# Ask for a generating function or an explicit expression of a triangle of positive integers

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 $$$$\label{C(n+1:n)-Explicit}\tag{PQ4} C_{n+1,n}=\frac{2n+3}{2}B\biggl(\frac{1}{2},n+2\biggr)-1 =\frac{(2n+2)!!}{(2n+1)!!}-1, \quad n\in\mathbb{N}_0=\{0,1,\dotsc\}.$$$$ where $$B(\alpha,\beta)$$ denotes the classical beta function. Therefore, by virtue of the formulas \eqref{C(n1)} and \eqref{C(n2)}, the recurrent relation \eqref{recur-c-C(n-k)-Four} can be inductively and recursively transformed to $$$$\label{Similar-Pascal-Rul}\tag{PR} C_{n+2,k}=C_{n+1,k}+C_{n+1,k-1}, \quad 1\le k\le n.$$$$

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 $$n\in\mathbb{N}$$?

It is common knowledge that the binomial coefficients $$\binom{n}{k}$$ satisfy Pascal's rule $$$$\binom{n+2}{k}=\binom{n+1}{k}+\binom{n+1}{k-1}.$$$$ This means that the sequence of binomial coefficients $$\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 $$$$\label{AB-Seq}\tag{AB} C_{n,k}=\sum_{j=0}^{k}\frac{(-1)^j}{2j+1}\binom{n}{k-j}, \quad 0\le k\le n-1.$$$$

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 $$$$\label{beta(m-j)}\tag{PQ8} C_{n+m,n}=\frac{1}{(2m-1)!!} \frac{(2n+2m)!!}{(2n+1)!!}-\sum_{j=0}^{m-1}\beta_{m,j}n^j, \quad m,n\in\mathbb{N}.$$$$ What is the explicit or closed-form expression of $$\beta_{m,j}$$ in \eqref{beta(m-j)} for $$0\le j\le m-1$$ and $$m\in\mathbb{N}$$?

• The first row is the sequence at oeis.org/A025547 Dec 13, 2022 at 4:06
• I think it will not help to just post some numbers without giving the method / definition how you compute those numbers. Dec 13, 2022 at 5:57
• Looks like the main diagonal is A129890(n)/A025549(n). Dec 13, 2022 at 7:11
• This should be closed if an explanation is not provided soon. Dec 13, 2022 at 8:23
• From your formulas, it looks like $$C_{n,k}=\sum_{j=0}^{k}\frac{(-1)^j}{2j+1}\binom{n}{k-j}.$$ Dec 14, 2022 at 20:18

The generating function: $${\cal C}(x,y) = \sum_{n,k\geq 0} C_{n,k} x^n y^{2k}$$ has the following explicit form: $${\cal C}(x,y) = \frac{\arctan(y)}{y(1-x(1+y^2))}.$$
UPDATED. Finally, we get the following formula for the coefficicnets $$\beta_{m,t}$$ in the formula for $$C_{n+m,n}$$: $$\beta_{m,t} = \sum_{j=t}^{m-1} \frac{(-1)^{j+m}}{(2(j-m)+1)\cdot j!}\sum_{\ell=t}^{j} s(j,\ell) \binom{\ell}{t} m^{\ell-t},$$ where $$s(\cdot,\cdot)$$ are Stirling numbers of 1st kind.
• Dear Max, using Mathematica 12, I tried to numerically verify your formula $$\sum_{j=0}^n \frac{(-1)^j}{(2j+1)(m+j)!}\sum_{\ell=t}^{m+j} s(m+j,\ell) \binom{\ell}{t} m^{\ell-t},$$ but found it seemingly wrong. Mar 27 at 13:54
• Dear Max, I substituted your formula into (PQ8) and took $m=1,2,3,4$, but I didn’t obtained (PQ4) to (PQ7). Mar 27 at 17:18
• The quantity $\beta_{m,j}$ should be independent of $n$, but your formula contains $n$ as a limit of a sum. Mar 27 at 17:29
• @qifeng618: Your representation for $C_{n+m,n}$ follows from complementation: $$\sum_{j=0}^n \ldots = \sum_{j=-m}^n \ldots - \sum_{j=-m}^{-1}\ldots,$$ where the first sum in the right-hand side gives a non-polynomial term, while the second sum is a polynomial in $n$ with coefficients $\beta_{m,t}$. Apr 4 at 15:24