Rewriting generating function in two variables

Question

Let $$ N_n(t) = \sum_{k \geq 0} N_{n,k}t^{k+1} $$ and $$ N(t,z) = 1 + \sum_{n \geq 1} N_n(t)z^n. $$ ($N_n(t)$ is a shift of the Narayana polynomial by $t$).

The polynomial $N_n(t)$ satisfies the following recurrence

$$ (n+1)N_n(t) = (2n-1)(1+t)N_{n-1}(t) - (n-2)(1-t)^2N_{n-2}(t). $$

This recurrence is supposed to correspond to

$$ (1-(1+t)z)N(t,z) = 1 + (t-1)z + z(2(1+t)z - (1-t)^2z^2 - 1) \frac{d}{dz}[N(t,z)] $$

according to the book I am reading. I have tried summing both sides of the recurrence over $n$ but I don't get it to match. May I please have some help deriving the identity above?

Answer 1

We consider the generating function \begin{align*} N(t,z)=1+\sum_{n\geq 1}N_n(t)z^n \end{align*} of the shifted Narayana polynomials $N_n(t) = \sum_{k \geq 0} N_{n,k}t^{k+1}$ with $N_{n,k}$ the Narayana numbers. They fulfil for $n\geq 2$ \begin{align*} (1+n)N_n(t)=(2n-1)(1+t)N_{n-1}(t)-(1-t)^2(n-2)N_{n-2}(t)\tag{1} \end{align*}

Here we show (1) corresponds with the differential equation \begin{align*} (1-(1+t)z)N(t,z) = 1 + (t-1)z + z\Big(2(1+t)z - (1-t)^2z^2 - 1\Big) \frac{d}{dz}[N(t,z)]\tag{2} \end{align*}

With $[z^n]$ denoting the coefficient of $z^n$ of a series and since \begin{align*} z\frac{d}{dz}N(t,z)=\sum_{n \geq 1} nN_n(t)z^n \end{align*}

we obtain for $n\geq 2$ from (2)

LHS: \begin{align*} [z^n]&(1-(1+t)z)N(t,z)\\ &=\left([z^{n}]-(1+t)[z^{n-1}]\right)\left(1+\sum_{n\geq 1}N_n(t)z^n\right)\\ &=N_n(t)-(1+t)N_{n-1}(t) \end{align*} RHS: \begin{align*} [z^n]&\left\{1 + (t-1)z + z\Big(2(1+t)z - (1-t)^2z^2 - 1\Big)\frac{d}{dz}[N(t,z)]\right\}\\ &=[z^n]\left\{z\Big(2(1+t)z - (1-t)^2z^2 - 1\Big) \frac{d}{dz}[N(t,z)]\right\}\\ &=\Big(2(1+t)[z^{n-1}] - (1-t)^2[z^{n-2}] - [z^n]\Big) \sum_{n \geq 1} nN_n(t)z^n\\ &=2(1+t)(n-1)N_{n-1}(t) - (1-t)^2(n-2)N_{n-2}(t) -nN_n(t) \\ \end{align*}

Equating LHS and RHS results in

\begin{align*} N_n(t)-(1+t)N_{n-1}(t)&=2(1+t)(n-1)N_{n-1}(t) - (n-2)(1-t)^2N_{n-2}(t) -nN_n(t)\\ (n+1)N_n(t)&=(2n-1)(1+t)N_{n-1}(t)-(n-2)(1-t)^2N_{n-2}(t) \end{align*} and the claim follows.