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\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:
$$L_{i,j}=\left \{ \begin{array}{ll}
0, & \text{if} \ \ 0\leqslant i<j\leqslant n;\\[0.2cm]
{i\choose j}, & \text{if} \ \ 0\leqslant j\leqslant i\leqslant n.
\end{array} \right.$$
We put $U(n)=L(n)^t$, where $A^t$ signifies the transpose of
matrix A. Moreover, a {\em lower Hessenberg matrix}
$H(n)=[H_{i,j}]_{0\leqslant i, j\leqslant n}$ is a square matrix
of order $n+1$, where $H_{i,j}=0$ whenever $j>i+1$ 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.

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\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
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\end{thebibliography}

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\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.

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\noindent (Concerned with sequences
\seqnum{A000032},
\seqnum{A000045},
\seqnum{A000129}, and
\seqnum{A001045}.)

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\vspace*{+.1in}
\noindent
Received December 3 2013;
revised version received March 2 2014. 
Published in {\it Journal of Integer Sequences}, March 26 2014.

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\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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