Context: I want to do partial fraction decomposition of a fraction of polynomials where the denominator is of the form $1 + aX + bX^2$.
Now to my concrete problem: Due to Bézout's identity or, more concretely, the Extended Euclidean Algorithm, we know that for any $n\in\mathbb{N}_{>0}$ and $c\in\mathbb{C}$, there exist polynomials $p(X), q(X) \in \mathbb C[X]$ such that $$p(X)(1 - c X)^n + q(X) (1 - \bar{c} X)^n = 1$$ Is there any simple closed form for these polynomials, perhaps in term of some well-known sequence of polynomials? Or a recursive formula for them?
For $n=1$, $p(X)$ and $q(X)$ are constant. We define $p_1 = \bar{c}/(\bar{c} - c)$ and $q_1 = -c/(\bar{c} - c)$. It can easily be seen that $p_1(1-cX)+q_1(1-\bar cX)=1$. Then $$ 1 = \left(p_1(1-cX)+q_1(1-\bar cX)\right)^{2n-1} \\ =\sum\limits_{k=0}^{2n-1} \binom{2n-1}{k}\left(p_1(1-cX)\right)^{2n-1-k}\left(q_1(1-\bar cX)\right)^{k} \\ =\sum\limits_{k=0}^{n-1} \binom{2n-1}{k}\left(p_1(1-cX)\right)^{2n-1-k}\left(q_1(1-\bar cX)\right)^{k} \\ +\sum\limits_{k=n}^{2n-1} \binom{2n-1}{k}\left(p_1(1-cX)\right)^{2n-1-k}\left(q_1(1-\bar cX)\right)^{k}\\ =\left(p_1(1-cX)\right)^{n}\sum\limits_{k=0}^{n-1} \binom{2n-1}{k}\left(p_1(1-cX)\right)^{n-1-k}\left(q_1(1-\bar cX)\right)^{k} \\ +\left(q_1(1-\bar cX)\right)^{n}\sum\limits_{k=0}^{n-1} \binom{2n-1}{k+n}\left(p_1(1-cX)\right)^{n-1-k}\left(q_1(1-\bar cX)\right)^{k}\\ =p_1^n(1-cX)^n\sum\limits_{k=0}^{n-1} \binom{2n-1}{k}\left(p_1(1-cX)\right)^{n-1-k}\left(q_1(1-\bar cX)\right)^{k} \\ +q_1^n(1-\bar cX)^n\sum\limits_{k=0}^{n-1} \binom{2n-1}{k}\left(p_1(1-cX)\right)^{k}\left(q_1(1-\bar cX)\right)^{n-1-k} $$ Such that you can use $$ p_n(X) = p_1^n\sum\limits_{k=0}^{n-1} \binom{2n-1}{k}\left(p_1(1-cX)\right)^{n-1-k}\left(q_1(1-\bar cX)\right)^{k} $$ and $$ q_n(X) = q_1^n\sum\limits_{k=0}^{n-1} \binom{2n-1}{k}\left(p_1(1-cX)\right)^{k}\left(q_1(1-\bar cX)\right)^{n-1-k} $$