/2},q^{-3/2},q^{-2},q^{-2}, \dots$ ? If so what's the problem? Or is the sequence of q factors more complicated? –  Pietro Majer Oct 20 '10 at 7:38
Maybe I haven't made myself clear enough. I believe the sequence of the $q$ factors is as you say (but I haven't computed more so I don't know), but I would have like to have an interpretation for those factors. Since those polynomials occurred naturally in a non-commutative extension of the determinant of the matrix I posted, I wondered if there was something like a quantum-determinant that would do the trick. –  Jean-Philippe Burelle Oct 20 '10 at 12:40

$$q^{-[n/2]/2}\sum_{k=0}^{[n/2]}(-1)^k \sum X_{j_1}\dots X_{j_{n-2k}},$$ where the second sum is over sequences $0=j_0,j_1,\dots ,j_ {n-2k},j_{n-2k+1}=n+1$ such that $j_0< j_1< \dots< j_ {n-2k}< j_{n-2k+1}$ and $j_{l+1}-j_l$ is odd for $l=0,\dots,n-2k$. Is not this a simple expression you are searching for?