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\begin{center}
\vskip 1cm{\LARGE\bf Symmetric Third-Order Recurring Sequences, \\
\vskip .1in
Chebyshev Polynomials, and Riordan Arrays} \vskip 1cm
\large
Paul Barry\\
School of Science\\
Waterford Institute of Technology\\
Ireland\\
\href{mailto:pbarry@wit.ie}{\tt pbarry@wit.ie} \\
\end{center}
\vskip .2 in


\begin{abstract}
We study a family of symmetric third-order recurring sequences with the aid of Riordan arrays and
Chebyshev polynomials. Formulas involving both
Chebyshev polynomials and Fibonacci numbers are established. The family of sequences defined by the product of consecutive
terms of the first family of sequences is also studied, and links to the
Chebyshev polynomials are again established, including continued fraction expressions. A multiplicative result is
established relating Chebyshev polynomials to sequences of doubled Chebyshev polynomials. Links to a special Catalan
related Riordan array are explored.
\end{abstract}

\section{Introduction}
In a posting to the \texttt{seqfan} mailing list, the forum for contributors to the OEIS \cite{SL1}, Richard Guy invited comments on the
`symmetric third-order recurring sequences'. This note explores some links between
these sequences and Riordan arrays \cite{SGWW, Spru}. Riordan arrays provide a convenient vehicle for exploring many
integer sequences, and can provide insight into the algebraic
relationships that exist between them. These can then often be translated into binomial identities.
\newline\newline
We can illustrate the context of this note by taking a simple example, based on the Fibonacci numbers $F_n$, where
$F_0=0$, $F_1=1$.
The sequence of ``golden rectangle numbers'' is the sequence of products $F_n F_{n+1}$. Thus we have (where we start at
index $1$)
\begin{center}\begin{tabular}{cccccccccc}
$n$ & & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $\ldots$ \\ \hline
$F_n$ & & $1$ & $1$ & $2$ & $3$ & $5$ & $8$ & $13$ & $\ldots$ \\
$F_{n+1}$ &  & $1$ & $2$ & $3$ & $5$ & $8$ & $13$ & $21$ & $\ldots$ \\ \hline
$F_nF_{n+1}$ & & $1$ & $2$ & $6$ & $15$ & $40$ & $104$ & $273$ & $\ldots$
\end{tabular}\end{center}
The generating function of the product sequence is given by
$$\frac{x}{1-2x-2x^2+x^3}=\frac{x}{(1+x)(1-3x+x^2)}.$$
Thus the product sequence satisfies a ``symmetric third-order recurrence''. It is also equal to an
alternating sum of the terms of the sequence with generating function
$$\frac{x}{1-3x+x^2}.$$
Thus $$F_n F_{n+1} = \sum_{k=0}^n (-1)^{n-k} F_{2k},$$ since
$F_{2n}$ is the sequence with g.f. $\frac{x}{1-3x+x^2}$. This motivates us to rearrange the table above in
a ``zig-zag'' fashion.
\begin{center}\begin{tabular}{cccccccccc}
$n$ & & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $\ldots$ \\ \hline
$v_{n+1}$ & & $1$ & $2$ & $2$ & $5$ & $5$ & $13$ & $13$ & $\ldots$ \\
$u_n $ &  & $1$ & $1$ & $3$ & $3$ & $8$ & $8$ & $21$ & $\ldots$ \\ \hline
$F_nF_{n+1}$ & & $1$ & $2$ & $6$ & $15$ & $40$ & $104$ & $273$ & $\ldots$
\end{tabular}\end{center}
Thus the sequence $F_n F_{n+1}$ has another factorization, involving doubled sequences.
The sequence $u_n$ or $1,1,3,3,8,8,\ldots$ has g.f. $$\frac{1+x}{1-3x^2+x^4},$$ while the sequence
$v_n$ or $1,1,2,2,5,5,\ldots$ has g.f.
$$\frac{(1+x)(1-x^2)}{1-3x^2+x^4}.$$ We note four points.
\begin{enumerate}
\item The ``base'' sequence $F_n$ has generating function
$$\frac{x}{1-x-x^2}=\frac{x(1+x-x^2)}{1-3x^2+x^4}.$$
\item The elements $F_{2n+2}$ of the sequence $1,3,8,13,\ldots$ are values of Chebyshev polynomials of the second
    kind\,:
$$F_{2n+2}=U_n(3/2).$$ Thus
$$F_{n+1}F_{n+2}=\sum_{k=0}^n (-1)^{n-k} U_k(3/2).$$
\item The sequence $1,2,5,13,\ldots$ represents the first differences of $1,3,8,21,\ldots$. Thus
the product sequence has another factorization $u_n v_{n+1}$ involving the doubling of a (Chebyshev) sequence and its
first differences.
\item We have
$$\sum_{i=0}^n F_iF_{i+1}=\sum_{i=0}^n\sum_{j=0}^i(-1)^{i-j}F_{2j}=\sum_{k=0}^{\lfloor \frac{n}{2}
\rfloor}F_{2(n-2k)}=\sum_{k=0}^{\lfloor \frac{n-1}{2} \rfloor}U_{n-2k-1}(3/2).$$
\end{enumerate}
The goal of this paper is to show that \emph{all} sequences with generating functions of the form
$$\frac{1}{1-kx-kx^2+x^4}$$ enjoy similar properties to those of the products of consecutive Fibonacci numbers: they are
all the product of consecutive terms of a related ``base'' sequence, and both sequences (base and product sequence) may be
expressed in terms of
Chebyshev polynomials. We employ the techniques of Riordan arrays to establish other interesting properties of these
sequences.
\newline\newline
\noindent The plan of the paper is as follows:
\begin{enumerate}
\item This Introduction
\item Notation and recall of results
\item The polynomial sequences $a(n;r)$, product sequences $b(n;r)$ and Riordan arrays
\item $a(n;r)$ and Chebyshev polynomials
\item $b(n;r)$ and Chebyshev polynomials
\item From $b(n;r)$ to $a(n;r)$
\item A Chebyshev product result
\item Additional Riordan array factorizations
\item A left inverse
\item The matrix $\left(\frac{1-x}{1+x^4},\frac{x}{1+x^2}\right)$
\item Appendix
\item Acknowledgements
\end{enumerate}
\section{Notation and recall of results}
For an integer sequence $a_n$, that is, an element of $\mathbf{Z}^\mathbf{N}$, the power series
$f(x)=\sum_{k=0}^{\infty}a_k x^k$ is called the \emph{ordinary generating function} or g.f. of the sequence.
$a_n$ is thus the coefficient of $x^n$ in this series. We denote this by
$a_n=[x^n]f(x)$. For instance, $F_n=[x^n]\frac{x}{1-x-x^2}$ is the $n$-th Fibonacci number, while
$C_n=[x^n]\frac{1-\sqrt{1-4x}}{2x}$ is the $n$-th Catalan number.
\newline\newline
\noindent
The \emph{Riordan group} \cite{SGWW}, \cite{Spru}, is a set of
infinite lower-triangular integer matrices, where each matrix is
defined by a pair of generating functions
$g(x)=1+g_1x+g_2x^2+\ldots$ and $f(x)=f_1x+f_2x^2+\ldots$ where
$f_1\ne 0$ \cite{Spru}. The associated matrix is the matrix whose
$i$-th column is generated by $g(x)f(x)^i$ (the first column being
indexed by 0). The matrix corresponding to the pair $g, f$ is
denoted by $(g, f)$ or $\cal{R}$$(g,f)$. The group law is then given
by
\begin{displaymath} (g, f)*(h, l)=(g(h\circ f), l\circ
f).\end{displaymath} The identity for this law is $I=(1,x)$ and the
inverse of $(g, f)$ is $(g, f)^{-1}=(1/(g\circ \bar{f}), \bar{f})$
where $\bar{f}$ is the compositional inverse of $f$.

A Riordan array of the form $(g(x),x)$, where $g(x)$ is the
generating function of the sequence $a_n$, is called the
\emph{sequence array} of the sequence $a_n$. Its general term is
$a_{n-k}$. Such arrays are also called \emph{Appell} arrays as they form the elements of the
Appell subgroup.
\newline\newline If $\mathbf{M}$ is the matrix $(g,f)$, and
$\mathbf{a}=(a_0,a_1,\ldots)'$ is an integer sequence with ordinary
generating function $\cal{A}$ $(x)$, then the sequence
$\mathbf{M}\mathbf{a}$ has ordinary generating function
$g(x)$$\cal{A}$$(f(x))$. The (infinite) matrix $(g,f)$ can thus be considered to act on the ring of
integer sequences $\mathbf{Z}^\mathbf{N}$ by multiplication, where a sequence is regarded as a
(infinite) column vector. We can extend this action to the ring of power series
$\mathbf{Z}[[x]]$ by
$$(g,f):\cal{A}(\mathnormal{x}) \longrightarrow \mathnormal{(g,f)}\cdot
\cal{A}\mathnormal{(x)=g(x)}\cal{A}\mathnormal{(f(x))}.$$
\begin{example} The so-called \emph{binomial matrix} $\mathbf{B}$ is the element
$(\frac{1}{1-x},\frac{x}{1-x})$ of the Riordan group. It has general
element $\binom{n}{k}$, and hence as an array coincides with Pascal's triangle. More generally, $\mathbf{B}^m$ is the
element $(\frac{1}{1-m x},\frac{x}{1-mx})$ of the Riordan group,
with general term $\binom{n}{k}m^{n-k}$. It is easy to show that the
inverse $\mathbf{B}^{-m}$ of $\mathbf{B}^m$ is given by
$(\frac{1}{1+mx},\frac{x}{1+mx})$.
\end{example}
\begin{example} If $a_n$ has generating function $g(x)$, then the generating function of the
sequence $$b_n=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} a_{n-2k}$$ is equal to
$$\frac{g(x)}{1-x^2}=\left(\frac{1}{1-x^2},x\right)\cdot g(x).$$
\end{example}
\noindent The row sums of the matrix $(g, f)$ have generating function
$$(g,f)\cdot \frac{1}{1-x}=\frac{g(x)}{1-f(x)}$$
 while the diagonal sums of $(g, f)$ (sums of left-to-right diagonals in the North East direction) have generating
function $g(x)/(1-xf(x))$. These coincide with the row sums of the ``generalized'' Riordan array $(g,xf)$. Thus the
Fibonacci numbers $F_{n+1}$ are the diagonal sums of the binomial matrix $\mathbf{B}$ given by
$\left(\frac{1}{1-x},\frac{x}{1-x}\right)$\,:
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0 & \ldots \\1 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 2
& 1 & 0 & 0 &
0 & \ldots \\ 1 & 3 & 3 & 1 & 0 & 0 & \ldots \\ 1 & 4 & 6
& 4 & 1 & 0 & \ldots \\1 & 5 & 10 & 10 & 5 & 1
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right)\end{displaymath} while they are the row sums of the
``generalized'' or ``stretched'' (using the nomenclature of \cite{CMS} ) Riordan array
$\left(\frac{1}{1-x},\frac{x^2}{1-x}\right)$\,:
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0 & \ldots \\1 & 0 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 1
& 0 & 0 & 0 &
0 & \ldots \\ 1 & 2 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 3 & 1
& 0 & 0 & 0 & \ldots \\1 & 4 & 3 & 0 & 0 & 0
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath}
We note that
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0 & \ldots \\1 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 2
& 1 & 0 & 0 &
0 & \ldots \\ 1 & 3 & 3 & 1 & 0 & 0 & \ldots \\ 1 & 4 & 6
& 4 & 1 & 0 & \ldots \\1 & 5 & 10 & 10 & 5 & 1
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right)\left(\begin{array}{c} 1\\
x\\
x^2\\
x^3\\
x^4\\
x^5\\
\vdots \end{array}\right)=\left(\begin{array}{c} 1\\
x+1\\
(x+1)^2\\
(x+1)^3\\
(x+1)^4\\
(x+1)^5\\
\vdots \end{array}\right)\end{displaymath} while a second application of this
matrix will send $x^n$ to $(x+2)^n$.
\newline\newline
\noindent To each proper Riordan matrix ($f_1 \ne 0$)
$$(g(x),f(x))=(g(x),xh(x))=(M_{n,k})_{n,k\ge 0}$$ there exist
\cite{ProdMat} two sequences
$\alpha=(\alpha_0,\alpha_1,\alpha_2,\ldots)(\alpha_0\ne 0)$
and
$\xi=(\xi_0,\xi_1,\xi_2,\ldots)$ such that \begin{enumerate}
\item Every
element in column $0$ can be expressed as a linear combination
of all the elements in the preceding row, the coefficients
being the elements
of the sequence $\xi$, i.e.
$$M_{n+1,0}=\xi_0M_{n,0}+\xi_1M_{n,1}+\xi_2M_{n,2}+\ldots$$
\item Every element $M_{n+1,k+1}$ not lying in
column $0$ or row $0$ can be expressed as a linear combination
of the elements of the preceding row, starting from the
preceding column on,
the coefficients being the elements of the sequence $\alpha$,
i.e. $$M_{n+1,k+1}=\alpha_0 M_{n,k}+\alpha_1
M_{n,k+1}+\alpha_2
M_{n,2}+\ldots$$ \end{enumerate}

\noindent The sequences $\alpha$ and $\xi$ are called the
$\alpha$-\emph{sequence} and the $\xi$-\emph{sequence} of the
Riordan matrix. It is
customary to use the same symbols $\alpha$ and $\xi$ as the
names of the corresponding generating functions. The functions
$g(x)$, $f(x)$,
$\alpha(x)$ and $\xi(x)$ are connected as follows\,: $$
f(x)=x\alpha(f(x)),\qquad
g(x)=\frac{M_{0,0}}{1-x\xi(f(x))}.$$
\noindent The $\alpha$-sequence is sometimes called the
$A$-\emph{sequence} of the array and then we write
$A(x)=\alpha(x)$.


Many interesting examples of Riordan arrays can be found in Neil Sloane's On-Line
Encyclopedia of Integer Sequences, \cite{SL1}, \cite{SL2}. Sequences are frequently referred to by their
OEIS number. For instance, the matrix $\mathbf{B}$ is \seqnum{A007318}.
\newline\newline
It is often the case that we work with ``generalized'' Riordan arrays, where we relax some of the conditions above. Thus
for instance \cite{CMS} discusses the notion of the
``stretched'' Riordan array. In this note, we shall encounter ``vertically stretched'' arrays of the form
$(g,h)$ where now $h_0=h_1=0$ with $h_2 \ne 0$. Such arrays are not invertible, but we may explore their left inversion.
In this context, standard Riordan arrays as described above are called
``proper'' Riordan arrays.
We note for instance that for any proper Riordan array $(g,f)$, its diagonal sums are just the row sums of the
vertically stretched array $(g,xf)$ and hence have g.f. $g/(1-xf)$.
\newline\newline
Each Riordan array $(g(x),f(x))$ has bi-variate generating function given by
$$\frac{g(x)}{1-yf(x)}.$$
For instance, the binomial matrix $\mathbf{B}$ has generating function
$$\frac{\frac{1}{1-x}}{1-y\frac{x}{1-x}}=\frac{1}{1-x(1+y)}.$$
\newline\newline
\noindent For a sequence $a_0, a_1, a_2, \ldots$ with g.f. $g(x)$, the ``aeration'' of the sequence is the sequence
$a_0, 0, a_1, 0, a_2, \ldots$ with interpolated zeros. Its g.f. is $g(x^2)$. The sequence $a_0, a_0, a_1, a_1, a_2,\ldots$
is
called the ``doubled'' sequence. It has g.f. $(1+x)g(x^2)$.
\newline\newline
The Chebyshev polynomials of the second kind, $U_n(x)$, are defined by
$$U_n(x)=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-k}{k}(-1)^k (2x)^{n-2k}$$
Alternatively,
$$U_n(x)=\sum_{k=0}^n \binom{\frac{n+k}{2}}{k}(-1)^{\frac{n-k}{2}}\frac{1+(-1)^{n-k}}{2}(2x)^k.$$
They satisfy the three-term recurrence
\begin{equation}U_{n+1}=2xU_n(x)-U_{n-1}(x), \quad U_0(x)=1,\quad U_1(x)=2x.\end{equation}
The generating function for $U_n(x)$ is given by
\begin{equation}\label{Chebgf}\sum_{k=0}^{\infty} U_n(x)y^n =\frac{1}{1-2xy+y^2}.\end{equation} We note that the
coefficient array of the
polynomials $U_n(x/2)$ which begin $1, x, x^2-1, x^3-2x, x^4-3x^2+1,\ldots,$ begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0 & \ldots \\0 & 1 & 0 & 0 & 0 & 0 & \ldots \\ -1 & 0
& 1 & 0 & 0 &
0 & \ldots \\ 0 & -2 & 0 & 1 & 0 & 0 & \ldots \\ 1 & 0 & -3
& 0 & 1 & 0 & \ldots \\0 & 3 & 0 & -4 & 0 & 1
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath} This is the Riordan array
$$\left(\frac{1}{1+x^2},\frac{x}{1+x^2}\right).$$ This follows since
$$\frac{\frac{1}{1+x^2}}{1-y\frac{x}{1+x^2}}=\frac{1}{1-xy+y^2}=\frac{1}{1-2\frac{x}{2}y+y^2}.$$
This matrix is \seqnum{A049310}.
Thus
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0 & \ldots \\0 & 1 & 0 & 0 & 0 & 0 & \ldots \\ -1 & 0
& 1 & 0 & 0 &
0 & \ldots \\ 0 & -2 & 0 & 1 & 0 & 0 & \ldots \\ 1 & 0 & -3
& 0 & 1 & 0 & \ldots \\0 & 3 & 0 & -4 & 0 & 1
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right)\left(\begin{array}{c} 1\\
x\\
x^2\\
x^3\\
x^4\\
x^5\\
\vdots \end{array}\right)=\left(\begin{array}{c} 1\\
x\\
x^2-1\\
x^3-2x\\
x^4-3x^2+1\\
x^5-4x^3+3x\\
\vdots \end{array}\right).\end{displaymath}
Similarly, the polynomials $U_n\left(\frac{x+2}{2}\right)$ have coefficient array given by
$$\left(\frac{1}{1+x^2}, \frac{x}{1+x^2}\right)\cdot\mathbf{B}^2=\left(\frac{1}{(1-x)^2},\frac{x}{(1-x)^2}\right).$$
In matrix terms, this gives us
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0 & \ldots \\2 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 3 & 4
& 1 & 0 & 0 &
0 & \ldots \\ 4 & 10 & 6 & 1 & 0 & 0 & \ldots \\ 5 & 20 & 21
& 8 & 1 & 0 & \ldots \\6 & 35 & 56 & 36 & 10 & 1
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right)\left(\begin{array}{c} 1\\
x\\
x^2\\
x^3\\
x^4\\
x^5\\
\vdots \end{array}\right)=\left(\begin{array}{c} 1\\
x+2\\
x^2+4x+3\\
x^3+6x^2+10x+4\\
x^4+8x^3+21x^2+20x+5\\
x^5+10x^4+36x^3+56x^2+35x+6\\
\vdots \end{array}\right).\end{displaymath} This coefficient matrix is \seqnum{A078812}, whose general term is
$\binom{n+k+1}{2k+1}$. The polynomials so defined are the Morgan-Voyce polynomials \cite{Morgan-Voyce}
$$M(n,x)=\sum_{k=0}^n \binom{n+k+1}{2k+1}x^k.$$
The Morgan-Voyce polynomials $m(n,x)$ are defined by
$$m(n,x)=\sum_{k=0}^n \binom{n+k}{2k} x^k=\sum_{k=0}^n \binom{n+k}{n-k}x^k.$$
The coefficient array for these polynomials, which has general term  $\binom{n+k}{2k}$, is the Riordan array
$$\left(\frac{1}{1-x},\frac{x}{(1-x)^2}\right).$$ This is \seqnum{A085478}. For this, we have
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0 & \ldots \\1 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 3
& 1 & 0 & 0 &
0 & \ldots \\ 1 & 6 & 5 & 1 & 0 & 0 & \ldots \\ 1 & 10 & 15
& 7 & 1 & 0 & \ldots \\1 & 15 & 35 & 28 & 9 & 1
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right)\left(\begin{array}{c} 1\\
x\\
x^2\\
x^3\\
x^4\\
x^5\\
\vdots \end{array}\right)=\left(\begin{array}{c} 1\\
x+1\\
x^2+3x+1\\
x^3+5x^2+6x+1\\
x^4+7x^3+15x^2+10x+1\\
x^5+9x^4+28x^3+35x^2+15x+1\\
\vdots \end{array}\right).\end{displaymath}
The Morgan-Voyce polynomials can be defined by a three-term recurrence as follows\,:
\begin{equation} m(0,x)=1,\quad m(1,x)=1+x,\quad m(n,x)=(x+2)m(n-1,x)-m(n-2,x) \end{equation}
and
\begin{equation} M(0,x)=1,\quad M(1,x)=2+x,\quad M(n,x)=(x+2)M(n-1,x)-M(n-2,x).\end{equation}
\noindent In the sequel, we shall use the notation $i=\sqrt{-1}$.
\section{The polynomial sequences $a(n;r)$, product sequences $b(n;r)$ and Riordan arrays}
We wish to study sequences of polynomials of the form
$$a(n;r)=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-k}{k}r^{\lfloor \frac{n-2k}{2} \rfloor},$$ with generating
function
$$g_a(x;r)=\frac{1+x-x^2}{1-(r+2)x^2+x^4}$$ and the related sequences
$b(n;r)$ with generating function
$$g_b(x;r)=\frac{1}{1-(r+1)x-(r+1)x^2+x^3}.$$
Examples of these sequences in the OEIS \cite{SL1} are given below.
\begin{center}
\begin{tabular}{|c|c|c|}\hline
$r$ & $a(n;r)$ & $b(n;r)$ \\\hline
0 & \seqnum{A157329} & \seqnum{A008619} \\\hline
1 & \seqnum{A000045}$(n+1)$ & \seqnum{A001654}$(n+1)$ \\\hline
2 & \seqnum{A002530}$(n+1)$ & \seqnum{A109437}$(n+1)$ \\\hline
3 & \seqnum{A136211}$(n+1)$ & \seqnum{A099025} \\\hline
4 & \seqnum{A041011} & \seqnum{A084158}$(n+1)$ \\\hline
5 & \seqnum{A152119}$(n+1)$ & \seqnum{A157335} \\\hline
\end{tabular}\end{center}
In each of the cases above, we have
$$b(n;r)=a(n;r)a(n+1;r).$$
Elements of the family $\{a(n;r)\}$ are images of the doubled sequences $1,1,r,r,r^2,r^2,\ldots$ with generating function
$\frac{1+x}{1-rx^2}$ by the Riordan array
$$\left(\frac{1}{1-x^2}, \frac{x}{1-x^2}\right)$$ (closely related to the Chebyshev polynomials). This is so since we
have
\begin{eqnarray*}\left(\frac{1}{1-x^2}, \frac{x}{1-x^2}\right)\cdot\frac{1+x}{1-rx^2}&=&
\frac{1}{1-x^2}\frac{1+\frac{x}{1-x^2}}{1-r\left(\frac{x}{1-x^2}\right)^2}\\
&=&\frac{1+x-x^2}{1-(r+2)x^2+x^4},\end{eqnarray*} the generating function of $a(n;r)$. An immediate consequence of this is
the fact that
\begin{eqnarray*} a(n;r)&=&\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-k}{k}r^{\lfloor \frac{n-2k}{2}\rfloor}\\
&=&\sum_{k=0}^n \binom{k}{n-k}r^{\lfloor \frac{2k-n}{2} \rfloor} \\
&=&\sum_{k=0}^n \binom{\frac{n+k}{2}}{k}\frac{1+(-1)^{n-k}}{2} r^{\lfloor \frac{k}{2} \rfloor}.\end{eqnarray*}
$a(n;r)$ thus represents the diagonal sums of the number triangle with general term $\binom{n}{k}r^{\lfloor \frac{n-k}{2}
\rfloor}$.
It is clear that $a(n;r)$ is a polynomial in $r$. We can express $a(n;r)$ directly in terms of ascending powers of $r$ by
noticing that the coefficient array of the polynomials $a(n;r)$, which begins
\begin{displaymath}\mathbf{\mathbf{A}}=\left(\begin{array}{ccccccc}
1 & 0 & 0 & 0 & 0 & 0 & \ldots \\1 &
0 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 1 & 0 & 0 & 0 & 0 & \ldots
\\
2 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 3 & 1 & 0 & 0 & 0 &
\ldots \\3 & 4 &
1 & 0 & 0 & 0 &\ldots\\ \vdots & \vdots & \vdots & \vdots &
\vdots & \vdots & \ddots\end{array}\right),\end{displaymath}
is the (generalized) Riordan array
$$\left(\frac{1+x-x^2}{(1-x^2)^2}, \frac{x^2}{(1-x^2)^2}\right)$$ with general term
$$\binom{\lfloor \frac{n+2k+1}{2} \rfloor}{\lfloor \frac{n-2k}{2} \rfloor}.$$ (A proof of this assertion is indicated in the Appendix). Thus we have
\begin{equation}a(n;r)=\sum_{k=0}^n \binom{\lfloor \frac{n+2k+1}{2} \rfloor}{\lfloor \frac{n-2k}{2}
\rfloor}r^k.\end{equation} 
This translates the algebraic identity
$$\left(\frac{1+x-x^2}{(1-x^2)^2}, \frac{x^2}{(1-x^2)^2}\right)\cdot \frac{1}{1-rx}=\frac{1+x-x^2}{1-(r+2)x^2+x^4}.$$
The family of polynomials $a(n;r)$ begins
$$1,1,r+1,r+2,r^2+3r+1,r^2+4r+3,r^3+5r^2+6r+1,\ldots$$
We note that if we define a family of polynomials $\mathcal{P}(n,x)$ as
$$\mathcal{P}(0,x)=1,\quad \mathcal{P}(1,x)=1, \quad \mathcal{P}(n,x)=\mathcal{P}(n-1,x)+x \mathcal{P}(n-2,x), \quad n>1$$
then
we have $$a(n,r)=r^{\lfloor \frac{n}{2} \rfloor} \mathcal{P}\left(n,\frac{1}{r}\right).$$
\noindent We further have
\begin{eqnarray*}
\frac{1+x-x^2}{1-(r+2)x^2+x^4}&=&\left(\frac{1}{1-x^2}, \frac{x}{1-x^2}\right)\cdot\frac{1+x}{1-rx^2}\\
&=&\left(\frac{1}{1-x^2}, \frac{x}{1-x^2}\right)\cdot (1+x, x^2) \cdot \frac{1}{1-rx}\\
&=& \left(\frac{1+x-x^2}{(1-x^2)^2},\frac{x^2}{(1-x^2)^2}\right)\cdot \frac{1}{1-rx}\end{eqnarray*}
where the last matrix is not invertible (due to the factor $(1+x,x^2)$).
\newline\newline
\noindent We can describe $a(n;r)$ in terms of the Morgan-Voyce polynomials as follows\,:
\begin{proposition} We have
$$a(2n;r)=m(n,r),\qquad a(2n+1;r)=M(n,r).$$
\end{proposition}
\begin{proof}
We have
\begin{eqnarray*}
a(2n;r)&=&\sum_{k=0}^n \binom{2n-k}{k}r^{n-k}\\
&=&\sum_{k=0}^n \binom{2n-(n-k)}{n-k}r^k\\
&=&\sum_{k=0}^n \binom{n+k}{n-k}r^k.\end{eqnarray*}
The case of $a(2n+1;r)$ follows in like manner.
\end{proof}
\begin{corollary}
$$a(n;r)=m(n/2,r)\frac{1+(-1)^n}{2}+M((n-1)/2,r)\frac{1-(-1)^n}{2}.$$
\end{corollary}
The main result of this paper is that the observation
$$b(n;r)=a(n;r)a(n+1;r)$$ holds in general.
\begin{theorem}\label{Thm_product} Let $a(n;r)$ be the sequence of polynomials with generating function
$$g_a(x;r)=\frac{1+x-x^2}{1-(r+2)x^2+x^4}.$$ Also let $b(n;r)$ be the sequence of polynomials with generating function
$$g_b(x;r)=\frac{1}{1-(r+1)x-(r+1)x^2+x^3}.$$ Then we have, for all $n,r \in \mathbb{Z}$,
$$b(n;r)=a(n;r)a(n+1;r).$$
\end{theorem}
\begin{proof}
We have
$$g_a(x;r)=\frac{1+x-x^2}{1-(r+2)x^2+x^4}=A(x)$$ so that
$$A(x)-(r+2)x^2A(x)+x^4A(x)=1+x-x^2.$$
Extracting the coefficient of $x^{n+4}$, we obtain
$$[x^{n+4}]A(x)-(r+2)[x^{n+2}]A(x)+[x^n]A(x)=[x^{n+4}](1+x-x^2)$$
\begin{equation}\label{Rec_1} A_{n+4}-(r+2)A_{n+2}+A_n=0. \end{equation}
By direct inspection, the first terms of the sequence $A_n=a(n;r)$ are
$${1,1,1+r,2+r,1+3r+r^2,3+4r+r^2,\ldots}.$$
We now set $B_n=A_nA_{n+1}$ and $T_n=A_{n}A_{n+3}$, and try to determine the corresponding generating functions.
From recurrence (\ref{Rec_1}), we have
\begin{eqnarray*}A_{n+4}&=&(r+2)A_{n+2}-A_n\\
A_{n+5}&=&(r+2)A_{n+3}-A_{n+1}.\end{eqnarray*}
Multiplying, we obtain
$$A_{n+4}A_{n+5}=(r+2)^2A_{n+2}A_{n+3}-(r+2)A_{n+1}A_{n+2}-(r+2)A_{n}A_{n+3}+A_nA_{n+1}.$$
This can be written as:
\begin{equation} \label{Rec_2} B_{n+4}=(r+2)^2B_{n+2}-(r+2)B_{n+1}-(r+2)T_n+B_n.\end{equation}
Now multiplying equation (\ref{Rec_1}) by $A_{n+3}$, we obtain
$$A_{n+3}A_{n+4}=(r+2)A_{n+2}A_{n+3}-A_nA_{n+3},$$ or
\begin{equation}\label{Rec_3} B_{n+3}=(r+2)B_{n+2}-T_n. \end{equation}
We can now use the method of coefficients \cite{Merlini_MC} to pass from the recurrences to the corresponding generating
functions
$B(x)$ and $T(x)$. In order to do so, we note that the coefficients of the sequence generated by $B(x)$ begin:
$${1,1+r,2+3r+r^2,2+7r+5r^2+r^3,\ldots}.$$
Equation (\ref{Rec_2}) becomes
\begin{equation}\begin{split} \frac{B(x)-1-(1+r)x-(2+3r+r^2)x^2-(2+7r+5r^2+r^3)x^3}{x^4}\\
=(r+2)\frac{B(x)-1-(1+r)x}{x^2}-(r+2)\frac{B(x)-1}{x}-(r+2)T(x)+B(x)\end{split}\end{equation}
while Equation (\ref{Rec_3}) becomes:
\begin{equation} \frac{B(x)-1-(1+r)x-(2+3r+r^2)x^2}{x^3}=(r+2)\frac{B(x)-1-(1+r)x}{x^2}-T(x).\end{equation}
We now have two linear equations for the two unknowns $B(x)$ and $T(x)$. Solving, we find that
$$B(x)=\frac{1}{1-(r+1)x-(r+1)x^2+x^3}=g_b(x;r)$$ as required.
\end{proof}
\begin{corollary}
$$T(x)=\frac{(r+2)-x}{1-(r+1)x-(r+1)x^2+x^3}$$ is the generating function of the product $a(n;r)a(n+3;r)$.
\end{corollary}
The author is indebted to an anonymous reviewer for the form of this proof.
\newline\newline
We note that this theorem establishes the binomial identity
\begin{equation}\sum_{j=0} ^k \binom{\lfloor \frac{n+1+2j}{2} \rfloor}{\lfloor \frac{n-2j}{2} \rfloor} \binom{\lfloor
\frac{n+2+2(k-j)}{2} \rfloor}{\lfloor \frac{n+1-2(k-j)}{2} \rfloor}=\sum_{j=0}^n
(-1)^{n-j}\binom{j+k+1}{2k+1}\end{equation} where the left hand side follows from the Cauchy form of the product of two
polynomials and the right hand side is an expression for the
general term of the $b(n;r)$ coefficient array $\left(\frac{1}{(1-x)(1-x^2)},\frac{x}{(1-x)^2}\right)$.
\section{$a(n;r)$ and Chebyshev polynomials}
We can express $a(n;r)$ in terms of Chebyshev polynomials. This is the content of the following proposition.
\begin{proposition}
\begin{equation} a(n;r)=i^n \bigg\{\frac{\sqrt{r}+1}{2\sqrt{r}}U_n\left(-i\frac{\sqrt{r}}{2}\right)+
\frac{\sqrt{r}-1}{2\sqrt{r}}U_n\left(i\frac{\sqrt{r}}{2}\right)\bigg\}.\end{equation}
\end{proposition}
\begin{proof}
We recall that the g.f. of $a(n;r)$ is $\frac{1+x-x^2}{1-(r+2)x^2+x^4}$. Now
\begin{eqnarray*}
\frac{1+x-x^2}{1-(r+2)x^2+x^4}&=&\frac{1-x^2}{1-(r+2)x^2+x^4}+\frac{x}{1-(r+2)x^2+x^4}\\
&=&\frac{1}{2}\bigg\{\frac{1}{1-\sqrt{r}x-x^2}+\frac{1}{1+\sqrt{r}x-x^2}\bigg\}
+\frac{1}{2\sqrt{r}}\bigg\{\frac{1}{1-\sqrt{r}x-x^2}-\frac{1}{1+\sqrt{r}x-x^2}\bigg\}\\
&=&\left(\frac{1}{2}+\frac{1}{2\sqrt{r}}\right)\frac{1}{1-\sqrt{r}x-x^2}+
\left(\frac{1}{2}-\frac{1}{2\sqrt{r}}\right)\frac{1}{1+\sqrt{r}x-x^2}\\
&=&\frac{\sqrt{r}+1}{2\sqrt{r}}\frac{1}{1-\sqrt{r}x-x^2}+
\frac{\sqrt{r}-1}{2\sqrt{r}}\frac{1}{1+\sqrt{r}x-x^2}.
\end{eqnarray*}
The assertion follows from this, given Eq. (\ref{Chebgf}).
\end{proof}
\begin{corollary}
\begin{equation}\label{corU}a(n;r)=i^n U_n\left(i\frac{\sqrt{r}}{2}\right)\bigg\{(-1)^n
\frac{\sqrt{r}+1}{2\sqrt{r}}+\frac{\sqrt{r}-1}{2\sqrt{r}}\bigg\}.\end{equation}
\end{corollary}
\begin{proof}
We have $U_n(-x)=(-1)^n U_n(x)$. The result follows from this.
\end{proof}
\noindent This now allows us to characterize the product $a(n;r)a(n+1;r)$ in terms of Chebyshev polynomials.
\begin{proposition}
\begin{equation}a(n;r)a(n+1;r)=i(-1)^{n+1}\frac{U_n\left(i\frac{\sqrt{r}}{2}\right)U_{n+1}\left(i\frac{\sqrt{r}}{2}\right)}{\sqrt{r}}.\end{equation}
\end{proposition}
\begin{proof}
\begin{eqnarray*}
a(n;r)a(n+1;r)&=&i(-1)^nU_n\left(i\frac{\sqrt{r}}{2}\right)U_{n+1}\left(i\frac{\sqrt{r}}{2}\right)\bigg\{(-1)^n\frac{\sqrt{r}+1}{2\sqrt{r}}+\frac{\sqrt{r}-1}{2\sqrt{r}}\bigg\}\cdot
\\&&\quad\bigg\{-(-1)^n\frac{\sqrt{r}+1}{2\sqrt{r}}+\frac{\sqrt{r}-1}{2\sqrt{r}}\bigg\}\\
&=&i(-1)^nU_n\left(i\frac{\sqrt{r}}{2}\right)U_{n+1}\left(i\frac{\sqrt{r}}{2}\right)\frac{1}{4r}(-(\sqrt{r}+1)^2+(\sqrt{r}-1)^2)\\
&=&i(-1)^nU_n\left(i\frac{\sqrt{r}}{2}\right)U_{n+1}\left(i\frac{\sqrt{r}}{2}\right)\frac{1}{4r}.-4\sqrt{r}\\
&=&i(-1)^{n+1}\frac{U_n\left(i\frac{\sqrt{r}}{2}\right)U_{n+1}\left(i\frac{\sqrt{r}}{2}\right)}{\sqrt{r}}.\end{eqnarray*}
\end{proof}
\noindent The following proposition provides a link between the polynomials $a(n;r)$ and continued fractions.
\begin{proposition}
\begin{equation}\label{cf} a(n+2;r)=a(n+1;r)\frac{r(1+(-1)^n)+1-(-1)^n}{2}+a(n;r)\end{equation}
\end{proposition}
\begin{proof}
We have
\begin{eqnarray*}
a(n+2;r)&=&i^{n+2}U_{n+2}\left(\frac{i\sqrt{r}}{2}\right)\bigg\{(-1)^{n+2}\frac{\sqrt{r}+1}{2\sqrt{r}}+\frac{\sqrt{r}-1}{2\sqrt{r}}\bigg\}\\
&=&-i^n\left(2i\frac{\sqrt{r}}{2}U_{n+1}\left(\frac{i\sqrt{r}}{2}\right)-U_n\left(\frac{i\sqrt{r}}{2}\right)\right)\bigg\{(-1)^{n+2}\frac{\sqrt{r}+1}{2\sqrt{r}}+\frac{\sqrt{r}-1}{2\sqrt{r}}\bigg\}\\
&=&-i^{n+1}\sqrt{r}U_{n+1}\left(\frac{i\sqrt{r}}{2}\right)\bigg\{(-1)^n\frac{\sqrt{r}+1}{2\sqrt{r}}+\frac{\sqrt{r}-1}{2\sqrt{r}}\bigg\}+\\
&&\qquad\qquad
i^nU_n\left(\frac{i\sqrt{r}}{2}\right)\bigg\{(-1)^n\frac{\sqrt{r}+1}{2\sqrt{r}}+\frac{\sqrt{r}-1}{2\sqrt{r}}\bigg\}\\
&=&a(n+2;r)\frac{-\sqrt{r}\bigg\{(-1)^n\frac{\sqrt{r}+1}{2\sqrt{r}}+\frac{\sqrt{r}-1}{2\sqrt{r}}\bigg\}}{\bigg\{(-1)^{n+1}\frac{\sqrt{r}+1}{2\sqrt{r}}+\frac{\sqrt{r}-1}{2\sqrt{r}}\bigg\}}+a(n;r)\\
&=&a(n+1;r)\frac{r(1+(-1)^n)+1-(-1)^n}{2}+a(n;r).
\end{eqnarray*}
\end{proof}
\noindent Inspired by \cite{Mansour}, we now define a family of rational functions $R(n,r)$ as follows\,:
\begin{displaymath}
R(n,r)=\left\{\begin{array}{ll}
1 & \text{if } n=0\\
1 & \text{if } n=1\\
\frac{1}{1+\frac{1}{r}R(n-1,r)} & \text{if } n>1
\end{array}\right.
\end{displaymath}
The first elements of the family are
$$1,1,\frac{r}{r+1},\frac{r+1}{r+2},\frac{r(r+2)}{r^2+3r+1},\frac{r^2+3r+1}{r^2+4r+3},\ldots$$
\begin{proposition}
$$a(n;r)=\textrm{Denominator}(R(n,r))$$
\end{proposition}
\begin{proof} Induction, using Eq. (\ref{cf}).
\end{proof}
\noindent Noting that
$$R(n,r)=\frac{i\sqrt{r}U_{n-1}\left(\frac{i\sqrt{r}}{2}\right)}{U_n\left(\frac{i\sqrt{r}}{2}\right)},\quad n>0, $$
we have
\begin{equation}a(n;r)=\textrm{Denominator}\left(\frac{i\sqrt{r}U_{n-1}\left(\frac{i\sqrt{r}}{2}\right)}{U_n\left(\frac{i\sqrt{r}}{2}\right)}\right).\end{equation}
Finally, we can use the expression
$$U_n(x)=2^n \prod_{k=1}^n \left(x-\cos\left(\frac{k \pi}{n+1}\right)\right)$$ and Corollary \textbf{\ref{corU}} to give
another characterization of $a(n;r)$.
\begin{proposition}
\begin{equation}a(n;r)=\prod_{k=1}^{\lfloor \frac{n}{2} \rfloor} \left(r+4\cos^2\left(\frac{k
\pi}{n+1}\right)\right).\end{equation}
\end{proposition}
\begin{proof}
We begin with the case $n=2m$. Then
\begin{eqnarray*}
i^n U_n\left(\frac{i\sqrt{r}}{2}\right)&=&i^n2^n\prod_{k=1}^n \left(\frac{i\sqrt{r}}{2}-\cos\left(\frac{k
\pi}{n+1}\right)\right)\\
&=&i^n \prod_{k=1}^n\left(i\sqrt{r}-2\cos\left(\frac{k \pi}{n+1}\right)\right)\\
&=&i^n \prod_{k=1}^m\left(i\sqrt{r}-2\cos\left(\frac{k \pi}{n+1}\right)\right)\left(i\sqrt{r}+2\cos\left(\frac{k
\pi}{n+1}\right)\right)\\
&=&i^n \prod_{k=1}^m\left(-r-4\cos^2\left(\frac{k \pi}{n+1}\right)\right)\\
&=&i^{2m} (-1)^m\prod_{k=1}^m\left(r+4\cos^2\left(\frac{k \pi}{n+1}\right)\right)\\
&=&\prod_{k=1}^{\lfloor \frac{n}{2} \rfloor}\left(r+4\cos^2\left(\frac{k \pi}{n+1}\right)\right).
\end{eqnarray*} Note that in the product expansion we have used a first with last, second with next-to-last, etc coupling.
We now turn to the case $n=2m+1$. We obtain
\begin{eqnarray*}
i^n U_n\left(\frac{i\sqrt{r}}{2}\right)&=&i^n2^n\prod_{k=1}^n \left(\frac{i\sqrt{r}}{2}-\cos\left(\frac{k
\pi}{n+1}\right)\right)\\
&=&i^n \prod_{k=1}^n\left(i\sqrt{r}-2\cos\left(\frac{k \pi}{n+1}\right)\right)\\
&=&i^n \bigg\{\prod_{k=1}^m\left(i\sqrt{r}-2\cos\left(\frac{k \pi}{n+1}\right)\right)\left(i\sqrt{r}+2\cos\left(\frac{k
\pi}{n+1}\right)\right)\bigg\}\cdot\\
& & \quad \quad \left(i\sqrt{r}-2\cos\left(\frac{(m+\frac{1}{2})\pi}{2m+1}\right)\right)\\
&=&i^n \prod_{k=1}^m\left(-r-4\cos^2\left(\frac{k
\pi}{n+1}\right)\right)\left(i\sqrt{r}-2\cos\left(\frac{\pi}{2}\right)\right)\\
&=&i^{2m+1} (-1)^m\prod_{k=1}^m\left(r+4\cos^2\left(\frac{k
\pi}{n+1}\right)\right)\left(i\sqrt{r}-2\cos\left(\frac{\pi}{2}\right)\right)\\
&=&\prod_{k=1}^{\lfloor \frac{n}{2} \rfloor}\left(r+4\cos^2\left(\frac{k \pi}{n+1}\right)\right)(-\sqrt{r}).
\end{eqnarray*}
The result follows now from the fact that
$$a(n;r)=i^n U_n\left(i\frac{\sqrt{r}}{2}\right)\bigg\{(-1)^n
\frac{\sqrt{r}+1}{2\sqrt{r}}+\frac{\sqrt{r}-1}{2\sqrt{r}}\bigg\}$$ where
$(-1)^n
\frac{\sqrt{r}+1}{2\sqrt{r}}+\frac{\sqrt{r}-1}{2\sqrt{r}}$ is the sequence
$$1,-\frac{1}{\sqrt{r}},1,-\frac{1}{\sqrt{r}},1,-\frac{1}{\sqrt{r}},\ldots.$$
\end{proof}
We note that the case $r=5$ of this result is noted in \seqnum{A152119} (Roger L. Bagula and Gary W. Adamson).
\section{$b(n;r)$ and Chebyshev polynomials}
We can also express the second sequence family $b(n;r)$ in terms of Chebyshev polynomials.
\begin{proposition}
\begin{equation} b(n;r)=\sum_{k=0}^n (-1)^{n-k}U_k\left(\frac{r+2}{2}\right).\end{equation}
\end{proposition}
\begin{proof} We have
\begin{eqnarray*}\left(\frac{1}{(1-x)(1-x^2)},\frac{x}{(1-x)^2}\right)\cdot
\frac{1}{1-rx}&=&\frac{1}{(1+x)(1-(r+2)x+x^2)}\\
&=&\frac{1}{1-(r+1)x-(r+1)x^2+x^3}\end{eqnarray*}
 as an easy calculation will verify. This implies that for the sequence $b(n;r)$ with generating function
 $$g_b(x;r)=\frac{1}{1-(r+1)x-(r+1)x^2+x^3}$$ we have
$$b(n;r)=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^{n-k}U_k\left(\frac{r+2}{2}\right).$$
\end{proof}
\begin{corollary} $b(n;r)$ is the first difference of the sequence with general term
$$w(n;r)=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}U_{n-2k}\left(\frac{r+2}{2}\right).$$
\end{corollary}
\begin{proof} We have
\begin{eqnarray*} b(n;r)&=&\sum_{k=0}^n (-1)^{n-k}U_k\left(\frac{r+2}{2}\right)\\
&=&U_n\left(\frac{r+2}{2}\right)-U_{n-1}\left(\frac{r+2}{2}\right)+U_{n-2}\left(\frac{r+2}{2}\right)-\ldots\\
&=&U_n\left(\frac{r+2}{2}\right)+U_{n-2}\left(\frac{r+2}{2}\right)+\ldots-U_{n-1}\left(\frac{r+2}{2}\right)-U_{n-3}\left(\frac{r+2}{2}\right)-\ldots\\
&=&\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}U_{n-2k}\left(\frac{r+2}{2}\right)-\sum_{k=0}^{\lfloor \frac{n-1}{2}
\rfloor}U_{n-2k-1}\left(\frac{r+2}{2}\right).\end{eqnarray*}
\end{proof}
\begin{corollary}
$$b(n;r)=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}\bigg\{
U_{n-2k+1}\left(\frac{r+2}{2}\right)-(r+1)U_{n-2k}\left(\frac{r+2}{2}\right)\bigg\}.$$
\end{corollary}
\begin{proof}
We use the Chebyshev recurrence
$$U_{n+1}=2xU_n(x)-U_{n-1}(x), \quad U_0(x)=1,\quad U_1(x)=2x.$$
\end{proof}
\noindent We note that the results above also follow directly from the observation that
$$g_b(x;r)=\frac{1}{(1+x)(1-(r+2)x+x^2)}=(1-x)\cdot\frac{1}{1-x^2}\cdot\frac{1}{1-(r+2)x+x^2}.$$
Now the general term of the Riordan array
$$\left(\frac{1}{(1-x)(1-x^2)},\frac{x}{(1-x)^2}\right)=\left(\frac{1+x}{(1-x^2)^2},\frac{x}{(1-x)^2}\right)=\left(\frac{1}{1+x},x\right)\left(\frac{1}{(1-x)^2},\frac{x}{(1-x)^2}\right)$$
can be expressed as
$$\sum_{j=0}^{n-k}(-1)^j \binom{n+k-j+1}{n-k-j}=\sum_{j=0}^{n-k}(-1)^j \binom{n+k-j+1}{2k+1}=\sum_{j=0}^n
(-1)^{n-j}\binom{j+k+1}{2k+1}.$$ This is \seqnum{A158909}, which has row sums given by
the product of consecutive Fibonacci numbers $F(n+1)F(n+2)$ and diagonal sums given by the Jacobsthal numbers
\seqnum{A001045}. The array begins
\begin{displaymath}\left(\begin{array}{ccccccc}
1 & 0 & 0 & 0 & 0 & 0 & \ldots \\1 &
1 & 0 & 0 & 0 & 0 & \ldots \\ 2 & 3 & 1 & 0 & 0 & 0 & \ldots
\\
2 & 7 & 5 & 1 & 0 & 0 & \ldots \\ 3 & 13 & 16 & 7 & 1 & 0 &
\ldots \\3 & 22 &
40 & 29 & 9 & 1 &\ldots\\ \vdots & \vdots & \vdots & \vdots &
\vdots & \vdots & \ddots\end{array}\right).\end{displaymath} We thus have
\begin{equation}b(n;r)=\sum_{k=0}^n \sum_{j=0}^{n-k}(-1)^j \binom{n+k-j+1}{n-k-j} r^k =\sum_{k=0}^n \sum_{j=0}^n
(-1)^{n-j}\binom{j+k+1}{2k+1}r^k.\end{equation}
The polynomials $b(n;x)$ begin
$$1, x+1, x^2+3x+2, x^3+5x^2+7x+2,x^4+7x^3+16x^2+13x+3,\ldots$$ We note that
\begin{eqnarray*}
1 &=& 1\\
x+1 &=& 1 \cdot (1+x)\\
x^2+3x+2 &=& (1+x)(2+x)\\
x^3+5x^2+7x+2 &=& (2+x)(1+3x+x^2)\\
x^4+7x^3+16x^2+13x+3&=& (1+3x+x^2)(3+4x+x^2)\ldots
\end{eqnarray*}
Alternative (and more esoteric) expressions can be found for $b(n;r)$ by taking other factorizations of the coefficient
array. For instance,
$$\left(\frac{1}{(1-x)(1-x^2)},\frac{x}{(1-x)^2}\right)=\left(\frac{1}{(1-x)^2},\frac{x}{(1-x)^2}\right)\cdot
\left(\frac{1+\sqrt{1+4x}}{2\sqrt{1+4x}},x\right)$$ leads to the expression
$$b(n;r)=\sum_{k=0}^n \sum_{j=0}^n \binom{n+j+1}{n-j}\binom{2j-2k-1}{j-k}(-1)^{j-k}r^k.$$
\section{From $b(n;r)$ to $a(n;r)$}
We now show that we can find a transformation from the sequences $b(n;r)$ with
generating function $$g_b(x;r)=\frac{1}{1-(r+1)x-(r+1)x^2+x^3}$$ to the sequences $a(n;r)$ with generating function
$g_a(x;r)=\frac{1+x-x^2}{1-(r+2)x^2+x^4}$. This follows since
\begin{eqnarray*} g_a(x;r)&=&\frac{1+x-x^2}{1-(r+2)x^2+x^4}\\&=&
\left(\frac{1+x-x^2}{(1-x^2)^2},\frac{x^2}{(1-x^2)^2}\right)\cdot \frac{1}{1-rx}\\
&=&\left(\frac{1+x-x^2}{(1-x^2)^2},\frac{x^2}{(1-x^2)^2}\right)\cdot
\left(\frac{1}{(1-x)(1-x^2)},\frac{x}{(1-x)^2}\right)^{-1}\cdot \frac{1}{1-(r+1)x-(r+1)x^2+x^3}\\
&=&\left(\frac{1+x-x^2}{(1-x^2)^2},\frac{x^2}{(1-x^2)^2}\right)\cdot (c(-x)^2+xc(-x)^4,1-c(-x))\cdot
\frac{1}{1-(r+1)x-(r+1)x^2+x^3}\\
&=& (1+x+x^3-x^4, x^2) \cdot \frac{1}{1-(r+1)x-(r+1)x^2+x^3}\\
&=& (1+x+x^3-x^4,x^2)\cdot g_b(x;r).\end{eqnarray*}
Here, we have used the notation
$$c(x)=\frac{1-\sqrt{1-4x}}{2x}$$ for the generating function of the Catalan numbers \seqnum{A000108}. \newline\newline
For instance, we have
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 & 0
& 0 & 0 & 0 & \ldots \\1 & 0 & 0 & 0 & 0 & 0 & \ldots \\ 0 & 1 & 0
& 0 & 0 & 0 & \ldots \\ 1 & 1 & 0 & 0 & 0 & 0 & \ldots \\ -1 & 0 &
1 & 0 & 0 & 0 & \ldots
\\0 & 1 & 1 & 0 & 0 & 0 &\ldots\\ \vdots & \vdots & \vdots & \vdots & \vdots
& \vdots &
\ddots\end{array}\right)\left(\begin{array}{c}1\\5\\30\\174\\1015\\5915\\\vdots\end{array}\right)=\left(\begin{array}{c}1\\1\\5\\6\\29\\35\\\vdots\end{array}\right)\end{displaymath}
which displays the sequence with g.f. $\frac{1+x-x^2}{1-6x^2+x^4}$ and general term
$$a(n;4)=\sum_{j=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-j}{j}4^{\lfloor \frac{n-2j}{2}\rfloor}$$ as the image of the
sequence with g.f. $\frac{1}{1-5x-5x^2+x^3}$ and general term $$b(n;4)=a(n;4)a(n+1;4).$$ We thus have the following
proposition.
\begin{proposition} The sequence $b(n;r)$ with generating function
$$\frac{1}{1-(r+1)x-(r+1)x^2+x^3}$$ is transformed into the sequence $a(n;r)$ with generating function
$\frac{1+x-x^2}{1-(r+2)x^2+x^4}$ and general term
$$a(n;r)=\sum_{j=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-j}{j}r^{\lfloor \frac{n-2j}{2}\rfloor}$$ by the (generalized)
Riordan array
$$(1+x+x^3-x^4, x^2).$$
\end{proposition}


We note that this transformation is not invertible. In terms of invertible transforms, we have the following result.
\begin{proposition} $a(n;r)$ is the image under the Riordan array $(1+x+x^3-x^4,x)$ of the aeration of
$b(n;r)$.
\end{proposition}
\begin{proof} We have
\begin{eqnarray*} \frac{1+x-x^2}{1-(r+2)x^2+x^4}&=&\frac{1+x+x^3-x^4}{(1+x^2)(1-(r+2)x^2+x^4)}\\
&=& (1+x+x^3-x^4,x)\cdot \frac{1}{1-(r+1)x^2-(r+1)x^4+x^6}.\end{eqnarray*}
\end{proof}
\begin{corollary} The aeration of $b(n;r)$ is the image of $a(n;r)$ under the Riordan array
$\left(\frac{1}{1+x+x^3-x^4}, x\right)$.
\end{corollary}
We now note that the Riordan array $\left(\frac{1}{1+x+x^3-x^4}, x\right)$ is the sequence array for
the sequence with general term
$$\sum_{j=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^{n-j} F(n-2j+1).$$ Hence we have
\begin{corollary}
\begin{equation}b(n;r)= \sum_{k=0}^{2n} \sum_{j=0}^{\lfloor \frac{2n-k}{2} \rfloor} (-1)^{2n-k-j}
F(2n-k-2j+1)a(k;r).\end{equation}
\end{corollary}
\section{A Chebyshev product result}
In this section, we observe that the polynomials
$$w(n;r)=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} U_{n-2k}\left(\frac{r+2}{2}\right)$$ exhibit
a multiplicative property. To this end, we let $U_n^d(x)$ represent a ``doubled'' Chebyshev polynomial, defined by
$$U_n^d(x)=U_{\frac{n}{2}}(x) \frac{1+(-1)^n}{2}+U_{\frac{n-1}{2}}(x) \frac{1-(-1)^n}{2}.$$
Clearly, $$U_n^d(x)=\sum_{k=0}^{\lfloor \frac{n}{4} \rfloor} \binom{\lfloor \frac{n}{2} \rfloor - k}{k} (-1)^k
(2x)^{\lfloor \frac{n}{2} \rfloor -2k}.$$ The generating function for the polynomials $U_n^d(x)$ is then given by
$$\frac{1+y}{1-2xy^2+y^4}.$$ The generating function of $\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}U_{n-2k}(x)$ is
similarly given by
$$\frac{1}{(1-y^2)(1-2x+y^2)}=\frac{1}{1-2xy+2xy^3-y^4}.$$
\begin{theorem} \label{Thm_cproduct}
We have
\begin{equation}\label{Chebcj}\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} U_{n-2k}(x)=U_n^d(x)U_{n+1}^d(x).\end{equation}
\end{theorem}
\begin{proof} The proof follows the same method as the proof of Theorem \ref{Thm_product}. For completeness, we give the
details.
We let
$$\frac{1+y}{1-2xy^2+y^4}=A(y)$$ so that
$$A(y)-2xy^2A(y)+y^4A(y)=1+y.$$
Extracting the coefficient of $y^{n+4}$, we obtain
$$[y^{n+4}]A(y)-2x[y^{n+2}]A(y)+[y^n]A(y)=[y^{n+4}](1+y)$$
\begin{equation}\label{Rec_c1} A_{n+4}-2xA_{n+2}+A_n=0. \end{equation}
By direct inspection, the first terms of the sequence $A_n=a(n;r)$ are
$${1,1,2x,2x,4x^2-1,4x^2-1,\ldots}.$$
We now set $B_n=A_nA_{n+1}$ and $T_n=A_{n}A_{n+3}$, and try to determine the corresponding generating functions.
From recurrence (\ref{Rec_c1}), we have
\begin{eqnarray*}A_{n+4}&=&2xA_{n+2}-A_n\\
A_{n+5}&=&2xA_{n+3}-A_{n+1}.\end{eqnarray*}
Multiplying, we obtain
$$A_{n+4}A_{n+5}=4x^2A_{n+2}A_{n+3}-2xA_{n+1}A_{n+2}-2xA_{n}A_{n+3}+A_nA_{n+1}.$$
This can be written as:
\begin{equation} \label{Rec_c2} B_{n+4}=4x^2B_{n+2}-2xB_{n+1}-2xT_n+B_n.\end{equation}
Now multiplying equation (\ref{Rec_c1}) by $A_{n+3}$, we obtain
$$A_{n+3}A_{n+4}=2xA_{n+2}A_{n+3}-A_nA_{n+3},$$ or
\begin{equation}\label{Rec_c3} B_{n+3}=2xB_{n+2}-T_n. \end{equation}
We can now use the method of coefficients to pass from the recurrences to the corresponding generating functions
$B(y)$ and $T(y)$. In order to do so, we note that the coefficients of the sequence generated by $B(y)$ begin:
$${1,2x,4x^2,2x(4x^2-1),\ldots}.$$
Equation (\ref{Rec_c2}) becomes
\begin{equation}\begin{split} \frac{B(y)-1-2xy-4x^2y^2-2x(4x^2-1)y^3}{y^4}\\
=4x^2\frac{B(y)-1-2xy}{y^2}-2x\frac{B(y)-1}{y}-2xT(y)+B(y)\end{split}\end{equation}
while Equation (\ref{Rec_c3}) becomes:
\begin{equation} \frac{B(y)-1-2xy-4x^2y^2}{y^3}=2x\frac{B(y)-1-2xy}{y^2}-T(y).\end{equation}
We now have two linear equations for the two unknowns $B(y)$ and $T(y)$. Solving, we find that
$$B(y)=\frac{1}{(1-y^2)(1-2xy+y^2)}$$ as required.
\end{proof}
\noindent We note that we also obtain
$$T(y)=\frac{2x-y}{(1-y^2)(1-2xy+y^2)}.$$
\noindent This theorem then implies that
\begin{equation}w(n;r)=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}
U_{n-2k}\left(\frac{r+2}{2}\right)=U_n^d\left(\frac{r+2}{2}\right)U_{n+1}^d\left(\frac{r+2}{2}\right).\end{equation}

\noindent An immediate consequence of this result is that we have
\begin{eqnarray*}
b(n;r)&=&\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} U_{n-2k}\left(\frac{r+2}{2}\right)-\sum_{k=0}^{\lfloor \frac{n-1}{2}
\rfloor} U_{n-2k-1}\left(\frac{r+2}{2}\right)\\
&=&U_n^d\left(\frac{r+2}{2}\right)U_{n+1}^d\left(\frac{r+2}{2}\right)-U_{n-1}^d\left(\frac{r+2}{2}\right)U_{n}^d\left(\frac{r+2}{2}\right)\\
&=&U_n^d\left(\frac{r+2}{2}\right)\left(U_{n+1}^d\left(\frac{r+2}{2}\right)-U_{n-1}^d\left(\frac{r+2}{2}\right)\right)
\end{eqnarray*}
This exhibits $b(n;r)$ as the product of a Chebyshev related polynomial and its first difference.
\section{Additional Riordan array factorizations}
We note that
$$\frac{1+x-x^2}{1-rx^2+x^4}=\left(\frac{1+x-x^2}{1+x^4},\frac{x^2}{1+x^4}\right)\cdot \frac{1}{1-rx}$$ and
hence
$$\frac{1+x-x^2}{1-(r+2)x^2+x^4}=\left(\frac{1+x-x^2}{1+x^4},\frac{x^2}{1+x^4}\right)\left(\frac{1}{1-x},\frac{x}{1-x}\right)^2
\cdot \frac{1}{1-rx}$$
Thus
$$\left(\frac{1+x-x^2}{(1-x^2)^2},\frac{x^2}{(1-x^2)^2}\right)=\left(\frac{1+x-x^2}{1+x^4},\frac{x^2}{1+x^4}\right)\left(\frac{1}{1-x},\frac{x}{1-x}\right)^2.$$
Here, $\left(\frac{1}{1-x},\frac{x}{1-x}\right)$ represents the standard binomial transformation.
We also have
\begin{eqnarray*}
(1+x+x^3-x^4,x^2)\cdot \left(\frac{1-x}{1-x^4},\frac{x}{1+x^2}\right)&=&
\left(\frac{1+x+x^3-x^4}{1+x^2+x^4+x^6},\frac{x^2}{1+x^4}\right)\\
&=&\left(\frac{1+x-x^2}{1+x^4},\frac{x^2}{1+x^4}\right).\end{eqnarray*}
A consequence of this last factorization is the fact that
$$(1+x+x^3-x^4,x^2)\cdot \frac{1}{1-(r-1)x-(r-1)x^2+x^3}=\frac{1+x-x^2}{1-rx^2+x^4}.$$
\section{A left inverse}
We now seek a left inverse \cite{CMS} for the array $(1+x+x^3-x^4,x^2)$. To this end we let
$t=\sqrt{x}$. Then
\begin{eqnarray*}
[x^n]\frac{\sqrt{x}^k}{1+\sqrt{x}+\sqrt{x}^3-\sqrt{x}^4}&=&[t^{2n}]\frac{t^k}{1+t+t^3-t^4}\\
&=&[t^{2n-k}]\frac{1}{1+t+t^3-t^4}\\
&=&[t^{2n-k}]\sum_{j=0}^{\infty} u_j x^j\\
&=& u_{2n-k}\end{eqnarray*}
where $$u_n=(-1)^n F(\lfloor \frac{n+2}{2} \rfloor) F(\lfloor \frac{n+3}{2} \rfloor).$$
We deduce from this that
\begin{eqnarray}b(n;r)&=&\sum_{k=0}^{2n}(-1)^k F(\lfloor \frac{2n-k+2}{2} \rfloor) F(\lfloor \frac{2n-k+3}{2}
\rfloor)\sum_{j=0}^{\lfloor \frac{k}{2} \rfloor} \binom{k-j}{j}r^{\lfloor \frac{k-2j}{2} \rfloor}\\
&=&\sum_{k=0}^{2n} (-1)^k F(n+1-\lfloor \frac{k}{2} \rfloor) F(n+1-\text{mod}(k,2)-\lfloor \frac{k}{2}
\rfloor)a(k;r).\end{eqnarray}
An alternative expression is given by
\begin{equation}b(n;r)=\sum_{k=0}^{2n}\sum_{j=0}^{\frac{2n-k}{3}}\sum_{m=0}^j
\binom{2n-k-2j-m}{j}\binom{j}{m}a(k;r).\end{equation}
In matrix terms, we have, taking $a(n;2)$ (\seqnum{A002530}), for example,
\begin{displaymath}\left(\begin{array}{ccccccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ldots \\
1 & -1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\ldots \\
4 & -2 & 1 & -1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\ldots \\
9 & -6 & 4 & -2 & 1 & -1 & 1 & 0 & 0 & 0 & 0\ldots \\
25 & -15 & 9 & -6 & 4 & -2 & 1 & -1 & 1 & 0 & 0\ldots \\
64 & -40 & 25 & -15 & 9 & -6 & 4 & -2 & 1 & -1 & 1\ldots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right)\left(\begin{array}{c} 1\\
1\\
3\\
4\\
11\\
15\\
41\\
56\\
153\\
209\\
571\\
\vdots \end{array}\right)=\left(\begin{array}{c} 1\\
3\\
12\\
44\\
165\\
615\\
\vdots \end{array}\right).\end{displaymath}
\section{The matrix $\left(\frac{1-x}{1-x^4},\frac{x}{1+x^2}\right)$ and its inverse}

We recall that

$$\left(\frac{1-x}{1-x^4},\frac{x}{1+x^2}\right)\cdot \frac{1}{1-(r+2)x}=\frac{1}{1-(r+1)x-(r+1)x^2+x^3}.$$ We have
$$\left(\frac{1-x}{1-x^4},\frac{x}{1+x^2}\right)=\left(\frac{1}{1+x},x\right)\cdot
\left(\frac{1}{1+x^2},\frac{x}{1+x^2}\right),$$ where the last matrix is
closely linked to the Chebyshev polynomials of the second kind. The matrix
$\left(\frac{1-x}{1-x^4},\frac{x}{1+x^2}\right)$ is \seqnum{A165620}. We let $T_{n,k}$ denote the general element of
$\left(\frac{1-x}{1-x^4},\frac{x}{1+x^2}\right)$. The general term of
$\left(\frac{1}{1+x},x\right)$ is $[k \le n](-1)^{n-k}$ while that of $\left(\frac{1}{1+x^2},\frac{x}{1+x^2}\right)$ is
given by
$$(-1)^{\frac{n-k}{2}}\frac{(1+(-1)^{n-k}}{2}\binom{\frac{n+k}{2}}{k}.$$ Thus we have
$$T_{n,k}=\sum_{j=0..n}(-1)^{n-j}(-1)^{\frac{j-k}{2}}\frac{1+(-1)^{j-k}}{2}\binom{\frac{j+k}{2}}{k}.$$
We can then express $b(n;r)$ as
\begin{eqnarray*} b(n;r)&=&\sum_{k=0}^n T_{n,k} (r+2)^k\\
&=&\sum_{k=0}^n T_{n,k} \sum_{j=0}^k \binom{k}{j}r^j 2^{k-j}\\
&=&\sum_{k=0}^n \sum_{l=0}^n (-1)^{n-l}(-1)^{\frac{l-k}{2}}\frac{1+(-1)^{l-k}}{2}\binom{\frac{l+k}{2}}{k}\sum_{j=0}^k
\binom{k}{j}r^j 2^{k-j}.\end{eqnarray*}

\noindent This could also have been deduced from the fact that
$$\left(\frac{1-x}{1-x^4},\frac{x}{1+x^2}\right)=\left(\frac{1}{(1-x)(1-x^2)},\frac{x}{(1-x)^2}\right)\cdot
\mathbf{B}^{-2}.$$
The row sums of $\left(\frac{1-x}{1-x^4},\frac{x}{1+x^2}\right)$ have g.f. $\frac{1}{1+x^3}$ while the diagonal sums are
the sequence $(-1)^n$.
The inverse of this matrix is given by
\begin{eqnarray*}(1+xc(x^2)+x^2c(x^2)^2+x^3c(x^2)^3,
xc(x^2))&=&\left(\frac{1+x-2x^2-(1+x)\sqrt{1-4x^2}}{2x^3},xc(x^2)\right)\\
&=&\left(\frac{c(x^2)+xc(x^2)-1}{x},xc(x^2)\right)\\
&=&(c(x^2)(1+xc(x^2)),xc(x^2))\\
&=&(1+xc(x^2),x)\cdot (c(x^2),xc(x^2)).\end{eqnarray*}
This matrix (\seqnum{A165621}) begins
\begin{displaymath}\mathbf{M}=\left(\begin{array}{ccccccc} 1 & 0 & 0 & 0
&
0 & 0 & \ldots \\1 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 1 & 1 &
0 & 0 & 0 &
\ldots \\ 2 & 2 & 1 & 1 & 0 & 0 & \ldots \\ 2 & 3 & 3 & 1 & 1
&
0 & \ldots \\5 & 5  & 4 & 4 & 1 & 1 &\ldots\\ \vdots & \vdots
&
\vdots &
\vdots & \vdots & \vdots &
\ddots\end{array}\right)\end{displaymath} where the first column is the sequence
$$1,1,1,2,2,5,5,14,14,42,\ldots$$ of doubled Catalan numbers, which has moment representation
$$\frac{1}{2\pi} \int_{-2}^2 x^n (1+x)\sqrt{4-x^2}dx$$ and Hankel transform \cite{Layman}
$$1, 0, -1, -1, 0, 1, 1,0,0,-1,1,\ldots$$ A number of interesting observations may be made about this matrix. Since
$g(x)=c(x^2)$, the $A$-sequence of this matrix has generating function $A(y)=1+y^2$. This then implies that
$$M_{n+1,k+1}=M_{n,k}+M_{n,k+2}.$$ The row sums of this matrix are
the central binomial coefficients $\binom{n+1}{\lfloor \frac{n+1}{2} \rfloor}$. The diagonal sums are \seqnum{A026008}.
The transform of the
sequence $1,0,1,0,1,0,\ldots$ by this matrix is $\binom{n}{\lfloor \frac{n}{2} \rfloor}$, \seqnum{A001405}. The transform
of $(-1)^n$ is the aerated Catalan numbers
$1,0,1,0,2,0,\ldots$, while the transform of $(-2)^n$ is a sequence whose Hankel transform  is $F(2n+1)$.
In general, the transform of the sequence with general term $r^n$ will have Hankel transform with g.f.
$\frac{1-x}{1+(r-1)x+x^2}$ and general term
$$\sum_{k=0}^n \binom{2n-k}{k}(-(r+1))^{n-k}.$$

\noindent It is interesting to take the Hankel transform of the columns of this matrix. We obtain a matrix which begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 & 0 & 0
&
0 & 0 & \ldots \\0 & 1 & 0 & 0 & 0 & 0 & \ldots \\ -1 & 0 & 1 &
0 & 0 & 0 &
\ldots \\ -1 & 0 & 0 & 1 & 0 & 0 & \ldots \\ 0 & -1 & 0 & 0 & 1
&
0 & \ldots \\1 & 0  & 0 & 0 & 0 & 1 &\ldots\\ \vdots & \vdots
&
\vdots &
\vdots & \vdots & \vdots &
\ddots\end{array}\right)\end{displaymath} (where we have multiplied the
columns by $(-1)^{\binom{k+1}{2}}$).

\section{Appendix}
In this section, we indicate a proof that the general term of the generalized Riordan array 
$$\left(\frac{1+x-x^2}{(1-x^2)^2}, \frac{x^2}{(1-x^2)^2}\right)$$ is given by 
$$ \binom{\lfloor \frac{n+2k+1}{2} \rfloor}{\lfloor \frac{n-2k}{2} \rfloor}.$$
We note first that 
$$\left(\frac{1+x-x^2}{(1-x^2)^2}, \frac{x^2}{(1-x^2)^2}\right)=(1+x-x^2,x)\left(\frac{1}{(1-x^2)^2}, \frac{x^2}{(1-x^2)^2}\right).$$
Now the Riordan array 
$$\left(\frac{1}{(1-x)^2}, \frac{x}{(1-x)^2}\right),$$ which is the unsigned version of \seqnum{A053122}, has 
general term 
$$\binom{n+k+1}{2k+1}.$$ Thus the matrix 
$$\left(\frac{1}{(1-x^2)^2}, \frac{x}{(1-x^2)^2}\right),$$ which represents an aeration of the matrix with general term 
$\binom{n+k+1}{2k+1}$, has general term 
$$\binom{\frac{n+3k+2}{2}}{2k+1}\frac{1+(-1)^{n-k}}{2}.$$
Considering now the generalized Riordan matrix 
$$\left(\frac{1}{(1-x)^2}, \frac{x^2}{(1-x)^2}\right)$$ is to effect the transformation $n \to n-k$ in the above formula.
We thus obtain that the general element of the generalized Riordan array
$$\left(\frac{1}{(1-x)^2}, \frac{x^2}{(1-x)^2}\right)$$ is given by 
$$\binom{\frac{n+2k+2}{2}}{2k+1}\frac{1+(-1)^{n-2k}}{2}=\binom{\frac{n}{2}+k+1}{2k+1}\frac{1+(-1)^n}{2}.$$
Thus the general term of 
$$\left(\frac{1+x-x^2}{(1-x^2)^2}, \frac{x^2}{(1-x^2)^2}\right)=(1+x-x^2,x)\left(\frac{1}{(1-x^2)^2}, \frac{x^2}{(1-x^2)^2}\right)$$ is given by the sum 
$$\binom{\frac{n}{2}+k+1}{2k+1}\frac{1+(-1)^n}{2}+\binom{\frac{n-1}{2}+k+1}{2k+1}\frac{1+(-1)^{n-1}}{2}-\binom{\frac{n-2}{2}+k+1}{2k+1}\frac{1+(-1)^{n-2}}{2}$$
or
$$\binom{\frac{n}{2}+k+1}{2k+1}\frac{1+(-1)^n}{2}+\binom{\frac{n}{2}+k+\frac{1}{2}}{2k+1}\frac{1-(-1)^n}{2}-\binom{\frac{n}{2}+k}{2k+1}\frac{1+(-1)^n}{2}$$
which simplifies to 
$$\frac{2(2k+1)}{n-2k}\binom{\frac{n}{2}+k+1}{2k+1}\frac{1+(-1)^n}{2}+\binom{\frac{n}{2}+k+\frac{1}{2}}{2k+1}\frac{1-(-1)^n}{2}.$$
By considering alternately the cases $n=2m$ and $n=2m+1$ we can establish that this last term coincides with 
$$ \binom{\lfloor \frac{n+2k+1}{2} \rfloor}{\lfloor \frac{n-2k}{2} \rfloor}.$$ 
\section{Acknowledgements} The author would like to thank the anonymous referees for their cogent suggestions which have
led to the current
expanded and hopefully clearer paper. In particular, the author acknowledges the input of one referee in alerting us to
the use of the method of coefficients used in the proofs of Theorems \ref{Thm_product} and \ref{Thm_cproduct}.
Acknowledgements are due also to Richard Guy, whose observations initiated this study.


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\bibitem{SL2} N. J. A.~Sloane, The On-Line Encyclopedia of Integer
Sequences, \emph{Notices of the AMS}, \textbf{50} (2003), pp.\, 912--915.


\bibitem{Spru} R. Sprugnoli, Riordan arrays and combinatorial sums,
\emph{Discrete Math.} \textbf{132} (1994) pp.\,267--290.

\bibitem{Morgan-Voyce} E. Weisstein, Morgan-Voyce Polynomials, from MathWorld--A Wolfram Web Resource.
\href{http://mathworld.wolfram.com/Morgan-VoycePolynomials.html}{\tt
http://mathworld.wolfram.com/Morgan-VoycePolynomials.html}, 2009


\end{thebibliography}

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\noindent 2000 {\it Mathematics Subject Classification}: Primary
11B83; Secondary  11A55, 11B37, 11Y55, 33C45.

\noindent \emph{Keywords:} Integer sequence, linear recurrence, Riordan arrays,
Chebyshev polynomials, continued fraction.

\bigskip
\hrule
\bigskip
\noindent (Concerned with sequences \seqnum{A000045},
\seqnum{A000108},
\seqnum{A001045},
\seqnum{A001405},
\seqnum{A001654},
\seqnum{A002530},
\seqnum{A008619},
\seqnum{A026008},
\seqnum{A041011},
\seqnum{A049310},
\seqnum{A053122},
\seqnum{A078812},
\seqnum{A084158},
\seqnum{A085478},
\seqnum{A099025},
\seqnum{A109437},
\seqnum{A136211},
\seqnum{A152119},
\seqnum{A157329},
\seqnum{A157335},
\seqnum{A158909},
\seqnum{A165620}, and
\seqnum{A165621}.)


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\vspace*{+.1in}
\noindent
Received February 27 2009;
revised versions received  March 30 2009; June 8 2009; December 3 2009.
Published in {\it Journal of Integer Sequences}, December 3 2009.

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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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