\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \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,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\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} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf More Determinant Representations for \\ \vskip .12in Sequences } \vskip 1cm \large Alireza Moghaddamfar \\ Department of Mathematics \\ K. N. Toosi University of Technology \\ P. O. Box 16315-1618 \\ Tehran \\ Iran \\ and \\ Research Institute for Fundamental Sciences (RIFS)\\ Tabriz\\ Iran\\ \href{mailto:moghadam@ipm.ir}{\tt moghadam@ipm.ir} \\ \href{mailto:moghadam@kntu.ac.ir}{\tt moghadam@kntu.ac.ir} \\ \ \\ Hadiseh Tajbakhsh\\ Department of Mathematics\\ K. N. Toosi University of Technology\\ P. O. Box 16315-1618\\ Tehran, Iran \end{center} \vskip .2 in \begin{abstract} In this paper, we will find some new families of infinite (integer) matrices whose entries satisfy a non-homogeneous recurrence relation and such that the sequence of their leading principal minors is a subsequence of the Fibonacci, Lucas, Jacobsthal, or Pell sequences. \end{abstract} \section{Introduction} Throughout this paper, unless noted otherwise, we will use the following notation. Let $\alpha=(\alpha_i)_{i\geqslant 0}$ and $\beta=(\beta_i)_{i\geqslant 0}$ be two arbitrary sequences starting with a common first term $\alpha_0=\beta_0$. We denote by $P_{\alpha,\beta}(n)$ the {\em generalized Pascal triangle} associated with the sequences $\alpha$ and $\beta$, which is introduced as follows. Actually, $P_{\alpha,\beta}(n)=[P_{i,j}]_{0\leqslant i, j\leqslant n}$ is a square matrix of order $n+1$ whose $(i,j)$-entry $P_{i,j}$ obeys the following rules: $$P_{i,0}=\alpha_i, \ \ P_{0,j}=\beta_j \ \ \text{for} \ \ i, j=0, 1, 2, \ldots, n,\ \ \text{and} \ \ P_{i,j}= P_{i,j-1}+P_{i-1,j} \ \ \text{for} \ \ 1\leqslant i, j\leqslant n.$$ We also denote by $T_{\alpha, \beta}(n)=[T_{i,j}]_{0\leqslant i, j\leqslant n}$ the {\em Toeplitz matrix} of order $n+1$ whose $(i,j)$-entry $T_{i,j}$ obeys the following rules: $$T_{i,0}=\alpha_i, \ \ T_{0,j}=\beta_j \ \ \text{for} \ \ i, j=0, 1, 2, \ldots, n,\ \ \text{and} \ \ T_{i,j}=T_{k, l} \ \ \text{if} \ \ i-j=k-l.$$ The {\em unipotent lower triangular matrix} $L(n)=[L_{i,j}]_{0\leqslant i, j\leqslant n}$ is again a square matrix of order $n+1$ with entries: $$L_{i,j}=\left \{ \begin{array}{ll} 0, & \text{if} \ \ 0\leqslant ii+1$ and $H_{i, i+1}\neq 0$ for some $i$, $0\leqslant i\leqslant n-1$. Given a matrix $A$, we denote by ${\rm R}_i(A)$ (resp., ${\rm C}_j(A)$) the row $i$ (resp., the column $j$) of $A$. We also denote by $A^{[1]}$ the submatrix obtained from $A$ by deleting the first column of $A$. Given a sequence $\varphi=(\varphi_i)_{i\geq 0}$, define the {\em binomial transform} of $\varphi$ to be the sequence $\hat{\varphi}=(\hat{\varphi}_i)_{i\geqslant0}$ with $${\hat{\varphi}}_i=\sum_{k=0}^i(-1)^{i+k}{i\choose k}\varphi_k.$$ The {\em Fibonacci sequence} (\seqnum{A000045} in \cite{integer}) is defined by the recurrence relation: $$F_0=0, \ F_1=1, \ \ \ F_n=F_{n-1}+F_{n-2} \ \ \text{for} \ \ n\geqslant 2.$$ The {\em Lucas sequence} (\seqnum{A000032} in \cite{integer}) is defined by the recurrence relation: $$L_0=2, \ L_1=1, \ \ \ L_n=L_{n-1}=L_{n-2} \ \ \text{for} \ \ n\geqslant 2.$$ The {\em Jacobsthal sequence} (\seqnum{A001045} in \cite{integer}) is defined by the recurrence relation: $$J_0=0, \ J_1=1, \ \ \ J_n=J_{n-1}+2J_{n-2} \ \ \text{for} \ \ n\geqslant 2.$$ The {\em Pell sequence} (\seqnum{A000129} in \cite{integer}) is defined by the recurrence relation: $$P_0=0, \ P_1=1, \ \ \ P_n=2P_{n-1}+P_{n-2} \ \ \text{for} \ \ n\geqslant 2.$$ Let $A=[A_{i, j}]_{i, j\geqslant 0}$ be an arbitrary infinite matrix. We denote the elementary row operation of type {\em three} by $O_{r, s}(\lambda)$, where $r\neq s$ and $\lambda$ a scalar, that is $$ {\rm {\rm R}}_k(O_{r, s}(\lambda)A)=\left \{ \begin{array}{ll} {\rm {\rm R}}_r(A)+\lambda {\rm {\rm R}}_s(A),& {\rm if} \ \ k=r; \\[0.2cm] {\rm {\rm R}}_k(A), & {\rm if} \ \ k\neq r. \end{array} \right.$$ The {\em $n$th leading principal minor} of $A$, denoted by $d_n(A)$, is defined as follows: $$d_n(A)=\det [A_{i, j}]_{0\leqslant i, j\leqslant n}, \ \ (n=0, 1, 2, 3, \ldots).$$ We put $D(A)=(d_n(A))_{n\geqslant 0}$. Two infinite matrices $A$ and $B$ are said to be {\em equimodular} if $D(A)=D(B)$. Given a sequence $\omega=(\omega_n)_{n\geqslant 0}$, a family $\{A_t| \ t\in I\}$ of equimodular matrices are said to be {\em $\omega$-equimodular} if $D(A_t)=\omega$ for all $t\in I$. We will denote the family of $\omega$-equimodular matrices by $\mathcal{A}_\omega$. The infinite matrices in $ \mathcal{A}_\omega$ are said to be {\em determinant representations of $\omega$}. Note that for {\em any} sequence $\omega=(\omega_n)_{n\geqslant 0}$, there is a determinant representation of $\omega$, in other words $\mathcal{A}_\omega\neq \emptyset$. Indeed, expanding along the last rows, it is easy to see that $$\left ( \begin{array}{crrrrl} \omega_0 & 1 & \ast & \ast & \ast & \cdots\\ -\omega_1 & 0 & 1 & \ast & \ast& \cdots\\ \omega_2 & 0 & 0 & 1 & \ast& \cdots\\ -\omega_3 & 0 & 0 & 0 & 1 & \cdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{array} \right )\in \mathcal{A}_\omega,$$ (see also Theorem 3.2 and the Remark after this theorem in \cite{yang}). Especially, there are many different determinant representations of $\omega$, when $\omega$ is a (sub-)sequence of Fibonacci, Lucas, Jacobsthal and Pell sequences. Some examples of such matrices can be found in \cite{mp, mtaj2}. In this paper, we are going to find some determinant representations of the sequences: $$\mathcal{F}=(F_{n+1})_{n\geqslant 0}, \ \ \mathcal{L}=(L_{n+1})_{n\geqslant 0}, \ \ \mathcal{J}=(J_{n+1})_{n\geqslant 0} \ \ \text{and} \ \ \mathcal{P}=(P_{n+1})_{n\geqslant 0}.$$ It is worthwhile to point out that we will use {\em non-homogeneous recurrence relations} to construct these determinant representations. In the sequel, we introduce a new family of (infinite) matrices $A(\infty)=[A_{i,j}]_{i, j\geqslant 0}$, whose entries obey a non-homogeneous recurrence relation. Actually, for two constants $u$ and $v$, and arbitrary sequences $\lambda=(\lambda_i)_{i\geqslant 0}$ and $\mu=(\mu_i)_{i\geqslant 0}$ with $\mu_0=0$, the first column and row of matrix $A(\infty)$ are the sequences $$(A_{i,0})_{i\geqslant 0}=(\lambda_0, \lambda_1, \lambda_2, \ldots, A_{i,0}=\lambda_i, \ldots),$$ and $$(A_{0,j})_{j\geqslant 0}=(\lambda_0, \lambda_0+u, \lambda_0+2u, \ldots, A_{0,j}=\lambda_0+ju, \ldots),$$ respectively, while the remaining entries $A_{i,j}$ $(i, j\geqslant 1)$ are obtained from the following non-homogeneous recurrence relation: $$A_{i,j}=A_{i,j-1}+A_{i-1,j}-\lambda_{i-1}+\mu_i-\mu_{i-1}+(j-1)(v-u), \ \ \ \ \ \ i, j\geqslant 1.$$ We denote by $A(n)$ the submatrix of $A(\infty)$ consisting of the entries in its first $n+1$ rows and columns. The matrix $A(3)$, for example, is then given by $$A(3)=\left (\begin{array}{ccccc} \lambda_0&\lambda_0+u&\lambda_0+2u&\lambda_0+3u\\[0.2cm] \lambda_1&\lambda_1+\mu_1+u&\lambda_1+2\mu_1+2u+v&\lambda_1+3\mu_1+3u+3v\\[0.2cm] \lambda_2&\lambda_2+\mu_2+u&\lambda_2+2\mu_2+\mu_1+2u+2v&\lambda_2+3\mu_2+3\mu_1+3u+7v\\[0.2cm] \lambda_3&\lambda_3+\mu_3+u&\lambda_3+2\mu_3+\mu_2+\mu_1+2u+3v&\lambda_3+3\mu_3+3\mu_2+4\mu_1+3u+12v\\[0.2cm] \end{array}\right).$$ Finally, the main result of this paper can be stated as follows:\\[0.3cm] {\bf Main Theorem.} \ {\em The matrix $A(n)$, $n\geqslant 0$, defined as above, satisfies the following statements: \begin{itemize} \item[{\rm (a)}] $A(n)=L(n)\cdot H(n)\cdot U(n)$, where $$H(n)=\left(\begin{array}{c|llcl} \hat{\lambda}_0 & u& 0& \cdots & 0 \\ \hline \hat{\lambda}_1& & & & \\ \hat{\lambda}_2 & & & & \\ \vdots & & & T_{(\hat{\mu}_1, \hat{\mu}_2, \hat{\mu}_3, \ldots), (\hat{\mu}_1, v, 0, 0, \ldots)}(n-1) & \\ \hat{\lambda}_{n} & & & & \\ \end{array}\right).$$\\ In particular, we have $\det(A(n))=\det(H(n))$. \item[{\rm (b)}] In the case when $u=v=1$ and $\lambda_i=(2^i-1)c+1$, we have the following statements: \subitem{\rm (b. 1)} if $\mu_i=\left(2^i+\frac{(i-2)(i+1)}{2}\right)c-\frac{i(i-3)}{2}$, then $\det(A(n))=F_{n+1}$. \subitem{\rm(b. 2)} if $\mu_i=\left(\frac{5\cdot 3^i }{4}-2^i-\frac{2i+1}{4}\right)c+\frac{5(3^i-1)}{4}+\frac{i}{2}$, then $\det(A(n))=L_{n+1}$. \subitem{\rm(b. 3)} if $\mu_i=i^2c-i^2+2i$, then $\det(A(n))=J_{n+1}$. \subitem{\rm(b. 4)} if $\mu_i=\left(2^{i+1}+\frac{(i+1)(i-4)}{2}\right)c+\frac{(5-i)i}{2}$, then $\det(A(n))=P_{n+1}$. \end{itemize}} As mentioned previously, we have obtained some determinant representations of the sequences: $$\mathcal{F}=(F_{n+1})_{n\geqslant 0}, \ \ \mathcal{L}=(L_{n+1})_{n\geqslant 0}, \ \ \mathcal{J}=(J_{n+1})_{n\geqslant 0} \ \ \text{and} \ \ \mathcal{P}=(P_{n+1})_{n\geqslant 0},$$ which are presented in the following: $$\begin{array}{ll} \left(\begin{array}{cccc}1&2&3&\cdots\\[0.3cm] c+1&2c+3&3c+6&\cdots\\[0.3cm] 3c+1&7c+3&12c+8&\cdots\\[0.3cm] \vdots&\vdots&\vdots&\ddots\end{array}\right)\in \mathcal{A}_{\mathcal{F}}, & \left(\begin{array}{cccc} 1&2&3&\cdots\\[0.3cm] c+1&2c+5&3c+10&\cdots\\[0.3cm] 3c+1&9c+13&16c+30& \cdots\\[0.3cm] \vdots&\vdots&\vdots&\ddots\end{array}\right)\in \mathcal{A}_{\mathcal{L}}, \\[2cm] \left(\begin{array}{cccc} 1 & 2 & 3 & \cdots\\[0.3cm] c+1 &2c+3 &3c+6 &\cdots\\[0.3cm] 3c+1 &7c+2 &12c+6 &\cdots\\[0.3cm] \vdots &\vdots &\vdots &\ddots \end{array}\right)\in \mathcal{A}_{\mathcal{J}} \ \ \text{and} & \left(\begin{array}{cccc} 1 & 2 & 3 & \cdots\\[0.3cm] c+1 &2c+4 &3c+8 &\cdots\\[0.3cm] 3c+1 &8c+5 &14c+13 &\cdots\\[0.3cm] \vdots &\vdots &\vdots &\ddots \end{array}\right)\in \mathcal{A}_{\mathcal{P}}. \end{array}$$ \section{Main results} As the first result of this paper, we consider the following theorem. \begin{theorem}\label{th1} For two arbitrary sequences $(\lambda_i)_{i\geqslant0}$ and $(\mu_i)_{i\geqslant 0}$, with $\mu_0=0$, and some integers $u$ and $v$, let $A(\infty)=[A_{i,j}]_{i,j\geqslant 0}$ be an infinite dimensional matrix whose entries are given by \begin{equation}\label{e1} A_{i,j}=A_{i,j-1}+A_{i-1,j}-\lambda_{i-1}+\mu_i-\mu_{i-1}+(j-1)(v-u), \ \ \ \ \ i, j\geqslant 1 \end{equation} and the initial conditions $A_{i, 0}=\lambda_i$ and $A_{0, i}=\lambda_0+iu$, $i\geqslant 0$. If $A(n)=[A_{i,j}]_{0\leqslant i, j\leqslant n}$, then we have \begin{equation}\label{e3} A(n)=L(n)\cdot H(n)\cdot U(n), \end{equation} where $$H(n)=\left(\begin{array}{c|llcl} \hat{\lambda}_0 & u& 0& \cdots & 0 \\ \hline \hat{\lambda}_1& & & & \\ \hat{\lambda}_2 & & & & \\ \vdots & & & T_{(\hat{\mu}_1, \hat{\mu}_2, \hat{\mu}_3, \ldots), (\hat{\mu}_1, v, 0, 0, \ldots)}(n-1) & \\ \hat{\lambda}_{n} & & & & \\ \end{array}\right).$$ \end{theorem} \begin{proof} First of all, we recall that the entries of $L(n)=[L_{i,j}]_{0\leqslant i,j\leqslant n}$ satisfy the following recurrence \begin{equation}\label{eee2} L_{i,j}=L_{i-1,j-1}+L_{i-1,j}, \ \ \ 1\leqslant i,j\leqslant n. \end{equation} Similarly, for the entries of $U(n)=[U_{i,j}]_{0\leqslant i, j\leqslant n}$ we have \begin{equation}\label{eee4} U_{i,j}=U_{i-1,j-1}+U_{i,j-1}, \ \ \ 1\leqslant i, j\leqslant n. \end{equation} In what follows, for convenience, we will let $A=A(n)$, $L=L(n)$, $H=H(n)$ and $U=U(n)$. Now, for the proof of the desired factorization we compute the $(i,j)$-entry of $L\cdot H\cdot U$, that is \begin{equation}\label{2009-26-2} (L\cdot H\cdot U)_{i,j}=\sum\limits_{r=0}^{n}\sum\limits_{s=0}^{n}L_{i,r}H_{r,s}U_{s,j}. \end{equation} In fact, we should establish $$\begin{array}{rcl}{\rm R}_0(L\cdot H\cdot U)&={\rm R}_0(A)=&(\lambda_0, \lambda_0+u, \ldots, \lambda_0+nu),\\[0.2cm] {\rm C}_0(L\cdot H\cdot U)&={\rm C}_0(A)=&(\lambda_0, \lambda_1, \ldots, \lambda_{n}),\end{array}$$ and finally, show that \begin{equation}\label{2009e4} (L\cdot H\cdot U)_{i,j}=(L\cdot H\cdot U)_{i-1,j-1}+(L\cdot H\cdot U)_{i-1,j}-\lambda_{i-1}+\mu_i-\mu_{i-1}+(j-1)(v-u), \end{equation} for $1\leqslant i, j\leqslant n$. Let us do the required calculations. Assume first that $i=0$. Then, we have $$(L\cdot H\cdot U)_{0,j}=\sum\limits_{r=0}^{n} \sum\limits_{s=0}^{n}L_{0,r}H_{r,s}U_{s,j}= \sum\limits_{s=0}^{n}H_{0,s}U_{s,j} = H_{0,0}U_{0,j}+H_{0,1}U_{1,j} = \lambda_0+ju,$$ and so ${\rm R}_0(L\cdot H\cdot U)={\rm R}_0(A)=(\lambda_0, \lambda_0+u, \ldots, \lambda_0+nu)$. Assume next that $j=0$. In this case, we obtain $$(L\cdot H\cdot U)_{i,0}=\sum\limits_{r=0}^{n} \sum\limits_{s=0}^{n}L_{i,r}H_{r,s}U_{s,0}= \sum\limits_{r=0}^{n}L_{i,r}H_{r,0}=\sum\limits_{r=0}^{n}{i \choose r}\hat{\lambda}_r = \lambda_i, $$ and hence we have ${\rm C}_0(L\cdot H\cdot U)={\rm C}_0(A)=(\lambda_0, \lambda_1, \ldots, \lambda_n)$. Finally, we must establish \eqref{2009e4}. Let us for the moment assume that $1\leqslant i, j\leqslant n$. In this case, we have \begin{equation}\label{end1} (L\cdot H\cdot U)_{i,j}= \sum\limits_{r=0}^{n}\sum\limits_{s=0}^{n}L_{i,r}H_{r,s}U_{s,j} =\sum\limits_{r=0}^{n}L_{i,r}H_{r,0}U_{0,j}+ \sum\limits_{r=0}^{n}\sum\limits_{s=1}^{n}L_{i,r}H_{r,s}U_{s,j}. \end{equation} Let $\Omega(i,j)=\sum\limits_{r=0}^{n}\sum\limits_{s=1}^{n}L_{i,r}H_{r,s}U_{s,j}$. Then, using \eqref{eee4}, we obtain \begin{equation}\label{end2} \begin{array}{lll} \Omega(i,j)& = & \sum\limits_{r=0}^{n}\sum\limits_{s=1}^{n}L_{i,r}H_{r,s}\big(U_{s-1,j-1}+U_{s,j-1}\big) = \sum\limits_{r=0}^{n} \sum\limits_{s=1}^{n}L_{i,r}H_{r,s}U_{s-1,j-1}+ \sum\limits_{r=0}^{n}\sum\limits_{s=1}^{n}L_{i,r}H_{r,s}U_{s,j-1} \\[0.4cm] & = & \sum\limits_{r=1}^{n}\sum\limits_{s=1}^{n}L_{i,r}H_{r,s}U_{s-1,j-1} +(L\cdot H\cdot U)_{i,j-1} +\sum\limits_{s=1}^{n}L_{i,0}H_{0,s}U_{s-1,j-1}- \sum\limits_{r=0}^{n}L_{i,r}H_{r,0}U_{0,j-1} \\ \end{array}\end{equation} For convenience, we write $\Theta(i,j)=\sum\limits_{r=1}^{n}\sum\limits_{s=1}^{n}L_{i,r}H_{r,s}U_{s-1,j-1}$. Now, we apply \eqref{eee2}, to get $$\begin{array}{lll} \Theta(i,j) & = & \sum\limits_{r=1}^{n}\sum\limits_{s=1}^{n}\big(L_{i-1,r-1}+L_{i-1,r}\big) H_{r,s}U_{s-1,j-1}\\[0.4cm] & = & \sum\limits_{r=1}^{n} \sum\limits_{s=1}^{n}L_{i-1,r-1}H_{r,s}U_{s-1,j-1}+\sum\limits_{r=1}^{n} \sum\limits_{s=1}^{n}L_{i-1,r}H_{r,s}U_{s-1,j-1}\\ & = & \sum\limits_{r=2}^{n} \sum\limits_{s=2}^{n}L_{i-1,r-1}H_{r,s}U_{s-1,j-1}+ \sum\limits_{r=1}^{n}L_{i-1,r-1}H_{r,1}U_{0,j-1} \\[0.4cm] & & +\sum\limits_{s=2}^{n}L_{i-1,0}H_{1,s}U_{s-1,j-1}+ \sum\limits_{r=1}^{n}\sum\limits_{s=1}^{n}L_{i-1,r}H_{r,s}U_{s-1,j-1}\\[0.4cm] & = & \sum\limits_{r=2}^{n}\sum\limits_{s=2}^{n}L_{i-1,r-1}H_{r,s}U_{s-1,j-1}+ \sum\limits_{r=1}^{n}L_{i-1,r-1}H_{r,1}U_{0,j-1} \\[0.4cm] & & +\sum\limits_{s=2}^{n}L_{i-1,0}H_{1,s}U_{s-1,j-1}+ \sum\limits_{r=1}^{n}\sum\limits_{s=1}^{n}L_{i-1,r}H_{r,s} \big(U_{s,j}-U_{s,j-1}\big) \ \ \big({\rm by \ } \eqref{eee4}\big)\\ & = & \sum\limits_{r=2}^{n}\sum\limits_{s=2}^{n}L_{i-1,r-1}H_{r-1,s-1} U_{s-1,j-1}+\sum\limits_{r=1}^{n}L_{i-1,r-1}H_{r,1}U_{0,j-1} \\[0.4cm] & & +\sum\limits_{s=2}^{n}L_{i-1,0}H_{1,s}U_{s-1,j-1}+ \sum\limits_{r=1}^{n}\sum\limits_{s=1}^{n}L_{i-1,r}H_{r,s}U_{s,j}\\[0.4cm] & & - \sum\limits_{r=1}^{n}\sum\limits_{s=1}^{n}L_{i-1,r}H_{r,s}U_{s,j-1} \ \ \ \ \ \big(\text{by the structure of} \ H\big) \\[0.4cm] & = & \sum\limits_{r=1}^{n}\sum\limits_{s=1}^{n}L_{i-1,r}H_{r,s}U_{s,j-1}+ \sum\limits_{r=1}^{n}L_{i-1,r-1}H_{r,1}U_{0,j-1}+\sum\limits_{s=2}^{n}L_{i-1,0}H_{1,s}U_{s-1,j-1} \\[0.4cm] & & +\sum\limits_{r=1}^{n}\sum\limits_{s=0}^{n}L_{i-1,r}H_{r,s}U_{s,j}- \sum\limits_{r=1}^{n}L_{i-1,r}H_{r,0}U_{0,j}-\sum\limits_{r=1}^{n}\sum\limits_{s=1}^{n}L_{i-1,r}H_{r,s}U_{s,j-1} \\[0.4cm] & & \big( {\rm note \ that \ } L_{i-1, n-1}=U_{n-1, j-1}=0 \big) \\[0.2cm] & = & \sum\limits_{r=1}^{n}L_{i-1,r-1}H_{r,1}U_{0,j-1}+ \sum\limits_{s=2}^{n}L_{i-1,0}H_{1,s}U_{s-1,j-1}+ \sum\limits_{r=0}^{n}\sum\limits_{s=0}^{n}L_{i-1,r}H_{r,s}U_{s,j}\\[0.4cm] & & -\sum\limits_{s=0}^{n}L_{i-1,0}H_{0,s}U_{s,j} - \sum\limits_{r=1}^{n}L_{i-1,r}H_{r,0}U_{0,j}\\[0.4cm] & = & \sum\limits_{r=1}^{n}L_{i-1,r-1}H_{r,1}U_{0,j-1}+ \sum\limits_{s=2}^{n}L_{i-1,0}H_{1,s}U_{s-1,j-1}+(L\cdot H\cdot U)_{i-1,j} \\[0.4cm] & & -\sum\limits_{s=0}^{n}L_{i-1,0}H_{0,s}U_{s,j}- \sum\limits_{r=1}^{n}L_{i-1,r}H_{r,0}U_{0,j} \ \ \big({\rm by \ } \ \eqref{2009-26-2}\big ).\\[0.4cm] \end{array}$$ By substituting this in \eqref{end2}, we obtain $$ \begin{array}{lll} \Omega(i,j)& = &(L\cdot H\cdot U)_{i,j-1}+(L\cdot H\cdot U)_{i-1,j} \\[0.2cm] & & +\sum\limits_{r=1}^{n}L_{i-1,r-1}H_{r,1}U_{0,j-1}+ \sum\limits_{s=2}^{n}L_{i-1,0}H_{1,s}U_{s-1,j-1}\\[0.4cm] & & -\sum\limits_{s=0}^{n}L_{i-1,0}H_{0,s}U_{s,j}- \sum\limits_{r=1}^{n}L_{i-1,r}H_{r,0}U_{0,j}\\[0.4cm] & & +\sum\limits_{s=1}^{n}L_{i,0}H_{0,s}U_{s-1,j-1}- \sum\limits_{r=0}^{n}L_{i,r}H_{r,0}U_{0,j-1}. \\[0.4cm] \end{array} $$ Finally, if the above expression is substituted in \eqref{end1} and the sums are put together, then we obtain $$\begin{array}{lll} (L\cdot H\cdot U)_{i,j}& =&(L\cdot H\cdot U)_{i-1,j}+(L\cdot H\cdot U)_{i,j-1}+\Psi(i,j), \end{array}$$ where $$\begin{array}{lll} \Psi(i,j)& := & \sum\limits_{r=0}^{n}L_{i,r}H_{r,0}U_{0,j}+ \sum\limits_{r=1}^{n}L_{i-1,r-1}H_{r,1}U_{0,j-1}+ \sum\limits_{s=2}^{n}L_{i-1,0}H_{1,s}U_{s-1,j-1} \\[0.4cm] &&-\sum\limits_{s=0}^{n}L_{i-1,0}H_{0,s}U_{s,j}-\sum\limits_{r=1}^{n}L_{i-1,r}H_{r,0}U_{0,j}+ \sum\limits_{s=1}^{n}L_{i,0}H_{0,s}U_{s-1,j-1}\\[0.4cm] &&-\sum\limits_{r=0}^{n}L_{i,r}H_{r,0}U_{0,j-1}. \\ \end{array}$$ However, by easy calculations one can show that $$ \begin{array}{l} \sum\limits_{r=0}^{n}L_{i,r}H_{r,0}U_{0,j}-\sum\limits_{r=0}^{n}L_{i,r}H_{r,0}U_{0,j-1} =0, \\[0.4cm] \sum\limits_{r=1}^{n}L_{i-1,r-1}H_{r,1}U_{0,j-1}=\sum\limits_{r=1}^{n}{i-1\choose r-1} \hat{\mu}_r=\sum\limits_{r=1}^{n}\left({i\choose r}-{i-1\choose r}\right)\hat{\mu}_r=\mu_i-\mu_{i-1},\\[0.4cm] \sum\limits_{r=1}^{n}L_{i-1,r}H_{r,0}U_{0,j}=\sum\limits_{r=0}^{n} \hat{\lambda}_r-\lambda_0=\lambda_{i-1}-\lambda_0,\\[0.4cm] \sum\limits_{s=2}^{n}L_{i-1,0}H_{1,s}U_{s-1,j-1} = (j-1)v,\\[0.4cm] \sum\limits_{s=0}^{n}L_{i-1,0}H_{0,s}U_{s,j}=\lambda_0+ju,\\[0.4cm] \sum\limits_{s=1}^{n}L_{i,0}H_{0,s}U_{s-1,j-1}= u, \end{array} $$ and so $$\Psi(i,j)=\mu_i-\mu_{i-1}-\lambda_{i-1}+(j-1)(v-u).$$ This completes the proof. \end{proof} Before stating the next result, we need to introduce some additional definitions. Let $\lambda=(\lambda_i)_{i\geqslant 0}$ and $\mu=(\mu_i)_{i\geqslant 0}$ be two arbitrary sequences. The {\em convolution} of $\lambda$ and $\mu$ is the sequence $\nu=(\nu_i)_{i\geqslant 0}$, where $$\nu_i=\sum_{k=0}^{i}\lambda_k\mu_{i-k}.$$ The {\em convolution matrix} associated with sequences $\lambda$ and $\mu$ is the infinite matrix $A(\infty)$ whose first column ${\rm C}_0(A(\infty))$ is $\lambda$ and whose $j$th column $(j=1, 2, \ldots)$ is the convolution of sequences ${\rm C}_{j-1}(A(\infty))$ and $\mu$. We say that the convolution matrix of the sequences $\lambda$ and $\lambda$ is the convolution matrix of the sequence $\lambda$. There are many well-known integer matrices which can be written as convolution matrices of some sequences. For instance, $U(\infty)$ is the convolution matrix of the sequences $(1, 0, 0, \ldots)$ and $(1, 1, 0, 0, \ldots)$ and $P_{(1, 1, \ldots),(1, 1, \ldots)}(\infty)$ is the convolution matrix of the sequence $(1, 1,\ldots)$. We will need the following technical result \cite[Theorem 3.1]{yang}. \begin{proposition}\label{prop1} Let $$A(x)=\sum_{n=1}^{\infty}a_nx^{n-1}, \ \ B(x)=\sum_{n=0}^{\infty}b_nx^n, \ \ V(x)=\sum_{n=0}^{\infty}v_nx^n \ \ \text{and} \ \ W(x)=\sum_{n=0}^{\infty}w_nx^n$$ be the generating functions for the sequences $(a_n)_{n\geqslant 1}$, $(b_n)_{n\geqslant 0}$, $(v_n)_{n\geqslant 0}$, and $(w_n)_{n\geqslant 0}$, respectively. Consider an infinite dimensional matrix of the following form: $$M(\infty)=\left(\begin{array}{c|ccc}b_0&v_0&v_0w_0&\cdots\\[0.2cm] b_1&v_1&v_0w_1+v_1w_0&\cdots\\[0.2cm] b_2&v_2&v_0w_2+v_1w_1+v_2w_0&\cdots\\[0.2cm] \vdots&\vdots&\vdots&\ddots\end{array}\right)$$ where ${\rm C}_0(M(\infty))=(b_0, b_1, \ldots)^t$ and $M(\infty)^{[1]}$ is the convolution matrix of the sequences $(v_i)_{i\geqslant 0}$ and $(w_j)_{j\geqslant 0}$. If \begin{equation}\label{eth1} A(W(x))=B(x)/V(x), \end{equation} then for any non-negative integer $n$, there holds $$\det(M(n))=(-1)^{n}v_0^{n+1}w_1^{n(n+1)/2}a_{n+1},$$ where $M(n)$ is the $(n+1)\times (n+1)$ upper left corner matrix of $M(\infty)$. \end{proposition} We are now in a position to prove the following theorem which is the second result of this paper. \begin{theorem}\label{th2} Let $A(n)$ be defined as in Theorem $\ref{th1}$ and let $c$ be a constant. In the case when $u=v=1$ and $\lambda_i=(2^i-1)c+1$, we have the following statements: \begin{itemize} \item[{\rm (a)}] if $\mu_i=\left(2^i+\frac{(i-2)(i+1)}{2}\right)c-\frac{i(i-3)}{2}$, then $\det(A(n))=F_{n+1}$. \item[{\rm (b)}] if $\mu_i=\left(\frac{5\cdot 3^i }{4}-2^i-\frac{2i+1}{4}\right)c+\frac{5(3^i-1)}{4}+\frac{i}{2}$, then $\det(A(n))=L_{n+1}$. \item[{\rm (c)}] if $\mu_i=i^2c-i^2+2i$, then $\det(A(n))=J_{n+1}$. \item[{\rm (d)}] if $\mu_i=\left(2^{i+1}+\frac{(i+1)(i-4)}{2}\right)c+\frac{(5-i)i}{2}$, then $\det(A(n))=P_{n+1}$. \end{itemize} \end{theorem} \begin{proof} Let $\mu=(\mu_i)_{i\geqslant 0}$ be a sequence with $\mu_0=0$ and let $c$ be a constant. Let $\lambda=(\lambda_i)_{i\geqslant 0}$ be a sequence with $\lambda_i=(2^i-1)c+1$. We consider the infinite matrices $A(\infty)=[A_{i,j}]_{i,j\geqslant 0}$ whose entries satisfy \begin{equation}\label{Fib} A_{i,j}=A_{i-1,j}+A_{i,j-1}-(2^i-1)c-1+\mu_i-\mu_{i-1} \ \ \ {\rm for} \ \ \ i, j\geqslant 1,\end{equation} with the initial conditions $A_{i, 0}=(2^i-1)c+1$ and $A_{0, i}=1+i$, $i\geqslant 0$. By Theorem \ref{prop1}, we observe that $$ A(n)=L(n)\cdot H(n)\cdot U(n),$$ where $$H(n)=\left(\begin{array}{c|llcl} 1 & 1& 0& \cdots & 0 \\ \hline c& & & & \\ c & & & & \\ \vdots & & & T_{(\hat{\mu}_1, \hat{\mu}_2, \hat{\mu}_3, \ldots), (\hat{\mu}_1, v, 0, 0, \ldots)}(n-1) & \\ c & & & & \\ \end{array}\right).$$ Evidently $\det(A(n))=\det(H(n))$, so it suffices to find $\det(H(n))$. From the structure of matrix $H(\infty)$, we have $${\rm C}_0(H(\infty))=(b_i)_{i\geqslant 0}=(1, c, c, \ldots)^t,$$ whose generating function is $$B(x)=\frac{1+(c-1)x}{1-x}.$$ \begin{itemize} \item[{\rm (a)}] Let $\mu_i=(2^i+\frac{(i-2)(i+1)}{2})c-\frac{i(i-3)}{2}$. In this case, we have the following infinite dimensional matrices: $$A(\infty)=\left(\begin{array}{ccccc}1&2&3&4&\cdots\\[0.3cm] c+1&2c+3&3c+6&4c+10&\cdots\\[0.3cm] 3c+1&7c+3&12c+8&18c+17&\cdots\\[0.3cm] 7c+1&17c+2&32c+8&53c+23&\cdots\\[0.3cm] \vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right),$$ and $$H(\infty)=\left(\begin{array}{c|llcl} 1 & 1& 0& \cdots & 0 \\ \hline c& & & & \\ c & & & T_{(c+1, 2c-1, c, c, \ldots), (c+1, 1, 0, 0, \ldots)}(\infty) & \\ \vdots & & & & \\ \end{array}\right).$$ Note that the submatrix $H(\infty)^{[1]}$ is the convolution of sequences $$(v_i)_{i\geqslant 0}=(1, c+1, 2c-1, c, c, \ldots), \ \ \ \ {\rm and} \ \ \ \ (w_i)_{i\geqslant 0}=(0, 1, 0, 0, 0, \ldots),$$ whose generating functions are $$V(x)=\frac{1+cx+(c-2)x^2-(c-1)x^3}{1-x} \ \ \ \ {\rm and} \ \ \ \ W(x)=x,$$ respectively. Plugging these generating functions into \eqref{eth1} yields $$A(W(x))=A(x)=\frac{\frac{1+(c-1)x}{1-x}}{\frac{1+cx+(c-2) x^2-(c-1)x^3}{1-x}}=1-x+2x^2-3x^3+\cdots+(-1)^{n}F_{n+1}x^n+\cdots,$$ and it follows by Proposition \ref{prop1} that $$\det(H(n))=(-1)^{n}v_0^{n+1}w_1^{n(n+1)/2}a_{n+1}=(-1)^{n}a_{n+1}=F_{n+1},$$ as required. \item[{\rm (b)}] Let $\mu_i=\left(\frac{5\cdot 3^i }{4}-2^i-\frac{2i+1}{4}\right)c+\frac{5(3^i-1)}{4}+\frac{i}{2}$. The infinite dimensional matrices created in this case are as follows: $$A(\infty)=\left(\begin{array}{ccccc} 1&2&3&4&\cdots\\[0.3cm] c+1&2c+5&3c+10&4c+16&\cdots\\[0.3cm] 3c+1&9c+13&16c+30& 24c+53& \cdots\\[0.3cm] 7c+1&31c+36& 62c+88& 101c+163&\cdots\\[0.3cm] \vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right),$$ and $$H(\infty)=\left(\begin{array}{c|llcl} 1 & 1& 0& \cdots & 0 \\ \hline c& & & & \\ c & & & T_{(c+3, 4c+5, 9c+10, 19c+20, \ldots), (c+3, 1, 0, 0, \ldots)}(\infty) & \\ \vdots & & & & \\ \end{array}\right).$$ Again, one can easily see that the submatrix $H(\infty)^{[1]}$ is the convolution of sequences $$(v_i)_{i\geqslant 0}=(1, c+3, 4c+5, 9c+10, 19c+20, \ \ldots),$$ (with general form $v_0=1$, $v_1=c+3$ and $v_i=(5\cdot 2^{i-2})(c+1)-c$ for $i\geqslant 2$), and $$(w_i)_{i\geqslant 0}=(0, 1, 0, 0, 0, \ldots).$$ The generating functions for these sequences are $$V(x)=\frac{(1+(c-1)x)(-x^2+x+1)}{(1-x)(1-2x)}, \ \ \ {\rm and} \ \ \ \ W(x)=x,$$ respectively. If $B(x)$, $V(x)$ and $W(x)$ are substituted in \eqref{eth1}, then we obtain $$A(W(x))=A(x)=\frac{\frac{1+(c-1)x}{1-x}}{\frac{(1+(c-1)x)(-x^2+x+1)} {(1-x)(1-2x)}}=1-3x+4x^2-7x^3+11x^4+\cdots+(-1)^{n}L_{n+1}x^n+\cdots,$$ and by Proposition \ref{prop1}, it follows that $$\det(H(n))=(-1)^{n}v_0^{n+1}w_1^{n(n+1)/2}a_{n+1}=(-1)^{n}a_{n+1}=L_{n+1},$$ as required. \item[{\rm (c)}] Let $\mu_i=i^2c-i^2+2i$. In this case, we have the following infinite dimensional matrices: $$A(\infty)=\left(\begin{array}{ccccc} 1 & 2 & 3 & 4 & \cdots\\[0.3cm] c+1 &2c+3 &3c+6 &4c+10 &\cdots\\[0.3cm] 3c+1 &7c+2 &12c+6 &18c+14 &\cdots\\[0.3cm] 7c+1 &16c-1 &30c+1 &50c+11 &\cdots\\[0.3cm] \vdots &\vdots &\vdots &\vdots &\ddots \end{array}\right)$$ and $$H(\infty)=\left(\begin{array}{c|llcl} 1 & 1& 0& \cdots & 0 \\ \hline c& & & & \\ c & & & T_{(c+1, 2c-2, 0, 0, \ldots), (c+1, 1, 0, 0, \ldots)}(\infty) & \\ \vdots & & & & \\ \end{array}\right).$$ Moreover, from the structure of $H(\infty)$, we see that the submatrix $H(\infty)^{[1]}$ is the convolution of sequences $$(v_i)_{i\geqslant 0}=(1, c+1, 2c-2, 0, 0, \ \ldots), \ \ \ \ {\rm and} \ \ \ \ (w_i)_{i\geqslant 0}=(0, 1, 0, 0, \ldots),$$ with generating functions $V(x)=1+(c+1)x+(2c-2)x^2$ and $W(x)=x$, respectively. Substituting the obtained generating functions in \eqref{eth1}, we obtain $$A(W(x))=A(x)=\frac{\frac{1+(c-1)x}{1-x}}{1+(c+1)x+(2c-2)x^2}=1-x+3x^2-5x^3+\cdots+(-1)^{n}J_{n+1}x^n+\cdots.$$ Therefore, it follows from Proposition \ref{prop1} that $$\det(H(n))=(-1)^{n}v_0^{n+1}w_1^{n(n+1)/2}a_{n+1}=(-1)^{n}a_{n+1}=J_{n+1},$$ as required. \item[{\rm (d)}] Let $\mu_i=\left(2^{i+1}+\frac{(i+1)(i-4)}{2}\right)c+\frac{(5-i)i}{2}$. This time, we will deal with the following matrices: $$A(\infty)=\left(\begin{array}{ccccc} 1 & 2 & 3 & 4 & \cdots\\[0.3cm] c+1 &2c+4 &3c+8 &4c+13 &\cdots\\[0.3cm] 3c+1 &8c+5 &14c+13 &21c+26 &\cdots\\[0.3cm] 7c+1 &21c+5 &41c+17 &68c+42 &\cdots\\[0.3cm] \vdots &\vdots&\vdots &\vdots &\ddots \end{array}\right),$$ and $$H(\infty)=\left(\begin{array}{c|llcl} 1 & 1& 0& \cdots & 0 \\ \hline c& & & & \\ c & & & T_{(c+2, 3c-1, 2c, 2c, \ldots), (c+2, 1, 0, 0, \ldots)}(\infty) & \\ \vdots & & & & \\ \end{array}\right).$$ In addition, the submatrix $H(\infty)^{[1]}$ of $H(\infty)$ is the convolution of sequences: $$(v_i)_{i\geqslant 0}=(1, c+2, 3c-1, 2c, 2c, \ \ldots) \ \ \ \ {\rm and} \ \ \ \ (w_i)_{i\geqslant 0}=(0, 1, 0, 0, \ldots).$$ Note that the generating functions of these sequences are $$V(x)=\frac{1+(1+c)x+(2c-3)x^2-(c-1)x^3}{1-x} \ \ \ {\rm and} \ \ \ W(x)=x.,$$ respectively. After having substituted these generating functions in \eqref{eth1}, we obtain $$A(W(x))=A(x)=\frac{\frac{1+(c-1)x}{1-x}}{\frac{1+(1+c)x+(2c-3)x^2-(c-1)x^3}{1-x}}=1-2x+5x^2-12x^3+\cdots+(-1)^{n}P_{n+1}x^n+\cdots$$ Now, by Proposition \ref{prop1}, we deduce that $$\det(H(n))=(-1)^{n}v_0^{n+1}w_1^{n(n+1)/2}a_{n+1}=(-1)^{n}a_{n+1}=P_{n+1},$$ as required. \end{itemize} This completes the proof. \end{proof} \section{Some remarks} In this section, we will explain how the sequences $(\lambda_i)_{i\geqslant 0}$ and $(\mu_i)_{i\geqslant 0}$ in Theorem \ref{th2}, are determined. Consider the following lower Hessenberg matrix $$H(\infty)=[H_{i, j}]_{i, j\geqslant 0}=\left (\begin{array}{cccccccc}h_{0,0}&h_{0,1}&0&0&0&\cdots\\[0.1cm] h_{1,0}&h_{1,1}&h_{1,2}&0&0&\cdots\\[0.1cm] h_{2,0}&h_{2,1}&h_{1,1}&h_{1,2}&0&\cdots\\[0.1cm] h_{3,0}&h_{3,1}&h_{2,1}&h_{1,1}&h_{1,2}&\cdots\\[0.1cm] h_{4,0}&h_{4,1}&h_{3,1}&h_{2,1}&h_{1,1}&\cdots\\[0.1cm] \vdots&\vdots&\vdots&\vdots&\vdots&\ddots\\ \end{array}\right).$$ Let $H(n)=[H_{i,j}]_{0\leqslant i, j\leqslant n}$, and let $d_n$ be the $n$th determinant of $H(n)$. In what follows, we show that the sequence of principal minors of $H(\infty)$, i.e., $D(H(\infty))=(d_n)_{n\geqslant 0}$, satisfies a recurrence relation. \begin{proposition}\label{prop-new-1} With the above notation, we have $$d_n=\left\{\begin{array}{ll} h_{0, 0}, & \text{if} \ \ n=0,\\[0.2cm] (-1)^n h_{0,1}(h_{1,2})^{n-1}h_{n,0}+ \sum\limits_{k=0}^{n-1}h_{n-k,1}(-h_{1,2})^{n-k-1}d_k, & \text{if} \ \ n\geqslant 1.\end{array}\right.$$ \end{proposition} \begin{proof} Obviously, $d_0=h_{0, 0}$. Hence, from now on we assume $n>1$. First, we apply the following row operations: $$\begin{array}{rcl} H_1(n) & = & \Big(\prod\limits_{i=1}^{n}O_{i,0}(\frac{-h_{i,1}}{h_{0,1}})\Big)H(n),\\[0.3cm] H_2(n) & = & \Big(\prod\limits_{i=1}^{n-1}O_{i+1,1}(\frac{-h_{i,1}}{h_{1,2}})\Big)H_1(n),\\[0.3cm] H_3(n) & =& \Big(\prod\limits_{i=1}^{n-2}O_{i+2,2}(\frac{-h_{i,1}}{h_{1,2}})\Big)H_2(n),\\[0.3cm] & \vdots & \\ H_{n}(n) & = & \Big(\prod\limits_{i=1}^{1}O_{i+(n-1),n-1}(\frac{-h_{i,1}}{h_{1,2}})\Big)H_{n-1}(n). \\[0.3cm] \end{array} $$ It is obvious that, step by step, the columns are ``emptied" until finally the following matrix $$H_{n}(n)=\left(\begin{array}{ccccccccc} \tilde{h}_{0,0}& h_{0,1}& 0& 0& 0& \cdots & 0 \\ \tilde{h}_{1,0} & 0 & h_{1,2} & 0 & 0 & \cdots & 0 \\ \tilde{h}_{2,0} & 0 & 0 & h_{1,2}& 0 & \cdots & 0\\ \tilde{h}_{3,0} & 0 & 0 & 0& h_{1,2} & \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots& \vdots\\ \tilde{h}_{n-1,0} & 0 & 0 &0 &0 & \cdots & h_{1,2} \\ \tilde{h}_{n,0} & 0 & 0& 0 & 0& \cdots & 0\\ \end{array}\right)_{(n+1)\times (n+1)},$$ is obtained, where \begin{equation}\label{e-2014-1} \tilde{h}_{i, 0}=\left\{\begin{array}{lll} h_{0,0}, & \text{if} & i=0;\\[0.2cm] h_{1,0}-\frac{h_{1,1}}{h_{0,1}}h_{0,0}, & \text{if} & i=1;\\[0.2cm] h_{i,0}-\frac{h_{i, 1 }}{h_{0,1}}h_{0,0}-\frac{1}{h_{1, 2}}\sum\limits_{k=1}^{i-1}h_{i-k,1}\tilde{h}_{k, 0}, & \text{if} & i\geqslant 2. \end{array}\right.\end{equation} Evidently, $d_n=\det(H_n(n))$. Expanding the determinant along the last row of $\det(H_n(n))$, we obtain \begin{equation}\label{e-2014-2} d_n=(-1)^n \tilde{h}_{n,0} h_{0,1}(h_{1,2})^{n-1}, \ \ \ (n\geqslant 1). \end{equation} Finally, after some simplification, it follows that $$\begin{array}{lll}d_n&=& (-1)^n \tilde{h}_{n,0} h_{0,1}(h_{1,2})^{n-1}\\[0.4cm] &=& (-1)^n h_{0,1}(h_{1,2})^{n-1}\Big[h_{n,0}-\frac{h_{n, 1 }}{h_{0, 1}}h_{0, 0}-\frac{1}{h_{1, 2}}\sum\limits_{k=1}^{n-1}h_{n-k,1}\tilde{h}_{k, 0}\Big] \ \ \ {\rm (by \ \eqref{e-2014-1})}\\[0.4cm] &=& (-1)^n h_{0,1}(h_{1,2})^{n-1}h_{n,0}+(-1)^{n+1} (h_{1,2})^{n-1}h_{n, 1 }h_{0, 0}+(-1)^{n+1} h_{0,1}(h_{1,2})^{n-2}\sum\limits_{k=1}^{n-1}h_{n-k,1}\tilde{h}_{k, 0} \\[0.4cm] &=& (-1)^n h_{0,1}(h_{1,2})^{n-1}h_{n,0}+(-1)^{n+1} (h_{1,2})^{n-1}h_{n, 1 }h_{0, 0}+ \sum\limits_{k=1}^{n-1}h_{n-k,1}(-h_{1,2})^{n-k-1}d_k \ \ \ \ {\rm (by \ \eqref{e-2014-2})} \\[0.4cm] &=& (-1)^n h_{0,1}(h_{1,2})^{n-1}h_{n,0}+ \sum\limits_{k=0}^{n-1}h_{n-k,1}(-h_{1,2})^{n-k-1}d_k.\\ \end{array}$$ and the result follows. \end{proof} In Proposition \ref{prop-new-1}, if we take $h_{0, 0}=h_{0, 1}=1$, $h_{1, 2}=1$, $h_{i, 0}=c$ and $h_{i, 1}=\hat{\mu}_i$ for $i\geqslant 1$, then we obtain $$d_n=\left\{\begin{array}{lll} 1, & \text{if} & n=0;\\[0.2cm] (-1)^nc+\sum\limits_{k=0}^{n-1}\hat{\mu}_{n-k}(-1)^{n-k-1}d_k, & \text{if} & n\geqslant 1.\end{array}\right.$$ Now, if $(d_n)_{n\geqslant 0}\in \{\mathcal{F}, \mathcal{L}, \mathcal{J}, \mathcal{P}\}$, then $$ \hat{\mu}_{n}=c+(-1)^{n-1}d_n+\sum\limits_{k=1}^{n-1}(-1)^{k+1}\hat{\mu}_{n-k}d_k,$$ from which we determine the sequence $(\hat{\mu}_i)_{i\geqslant 1}$. Now, we form $$H(\infty)=\left(\begin{array}{c|llcl} 1 & 1& 0& \cdots & \\ \hline c& & & & \\ c & & & & \\ \vdots & & & T_{(\hat{\mu}_1, \hat{\mu}_2, \hat{\mu}_3, \ldots), (\hat{\mu}_1, 1, 0, 0, \ldots)}(\infty) & \\ c & & & & \\ \end{array}\right).$$ Finally, the sequences $(\lambda_i)_{i\geqslant 0}$ and $(\mu_i)_{i\geqslant 0}$ are determined by the equation $A(n)=L(n)\cdot H(n)\cdot U(n)$. \section{Acknowledgement} Our special thanks go to the Research Institute for Fundamental Sciences, Tabriz, Iran, for having sponsored this paper. \begin{thebibliography}{9} \bibitem{mp} A. R. Moghaddamfar and S. M. H. Pooya, Generalized Pascal triangles and Toeplitz matrices, {\em Electron. J. Linear Algebra} {\bf 18} (2009), 564--588. \bibitem{mtaj2} A. R. Moghaddamfar and H. Tajbakhsh, Lucas sequences and determinants, {\em Integers}, {\bf 12} (2012), 21--51. \bibitem{integer} N. J. A. Sloane, {\em The On-Line Encyclopedia of Integer Sequences}. Published electronically at \url{http://oeis.org}, 2013. \bibitem{yang} Y. Yang and M. Leonard, Evaluating determinants of convolution-like matrices via generating functions, {\em Int. J. Inf. Syst. Sci.}, {\bf 3} (2007), 569--580. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11C20; Secondary 15B36, 10A35. \noindent \emph{Keywords: } determinant, generalized Pascal triangle, Toeplitz matrix, matrix factorization, convolution matrix, lower Hessenberg matrix. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A000032}, \seqnum{A000045}, \seqnum{A000129}, and \seqnum{A001045}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received December 3 2013; revised version received March 2 2014. Published in {\it Journal of Integer Sequences}, March 26 2014. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document} .