**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}\label{AB-Seq}\tag{AB} 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**

Except the sequence of the binomial coefficients $\binom{n}{k}$ and the sequence in \eqref{AB-Seq}, are there any more solutions 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{3n+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 $m\in\mathbb{N}$?

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