I encountered the following triangle of positive integers:

|$c_{n,k}$ | $n=1$ | $n=2$ | $n=3$ | $n=4$ | $n=5$ | $n=6$ | $n=7$ | $n=8$|
|:------:|:------:|:------:|:------:|:------:|:------:|:------:|:------:|:------:|
|$k=0$ | $1$ | $3$ | $15$ | $105$ | $315$ | $3465$ | $45045$ | $45045$|
|$k=1$ || $5$ | $40$ | $385$ | $1470$ | $19635$ | $300300$ | $345345$|
|$k=2$ ||| $33$ | $511$ | $2688$ | $45738$ | $849849$ | $1150149$|
|$k=3$ |||| $279$ | $2370$ | $55638$ | $1317888$ | $2167737$|
|$k=4$ ||||| $965$ | $36685$ | $1200199$ | $2518087$|
|$k=5$ ||||| | $11895$ | $631540$ | $1831739$|
|$k=6$ ||||| | | $169995$ | $801535$|
|$k=7$ ||||| | | | $184331$|

Let $C_{n,k}=\frac{c_{n,k}}{c_{n,0}}$ for $0\le k\le n-1$ and $n\in\mathbb{N}$. These numbers satisfy the following recurrent relations
\begin{gather}
C_{n+2,0}-C_{n+1,0}=0, \label{recur-c-C(n-k)-One}\\
(2n+3)C_{n+2,1}-(2n+3)C_{n+1,1}-(4n+5)C_{n+1,0}+2(n+1)C_{n,0}=0, \label{recur-c-C(n-k)-Two}\\
(2n+3)C_{n+2,n+1}-(4n+5)C_{n+1,n}+2(n+1)C_{n,n-1}=0, \label{recur-c-C(n-k)-Three}
\end{gather}
and
\begin{equation}\label{recur-c-C(n-k)-Four}
(2n+3)(C_{n+2,k}-C_{n+1,k}-C_{n+1,k-1})
=2(n+1)(C_{n+1,k-1}-C_{n,k-1}-C_{n,k-2})
\end{equation}
for $2\le k\le n$ and $n\in\mathbb{N}=\{1,2,3,\dotsc\}$.

It is not difficult to obtain
\begin{gather}
C_{n,0}=1, \quad n\in\mathbb{N}_0=\{0,1,2,\dotsc\},\\
C_{n,1}=\frac{3n-1}{3}, \quad n\ge2,\\
C_{n,2}=\frac{15 n^2-25 n+6}{30}, \quad n\ge3,\\
C_{n,3}=\frac{35n^3-140n^2+147n-30}{210}, \quad n\ge4,
\end{gather}
and
\begin{equation}\label{C(n+1:n)-Explicit}
C_{n,n-1}=\frac{2n+1}{2}B\biggl(\frac{1}{2},n+1\biggr)-1, \quad n\in\mathbb{N},
\end{equation}
where $B(\alpha,\beta)$ denotes the classical beta function.

Can one find an explicit expression of the sequence $c_{n,k}$ for $0\le k\le n-1$ and $n\in\mathbb{N}$?

What is the generating function of the sequence $c_{n,k}$ for $0\le k\le n-1$ and $n\in\mathbb{N}$?