\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{amscd} \usepackage{graphicx} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics} \usepackage{latexsym} \usepackage{epsf} \usepackage{breakurl} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{https://oeis.org/#1}{\rm \underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{problem}[theorem]{Problem} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \def\modd#1#2{#1\ \mbox{\rm (mod}\ #2\mbox{\rm )}} \begin{center} \vskip 1cm{\LARGE\bf A Generalization of the Pascal and \\ \vskip .1in Stirling Matrices} \vskip 1cm \begin{center} \Large A. R. Moghaddamfar\\ Faculty of Mathematics\\ K. N. Toosi University of Technology\\ Tehran, P. O. Box 16765--3381\\ Iran\\ \href{mailto:moghadam@kntu.ac.ir}{\tt moghadam@kntu.ac.ir} \\ \ \\ Navid Salehy\\ Department of Mathematics\\ University of New Orleans\\ New Orleans, LA 70148 \\ USA\\ \href{mailto:ssalehy@uno.edu}{\tt ssalehy@uno.edu}\\ \ \\ Nima Salehy\\ Department of Mathematics and Statistics\\ Louisiana Tech University\\ Ruston, LA 71272\\ USA\\ \href{mailto:nsalehy@latech.edu}{\tt nsalehy@latech.edu}\\ \end{center} \end{center} \vskip .2 in \begin{abstract} The purpose of this article is threefold. First, we introduce a natural generalization of the Pascal and Stirling matrices and investigate its LU decomposition and determinant. Second, we study several well-known integer sequences arising from Pascal's triangle and present a systematic method for recovering these sequences using the Pascal matrix and the exponential function. Third, we obtain determinant representations for linear recurrence relations with constant coefficients. \end{abstract} \section{Introduction, background and definitions}\label{Sec1} We begin by recalling the standard definitions and notation for the Pascal and Stirling matrices. We then introduce shifted versions of these matrices and a natural generalization that will be studied throughout the article. \subsection{The classical Pascal matrix} One of the most well-known numerical triangles in mathematics is Pascal's triangle. It is named after the 17th-century French mathematician Blaise Pascal (1623--1662), although Pascal himself referred to it as the ``Arithmetical Triangle.'' It should be noted, however, that Pascal was not the first to discover this triangle. It had been known and used by mathematicians in Persia, China, India, and elsewhere for centuries before Pascal \cite{Edw}. Pascal's triangle is shown in Fig.~\ref{fig1}. It is constructed as follows: the top entry and the two edges of the triangle are filled with $1$'s, and each remaining entry is obtained by adding the two entries directly above it in the preceding row. \begin{figure}[H] $$\begin{array}{cccccccccccccccc} && & & & & & 1 & & & & & & & & \\ && & & & & 1 & & 1 & & & & & & & \\ && & & & 1& & 2 & & 1 & & & & & & \\ && & & 1 & & 3 & & 3 & & 1& & & & & \\ && & 1& & 4 & & 6 & & 4 & & 1& & & & \\ && 1& & 5 & & 10 & & 10 & & 5 & & 1& & & \\ &1 & & 6& & 15& & 20 & & 15 & & 6 & & 1& & \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \end{array}$$ \caption{Pascal's triangle.}\label{fig1} \end{figure} Here, we are interested in square arrays of numbers derived from Pascal's triangle, which can be constructed in various ways. We will mention some of these methods subsequently. Given a positive integer $n$, the $n\times n$ {\em classical Pascal matrix} $P(n)=[P_{i, j}]_{1\leq i, j\leq n}$ is defined by \[P_{i, j}=\binom{i-1}{j-1}, \qquad 1\leq i, j\leq n,\] where $\binom{i-1}{j-1}=0$ whenever $j>i$. Equivalently, the entries of $P(n)$ can be defined recursively by \begin{displaymath} P_{i, j}= \begin{cases} 1, & \text{if $i\geq 1$, $j=1$;}\\ 0, & \text{if $i=1$, $j>1$;}\\ P_{i-1, j-1}+P_{i-1, j}, & \text{if $i, j\geq 2$.} \end{cases} \end{displaymath} For example, when $n=5$, we have \[P(5)=\begin{pmatrix} 1&.&.&.&.\\ 1&1&.&.&.\\ 1&2&1&.&.\\ 1&3&3&1&.\\ 1&4&6&4&1 \end{pmatrix}.\] \subsection{Stirling numbers and matrices} Stirling numbers are divided into two classes: Stirling numbers of the first kind and Stirling numbers of the second kind. In this article, we consider only Stirling numbers of the second kind. For positive integers $n$ and $k$ with $1\leq k\leq n$, the {\em Stirling number of the second kind}, denoted by $S(n, k)$, is defined by \[ x^n=\sum_{k=1}^{n} S(n, k)x(x-1)(x-2)\cdots(x-k+1). \] In particular, $S(n, n)=1$. We use the convention that $S(n, k)=0$ whenever $k>n$. The Stirling numbers of the second kind satisfy the recurrence relation \begin{equation}\label{eq-s2} S(n,k)=S(n-1,k-1)+kS(n-1,k),\qquad n,k\geq 2. \end{equation} Similarly, we define the $n\times n$ (lower) triangular matrix $S(n)=[S_{i, j}]_{1\leq i, j\leq n}$ with $S_{i, j}=S(i, j)$, which is called the {\em Stirling matrix of the second kind}. The recurrence relation \eqref{eq-s2} yields the following recursive description of $S(n)$. \begin{displaymath} S_{i, j}=\begin{cases} 1, & \text{if $i\geq 1, j=1$;}\\ 0, & \text{if $i=1, j>1$;}\\ S_{i-1, j-1}+jS_{i-1, j}, & \text{if $i, j\geq 2$.} \end{cases}\end{displaymath} For instance, when $n=5$, the Stirling matrix $S(5)$ has the following form: $$S(5)=\left(\begin{matrix} 1& . & . & . & .\\ 1 & 1 & . & . & .\\ 1 & 3 & 1 & . & .\\ 1 & 7 & 6 & 1 & .\\ 1 & 15 & 25 & 10 & 1 \\ \end{matrix}\right).$$ \subsection{Shifted Pascal and Stirling matrices} The $n\times n$ {\em shifted Pascal and Stirling matrices}, $\mathsf{P}(n)=[\mathsf{P}_{i, j}]_{1\leq i, j\leq n}$ and $\mathsf{S}(n)=[\mathsf{S}_{i, j}]_{1\leq i, j\leq n}$, respectively, are defined by \begin{displaymath}\mathsf{P}_{i, j}=\begin{cases} 1, & \text{if $i\geq 1, j=1$;}\\ 1, & \text{if $i=1, j>1$;}\\ \mathsf{P}_{i, j-1}+\mathsf{P}_{i-1, j}, & \text{if $i, j\geq 2$,} \end{cases}\end{displaymath} and \begin{displaymath}\mathsf{S}_{i, j}=\begin{cases} 1, & \text{if $i\geq 1, j=1$;}\\ 1, & \text{if $i=1, j>1$;}\\ \mathsf{S}_{i, j-1}+j\mathsf{S}_{i-1, j}, & \text{if $i, j\geq 2$.} \end{cases}\end{displaymath} For instance, for $n=5$, these matrices have the following form: $$\mathsf{P}(5)=\left(\begin{matrix}1&1 & 1& 1& 1\\ 1 & 2 & 3 & 4 & 5 \\ 1 & 3 & 6 & 10 & 15 \\ 1 & 4 & 10 & 20 & 35 \\ 1 & 5 & 15 & 35 & 70 \\ \end{matrix}\right) \quad \text{and} \quad \mathsf{S}(5)=\left(\begin{matrix}1&1 & 1& 1& 1\\ 1 & 3 & 6 & 10 & 15 \\ 1 & 7 & 25 & 65 & 140 \\ 1 & 15 & 90 & 350 & 1050 \\ 1 & 31 & 301 & 1701 & 6951 \\ \end{matrix}\right).$$ It is well known that \[\mathsf{P}(n)=P(n)P(n)^t,\] for every positive integer $n$; see, for example, \cite{EdSt}. The relationship between the shifted Pascal and Stirling matrices has also been investigated in several works; see, for example, \cite{MaGu,SpZi}. Motivated by these constructions, we introduce the following generalizations. \subsection{Generalizations} Let $\alpha=(\alpha_j)_{j\geq 1}$ be an arbitrary sequence, and define the matrices $S_\alpha (n)=[\widehat{S}_{i, j}]_{1\leq i, j\leq n}$ and $\mathsf{S}_\alpha (n)=[\widehat{\mathsf{S}}_{i, j}]_{1\leq i, j\leq n}$ by \begin{equation}\label{a-1}\widehat{S}_{i, j}=\begin{cases} \alpha_1, & \text{if $i\geq 1, j=1$;}\\ 0, & \text{if $i=1, j>1$;}\\ \widehat{S}_{i-1, j-1}+\alpha_j\widehat{S}_{i-1, j}, & \text{if $i, j\geq 2$,} \end{cases}\end{equation} and \begin{displaymath} \widehat{\mathsf{S}}_{i, j}=\begin{cases} \alpha_1, & \text{if $i\geq 1, j=1$;}\\ \alpha_1, & \text{if $i=1, j>1$;}\\ \widehat{\mathsf{S}}_{i, j-1}+\alpha_j\widehat{\mathsf{S}}_{i-1, j}, & \text{if $i, j\geq 2$.} \end{cases}\end{displaymath} For example, the matrices $S_\alpha (n)$ and $\mathsf{S}_\alpha (n)$ corresponding to $n=3$ are $$S_\alpha (3)=\left(\begin{matrix} \alpha_1 & . & .\\ \alpha_1 & \alpha_1 & .\\ \alpha_1 & \alpha_1+\alpha_2\alpha_1 & \alpha_1 \end{matrix}\right),$$ and $$\mathsf{S}_\alpha (3)=\left(\begin{matrix} \alpha_1 & \alpha_1 & \alpha_1 \\ \alpha_1 & \alpha_1+\alpha_2\alpha_1 & \alpha_1+\alpha_2\alpha_1+\alpha_3\alpha_1 \\ \alpha_1 & \alpha_1+\alpha_2\alpha_1+\alpha_2^2\alpha_1 & \alpha_1+\alpha_2\alpha_1+\alpha_2^2\alpha_1+\alpha_3\alpha_1+\alpha_3\alpha_2\alpha_1+\alpha_3^2\alpha_1 \end{matrix}\right).$$ Since \[S_{(\alpha_1,\alpha_2,\alpha_3,\ldots)}(n)=\alpha_1S_{(1,\alpha_2,\alpha_3,\ldots)}(n),\] and \[\mathsf{S}_{(\alpha_1,\alpha_2,\alpha_3,\ldots)}(n)=\alpha_1\mathsf{S}_{(1,\alpha_2,\alpha_3,\ldots)}(n),\] we may, without loss of generality, restrict our attention to sequences $\alpha=(\alpha_j)_{j\geq1}$ satisfying $\alpha_1=1$. Observe that when $\alpha=(1, 1, 1,\ldots)$, we have \[ S_\alpha(n)=P(n) \qquad\text{and}\qquad \mathsf{S}_\alpha(n)=\mathsf{P}(n). \] On the other hand, when $\alpha=(1, 2, 3, \ldots)$, we have \[S_\alpha(n)=S(n) \qquad\text{and}\qquad \mathsf{S}_\alpha(n)=\mathsf{S}(n). \] Thus, $S_\alpha(n)$ provides a natural common generalization of the Pascal and Stirling matrices, while $\mathsf{S}_\alpha(n)$ provides a corresponding generalization of their shifted versions. In this article, for an arbitrary sequence $\alpha=(\alpha_j)_{j\geq 1}$ with $\alpha_1=1$, we investigate the $n\times n$ matrix $\mathsf{S}_\alpha (n)$. In particular, we obtain an LU decomposition of it as $\mathsf{S}_\alpha (n)=S_\alpha (n) \cdot \mathsf{U}_\alpha(n)$, where $\mathsf{U}_\alpha(n)$ is an upper triangular matrix associated with $\alpha$ (defined by a recurrence relation in Section \ref{Sec2}), and find its determinant. This result generalizes the corresponding results in \cite{EdSt, Kr, MoSaSa}. Since $S_\alpha (n)$ is a lower triangular matrix with 1's on the diagonal, to find the determinant of the matrix $\mathsf{S}_\alpha (n)$, we must explicitly specify the diagonal entries of $\mathsf{U}_\alpha(n)$. For this purpose, we use complete homogeneous symmetric polynomials, which we will introduce below. \subsection{The complete homogeneous symmetric polynomials} {\em The complete homogeneous symmetric polynomials} $h_k(x_1,\ldots,x_n)$, $k\geq 0$, are defined by $h_0(x_1, x_2, \ldots, x_n)=1$, and $$h_k(x_1, x_2, \ldots, x_n)=\sum_{1\leq i_1\leq \cdots \leq i_k\leq n} x_{i_1} x_{i_2} \cdots x_{i_k},$$ or equivalently $$h_k(x_1, x_2, \ldots, x_n)=\sum_{\substack{l_1+l_2+\cdots +l_n=k\\ l_1, l_2, \ldots, l_n\geq 0}} x_1^{l_1} x_2^{l_2} \cdots x_n^{l_n},$$ for all $k\geq 1$. Thus, we have \begin{align*} h_1(x_1, x_2, \ldots, x_n) &=x_1+x_2+\cdots +x_n,\\ h_2(x_1, x_2, \ldots, x_n)&=x_1^2+x_1x_2+\cdots +x_1x_n+x_2^2+x_2x_3+\cdots+x_{n-1}x_n+x_n^2,\\ & \vdots \\ h_n(x_1, x_2, \ldots, x_n)&=\sum_{\substack{l_1+l_2+\cdots +l_n=n\\ l_1, l_2, \ldots, l_n\geq 0}} x_1^{l_1} x_2^{l_2} \cdots x_n^{l_n}. \end{align*} We now define \cite{SpZi} the infinite lower triangular matrix $G(x_1, x_2, \ldots)=[G_{i, j}]_{i, j\geq 1}$, where \begin{displaymath} G_{i, j}=\begin{cases} 0, & \text{if $i2$. We distinguish several cases. First, if $i=1$, then \[\mathsf{U}_{1,j}=1=h_0(\alpha_1,\ldots,\alpha_j),\] for every $j>1$. Next, suppose that $i=2$ and $j>2$. It follows from \eqref{u-1-2} that \[\begin{aligned} \mathsf{U}_{2,j} &=\mathsf{U}_{2,j-1} +\alpha_1\mathsf{U}_{1,j-1} +(\alpha_j-\alpha_1)\mathsf{U}_{1,j}\\ &=\mathsf{U}_{2,j-1}+\alpha_j. \end{aligned} \] By the inductive hypothesis, \[\mathsf{U}_{2,j-1}=h_1(\alpha_2,\ldots,\alpha_{j-1})=\alpha_2+\cdots+\alpha_{j-1},\] and therefore \[\mathsf{U}_{2,j} =\alpha_2+\cdots+\alpha_{j-1}+\alpha_j =h_1(\alpha_2,\ldots,\alpha_j).\] We next consider the diagonal case $i=j\geq 2$. Once again, it follows from \eqref{u-1-2} that \begin{equation}\label{newi=j} \mathsf{U}_{i, i} =\mathsf{U}_{i, i-1} +\alpha_{i-1}\mathsf{U}_{i-1,i-1} +(\alpha_i-\alpha_{i-1})\mathsf{U}_{i-1,i}. \end{equation} Since $\mathsf{U}_{\alpha}(n)$ is upper triangular, $\mathsf{U}_{i, i-1}=0$. Also, since $(i-1)+(i-1)<2i$ and $(i-1)+i<2i$, the inductive hypothesis gives \[\mathsf{U}_{i-1,i-1}=\alpha_{i-1}^{i-2} \quad \text{and} \quad \mathsf{U}_{i-1,i} =h_{i-2}(\alpha_{i-1},\alpha_i).\] Substituting these into \eqref{newi=j} implies that \[\mathsf{U}_{i, i}=\alpha_{i-1}^{i-1}+(\alpha_i-\alpha_{i-1}) h_{i-2}(\alpha_{i-1},\alpha_i).\] On the other hand, using the standard identity \[(y-x)h_r(x, y)=y^{r+1}-x^{r+1},\] we obtain \[(\alpha_i-\alpha_{i-1})h_{i-2}(\alpha_{i-1},\alpha_i)=\alpha_i^{i-1}-\alpha_{i-1}^{i-1}.\] Hence \[\mathsf{U}_{i, i}=\alpha_i^{i-1}=h_{i-1}(\alpha_i),\] as required. It remains to consider the case $j>i\geq 3$. By the inductive hypothesis, we have \[\begin{aligned} \mathsf{U}_{i,j-1} &=h_{i-1}(\alpha_i,\ldots,\alpha_{j-1}),\\ \mathsf{U}_{i-1,j-1} &=h_{i-2}(\alpha_{i-1},\ldots,\alpha_{j-1}),\\ \mathsf{U}_{i-1,j} &=h_{i-2}(\alpha_{i-1},\ldots,\alpha_j). \end{aligned}\] We also use the following standard identities for complete homogeneous symmetric polynomials: \[h_k(x_1, \ldots, x_m)=h_k(x_1, \ldots, x_{m-1})+x_mh_{k-1}(x_1, \ldots, x_m),\] and \[h_k(x_1, \ldots, x_m)=x_1h_{k-1}(x_1, \ldots, x_m)+h_k(x_2, \ldots, x_m).\] Using \eqref{u-1-2} and the two identities above, we obtain \[\begin{aligned} \mathsf{U}_{i, j}= & h_{i-1}(\alpha_i,\ldots,\alpha_{j-1}) +\alpha_{i-1}h_{i-2}(\alpha_{i-1}, \ldots, \alpha_{j-1})+(\alpha_j-\alpha_{i-1})h_{i-2}(\alpha_{i-1}, \ldots, \alpha_j)\\ = &h_{i-1}(\alpha_i, \ldots, \alpha_{j-1}) +\alpha_{i-1}h_{i-2}(\alpha_{i-1}, \ldots, \alpha_{j-1})\\ &+\alpha_j\left[\alpha_{i-1}h_{i-3}(\alpha_{i-1},\ldots,\alpha_j) +h_{i-2}(\alpha_i,\ldots,\alpha_j)\right] \\ & -\alpha_{i-1} \left[ h_{i-2}(\alpha_{i-1},\ldots,\alpha_{j-1}) +\alpha_j h_{i-3}(\alpha_{i-1}, \ldots, \alpha_j) \right]\\ =&h_{i-1}(\alpha_i,\ldots,\alpha_{j-1}) +\alpha_jh_{i-2}(\alpha_i,\ldots,\alpha_j)\\ =&h_{i-1}(\alpha_i,\ldots,\alpha_j). \end{aligned}\] This completes the induction. Finally, since $\mathsf{U}_{\alpha}(n)$ is upper triangular, its determinant is the product of its diagonal entries. Thus \[\det\mathsf{U}_{\alpha}(n)= \prod_{i=1}^n\mathsf{U}_{i, i}=\prod_{i=1}^n\alpha_i^{i-1}.\] The proof is now complete. \end{proof} We are now ready to state the main result of this section. \begin{theorem} Let $\alpha=(\alpha_j)_{j\geq 1}$ be an arbitrary sequence with $\alpha_1=1$, and let $n$ be a positive integer. Then, we have the following matrix decomposition: \begin{equation}\label{decom1} \mathsf{S}_\alpha (n)=S_\alpha (n) \cdot \mathsf{U}_\alpha(n).\end{equation} Additionally, we obtain $$\det \mathsf{S}_\alpha (n)=\prod_{j=1}^{n} \alpha_j^{j-1}.$$ \end{theorem} Before proving this theorem, it is useful to highlight some special cases: \begin{itemize} \item[{\rm (a)}] If $\alpha=(1, 1, 1, \ldots)$, then $\mathsf{S}_\alpha (n)=\mathsf{P}(n)$, $S_\alpha (n)=P(n)$ and $\mathsf{U}_\alpha(n)=P(n)^t$, and we get $\mathsf{P}(n)=P(n)\cdot P(n)^t$ and $\det \mathsf{P}(n)=1$ \cite{EdSt}. \item[{\rm (b)}] If $\alpha=(1, 2, 3, \ldots)$, then $\mathsf{S}_\alpha (n)=\mathsf{S}(n)$ and $S_\alpha (n)=S(n)$, and we get $\mathsf{S}(n)=S(n)\cdot \mathsf{U}_\alpha(n)$ and $\det \mathsf{S}(n)=\prod_{j=1}^{n}j^{j-1}$ \cite{Kr, MoSaSa}. \end{itemize} \begin{proof} As a convenient notation, let us write $\mathsf{S}=\mathsf{S}_\alpha (n)$, $S=S_\alpha (n)$, and $\mathsf{U}=\mathsf{U}_\alpha(n)$. In order to prove \eqref{decom1}, namely $\mathsf{S}=S \cdot \mathsf{U}$, therefore, it suffices to show that their entries correspond to the same recurrence relations satisfying the same initial conditions, that is $$(S \cdot \mathsf{U})_{1, j}=\widehat{\mathsf{S}}_{1, j} \quad \text{and} \quad (S \cdot \mathsf{U})_{i, 1}=\widehat{\mathsf{S}}_{i, 1},$$ for every $1\leq i, j\leq n$, and \begin{equation}\label{relation-MSU} (S \cdot \mathsf{U})_{i, j} = (S \cdot \mathsf{U})_{i, j-1} + \alpha_j (S \cdot \mathsf{U})_{i-1, j}, \end{equation} for every $2 \leq i, j \leq n$. First of all, using simple calculations, we have $$(S \cdot \mathsf{U})_{1, j}=\sum_{k=1}^{n} \widehat{S}_{1, k} \mathsf{U}_{k, j}=\widehat{S}_{1, 1}\mathsf{U}_{1, j}=1\cdot 1=1=\widehat{\mathsf{S}}_{1, j},$$ and $$(S \cdot \mathsf{U})_{i, 1}=\sum_{k=1}^{n} \widehat{S}_{i, k}\mathsf{U}_{k, 1}=\widehat{S}_{i, 1}\mathsf{U}_{1, 1}=1\cdot 1=1=\widehat{\mathsf{S}}_{i, 1},$$ as required. Now, to verify \eqref{relation-MSU}, we proceed as follows. Let us, for the moment, assume that $2 \leq i, j \leq n$. Since \begin{align*} (S \cdot \mathsf{U})_{i, j}&= \sum\limits_{k=1}^{n} \widehat{S}_{i, k}\mathsf{U}_{k, j} =\widehat{S}_{i, 1}\mathsf{U}_{1, j}+\sum\limits_{k=2}^{n} \widehat{S}_{i, k}\mathsf{U}_{k, j}\\ &= \widehat{S}_{i, 1}\mathsf{U}_{1, j-1}+\sum\limits_{k=2}^{n} \widehat{S}_{i, k}\big[\mathsf{U}_{k, j-1} +\alpha_{k-1} \mathsf{U}_{k-1, j-1} + (\alpha_j-\alpha_{k-1}) \mathsf{U}_{k-1, j}\big]\\ &=(S\cdot \mathsf{U})_{i, j-1} +\sum\limits_{k=2}^{n} \widehat{S}_{i, k}\big[\alpha_{k-1} \mathsf{U}_{k-1, j-1} + (\alpha_j-\alpha_{k-1}) \mathsf{U}_{k-1, j}\big], \end{align*} it suffices to show that \begin{equation}\label{last-eq} \sum_{k=2}^{n} \widehat{S}_{i, k}\big[\alpha_{k-1} \mathsf{U}_{k-1, j-1} + (\alpha_j-\alpha_{k-1}) \mathsf{U}_{k-1, j}\big]=\alpha_j (S \cdot \mathsf{U})_{i-1, j}.\end{equation} To do this, we note that using \eqref{a-1}, the sum on the left-hand side of \eqref{last-eq} is equal to \begin{align*} & \sum\limits_{k=2}^{n} \big[\widehat{S}_{i-1, k-1}+\alpha_k \widehat{S}_{i-1, k}\big]\big[\alpha_{k-1} \mathsf{U}_{k-1, j-1} + (\alpha_j-\alpha_{k-1}) \mathsf{U}_{k-1, j}\big]\\ &=\alpha_1 \widehat{S}_{i-1, 1}\mathsf{U}_{1, j-1}+\sum\limits_{k=3}^{n}\alpha_{k-1} \widehat{S}_{i-1, k-1}\mathsf{U}_{k-1, j-1} \\ & \quad + (\alpha_j-\alpha_{1})\widehat{S}_{i-1, 1} \mathsf{U}_{1, j}+ \sum\limits_{k=3}^{n} (\alpha_j-\alpha_{k-1})\widehat{S}_{i-1, k-1} \mathsf{U}_{k-1, j}\\ &\quad +\sum\limits_{k=2}^{n} \alpha_k\alpha_{k-1} \widehat{S}_{i-1, k}\mathsf{U}_{k-1, j-1}+ \sum\limits_{k=2}^{n} \alpha_k(\alpha_j-\alpha_{k-1}) \widehat{S}_{i-1, k} \mathsf{U}_{k-1, j}\\ &=\alpha_j \widehat{S}_{i-1, 1}\mathsf{U}_{1, j} +\sum\limits_{k=2}^{n-1}\alpha_{k} \widehat{S}_{i-1, k}\mathsf{U}_{k, j-1}+ \sum\limits_{k=2}^{n-1} (\alpha_j-\alpha_{k})\widehat{S}_{i-1, k} \mathsf{U}_{k, j}\\ &\quad +\sum\limits_{k=2}^{n-1} \alpha_k\alpha_{k-1} \widehat{S}_{i-1, k}\mathsf{U}_{k-1, j-1}+ \sum\limits_{k=2}^{n-1} \alpha_k(\alpha_j-\alpha_{k-1}) \widehat{S}_{i-1, k} \mathsf{U}_{k-1, j}\\ & \quad (\text{note that} \quad \mathsf{U}_{1, j-1} =\mathsf{U}_{1, j} \quad \text{and} \quad \widehat{S}_{i-1, n}=0) \end{align*} \begin{align*} &=\alpha_j \widehat{S}_{i-1, 1}\mathsf{U}_{1, j} +\sum\limits_{k=2}^{n-1}\widehat{S}_{i-1, k}\Big(\alpha_{k} \big[\mathsf{U}_{k, j-1}+\alpha_{k-1}\mathsf{U}_{k-1, j-1} +(\alpha_j-\alpha_{k-1}) \mathsf{U}_{k-1, j}\big]\\ &\quad + (\alpha_j-\alpha_{k})\mathsf{U}_{k, j}\Big)\\ &=\alpha_j \widehat{S}_{i-1, 1}\mathsf{U}_{1, j} +\sum\limits_{k=2}^{n-1}\widehat{S}_{i-1, k}\big[\alpha_{k} \mathsf{U}_{k, j}+ (\alpha_j-\alpha_{k})\mathsf{U}_{k, j}\big] \quad (\text{by}\quad\eqref{u-1-2} ) \\ &=\alpha_j \sum\limits_{k=1}^{n-1}\widehat{S}_{i-1, k} \mathsf{U}_{k, j}=\alpha_j \sum\limits_{k=1}^{n}\widehat{S}_{i-1, k} \mathsf{U}_{k, j}=\alpha_j (S \cdot \mathsf{U})_{i-1, j}, \quad (\text{note that}\quad\widehat{S}_{i-1, n}=0) \end{align*} as required. The first part of the theorem is now complete. To prove the second statement, using the matrix decomposition \eqref{decom1}, Lemma \ref{lm1} and the fact that $\det S_\alpha(n)=1$, we observe that $$\det \mathsf{S}_\alpha (n)= \det (S_\alpha (n) \cdot \mathsf{U}_\alpha(n))=\det S_\alpha (n) \det \mathsf{U}_\alpha(n)=\prod_{j=1}^{n} \alpha_j^{j-1},$$ and the proof is complete. \end{proof} \section{Sequences associated with Pascal's triangle}\label{Sec3} In this section, let $d^{(k)}$ denote the $k$th diagonal of Pascal's triangle (see Fig.~\ref{fig1}). Hence, $$d^{(k)}=(d^{(k)}_i)_{i\geq 0}=\left({k+i\choose i}\right)_{i\geq 0} \quad (k=0, 1, 2, \ldots).$$ For instance, we have \begin{table}[H] \begin{center} \begin{tabular}{l|ll} \hline $k$ & $d^{(k)}$ & OEIS$\#$ \\ \hline $0$ & $(1, 1, 1, 1, 1, \ldots)$ & \seqnum{A000012} \\ $1$ & $(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \ldots)$ & \seqnum{A000027} \\ $2$ & $(1, 3, 6, 10, 15, 21, 28, 36, 45, 55, \ldots)$ & \seqnum{A000217} \\ $3$ & $(1, 4, 10, 20, 35, 56, 84, 120, 165, 220, \ldots)$ & \seqnum{A000292} \\ $4$ & $(1, 5, 15, 35, 70, 126, 210, 330, 495, 715, \ldots)$ & \seqnum{A000332} \\ \hline \end{tabular} \end{center} \caption{The $k$th diagonal of Pascal's triangle for $k=0, 1, 2, 3, 4$.} \label{T1} \end{table} In the sequel, we introduce some lower triangular matrices that are constructed using $d^{(k)}$: $$\mathsf{L}_{d^{(k)}}(\infty)=\left(\begin{matrix} \mathbf{0}_{k\times \infty}\\ \operatorname{diag}d^{(k)} \end{matrix}\right), \quad k=0, 1, 2, \ldots,$$ for example, for $k=0, 1, 2, 3, 4$, we have $\mathsf{L}_{d^{(0)}}(\infty)=I_{\infty}$, $$\begin{array}{ll}\mathsf{L}_{d^{(1)}}(\infty)=\left(\begin{matrix} . & . &. &. &. &. & \cdots\\ 1 &. &. &. &. &. & \cdots\\ . & 2 &. &. &. &. & \cdots\\ . &. & 3 &. &. &. & \cdots\\ . &. &. & 4 &. &. & \cdots\\ . &. &. &. & 5 &. & \cdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{matrix}\right), & \mathsf{L}_{d^{(2)}}(\infty)=\left(\begin{matrix} . & . & . & . & . & . & \cdots\\ . & . & . & . & . & . & \cdots\\ 1 & . & . & . & . & . & \cdots\\ . & 3 & . & . & . & . & \cdots\\ . & . & 6 & . & . & . & \cdots\\ . & . & . & 10 & . & . & \cdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{matrix}\right)\\[2cm] \mathsf{L}_{d^{(3)}}(\infty)=\left(\begin{matrix} . & . & . & . & . & . & \cdots\\ .& . & . & . & . & . & \cdots\\ . & . & . & . & . & . & \cdots\\ 1 & . & . & . & . & . & \cdots\\ . & 4 & . & . & . & . & \cdots\\ . & . & 10 & . & . & . & \cdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{matrix}\right), & \mathsf{L}_{d^{(4)}}(\infty)=\left(\begin{matrix} . & . & . & . & . & . & \cdots\\ . & . & . & . & . & . & \cdots\\ . & . & . & . & . & . & \cdots\\ . & . & . & . & . & . & \cdots\\ 1 & . & . & .& . & . & \cdots\\ . & 5 & . & . & . & . & \cdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{matrix}\right).\end{array}$$ \begin{theorem}\label{th3} Let $k\geq 1$ be an integer. With the notation above, we have the following \begin{equation}\label{eee-7} \exp(\mathsf{L}_{d^{(k)}}(n))=T_{\alpha^{(k)}, \beta} (n)\odot \sum\limits_{i=0}^{\infty} \mathsf{L}_{d^{(ki)}}(n), \end{equation} where $\alpha^{(k)}=(\alpha^{(k)}_i)_{i\geq 1}$ is the sequence given by the recurrence $$\alpha^{(k)}_i=\binom{i-2}{k-1}\alpha^{(k)}_{i-k}, \quad i>k,$$ and the initial conditions $\alpha^{(k)}_1=1$, $\alpha^{(k)}_i=0$, $i=2, 3, \ldots, k$, and $\beta=(1, 0, 0, 0, \ldots)$. \end{theorem} Before we proceed with the proof of Theorem \ref{th3}, it seems appropriate to mention a routine observation. Given an arbitrary sequence $\gamma=(\gamma_i)_{i\geq 1}$ and a positive integer $k$, the infinite matrix $\mathsf{L}_{k, \gamma}(\infty)=[L_{i, j}]_{i, j\geq 1}$ is defined so that \begin{displaymath} L_{i, j}=\begin{cases} \gamma_j, & \text{if $i-j=k$;}\\ 0, & \text{otherwise}, \end{cases}\end{displaymath} or equivalently $$\mathsf{L}_{k, \gamma}(\infty)=\left(\begin{matrix} \mathbf{0}_{k\times \infty}\\ \operatorname{diag} \gamma \end{matrix}\right), \quad k\geq 1.$$ Thus, for example, we have $$\mathsf{L}_{2, \gamma}(6)=\left(\begin{matrix} . & . &. &. &. & .\\ . & . &. &. &. & .\\ {\gamma_1} &. &. &. &. & .\\ . & {\gamma_2} &. &. &. & .\\ . &. & {\gamma_3} &. &. & .\\ . &. &. & {\gamma_4} &. & . \end{matrix}\right).$$ Let $\exp(\mathsf{L}_{k, \gamma}(\infty))=[E_{i, j}]_{i, j\geq 1}$. It is easy to check that \begin{equation}\label{eij} E_{i, j}=\begin{cases} 0, & \text{if $ij, \quad i-j\not\equiv 0\pmod{k}$;}\\ \frac{\gamma_j\gamma_{j+k}\gamma_{j+2k}\cdots \gamma_{i-k}}{\left(\frac{i-j}{k}\right)!}, & \text{if $i>j, \quad i-j\equiv 0\pmod{k}$.} \end{cases}\end{equation} \begin{proof} Fix an integer $k\geq 1$, and let $\gamma_i=d^{(k)}_{i-1}=\binom{i-1+k}{i-1}$. We first claim that the sequence $(E_{i, 1})_{i\geq 1}$ (the first column of $\exp(\mathsf{L}_{d^{(k)}}(\infty))$) is equal to the sequence $\alpha^{(k)}$, and for this purpose, it is enough to check by induction on $i$ that $$E_{i, 1}=\binom{i-2}{k-1}E_{i-k, 1} \quad (i>k),$$ and $E_{1, 1}=1$, $E_{2, 1}=E_{3, 1}=\ldots=E_{k, 1}=0$. Indeed, for $i>k$ and $i-1\equiv 0\pmod{k}$, we observe by the inductive hypothesis that \begin{align*}E_{i, 1}&=\frac{\gamma_1\gamma_{1+k} \gamma_{1+2k} \cdots \gamma_{i-2k} \gamma_{i-k}}{\left(\frac{i-1}{k}\right)!}\\ &=\frac{\gamma_1\gamma_{1+k} \gamma_{1+2k}\cdots \gamma_{i-2k}}{\left(\frac{(i-k)-1}{k}\right)!}\cdot \frac{\gamma_{i-k}}{\frac{i-1}{k}}\\&=E_{i-k, 1}\cdot \frac{{i-1\choose k}}{\frac{i-1}{k}}\\&={i-2\choose k-1}E_{i-k, 1}. \end{align*} Next, we show that if $r-s=i-1$ and $r-s\equiv 0\pmod{k}$, then \begin{equation}\label{ersi1} E_{r, s}=E_{i, 1}\cdot \binom{r-1}{i-1}. \end{equation} In fact, we have \begin{align*} E_{r, s}&=\frac{\gamma_s\gamma_{s+k}\cdots \gamma_{r-k}}{\left(\frac{r-s}{k}\right)!}=\frac{\gamma_1\gamma_{1+k}\cdots \gamma_{i-k}}{\left(\frac{i-1}{k}\right)!} \cdot \frac{\gamma_s\gamma_{s+k}\cdots \gamma_{r-k}}{\gamma_1\gamma_{1+k}\cdots \gamma_{i-k}}\\&=E_{i, 1} \cdot \frac{\gamma_s\gamma_{s+k}\cdots \gamma_{r-k}}{\gamma_1\gamma_{1+k}\cdots \gamma_{i-k}}\\ &=E_{i, 1}\cdot \frac{\binom{s-1+k}{s-1}\binom{s-1+2k}{s-1+k}\binom{s-1+3k}{s-1+2k}\cdots \binom{r-1}{r-1-k}}{{0+k\choose 0}{2k\choose k}{3k\choose 2k}\cdots{i-1\choose i-1-k}}\\&=E_{i, 1}\cdot \frac{(r-1)!}{(s-1)!(i-1)!}\\ & =E_{i, 1}\cdot \frac{(r-1)!}{(r-i)!(i-1)!}\\&=E_{i, 1}\cdot \binom{r-1}{i-1}. \end{align*} Finally, it follows from the fact that $(E_{i, 1})_{i\geq 1}=\alpha^{(k)}$, and \eqref{eij}, \eqref{ersi1} that $$\exp(\mathsf{L}_{d^{(k)}}(n))=T_{\alpha^{(k)}, \beta} (n)\odot \sum\limits_{i=0}^{\infty} \mathsf{L}_{d^{(ki)}}(n),$$ and the proof is complete. \end{proof} \begin{example} We have {\footnotesize \begin{align*}\footnotesize \exp(\left(\begin{matrix} . & . & . & . & . & . & . & . & . & .\\ . & . & . & . & . & . & . & . & . & .\\ 1 & . & . & . & . & . & . & . & . & .\\ . & 3 & . & . & . & . & . & . & . & .\\ . & . & 6 & . & . & . & . & . & . & .\\ . & . & . & 10 & . & . & . & . & . & .\\ . & . & . & . & 15 & . & . & . & . & .\\ . & . & . & . & . & 21 & . & . & . & .\\ . & . & . & . & . & . & 28 & . & . & .\\ . & . & . & . & . & . & . & 36 & . & . \end{matrix}\right))=\left(\begin{matrix} 1 & . & . & . & . & . & . & . & . & .\\ . & 1 & . & . & . & . & . & . & . & .\\ 1 & . & 1 & . & . & . & . & . & . & .\\ . & 3 & . & 1 & . & . & . & . & . & .\\ 3 & . & 6 & . & 1 & . & . & . & . & .\\ . & 15 & . & 10 & . & 1 & . & . & . & .\\ 15 & . & 45 & . & 15 & . & 1 & . & . & .\\ . & 105 & . & 105 & . & 21 & . & 1 & . & .\\ 105 & . & 420 & . & 210 & . & 28 & . & 1 & .\\ . & 945 & . & 1260 & . & 378 & . & 36 & . & 1 \end{matrix}\right)\\[0.5cm] =\left(\begin{matrix} 1 & . & . & . & . & . & . & . & . & .\\ . & 1 & . & . & . & . & . & . & . & .\\ 1 & . & 1 & . & . & . & . & . & . & .\\ . & 1 & . & 1 & . & . & . & . & . & .\\ 3 & . & 1 & . & 1 & . & . & . & . & .\\ . & 3 & . & 1 & . & 1 & . & . & . & .\\ 15 & . & 3 & . & 1 & . & 1 & . & . & .\\ . & 15 & . & 3 & . & 1 & . & 1 & . & .\\ 105 & . & 15 & . & 3 & . & 1 & . & 1 & .\\ . & 105 & . & 15 & . & 3 & . & 1 & . & 1 \end{matrix}\right) \odot \left(\begin{matrix} 1 & . & . & . & . & . & . & . & . & .\\ . & 1 & . & . & . & . & . & . & . & .\\ 1 & . & 1 & . & . & . & . & . & . & .\\ . & 3 & . & 1 & . & . & . & . & . & .\\ 1 & . & 6 & . & 1 & . & . & . & . & .\\ . & 5 & . & 10 & . & 1 & . & . & . & .\\ 1 & . & 15 & . & 15 & . & 1 & . & . & .\\ . & 7 & . & 35 & . & 21 & . & 1 & . & .\\ 1 & . & 28 & . & 70 & . & 28 & . & 1 & .\\ . & 9 & . & 84 & . & 126 & . & 36 & . & 1 \end{matrix}\right). \end{align*}} \end{example} Given a formal power series \[f(t)=\sum_{r=0}^{\infty}a_rt^r,\] over a field of characteristic zero, define the lower triangular matrix \[M(f)=[M(f)_{i, j}]_{i, j\geq1}\] by \begin{displaymath} M(f)_{i, j}= \begin{cases} \dfrac{(i-1)!}{(j-1)!}[t^{i-j}]f(t),& i\geq j,\\[0.2cm] 0,&ik),\end{equation} where $c_1, c_2, \ldots, c_k$ are constants, and $c_k\neq 0$. It follows from the general recursion theorem that for any $k$ initial conditions $\alpha_1, \alpha_2, \ldots, \alpha_k$, there exists a unique sequence $(\alpha_n)_{n\geq 1}$ that satisfies \eqref{e999} with the same initial conditions. Let $(\beta_n)_{n\geq 1}$ be a sequence defined as $\beta_1=\beta_2=\ldots=\beta_{k-1}=0$, $\beta_k=1$, and \begin{equation}\label{e2} \beta_n-c_1\beta_{n-1}-c_2\beta_{n-2}-\cdots-c_k\beta_{n-k}=0 \quad (n>k).\end{equation} Consider the $n\times n$ upper Hessenberg-Toeplitz matrix $$H_n(-1; c_1, c_2, \ldots, c_k)=[H_{i, j}]_{1\leq i, j\leq n},$$ where \begin{displaymath} H_{i, j}=\begin{cases} c_{j-i+1}, & \text{if $0\leq j-i\leq k-1$;}\\ -1, & \text{if $j-i=-1$;}\\ 0, & \text{otherwise.}\\ \end{cases}\end{displaymath} For instance, for $n=5$ and $k=3$, the corresponding matrix is $$H_5(-1; c_1, c_2, c_3)=\left(\begin{matrix} c_1 & c_2& c_3 & . & .\\ -1& c_1 & c_2& c_3 & .\\ . & -1& c_1 & c_2& c_3 \\ . & . & -1& c_1 & c_2\\ . & . & . & -1& c_1\\ \end{matrix}\right). $$ Note that the rows of $H_n(-1; c_1, c_2, \ldots, c_k)$ represent the negatives of the coefficients of the difference equation \eqref{e2} satisfied by the sequence $(\beta_n)_{n\geq 1}$. Let $D_{n, k}=\det H_n(-1; c_1, c_2, \ldots, c_k)$. For $n>k$, using expansion along the last column, we obtain \begin{equation}\label{e3} D_{n, k}=\sum_{j=1}^{k} c_jD_{n-j, k}. \end{equation} If we define \begin{equation}\label{e4} D_{1-k, k}=D_{2-k, k}=\ldots=D_{-2, k}=D_{-1, k}=0; \quad D_{0, k}=1, \end{equation} then \eqref{e3} remains valid for $n\geq 1$. Hence, for fixed $k$, it is evident that both $D_{n, k}$ and $(\beta_n)_{n\geq 1}$ obey the identical recurrence relation of order $k$. Consequently, $D_{n, k}=\beta_{n+k}$. Therefore, the subsequent theorem has been established. \begin{theorem}\label{th22} With the above notation, we have $D_{n, k}=\beta_{n+k}$ for every $n\geq 1$. \end{theorem} Now, let us give two examples to illustrate the result of Theorem \ref{th22}: (a) {\em The Berstel sequence} (\seqnum{A007420} in \cite{OEIS}). Suppose $(b_n)_{n\geq 1}$ is Berstel's ternary recurrence sequence defined by $$b_1=b_2=0, \quad b_3=1, \quad b_n=2b_{n-1}-4b_{n-2}+4b_{n-3}, \quad n>3.$$ Then, the following determinant representation of $(b_{n+3})_{n\geq 1}$ follows from Theorem \ref{th22}, $$b_{n+3}=\left|\begin{array}{cccccc} 2 & -4 & 4 & . & & .\\ -1& 2 & -4 & 4 & \ddots & .\\ . & -1& 2 & -4& \ddots & .\\ . & . & -1 & 2 & \ddots & 4\\ & \ddots & \ddots & \ddots & \ddots & -4\\ . & . & . & & -1& 2 \\ \end{array}\right|_{n\times n}. $$ (b) The {\em $k$-generalized Fibonacci sequence}. The $k$-generalized Fibonacci sequence $(F^{(k)}_{n})_{n\geq 1}$ satisfies the $k$-th order recurrence $$F^{(k)}_{n}=F^{(k)}_{n-1}+F^{(k)}_{n-2}+\cdots+F^{(k)}_{n-k}, \quad n>k,$$ with initial conditions $$F^{(k)}_{1}=F^{(k)}_{2}=\ldots=F^{(k)}_{k-1}=0, \quad F^{(k)}_k=1.$$ We recall that these numbers are also called Fibonacci $k$-sequences, generalized Fibonacci numbers of order $k$, Fibonacci $k$-step numbers, and $k$-bonacci numbers (see also \cite{Lee-Lee, Miles}). The original Fibonacci sequence $(F_{i-1})_{i\geq 1}$ (\seqnum{A000045} in \cite{OEIS}) can be obtained by putting $k=2$: $$F^{(2)}_1=0, \quad F^{(2)}_2=1, \quad F^{(2)}_{n}=F^{(2)}_{n-1}+F^{(2)}_{n-2}, \quad n>2.$$ Using Theorem \ref{th22}, for instance, $(F^{(5)}_{n+5})_{n\geq 1}$ has the following determinant representation: $$F^{(5)}_{n+5}=\left|\begin{array}{cccccccc} 1 & 1 & 1 & 1& 1 & . & . &. \\ -1& 1 & 1& 1 & 1 & 1 & \ddots & . \\ . & -1& 1& 1 & 1& 1 & \ddots & . \\ . & . & -1& 1 & 1 & 1 & \ddots & 1 \\ . & . & . & -1 & 1 & 1 & \ddots & 1\\ . & . & . & . & -1 & 1 & \ddots & 1 \\ . & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & 1 \\ . & . & . & . & . & . & -1 & 1\\ \end{array}\right|_{n\times n}. $$ \section{Acknowledgments} The authors would like to express their gratitude to the referee(s) for carefully reading the manuscript and providing valuable suggestions and comments. \begin{thebibliography}{99} \bibitem{EdSt} A. Edelman and G. Strang, Pascal matrices, {\em Amer. Math. Monthly} {\bf 111} (2004), 189--197. \bibitem{Edw} A. W. F. Edwards, {\em Pascal's Arithmetical Triangle: The Story of a Mathematical Idea}, Charles Griffin, 1987 and Johns Hopkins University Press, Baltimore, MD, 2002. \bibitem{Kr} C. Krattenthaler, Advanced determinant calculus: A complement, {\em Linear Algebra Appl.} {\bf 411} (2005) 68--166. \bibitem{Lee-Lee} G. Y. Lee and S. G. Lee, A note on generalized Fibonacci numbers, {\em Fibonacci Quart.} {\bf 33} (1995), 273--278. \bibitem{Lind} D. A. Lind, A determinant involving generalized binomial coefficients, {\em Fibonacci Quart.} {\bf 9} (1971), 113--119. \bibitem{MaGu} P. Maltais and T. A. Gulliver, Pascal matrices and Stirling numbers, {\em Appl. Math. Lett.} {\bf 11} (1998), 7--11. \bibitem{Miles} E. P. Miles, Jr., Generalized Fibonacci numbers and associated matrices, {\em Amer. Math. Monthly} {\bf 67} (1960), 745--752. \bibitem{MoSaSa} A. R. Moghaddamfar, S. Navid Salehy, and S. Nima Salehy, The determinants of matrices with recursive entries, {\em Linear Algebra Appl.} {\bf 428} (2008), 2468--2481. \bibitem{MSS} A. R. Moghaddamfar, S. Navid Salehy, and S. Nima Salehy, Determinant representations of sequences: a survey, {\em Spec. Matrices} {\bf 1} (2013) 46--60. \bibitem{OEIS} N. J. A. Sloane et al., The On-Line Encyclopedia of Integer Sequences, 2023. Available at \url{https://oeis.org}. \bibitem{SpZi} M. Z. Spivey and A. M. Zimmer, Symmetric polynomials, Pascal matrices, and Stirling matrices, {\em Linear Algebra Appl.} {\bf 428} (2008), 1127--1134. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2020 {\it Mathematics Subject Classification}: Primary 11C20; Secondary 15A15, 15A23. \noindent \emph{Keywords:} Stirling number of the second kind, Stirling matrix of the second kind, Pascal matrix, complete homogeneous symmetric polynomial. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A000012}, \seqnum{A000027}, \seqnum{A000045}, \seqnum{A000085}, \seqnum{A000110}, \seqnum{A000217}, \seqnum{A000258}, \seqnum{A000292}, \seqnum{A000307}, \seqnum{A000332}, \seqnum{A000357}, \seqnum{A000405}, \seqnum{A000984}, \seqnum{A001669}, \seqnum{A001680}, \seqnum{A001681}, \seqnum{A007420}, \seqnum{A033306}, \seqnum{A081624}, \seqnum{A081629}, \seqnum{A081697}, \seqnum{A081740}, \seqnum{A105479}, \seqnum{A105480}, and \seqnum{A123023}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received June 3 2024; revised versions received June 4 2024; June 11 2026; September 22 2026; September 23 2026. Published in {\it Journal of Integer Sequences}, September 25 2026. \bigskip \hrule \bigskip \noindent Return to \href{https://cs.uwaterloo.ca/journals/JIS/}{Journal of Integer Sequences home page}. \vskip .1in \end{document}