Let $c(x)=\frac{1-\sqrt{1-4x}}{2x}$ be the generating function of the Catalan numbers.
It satisfies $$\frac{1}{c(x)^k}+x^k c(x)^k=L_k(1,-x),$$
where $L_n(x,s)$ denote the Lucas polynomials defined by $ L_n(x,s)=x L_{n-1}(x,s)+s L_{n-2}(x,s)$ with initial values $ L_0(x,s)=2$ and $ L_(x,s)=x.$
This follows from the well-known und easily verified identity $L_n(x+y,-xy)=x^n+y^n,$ because $\frac{1}{c(x)}=1-xc(x).$
I am interested in a generalization for the Narayana polynomials $C_n(t)=\sum_{k=0}^{n-1}\binom{n}{k}\binom{n-1}{k}\frac{t^k}{k+1}.$
In the literature there occur two different generating functions, $c_0(x,t)=\sum_{n\geq0}C_n(t)x^n$ and $c_1(x,t)=1+\sum_{n\geq1}tC_n(t)x^n$.
Let us define analogs of $c(x)^k$ by $c(x,t)^{(2k)}= c(x,t)^{(2k-1)}c_1(x,t)$ and
$c(x,t)^{(2k+1)}= c(x,t)^{(2k)}c_0(x,t)$ with $c(x,t)^{(0)}= 1$ and call their coefficients convolution powers of the Narayana numbers.
Computations suggest that
$$\frac{1}{c(x,t)^{(k)}}+t^{\lfloor{\frac{k+1}{2}}\rfloor}x^k c(x,t)^{(k)}=h_k(x,t)$$
for some polynomials $ h_k(x,t)$ of degree ${\lfloor{\frac{k+1}{2}}\rfloor}$.
For even $k$ this follows as above and gives $h_{2k}(x,t)=L_k(1-(1+t)x,-t x^2).$
The first $h_k(x,t)$ are $h_1(x,t)=1+(t-1)x$, $h_2(x,t)=1-(1+t)x$, $h_3(x,t)=1-(2+t)x+(1-t)x^2$, $h_4(x,t)=1-2(1+t)x+(1+t^2)x^2$, $h_5(x,t)=1-(3+2t)x+(3+t+t^2)x^2+(1-t)x^3$.
Any suggestions for a proof for odd $k$?
Best Answer
First off, it should be added that $C_0(t)=1$, which does not follow from the given formula.
Let's show that $$h_{2k+1}(x,t) = L_{2k}(1-(1+t)x,-tx^2)tx + L_{2k+2}(1-(1+t)x,-tx^2).$$
First notice that $c_0(x,t)$ satisfies the identity $1 - (1+(t-1)x)c_0(x,t) + txc_0(x,t)^2=0$ and that $c_1(x,t) = 1 + (c_0(x,t)-1)t$.
We have $c(x,t)^{(k)} = c_0(x,t)^{\lceil k/2\rceil} c_1(x,t)^{\lfloor k/2\rfloor}$. So, our goal is to evaluate $$h_{2k+1}(x,t) = a + b,$$ where $$a:=\frac1{c_0(x,t)^{k+1} c_1(x,t)^k},\quad b:=t^{k+1}x^{2k+1}c_0(x,t)^{k+1} c_1(x,t)^k.$$
From the known formulae for $h_{2k}(x,t)$ and $h_{2k+2}(x,t)$, we get the following system of linear equations: $$\begin{cases} c_0(x,t)\cdot a + \frac1{txc_0(x,t)}\cdot b = L_{2k}(1-(1+t)x,-tx^2),\\ \frac1{c_1(x,t)}\cdot a + xc_1(x,t)\cdot b = L_{2k+2}(1-(1+t)x,-tx^2). \end{cases} $$ Solving it for $a$ and $b$, we deduce the formula for $a+b$ given above.
PS. Computation of manageable formulae for $a$ and $b$ is a bit tedious, but knowing the target formula, we can take a linear combination $$L_{2k}(1-(1+t)x,-tx^2)tx + L_{2k+2}(1-(1+t)x,-tx^2) = \left(c_0(x,t)tx + \frac1{c_1(x,t)}\right)\cdot a + \left(\frac1{c_0(x,t)} + xc_1(x,t)\right)\cdot b,$$ and it just remains to verify that the coefficients of $a$ and $b$ both equal $1$.