Let $c(n,k)$ denote the number of Dyck paths of semilength $n$ which are contained in the strip $0 \leq y \leq 2k + 1.$

They satisfy the recursion $\sum_{j=0}^{k+1}(-1)^j \binom{2k+2-j}{j}c(n-j,k)=0$ for $n>k.$

We can extend the sequence to negative $n$ such that this recursion holds for all $n \in \mathbb{Z}.$

I am interested in the generating function of the sequence ${\left( {c( - n,k)} \right)_{n \geq 0}}.$

It is well known that $\sum\limits_{n \geq 0} {c(n,k){x^n}} = \frac{{{F_{2k + 1}}( - x)}}{{{F_{2k + 2}}( - x)}}$ if by ${F_n}(x) = \sum\limits_{j = 0}^{\left\lfloor {\frac{n}{2}} \right\rfloor } \binom{n-j}{j} x^j $ we denote the Fibonacci polynomials which satisfy ${F_n}(x) = {F_{n - 1}}(x) + x{F_{n - 2}}(x)$ with initial values $F_0(x)=F_1(x)=1.$

Computations for small $k$ suggest that $\sum\limits_{n \geq 0} {c( - n,k){x^n}} = - \frac{1}{x}\frac{{{F_{2k}}( - \frac{1}{x})}}{{{F_{2k + 2}}( - \frac{1}{x})}}.$ As mentioned in OEIS A080937 and A038213 for $n=2$ this result is due to Michael Somos.

These generating functions imply that $c(n,k)$ satisfies the recursion for $\left| n \right| > k.$

But to show that $c(-n,k)$ is the looked for extension we need the recursion for all $n$. Any idea how to do this?