Rewriting generating function in two variables 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?
 A: 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.

