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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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    $\begingroup$ The first row is the sequence at oeis.org/A025547 $\endgroup$
    – qifeng618
    Dec 13, 2022 at 4:06
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    $\begingroup$ I think it will not help to just post some numbers without giving the method / definition how you compute those numbers. $\endgroup$ Dec 13, 2022 at 5:57
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    $\begingroup$ Looks like the main diagonal is A129890(n)/A025549(n). $\endgroup$ Dec 13, 2022 at 7:11
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    $\begingroup$ This should be closed if an explanation is not provided soon. $\endgroup$ Dec 13, 2022 at 8:23
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    $\begingroup$ From your formulas, it looks like $$C_{n,k}=\sum_{j=0}^{k}\frac{(-1)^j}{2j+1}\binom{n}{k-j}.$$ $\endgroup$ Dec 14, 2022 at 20:18

1 Answer 1

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The generating function: $${\cal C}(x,y) = \sum_{n,k\geq 0} C_{n,k} x^n y^{2k}$$ has the following explicit form: $${\cal C}(x,y) = \frac{\arctan(y)}{y(1-x(1+y^2))}.$$

For "one more problem", you may like to check a paper on the Graham–Knuth–Patashnik recurrence.

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.

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  • $\begingroup$ 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. $\endgroup$
    – qifeng618
    Mar 27 at 13:54
  • $\begingroup$ @qifeng618: please elaborate on what you think is wrong. $\endgroup$ Mar 27 at 16:18
  • $\begingroup$ Dear Max, I substituted your formula into (PQ8) and took $m=1,2,3,4$, but I didn’t obtained (PQ4) to (PQ7). $\endgroup$
    – qifeng618
    Mar 27 at 17:18
  • $\begingroup$ The quantity $\beta_{m,j}$ should be independent of $n$, but your formula contains $n$ as a limit of a sum. $\endgroup$
    – qifeng618
    Mar 27 at 17:29
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    $\begingroup$ @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 $\beta_{m,t}$. $\endgroup$ Apr 4 at 15:24

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