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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 $k$?

Johann Cigler
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  • Could you list the first few $h_k(x,t)$? (There are all sorts of polynomial sequences in the OEIS related to the Catalan numbers.) // Btw, the arguments $(1-(1+t)x,-tx^2)$ are related to the o.g.f. for the bivariate Narayana polynomials given in my MO-Q https://mathoverflow.net/questions/465989/generalized-puiseux-series-for-diagonal-reflections-of-the-curves-y-fracx with $m =1$, $a=-1$, and $b=-t$. – Tom Copeland Mar 07 '24 at 15:33
  • For t=1, probably related to OEIS A132460 and A034807 in which the refs https://www.fq.math.ca/Scanned/38-5/lang.pdf and https://arxiv.org/pdf/2304.12937.pdf have Binet formulas. A third paper with a generalized Binet formula is http://math.colgate.edu/%7Eintegers/w69/w69.pdf. – Tom Copeland Mar 07 '24 at 17:00
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    In case links should break, these are "Improved Formula for the Multi-Section of the Linear Three-Term Recurrence Sequence" by Gary Detlefs and Wolfdieter Lang, "On polynomials related to powers of the generating functions of Catalan numbers" by Lang, and "Irreducibility of generalized Fibonacci polynomials" by Rigoberto Florez and J. C. Saunders. – Tom Copeland Mar 07 '24 at 17:13

1 Answers1

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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 $1$.

Max Alekseyev
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