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?



Number of bounded Dyck paths with "negative length"
If $f(n)$ satisfies a linear recurrence with constant coefficients for all $n\in \mathbb{Z}$ and we set $F(x)=\sum\_{n\geq 0} f(n)x^n$, then $\sum\_{n\geq 1}f(-n)x^n = -F(1/x)$ (as rational functions). See *Enumerative Combinatorics*, vol. 1, second ed., Prop. 4.2.3.


**Addendum.** Using Exercise 3.66(d) in *Enumerative Combinatorics*,
vol. 1, second ed., it is not hard to show that $c(-n,k)$ is equal
to the number of sequences $(a\_1,a\_2,\dots,a\_{2n-1})$ of positive
integers satisfying $1\leq a\_i\leq k+1$ and $a\_1\leq a\_2 \geq a\_3 \leq
 a\_4 \geq \cdots\geq a\_{2n-1}$.


