Preliminaries
I encountered the following triangle of positive integers:
| $c_{n,k}$ |
$n=1$ |
$n=2$ |
$n=3$ |
$n=4$ |
$n=5$ |
$n=6$ |
$n=7$ |
$n=8$ |
| $k=0$ |
$1$ |
$3$ |
$15$ |
$105$ |
$315$ |
$3465$ |
$45045$ |
$45045$ |
| $k=1$ |
|
$5$ |
$40$ |
$385$ |
$1470$ |
$19635$ |
$300300$ |
$345345$ |
| $k=2$ |
|
|
$33$ |
$511$ |
$2688$ |
$45738$ |
$849849$ |
$1150149$ |
| $k=3$ |
|
|
|
$279$ |
$2370$ |
$55638$ |
$1317888$ |
$2167737$ |
| $k=4$ |
|
|
|
|
$965$ |
$36685$ |
$1200199$ |
$2518087$ |
| $k=5$ |
|
|
|
|
|
$11895$ |
$631540$ |
$1831739$ |
| $k=6$ |
|
|
|
|
|
|
$169995$ |
$801535$ |
| $k=7$ |
|
|
|
|
|
|
|
$184331$ |
The first row $c_{n,0}$ for $n\in\mathbb{N}=\{1,2,\dotsc\}$ is perhaps the sequence at https://oeis.org/A025547. The other positive integers $c_{n,k}$ are defined as follows.
Let $C_{n,k}=\frac{c_{n,k}}{c_{n,0}}$ for $0\le k\le n-1$ and $n\in\mathbb{N}$. These real numbers $C_{n,k}$ satisfy
$$
C_{n,0}=1, \quad n\in\mathbb{N}
$$
and the following recurrent relations
\begin{gather}
C_{n+2,1}-C_{n+1,1}=1, \\
(2n+3)(C_{n+2,n+1}-C_{n+1,n})=2(n+1)(C_{n+1,n}-C_{n,n-1}), \\
(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}). \label{recur-c-C(n-k)-Four} \tag{PQ}
\end{gather}
It is not difficult to obtain
\begin{gather}
C_{n,1}=\frac{3n-1}{3}, \quad n\ge2,\label{C(n1)}\tag{PQ1}\\
C_{n,2}=\frac{15 n^2-25 n+6}{30}, \quad n\ge3,\label{C(n2)}\tag{PQ2}\\
C_{n,3}=\frac{35n^3-140n^2+147n-30}{210}, \quad n\ge4,\label{C(n3)}\tag{PQ3}
\end{gather}
and
\begin{equation}\label{C(n+1:n)-Explicit}\tag{PQ4}
C_{n+1,n}=\frac{2n+3}{2}B\biggl(\frac{1}{2},n+2\biggr)-1
=\frac{(2n+2)!!}{(2n+1)!!}-1, \quad n\in\mathbb{N}_0=\{0,1,\dotsc\}.
\end{equation}
where $B(\alpha,\beta)$ denotes the classical beta function. Therefore, by virtue of the formulas \eqref{C(n1)} and \eqref{C(n2)}, the recurrent relation \eqref{recur-c-C(n-k)-Four} can be inductively and recursively transformed to
\begin{equation}\label{Similar-Pascal-Rul}\tag{PR}
C_{n+2,k}=C_{n+1,k}+C_{n+1,k-1}, \quad 1\le k\le n.
\end{equation}
Two Problems
(1) Can one find an explicit expression of the sequence $c_{n,k}$ for $0\le k\le n-1$ and $n\in\mathbb{N}$?
(2) What is the generating function of the sequence $c_{n,k}$ for $0\le k\le n-1$ and $n\in\mathbb{N}$?
About Pascal's rule
It is common knowledge that the binomial coefficients $\binom{n}{k}$ satisfy Pascal's rule
\begin{equation}
\binom{n+2}{k}=\binom{n+1}{k}+\binom{n+1}{k-1}.
\end{equation}
This means that the sequence of binomial coefficients $\binom{n}{k}$ is a solution to the recurrent relation \eqref{Similar-Pascal-Rul}.
As Alexander Burstein commented below, another solution to the recurrent relation \eqref{Similar-Pascal-Rul}, satisfying \eqref{C(n1)}, \eqref{C(n2)}, \eqref{C(n3)}, and \eqref{C(n+1:n)-Explicit}, is
\begin{equation}
C_{n,k}=\sum_{j=0}^{k}\frac{(-1)^j}{2j+1}\binom{n}{k-j}, \quad 0\le k\le n-1.
\end{equation}
One more problem
What is the general solution to the recurrent relation \eqref{Similar-Pascal-Rul}?
Alternative form of $C_{n,k}$
Similar to \eqref{C(n+1:n)-Explicit}, the following expressions are also valid:
\begin{align}
C_{n+2,n}&=\frac{1}{3}\frac{(2n+4)!!}{(2n+1)!!}-\frac{n+5}{3}, \label{C(n+2+n)-Explicit}\tag{PQ5}\\
C_{n+3,n}
&=\frac{1}{15}\frac{(2n+6)!!}{(2n+1)!!}-\frac{15 n^2+65 n+66}{30}, \label{C(n+3:n)-Explicit}\tag{PQ6}\\
C_{n+4,n}
&=\frac{1}{105}\frac{(2n+8)!!}{(2n+1)!!} -\frac{35 n^3+280 n^2+707 n+558}{210} \label{C(n+4:n)-Form}\tag{PQ7}
\end{align}
for $n\in\mathbb{N}_0$. I guess that
\begin{equation}\label{beta(m-j)}\tag{PQ8}
C_{n+m,n}=\frac{1}{(2m-1)!!} \frac{(2n+2m)!!}{(2n+1)!!}-\sum_{j=0}^{m-1}\beta_{m,j}n^j, \quad m,n\in\mathbb{N}.
\end{equation}
What is the explicit or closed-form expression of $\beta_{m,j}$ in \eqref{beta(m-j)} for $0\le j\le m-1$ and $m\in\mathbb{N}$?
