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Your sequence $p_n (u,v)$ can be defined by $p_n (u,v) = vp_{n - 1} (u,v) - up_{n - 2} (u,v) + p_{n - 3} (u,v)$ with initial values $p_0 (u,v) = 1,p_1 (u,v) = v,p_2 (u,v) = v^2 - u.$ In my above remark I have overlooked a term in the fifth polynomial. This is now the same as the formula given by ARupinski.

Added later: Extend the sequence $p_n (u,v)$ to negative indices by $p_{ - 1} (u,v) = p_{ - 2} (u,v) = 0$ and $p_{ - n} (u,v) = p_{ n - 3} (v,u)$ for $n >2 .$ Define a new sequence of polynomials $r_n (u,v)$ by the same recurrence and initial values $r_{ - 1} (u,v) = 1,r_0 (u,v) = 0,r_1 (u,v) = - u.$ Extend it to negative values by $r_{ - n} (u,v) = r_{n - 2} (v,u).$

Let $A$ be the matrix with rows $(0,1,0),(0,0,1),(1, - u,v).$

Then $A^n$ is the matrix with the following rows: $\left( {p_{n - 3 + j} (u,v),r_{n - 2 + j} (u,v),p_{n - 2 + j} (u,v} \right)$ for $0 \le j \le 2.$

It seems that the sequence $r_n (u,v)$ or the sequence $r_{2n} (u,v)/(1 - uv)$ has analogous properties with respect to the zeroes. Is it also related to the group representation?

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Your sequence $p_n (u,v)$ can be defined by $p_n (u,v) = vp_{n - 1} (u,v) - up_{n - 2} (u,v) + p_{n - 3} (u,v)$ with initial values $p_0 (u,v) = 1,p_1 (u,v) = v,p_2 (u,v) = v^2 - u.$ In my above remark I have overlooked a term in the fifth polynomial. This is now the same as the formula given by ARupinski.