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\begin{center}
\vskip 1cm{\LARGE\bf
Hankel Transforms of Linear Combinations \\
\vskip .1in
of Catalan Numbers}

\vskip 1cm
\large
Michael Dougherty, Christopher French, Benjamin Saderholm, and Wenyang Qian\\
Department of Mathematics and Statistics\\
Grinnell College\\
Grinnell, IA 50112\\
USA\\
\href{mailto:doughert@grinnell.edu}{\tt doughert@grinnell.edu}\\
\href{mailto:frenchc@grinnell.edu}{\tt frenchc@grinnell.edu}\\
\href{mailto:saderhol@grinnell.edu}{\tt saderhol@grinnell.edu}\\
\href{mailto:qianwney@grinnell.edu}{\tt qianweny@grinnell.edu}\\
\end{center}

\vskip .2in

\begin{abstract}
Cvetkovi\'c, Rajkovi\'c, and Ivkovi\'c proved that the Hankel transform
of the sequence of sums of adjacent Catalan numbers is the bisection of the sequence
of Fibonacci numbers.  Here, we find recurrence relations for the Hankel
transform of more general linear combinations of Catalan numbers, involving
up to four adjacent Catalan numbers, with arbitrary coefficients.
Using these, we make certain conjectures about the recurrence relations satisfied
by the Hankel transform of more extended linear combinations.
\end{abstract}


\section{Introduction}\label{sec:intro}

Given a sequence of integers $\{a_n\}_{n=0}^\infty$, the Hankel matrix of the sequence is the infinite matrix
$H$ whose $(i,j)$ entry is $a_{i+j}$.  Here, we allow $i$ and $j$ to be $0$, so our rows and columns are numbered
beginning with $0$.  That is, 
\[H=\left(\begin{matrix} a_0 &amp; a_1 &amp; a_2 &amp; \cdots \\
a_1 &amp; a_2 &amp; a_3 &amp; \cdots \\
a_2 &amp; a_3 &amp; a_4 &amp; \cdots \\
a_3 &amp; a_4 &amp; a_5 &amp; \cdots \\
\vdots &amp; \vdots &amp; \vdots &amp; \ddots
\end{matrix}\right).\]
In general, if $A$ is an infinite matrix, with row and column indices beginning at $0$, and
$n\geq 0$ is an integer, we let $A(n)$ denote the upper-left $(n+1)\times (n+1)$ submatrix of $A$.
Thus, $H(0)$ is the $1\times 1$ matrix $(a_0)$, $H(1)$ is the $2\times 2$-matrix
\[H(1)=\left(\begin{matrix} a_0 &amp; a_1 \\ a_1 &amp; a_2 \end{matrix}\right),\]
and
\[H(2)=\left(\begin{matrix} a_0 &amp; a_1 &amp; a_2 \\ a_1 &amp; a_2 &amp; a_3 \\ a_2 &amp; a_3 &amp; a_4 \end{matrix}\right).\]
We let $h_n=\det(H(n))$.  Then the sequence $\{h_n\}_{n=0}^\infty$ is called the {\it Hankel transform}
of the sequence $\{a_n\}_{n=0}^\infty$.

The Hankel transform is particularly interesting when applied to the Catalan numbers.  Recall
that the Catalan numbers $c_n$ are defined by the recurrence relation
\[c_{n+1}=c_nc_0+c_{n-1}c_1+c_{n-2}c_2+\cdots +c_0c_n\]
where $c_0=1$.  It is well-known that the Hankel transform of the sequence of
Catalan numbers is the sequence of all $1$'s.  Mays and Wojciechowski \cite{MW} found
the Hankel transform of the sequences obtained by removing $1$, $2$, or $3$ terms
from the beginning of the Catalan number sequence.  If we remove $1$ term, then
the Hankel transform is a sequence of $1$'s; if we remove $2$ terms, the Hankel
transform is the sequence $\{n+2\}_{n=0}^\infty$;
if we remove $3$ terms, the Hankel transform
is the sequence $\left\lbrace\frac{(n+2)(n+3)(2n+5)}{6}\right\rbrace_{n=0}^\infty$.
(Note that our $n$ is $k-1$ in \cite{MW}.)
Cvetkovi\'c, Rajkovi\'c, and Ivkovi\'c \cite{CRI} proved that the Hankel transform of the sequence $\{c_n+c_{n+1}\}$ is the sequence
$2, 5, 13, 34, 89, \cdots$ consisting of every other Fibonacci number.  Benjamin,
Cameron, Quinn, and Yerger \cite{BCQY} also reproved this result using a more
combinatorial approach.  Notice that, in particular, the Hankel transform
of the sequence $\{c_n+c_{n+1}\}$
satisfies the homogeneous linear recurrence relation
\begin{equation}\label{eqn:131}
h_{n+2}-3h_{n+1}+h_n=0.
\end{equation}

In this paper, we take up the question of finding recurrence relations satisfied by more general linear combinations
of Catalan numbers.  In particular, we ask

\begin{question}\label{question} Suppose given $k+1$ integers $\alpha_0, \alpha_1, \alpha_2, \cdots, \alpha_k$.  Let $\{h_n\}$
denote the Hankel transform of the sequence whose $n^{\text th}$ term is $\sum_{i=0}^k \alpha_i c_{n+i}$.
What homogeneous linear recurrence relation, if any, is satisfied by the sequence $h_n$?
\end{question}

For example, in the case where $k=1$ and $\alpha_0=\alpha_1=1$, we know that $h_n$ satisfies the recurrence relation
of Equation \ref{eqn:131}.  Of course, if $k=0$ and $\alpha_0=1$, then $\{h_n\}$ is just the Hankel transform
of the sequence of Catalan numbers, so $h_n=1$ for all $n$.
From this, it is easy to see that when $k=0$ and $\alpha_0$ is arbitrary,
we have $h_n=\alpha_0^{n+1}$.  Thus, the sequence $\{h_n\}$ satisfies the first-order homogeneous linear
recurrence relation \[h_{n+1}-\alpha_0h_n=0.\]

Our next three theorems, which we will prove in this paper,
answer Question \ref{question} when $k=1$, $k=2$, and $k=3$.  For ease of reading,
we adjust our
notation, replacing $\alpha_0$, $\alpha_1$, $\alpha_2$, and $\alpha_3$ by $\alpha$, $\beta$, $\gamma$, $\delta$.

\begin{theorem}\label{thm:1} The Hankel transform of the sequence $\{\alpha c_n+\beta c_{n+1}\}$ satisfies
the second-order homogeneous linear recurrence relation
\[h_{n+2}-(\alpha+2\beta)h_{n+1}+\beta^2 h_n=0.\]
\end{theorem}

\begin{theorem}\label{thm:2} The Hankel transform of the sequence $\{\alpha c_n+\beta c_{n+1}+\gamma c_{n+2}\}$
satisfies the fourth-order homogeneous linear recurrence relation

\begin{eqnarray*}
h_{n+4}+(-\alpha-2\beta-4\gamma)h_{n+3}+
(\beta^2-2\alpha\gamma+4\beta\gamma
+6\gamma^2)h_{n+2}\\
+(-\alpha-2\beta-4\gamma)\gamma^2 h_{n+1}+\gamma^4h_n=0.
\end{eqnarray*}

\end{theorem}


\begin{theorem}\label{thm:3} The Hankel transform of the sequence $\{\alpha c_n+\beta c_{n+1}+\gamma c_{n+2}+\delta c_{n+3}\}$
satisfies the eighth-order homogeneous linear recurrence relation

{\scriptsize
\begin{eqnarray*}
h_{n+8}\\
+
(-\alpha-2\beta-4\gamma-8\delta)h_{n+7}\\
+(\beta^2-2\alpha\gamma+4\beta\gamma+6\gamma^2-12\alpha\delta+4\beta\delta+24\gamma\delta+28\delta^2)h_{n+6}\\
+(-\alpha\gamma^2-2\beta\gamma^2-4\gamma^3+2\alpha\beta\delta+4\beta^2\delta-4\alpha\gamma\delta
-24\gamma^2\delta-7\alpha\delta^2+2\beta\delta^2-60\gamma\delta^2-56\delta^3)h_{n+5}\\
+(\gamma^4-4\beta\gamma^2\delta+8\gamma^3\delta+\alpha^2\delta^2+4\alpha\beta\delta^2
+6\beta^2\delta^2+12\alpha\gamma\delta^2-8\beta\gamma\delta^2+36\gamma^2\delta^2+
40\alpha\delta^3-8\beta\delta^3
+80\gamma\delta^3+70\delta^4)h_{n+4}\\
+(-\alpha\gamma^2-2\beta\gamma^2-4\gamma^3+2\alpha\beta\delta+4\beta^2\delta-4\alpha\gamma\delta
-24\gamma^2\delta-7\alpha\delta^2+2\beta\delta^2-60\gamma\delta^2-56\delta^3)\delta^2 h_{n+3}\\
+(\beta^2-2\alpha\gamma+4\beta\gamma+6\gamma^2-12\alpha\delta+4\beta\delta+24\gamma\delta+28\delta^2)
\delta^4 h_{n+2}\\
+(-\alpha-2\beta-4\gamma-8\delta)\delta^6 h_{n+1}\\
+\delta^8 h_n=0.
\end{eqnarray*}
}
\end{theorem}

Of course, Theorem \ref{thm:2} follows from Theorem \ref{thm:3} by setting $\delta$ equal to $0$,
and Theorem \ref{thm:1} follows from Theorem \ref{thm:2} by setting $\gamma=0$.  Notice also 
that the result of Cvetkovi\'c, Rajkovi\'c, and Ivkovi\'c \cite{CRI} follows from Theorem \ref{thm:1} by setting $\alpha=\beta=1$.  In this
case, it is trivial to check that the initial two terms of the Hankel transform are $2$ and $5$, and
then the recurrence relation implies the rest.

While the recurrence relations given above quickly grow in complexity as $k$ increases,
there are several features of these recurrence relations
that seem interesting.  On the basis of these features, we make the following conjectures.
First, we note that the order of the recurrence relations doubles each time
we increase $k$ by $1$.

\begin{conjecture}\label{con:1} The Hankel transform of the sequence whose $n^{\text th}$ term is 
\[\sum_{i=0}^k \alpha_i c_{n+i}\]
satisfies a homogeneous linear recurrence relation of order $2^k$.
\end{conjecture}

We also notice that the coefficient of $h_{n+2^k-1}$ is of a particularly simple form.

\begin{conjecture} The coefficient of $h_{n+2^k-1}$ in the recurrence relation of Conjecture \ref{con:1} is
\[-(\alpha_0+2\alpha_1+4\alpha_2+\cdots+2^k\alpha_k).\]
\end{conjecture}

Finally, we note that the coefficients of $h_{n+2^k-r}$ and $h_{n+r}$ seem to be related.  For example,
in Theorem \ref{thm:2}, the coefficient of $h_{n+1}$ is the product of $\gamma^2$ and
the coefficient of $h_{n+3}$.  Likewise, the coefficient of $h_n$ is the product of $\gamma^4$
and the coefficient of $h_{n+4}$.

\begin{conjecture} When $r&lt;2^{k-1}$, the coefficient of $h_{n+r}$ in the recurrence relation of Conjecture \ref{con:1} is equal
to the product of $\alpha_k^{2^k-2r}$ and the coefficient of $h_{n+2^k-r}$.
\end{conjecture}

Using Theorem \ref{thm:2}, we can show that certain sequences in the OEIS \cite{OEIS}
are Hankel transforms of linear
combinations of Catalan numbers.  For example, if we set $\alpha=1, \beta=1, \gamma=1$, we get the sequence
\[4, 20, 111, 624, 3505, 19676, 110444, 619935, 3479776, 19532449, \ldots\]
which is sequence \seqnum{A014523} in the OEIS, the number of Hamiltonian paths in a $4\times (2n+1)$ grid
(see also \cite{CK}).    To see this, we simply observe from the OEIS that sequence \seqnum{A014523} satisfies the recurrence
relation \[a_{n+4}-7a_{n+3}+9a_{n+2}-7a_{n+1}+a_n=0.\]
By Theorem \ref{thm:2}, this is the same as the recurrence relation satisfied
by the Hankel transform of $c_n+c_{n+1}+c_{n+2}$.  One now simply observes that the first four terms
of the two sequences coincide!

Likewise, if we set $\alpha=1, \beta=2, \gamma=1$, we get the sequence
\[5, 30, 199, 1355, 9276, 63565, 435665, 2986074, 20466835, 140281751, \ldots\]
which is sequence \seqnum{A103433} in the OEIS, namely the sequence
$\sum_{i=0}^n F_{2i-1}^2$ where $F_k$ denotes the $k^{\text th}$ Fibonacci number.
To see this, we note from the OEIS that sequence \seqnum{A103433} has generating function
\[G(x)=\frac{x(1-4x+x^2)}{(1-7x+x^2)(1-x)^2}=\frac{x-4x^2+x^3}{x^4-9x^3+16x^2-9x+1}.\]
From here, it is easy to check that the sequence satisfies the recurrence relation
\[h_{n+4}-9h_{n+3}+16h_{n+2}-9h_{n+1}+h_n=0.\] By Theorem \ref{thm:2}, 
this is the same as the recurrence
relation satisfied by the Hankel transform of $c_n+2c_{n+1}+c_{n+2}$.  Again, one simply observes
that the first four terms of the two sequences coincide.

Our paper is organized as follows.  In Section \ref{sec:CNHM}, we state some basic
results concerning Hankel matrices and shifts of Catalan numbers. These results
are proved in Section \ref{sec:proofs}.  Using only
these basic results, we prove Theorem \ref{thm:1} in
Section \ref{sec:CNHM}.  For Theorems \ref{thm:2} and \ref{thm:3} we need somewhat
more sophisticated arguments.  To prepare for these, we state and prove two
easy lemmas about recurrence relations in Section \ref{sec:recurrence}.  The work of Section
\ref{sec:CNHM} shows that we can simplify our study of the Hankel matrices
of linear combinations of Catalan numbers by investigating a class of matrices
which we call hyperdiagonal.  We carry out this investigation in Section
\ref{sec:hyper}, and use the results from there to finish off the proofs of Theorems
\ref{thm:2} and \ref{thm:3} in Section \ref{sec:end}.


\section{Catalan Numbers and Hankel Matrices}\label{sec:CNHM}

In this section, we state some results about the Hankel matrices of shifts of the sequence of Catalan numbers;
we will prove these results in the next section.  Using
these results, we give an easy proof of Theorem \ref{thm:1}.  To begin, we need some definitions.

\begin{definition} Let $C$ be the Hankel matrix of the sequence of Catalan numbers, so $C_{i,j}=c_{i+j}$,
where $i,j$ are non-negative integers.  Let $C\{k\}$ denote the Hankel matrix of the sequence
of Catalan numbers with the first $k$ terms removed, so $C\{k\}_{i,j}=c_{i+j+k}$.  In particular,
$C=C\{0\}$.
\end{definition}

\begin{definition}\label{def:LSUC} Let $L$ be the infinite matrix defined by
$L_{i,j}=(-1)^{i+j}{i+j\choose i-j}$.
(When $i&lt;j$, $L_{i,j}=0$.)  Let $U=L^T$, so $U_{i,j}=(-1)^{i+j}{i+j\choose j-i}$.
\end{definition}

\begin{notn}\label{notn:upperleft} For an infinite matrix $A$ and an integer $n\geq 0$,
let $A(n)$ denote the upper-left $(n+1)\times (n+1)$ submatrix of $A$.
\end{notn}

In the next section, we will prove

\begin{proposition}\label{prop:LC1U}
The matrix product $LCU$ is the identity matrix.  The matrix $LC\{1\}U$ is tridiagonal
with \[(LC\{1\}U)_{i,j}=\begin{cases} 1, &amp; \text{ if } |i-j|=1\text{ or }i=j=0;
\\ 2, &amp; \text{ if } i=j\geq 1; \\ 0, &amp; \text{otherwise.}
\end{cases}\]
\end{proposition}

That is, $LC\{1\}U$ takes the form
\[\left(\begin{matrix}
1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots\\
1 &amp; 2 &amp; 1 &amp; 0 &amp; 0 &amp; \cdots \\
0 &amp; 1 &amp; 2 &amp; 1 &amp; 0 &amp; \cdots \\
0 &amp; 0 &amp; 1 &amp; 2 &amp; 1 &amp; \cdots \\
0 &amp; 0 &amp; 0 &amp; 1 &amp; 2 &amp; \cdots \\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots
\end{matrix}\right).\]

Assuming this result, we now obtain an elementary proof of Theorem \ref{thm:1}.

\begin{proof}

Let $\{h_n\}$ be the Hankel transform of the sequence $\alpha c_n+\beta c_{n+1}$.
The Hankel matrix of this sequence is $\alpha C+\beta C\{1\}$.
Recall that for an infinite matrix $A$, $A(n)$ denotes the upper-left $(n+1)\times (n+1)$
submatrix of $A$.  Then, we have \[h_n=\det((\alpha C+\beta C\{1\})(n))\]
Since $L(n)$ and $U(n)$ are triangular $(n+1)\times (n+1)$ matrices with $1$'s on the diagonal, they both
have determinant $1$, so we now have
\[h_n=\det(L(n)(\alpha C+\beta C\{1\})(n)U(n))\]
\[=\det((L(\alpha C+\beta C\{1\})U)(n))=\det((\alpha LCU+\beta LC\{1\}U)(n)).\]
By Proposition \ref{prop:LC1U}, $\alpha LCU+\beta LC\{1\}U$ is equal to
\[\left(\begin{matrix}
\beta+\alpha &amp; \beta &amp; 0 &amp; 0 &amp; 0 &amp; \cdots\\
\beta &amp; 2\beta+\alpha &amp; \beta &amp; 0 &amp; 0 &amp; \cdots \\
0 &amp; \beta &amp; 2\beta+\alpha &amp; \beta &amp; 0 &amp; \cdots \\
0 &amp; 0 &amp; \beta &amp; 2\beta+\alpha &amp; \beta &amp; \cdots \\
0 &amp; 0 &amp; 0 &amp; \beta &amp; 2\beta+\alpha &amp; \cdots \\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots
\end{matrix}\right).\]
Thus, \[h_{n+2}=\left|\begin{matrix}
\beta+\alpha &amp; \beta &amp; \cdots &amp; 0 &amp; 0 &amp; 0 \\
\beta &amp; 2\beta+\alpha &amp; \cdots &amp; 0 &amp; 0 &amp; 0 \\
\vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp; \vdots \\
0 &amp; 0 &amp; \cdots &amp; 2\beta+\alpha &amp; \beta &amp; 0 \\
0 &amp; 0 &amp; \cdots &amp; \beta &amp; 2\beta+\alpha &amp; \beta \\
0 &amp; 0 &amp; \cdots &amp; 0 &amp; \beta &amp; 2\beta+\alpha \\
\end{matrix}\right|\]
where the given matrix has $n+3$ rows and $n+3$ columns.
Expanding along the last column gives
\[(2\beta+\alpha)\left|\begin{matrix}
\beta+\alpha &amp; \beta &amp; \cdots &amp; 0 &amp; 0  \\
\beta &amp; 2\beta+\alpha &amp; \cdots &amp; 0 &amp; 0  \\
\vdots &amp; \vdots &amp; \ddots &amp; \vdots \\
0 &amp; 0 &amp; \cdots &amp; 2\beta+\alpha &amp; \beta \\
0 &amp; 0 &amp; \cdots &amp; \beta &amp; 2\beta+\alpha  \\
\end{matrix}\right|-\beta
\left|\begin{matrix}
\beta+\alpha &amp; \beta &amp; \cdots &amp; 0  &amp; 0 \\
\beta &amp; 2\beta+\alpha &amp; \cdots &amp; 0 &amp; 0  \\
\vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp; \vdots  \\
0 &amp; 0 &amp; \cdots &amp; 2\beta+\alpha &amp; \beta  \\
0 &amp; 0 &amp; \cdots &amp; 0 &amp; \beta  \\
\end{matrix}\right|.\]
Expanding the right-hand matrix along the bottom row now yields
\[(2\beta+\alpha)\left|\begin{matrix}
\beta+\alpha &amp; \beta &amp; \cdots &amp; 0 &amp; 0  \\
\beta &amp; 2\beta+\alpha &amp; \cdots &amp; 0 &amp; 0  \\
\vdots &amp; \vdots &amp; \ddots &amp; \vdots \\
0 &amp; 0 &amp; \cdots &amp; 2\beta+\alpha &amp; \beta \\
0 &amp; 0 &amp; \cdots &amp; \beta &amp; 2\beta+\alpha  \\
\end{matrix}\right|-\beta^2
\left|\begin{matrix}
\beta+\alpha &amp; \beta &amp; \cdots  &amp; 0 \\
\beta &amp; 2\beta+\alpha &amp; \cdots &amp; 0  \\
\vdots &amp; \vdots &amp; \ddots &amp; \vdots   \\
0 &amp; 0 &amp; \cdots &amp; 2\beta+\alpha  \\
\end{matrix}\right|\]
\[=(2\beta+\alpha)h_{n+1}-\beta^2 h_n.\]
Thus, $h_{n+2}-(2\beta+\alpha)h_{n+1}+\beta^2 h_n=0$ for all $n\geq 0$.
\end{proof}

In the next section, we will also prove the following Proposition:

\begin{proposition}\label{prop:LCKU2} When $i+j\geq k\geq 0$, $(LC\{k\}U)_{i,j}={2k\choose k+i-j}$.
\end{proposition}

Again, ${n\choose m}$ is defined to be $0$ when $m&lt;0$ or $m&gt;n$.
We will later use this proposition in the proofs of Theorems \ref{thm:2} and
\ref{thm:3}.  As an example, $LC\{2\}U$ takes the form
\[\left(\begin{matrix}
2 &amp; 3 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots\\
3 &amp; 6 &amp; 4 &amp; 1 &amp; 0 &amp; 0 &amp; \cdots \\
1 &amp; 4 &amp; 6 &amp; 4 &amp; 1 &amp; 0 &amp; \cdots \\
0 &amp; 1 &amp; 4 &amp; 6 &amp; 4 &amp; 1 &amp; \cdots \\
0 &amp; 0 &amp; 1 &amp; 4 &amp; 6 &amp; 4 &amp; \cdots \\
0 &amp; 0 &amp; 0 &amp; 1 &amp; 4 &amp; 6 &amp; \cdots \\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots
\end{matrix}\right).\]
The $3$ entries in the top left (the $2$ and the two $3$'s) are not given by Proposition
\ref{prop:LCKU2}, but can easily be computed directly.
Likewise, $LC\{3\}U$ takes the form
\[\left(\begin{matrix}
5 &amp; 9 &amp; 5 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp;\cdots\\
9 &amp; 19 &amp; 15 &amp; 6 &amp; 1 &amp; 0 &amp; 0 &amp; \cdots \\
5 &amp; 15 &amp; 20 &amp; 15 &amp; 6 &amp; 1 &amp; 0 &amp;\cdots \\
1 &amp; 6 &amp; 15 &amp; 20 &amp; 15 &amp; 6 &amp; 1 &amp; \cdots \\
0 &amp; 1 &amp; 6 &amp; 15 &amp; 20 &amp; 15 &amp; 6 &amp; \cdots \\
0 &amp; 0 &amp; 1 &amp; 6 &amp; 15 &amp; 20 &amp; 15 &amp; \cdots \\
0 &amp; 0 &amp; 0 &amp; 1 &amp; 6 &amp; 15 &amp; 20  &amp; \cdots \\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots
\end{matrix}\right).\]
Again, the six entries in the top left (the three $5$'s, the two $9$'s, and the $19$)
are not given by Proposition \ref{prop:LCKU2}, but can easily be computed directly.



\section{Proofs of Propositions \ref{prop:LC1U} and \ref{prop:LCKU2}}\label{sec:proofs}

In this section, as the title suggests, we prove Propositions \ref{prop:LC1U} and \ref{prop:LCKU2}.
The main technical tool in these proofs is the following Lemma.  The second author used this idea previously \cite{CF} having found it originally in a paper by Bajunaid,
Cohen, Colonna, and Signman \cite{BCCS}.

\begin{lemma}\label{lem:genfun}
Let $g(z)$ be the power series $\sum_{n=0}^\infty c_k z^k$.  Then, $g(z(1-z))=\sum_{k=0}^\infty z^k$.
\end{lemma}

\begin{proof}
By definition of the Catalan numbers,
\[g(z)^2=\sum_{k=0}^\infty \left(\sum_{l=0}^k c_lc_{k-l}\right) z^k
=\sum_{k=0}^\infty c_{k+1}z^k.\]
Thus, \[zg(z)^2=\sum_{k=0}^\infty c_{k+1}z^{k+1}=\sum_{k=1}^\infty c_kz^k=g(z)-1.\]
We may substitute $z(1-z)$ into this equation to obtain
\[z(1-z)g(z(1-z))^2=g(z(1-z))-1.\]
Multiplying by $z(1-z)$ yields
\[\left(z(1-z)g(z(1-z))\right)^2=\left(z(1-z)g(z(1-z))\right)-z(1-z).\]
If we let $h(z)=z(1-z)g(z(1-z))$, then the above equation implies
$h(z)^2-h(z)+z(1-z)=0,$ so that $(h(z)-z)(h(z)-(1-z))=0$.  Since the ring of power
series is an integral domain, this implies $h(z)=z$ or $h(z)=1-z$.  But we know
$h(0)=0$, so $h(z)=z$, whence $z(1-z)g(z(1-z))=z$.  This implies
$g(z(1-z))=\frac{1}{1-z}=\sum_{k=0}^\infty z^k$.
\end{proof}

\begin{corollary}\label{cor:genfun} Suppose $i\geq t\geq 1$.  Then, we have
\[\sum_{k=i-t}^{2i-t} (-1)^{k+t} {k+t\choose 2i-t-k} c_k=0.\]
For any $i\geq 0$, we have
\[\sum_{k=i}^{2i} (-1)^{k} {k\choose 2i-k} c_k=1,\]
and
\[\sum_{k=i+1}^{2i+1} (-1)^{k-1} {k-1\choose 2i+1-k} c_k=2i+1.\]
\end{corollary}

\begin{proof}
First, we have \[(1-z)^tg(z(1-z))=\sum_{k=0}^\infty c_k z^k(1-z)^{k+t}
=\sum_{k=0}^\infty \sum_{m=0}^{k+t} (-1)^m {k+t\choose m} c_k z^{k+m}.\]
For an integer $i\geq 0$, the coefficient of $z^{2i-t}$ is
\[\sum_{k=i-t}^{2i-t} (-1)^{k+t} {k+t\choose 2i-t-k} c_k.\]
On the other hand, by Lemma \ref{lem:genfun},
$(1-z)^tg(z(1-z))=(1-z)^{t-1}$, so if $i\geq t$, then $2i-t\geq t$, so the
coefficient of $i-t$ is $0$.

For the second formula, we have \[g(z(1-z))=\sum_{k=0}^\infty c_k z^k(1-z)^k
=\sum_{k=0}^\infty \sum_{m=0}^k (-1)^m {k\choose m} c_k z^{k+m}.\]
For an integer $i\geq 0$, the coefficient of $z^{2i}$ is
\[\sum_{k=i}^{2i} (-1)^{k} {k\choose 2i-k} c_k.\]
By Lemma \ref{lem:genfun}, this coefficient is equal to $1$.

Finally, we have \[\frac{g(z(1-z))-1}{1-z}=\sum_{k=1}^\infty c_k z^k(1-z)^{k-1}
=\sum_{k=1}^\infty \sum_{m=0}^{k-1} (-1)^m {k-1\choose m} c_k z^{k+m}.\]
For an integer $i\geq 0$, the coefficient of $z^{2i+1}$ is
\[\sum_{k=i+1}^{2i+1} (-1)^{k-1} {k-1\choose 2i+1-k} c_k.\]
But by Lemma \ref{lem:genfun} again,
\[\frac{g(z(1-z))-1}{1-z}=\frac{1/(1-z) -1}{1-z}=\frac{z}{(1-z)^2}=
\sum kz^k,\] so the coefficient of $z^{2i+1}$ is $2i+1$.

\end{proof}



\begin{lemma}\label{lem:LC} The product $LC$ is an upper triangular matrix,
with $LC_{i,i}=1$ for all $i\geq 0$, and
$LC_{i,i+1}=2i+1$.
\end{lemma}

\begin{proof} First, we observe that since $L$ is lower triangular,
\[(LC)_{i,j}=\sum_{k=0}^\infty L_{i,k}C_{k,j}=\sum_{k=0}^i L_{i,k}C_{k,j},\]
so that the matrix product $LC$ is well-defined.  Now, by definition of $L$ and $C$,
\[(LC)_{i,j}=\sum_{k=0}^i (-1)^{i+k} {i+k\choose i-k}c_{k+j}
=\sum_{k=j}^{i+j} (-1)^{i-j+k} {i-j+k\choose i+j-k} c_k.\]
When $j&lt;i$, let $t=i-j$.  Then $i\geq t\geq 1$, and the sum becomes
\[\sum_{k=i-t}^{2i-t} (-1)^{k+t} {k+t\choose 2i-t-k} c_k,\]
which is $0$ by Corollary \ref{cor:genfun}.  The other two statements similarly
follow from Corollary \ref{cor:genfun}.
\end{proof}

We are now ready to prove Proposition \ref{prop:LC1U}.

\begin{proof} (of Proposition \ref{prop:LC1U}).
First, $LC$ and $U$ are both upper triangular with $1$'s on the diagonal,
so $LCU$ is upper triangular with $1$'s on the diagonal.  Also, $(LCU)^t=U^tC^tL^t=LC^tU$.
Since $C$ is symmetric, this is $LCU$.  Thus, $LCU$ is also symmetric, so $LCU$ is the identity.

Now, $(LC\{1\})_{i,j}=0$ if $i&gt;j+1$.  Since $U_{i,j}=0$ for $i&gt;j$, it follows easily that
$(LC\{1\}U)_{i,j}=0$ whenever $i&gt;j+1$.  Moreover, when $i\geq 1$, we have
\[(LC\{1\}U)_{i,i-1}=\sum_{k=0}^\infty (LC\{1\})_{i,k}U_{k,i-1}\]
\[=\sum_{k=i-1}^{i-1} (LC\{1\})_{i,k} U_{k,i-1}=(LC\{1\})_{i,i-1}U_{i-1,i-1}=LC_{i,i}U_{i-1,i-1}.\]
By Lemma \ref{lem:LC}, this is $1$.  When $i\geq 1$, we have
\[(LC\{1\}U)_{i,i}=\sum_{k=0}^\infty (LC\{1\})_{i,k}U_{k,i}\]
\[=\sum_{k=i-1}^i (LC\{1\})_{i,k} U_{k,i}=(LC\{1\})_{i,i-1}U_{i-1,i}+(LC\{1\})_{i,i}U_{i,i}\]
\[=(LC)_{i,i}U_{i-1,i}+LC_{i,i+1}U_{i,i}.\]
By Lemma \ref{lem:LC} and the definition of $U$, this is
$-(2i-1)+2i+1=2$.  Likewise, $(LC\{1\}U)_{0,0}=(LC\{1\})_{0,0}U_{0,0}=(LC)_{0,1}U_{0,0}=1$.
The rest of the claim follows since $LC\{1\}U$ is symmetric.
\end{proof}

We now turn to considering $LC\{k\}U$.  The following definition will be useful.

\begin{definition}
Let $S$ be the infinite matrix defined by $S_{i,j}=\delta_{i,j+1}$, where $\delta$ is the Kronecker 
delta symbol.
\end{definition}

We note that if $A$ is an infinite matrix, then $AS$ is the matrix obtained by removing the first
column of $A$.  In particular, if $A$ is the Hankel matrix of a sequence $\{a_i\}_{i=0}^\infty$,
then $AS$ is the Hankel matrix of the sequence $\{a_i\}_{i=1}^\infty$.  Thus, $CS^k=C\{k\}$.

\begin{proposition}\label{prop:LCKU} For any $k\geq 0$, we have $(LC\{1\}U)^k=LC\{k\}U$.
\end{proposition}

\begin{proof} We have seen already in Proposition \ref{prop:LC1U} 
that $LCU=I$, so $LC\{0\}U=(LC\{1\}U)^0$.  It is easy to check
that if $A$ and $B$ are both upper triangular, then $(AB)(n)=A(n)B(n)$.
Thus, $(LC)(n)U(n)$ is the $n+1\times n+1$ identity matrix.  Thus, $LC(n)=U(n)^{-1}$,
so $(ULC)(n)=U(n)(LC)(n)$ is the $n+1\times n+1$ identity matrix for each $n$.  Therefore,
$ULC$ is the infinite identity matrix.  Thus,
\[(LC\{1\}U)=(LCSU)^k=LCS(ULCS)^{k-1}U\]
\[=LCS(S^{k-1})U=LCS^kU=LC\{k\}U.\]
\end{proof}


We now can prove Proposition \ref{prop:LCKU2}.

\begin{proof} (of Proposition \ref{prop:LCKU2})
We prove this by induction on $k$.  The cases when $k=0$
or $k=1$ follow easily from Proposition \ref{prop:LC1U}.  Now, suppose the claim
holds when $k=t\geq 1$.  Then by Proposition \ref{prop:LCKU} we have
\[LC\{t+1\}U=(LC\{1\}U)^{t}(LC\{1\}U)=(LC\{t\}U)(LC\{1\}U).\]
Thus, we have
\[(LC\{t+1\}U)_{i,j}=\sum_{l} (LC\{t\}U)_{i,l}(LC\{1\}U)_{l,j}.\]

If $j\geq 1$, then by Proposition \ref{prop:LC1U}, we get
\[(LC\{t+1\}U)_{i,j}=(LC\{t\}U)_{i,j-1}+2(LC\{t\}U)_{i,j}+(LC\{t\}U)_{i,j+1}.\]
Now, assuming $i+j\geq t+1$, we have $i+(j-1)\geq t$, so by the inductive
hypothesis, the above sum is
\[{2t\choose t+i-j+1}+2{2t\choose t+i-j}+{2t\choose t+i-j-1}\]
\[=\left({2t\choose t+i-j+1}+{2t\choose t+i-j}\right)+\left({2t\choose t+i-j}+
{2t\choose t+i-j-1}\right) \]
\[={2t+1\choose t+i-j+1}+{2t+1\choose t+i-j}={2(t+1)\choose (t+1)+i-j},\]
as needed.

On the other hand, if $j=0$, then by Proposition \ref{prop:LC1U}, we get
\[(LC\{t+1\}U)_{i,j}=\sum_{l} (LC\{t\}U)_{i,l}(LC\{1\}U)_{l,j}
=(LC\{t\}U)_{i,0}+(LC\{t\}U)_{i,1}.\]
Now, since we may assume $i+j\geq t+1$, we have $i\geq t+1$, so
by the inductive hypothesis, the above term is
${2t\choose t+i}+{2t\choose t+i-1}={2t+1\choose t+i}$.  If $i=t+1$,
then \[{2t+1\choose t+i}=1={2(t+1)\choose (t+1)+i},\] while if $i&gt;t+1$,
then \[{2t+1\choose t+i}=0={2(t+1)\choose (t+1)+i},\] as needed.
\end{proof}


\section{Recurrence Relations}\label{sec:recurrence}

In this section, we make two observations about solutions to recurrence relations.  The second observation,
Lemma \ref{lem:characteristic},
is an easy application of the Cayley-Hamilton theorem to recurrence relations.  The first observation,
Lemma \ref{lem:shorter}, describes a technique
for showing that a given solution of a recurrence relation in fact satisfies a recurrence relation of lower order.
An example will help to illustrate this idea.

\begin{example}
Suppose a sequence $\{x_n\}$ satisfies the recurrence relation \[a_{n+3}-2a_{n+2}+a_n=0,\] and suppose
that $x_2=x_1+x_0$.  Then the sequence $\{x_n\}$ satisfies the recurrence relation \[a_{n+2}-a_{n+1}-a_n=0.\]
To see this, define a new sequence $\{y_n\}$ by $y_n=x_{n+2}-x_{n+1}-x_n$.  Then 
\[y_{n+1}-y_n=(x_{n+3}-x_{n+2}-x_{n+1})-(x_{n+2}-x_{n+1}-x_n)\]
\[=x_{n+3}-2x_{n+2}+x_n=0,\]
so $y_{n+1}=y_n$ for all $n$.  At the same time,
$y_0=x_2-x_1-x_0=0$.  Thus, $y_n=0$ for all $n$, so that $x_{n+2}-x_{n+1}-x_n=0$
for all $n$.
\end{example}

To generalize the example above, recall that a recurrence relation
\[c_ka_{n+k}+c_{k-1}a_{n+k-1}+\cdots+c_1a_{n+1}+c_0a_n=0\]
has an associated characteristic polynomial
\[p(x)=c_kx^k+c_{k-1}x^{k-1}+\cdots +c_1x+c_0.\]

In the following lemma, we suppose given two polynomials of degree $k_1$ and $k_2$ respectively:
\[p_1(x)=\sum_j c_{j,1} x^j,\quad p_2(x)=\sum_j c_{j,2} x^j.\]
Here, the summations are over all integers, but $c_{j,1}=0$ for all $j&gt;k_1$ and all $j&lt;0$, while
$c_{j,2}=0$ for all $j&gt;k_2$ and all $j&lt;0$.
We let \[q(x)=p_1(x)\cdot p_2(x)=\sum_j \left(\sum_l c_{l,1}c_{j-l,2}\right)x^j.\]

\begin{lemma}\label{lem:shorter} Suppose given a sequence $\{x_n\}$ such that for some positive
integer $m$, we have
\begin{enumerate}
	\item $\{x_n\}$ satisfies the recurrence relation with characteristic polynomial
$q(x)$ for $n\geq m$, and
	\item For $0\leq n\leq m+k_1-1$, we have $\sum_j c_{j,2}x_{n+j}=0$.  That is, for $0\leq n\leq m+k_1-1$,
$\{x_n\}$ satisfies the recurrence relation with characteristic polynomial $p_2(x)$.
\end{enumerate}
Then $\{x_n\}$ satisfies the recurrence relation with characteristic polynomial $p_2(x)$ for all $n\geq 0$.
\end{lemma}

\begin{proof} Let $y_n=\sum_j c_{j,2} x_{n+j}$.  Then \[\sum_l c_{l,1} y_{n+l}=\sum_l \sum_j  c_{l,1}c_{j,2} x_{n+l+j}
=\sum_j\left(\sum_l c_{l,1}c_{j-l,2}\right) x_{n+j}.\]  By our first hypothesis, this last term is $0$ whenever $n\geq m$,
so $\{y_n\}$ satisfies the $k_1^{\text{th}}$-order homogeneous linear
recurrence relation with characteristic polynomial $p_1(x)$ for $n\geq m$.  By our second hypothesis, $y_n=0$ for
$0\leq n\leq m+k_1-1$, which includes the $k_1$ consecutive values of $n$ beginning at $m$.  It follows
by induction that $y_n=0$ for all $n\geq 0$.  Thus,
$\{x_n\}$ satisfies the recurrence relation with characteristic polynomial $p_2(x)$.
\end{proof}

Now, we turn to our application of the Cayley-Hamilton theorem.
Consider a sequence $\{\vec v_n\}$ of vectors in $\mathbb R^m$; we
let $v_n^i$ denote the $i^\text{th}$ component of $\vec v_n$.

\begin{lemma}\label{lem:characteristic} Suppose
$\{\vec v_n\}$ satisfies the recurrence relation
$\vec v_{n+1}=A\vec v_n$, where $A$ is an $m\times m$ matrix.  Then for each $i\in \{1, 2, \ldots, m\}$,
the sequence $v_n^i$ satisfies a recurrence relation whose characteristic polynomial is equal to the
characteristic polynomial of the matrix $A$, $\det(\lambda I-A)$.
\end{lemma}

\begin{proof}
Let $p(\lambda)=\det(\lambda I-A)=c_m\lambda^m+c_{m-1}\lambda^{m-1}+\cdots+c_0$.  Then by the
Cayley-Hamilton theorem, $p(A)=0$,
so for any $n$, \[c_mA^m\vec v_n+c_{m-1}A^{m-1}\vec v_n+\cdots +c_0\vec v_n=\vec 0.\]  Since $\vec v_{n+1}=A\vec v_n$
for all $n$, this yields
\[c_m\vec v_{n+m}+c_{m-1}\vec v_{n+m-1}+\cdots +c_0\vec v_n=\vec 0.\]
Now, taking the $i^\text{th}$ component of the above equation yields the result.
\end{proof}


\section{Hyperdiagonal matrices}\label{sec:hyper}
Just as Proposition \ref{prop:LC1U} allowed us to study the Hankel transform
of the sequence $\alpha c_n+\beta c_{n+1}$ in the proof of Theorem \ref{thm:1},
so Proposition \ref{prop:LCKU2} allows us to investigate the Hankel transforms
of more complicated linear combinations of Catalan numbers.  Proposition
\ref{prop:LCKU2} motivates studying a class of matrices which we now describe.

\begin{definition}\label{def:hyperdiagonal} An infinite matrix $M$ is {\it $k$-hyperdiagonal} if $M_{i,j}=0$
whenever $|j-i|&gt;k$, and for $j+i\geq k$, $M_{i,j}$ only depends on $|j-i|$.
We let $m_t$ and $m_{-t}$ denote the value of $M_{i,j}$ when $|j-i|=t$ and $j+i\geq k$
(so $m_t=0$ when $t&gt;k$ or $t&lt;-k$.)
\end{definition}

\begin{example}\label{ex:k=123}
By Proposition \ref{prop:LCKU2}, the matrix $LC\{k\}U$  is $k$-hyperdiagonal,
with $m_t={2k\choose k-t}={2k\choose k+t}$.
\end{example}

In this section, we will investigate the determinants of the sequence of
upper left $(n+1)\times (n+1)$ submatrices of a $k$-hyperdiagonal matrix $M$.

\begin{notn} Suppose $M$ is a $k$-hyperdiagonal infinite matrix, and
let $\vec r$ denote the integer-valued vector $\langle r_1, r_2, \cdots, r_k\rangle$, where
$-k&lt;r_1&lt;r_2&lt;\cdots&lt;r_k\leq k$.
Let $M(n)_{\vec r}$ denote the
$(n+1)\times (n+1)$ matrix whose $(i,j)$ entry is given by
\[\begin{cases} M_{i,j}, &amp; 0\leq i\leq n \text{ and }0\leq j&lt; \max\{0,n-k+1\}; \\
M_{i,n+r_{j-(n-k)}}, &amp; 0\leq i\leq n \text{ and }\max\{0,n-k+1\}\leq j\leq n.\end{cases}\]
\end{notn}

Thus, when $n\geq k$, $M(n)_{\vec r}$ is obtained by removing all rows
of $M$ past the $n^\text{th}$ row, retaining the $0^\text{th}$ through $(n-k)^{\text{th}}$
columns, removing all columns past the $(n+k)^\text{th}$ column, and
retaining the $(n+r_i)^{\text{th}}$ columns for $1\leq i\leq k$.

\begin{remark}\label{rem:case} If $\vec r=\langle 1-k, 2-k, 3-k, \ldots, 0\rangle$, then
$M(n)_{\vec r}=M(n)$, where we recall from Notation \ref{notn:upperleft} that
$M(n)$ denotes is the upper left $(n+1)\times (n+1)$-submatrix of $M$.
Indeed, in this case, $r_i=i-k$, so $n+r_{j-(n-k)}=j$, whence
$M_{i,n+r_{j-(n-k)}}=M_{i,j}$.
\end{remark}

\begin{example} Suppose $k=2$ and \[M=\left(\begin{matrix} 
* &amp; * &amp; x &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\
* &amp; z &amp; y &amp; x &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\
x &amp; y &amp; z &amp; y &amp; x &amp; 0 &amp; 0 &amp; \cdots \\
0 &amp; x &amp; y &amp; z &amp; y &amp; x &amp; 0 &amp; \cdots \\
0 &amp; 0 &amp; x &amp; y &amp; z &amp; y &amp; x &amp; \cdots \\
0 &amp; 0 &amp; 0 &amp; x &amp; y &amp; z &amp; y &amp; \cdots \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; x &amp; y &amp; z &amp; \cdots \\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots
\end{matrix}\right).\]
Then \[M(6)_{\langle 0,2\rangle}=\left(\begin{matrix}
* &amp; * &amp; x &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\
* &amp; z &amp; y &amp; x &amp; 0 &amp; 0 &amp; 0 \\
x &amp; y &amp; z &amp; y &amp; x &amp; 0 &amp; 0  \\
0 &amp; x &amp; y &amp; z &amp; y &amp; 0 &amp; 0  \\
0 &amp; 0 &amp; x &amp; y &amp; z &amp; x &amp; 0  \\
0 &amp; 0 &amp; 0 &amp; x &amp; y &amp; y &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; x &amp; z &amp; x  \\
\end{matrix}\right).\]
Here, we have retained columns numbered $0$ through $4$, together with columns numbered
$6=6+0$ and $8=6+2$.  As another example, if we retain columns numbered $5=6+(-1)$ and $7=6+1$,
we get
\[M(6)_{\langle -1,1\rangle}=\left(\begin{matrix}
* &amp; * &amp; x &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\
* &amp; z &amp; y &amp; x &amp; 0 &amp; 0 &amp; 0 \\
x &amp; y &amp; z &amp; y &amp; x &amp; 0 &amp; 0  \\
0 &amp; x &amp; y &amp; z &amp; y &amp; x &amp; 0  \\
0 &amp; 0 &amp; x &amp; y &amp; z &amp; y &amp; 0  \\
0 &amp; 0 &amp; 0 &amp; x &amp; y &amp; z &amp; x \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; x &amp; y &amp; y  \\
\end{matrix}\right).\]
\end{example}

Observe that the determinant of $M(6)_{\langle 0, 2\rangle}$ is
equal to the product of $x$ and the determinant of $M(5)_{\langle -1, 1\rangle}$.  Likewise,
the determinant of $M(6)_{\langle -1, 1\rangle}$ is
\[x\left(\det(M(5)_{\langle 0,2 \rangle})\right)
-y\left(\det(M(5)_{\langle -1, 2\rangle})\right)
+y\left(\det(M(5)_{\langle -1, 0\rangle})\right).\]
As we will shortly show, the determinant of $M(n)_{\vec r}$ can be
expressed as a linear combination of determinants of the form
$M(n-1)_{\vec s}$ for various vectors $\vec s$.  We now describe how the
needed vectors $\vec s$ can be obtained from $\vec r$.

\begin{notn} 
Suppose $0\leq l\leq k$, and suppose given a vector $\vec r$ with \[-k&lt;r_1&lt;r_2&lt;\cdots&lt;r_k\leq k.\]
If $r_k&lt;k$ or $l=k$, let $\delta_l\vec r$ be the vector obtained by inserting $-k$ as a $0^\text{th}$
term, removing the $l^\text{th}$ term, and then adding one to all the components of $\vec r$.
If $r_k=k$ and $l&lt;k$, then $\delta_l\vec r$ is undefined.
\end{notn}

Thus, if $l=k$, then
\[\delta_l\vec r=\langle -k+1, r_1+1, r_2+1, \cdots, r_{k-1}+1\rangle.\]
When $1\leq l\leq k-1$, and $r_k&lt;k$, we have
\[\delta_l\vec r=\langle -k+1, r_1+1, r_2+1, \cdots, r_{l-1}+1,r_{l+1}+1, \cdots, r_k+1\rangle.\]
If $l=0$ and $r_k&lt;k$, then
\[\delta_0\vec r=\langle r_1+1, r_2+1, \cdots, r_k+1\rangle.\]

\begin{lemma}\label{lem:remove} Suppose $0\leq l\leq k$, $n\geq k-l$, and $\vec r$ is a vector
with \[-k&lt;r_1&lt;r_2&lt;\cdots&lt;r_k\leq k.\]
If $r_k=k$ and $l&lt;k$, then removing the $n^\text{th}$ row and the
$(n-k+l)^\text{th}$ column of $M(n)_{\vec r}$ yields a matrix with a column of all zeros.
If $r_k&lt;k$ or $l=k$, then removing the $n^\text{th}$ row and the
$(n-k+l)^\text{th}$ column of $M(n)_{\vec r}$ yields $M(n-1)_{\delta_l \vec r}$.
\end{lemma}

\begin{proof} 
First, if $r_k=k$ and $l&lt;k$, then for $0\leq i\leq n-1$, the $(i,n)$-entry of $M(n)_{\vec r}$
is $M_{i,n+k}$, which is $0$ since $n+k-i&gt;k$ and $M$ is $k$-hyperdiagonal.
Thus, since $n-k+l&lt;n$, it follows that removing the $n^{\text{th}}$ row
and $(n-k+l)^\text{th}$ of $M(n)_{\vec r}$ yields a matrix whose last column is all $0$'s. 

Now, suppose $r_k&lt;k$ or $l=k$.  Let $M'$ denote the $n\times n$ matrix obtained
by removing the $n^\text{th}$ row and the
$(n-k+l)^\text{th}$ column from $M(n)_{\vec r}$.  Then $M'_{i,j}$
is equal to the $(i,j)$ entry of $M(n)_{\vec r}$ when $0\leq j&lt;n-k+l$ and is equal
to the $(i,j+1)$ entry of $M(n)_{\vec r}$ when $n-k+l\leq j\leq n-1$.  Thus,
\[M'_{i,j}=\begin{cases}
M_{i,j}, &amp; \text{ if } 0\leq i\leq n-1\text{ and } 0\leq j&lt;\max\{0,n-k+1\}; \\
M_{i,n+r_{j-(n-k)}}, &amp; \text{ if } 0\leq i\leq n-1\text{ and } \max\{0,n-k+1\}\leq j&lt;n-k+l; \\
M_{i,n+r_{j+1-(n-k)}}, &amp; \text{ if } 0\leq i\leq n-1\text{ and }n-k+l\leq j\leq n-1.\end{cases}\]
Now, for a sequence $-k&lt;s_1&lt;s_2&lt;\cdots&lt;s_k\leq k$, the $(i,j)$ entry of the $n\times n$ matrix
$M(n-1)_{\vec s}$ is
\[\begin{cases} M_{i,j}, &amp; \text{ if } 0\leq i\leq n-1\text{ and } 0\leq j&lt; \max\{0,n-k\}; \\
M_{i,(n-1)+s_{j-(n-1-k)}}, &amp; \text{ if } 0\leq i\leq n-1\text{ and }\max\{0,n-k\}\leq j\leq n-1.
\end{cases}\]
When $s_1=-k+1$ and $j=n-k$, $M_{i,(n-1)+s_{j-(n-1-k)}}=M_{i,j}$, so the above can be rewritten as
\[\begin{cases} M_{i,j}, &amp; \text{ if } 0\leq i\leq n-1\text{ and }0\leq j&lt;\max\{0,n-k+1\}; \\
M_{i,(n-1)+s_{j-(n-1-k)}}, &amp; \text{ if } 0\leq i\leq n-1\text{ and }\max\{0,n-k+1\}&lt;j\leq n-1.
\end{cases}\]
If we set $\vec s=\delta_l\vec r$, then we see that $M(n-1)_{\vec s}=M'$, as needed.
\end{proof}



\begin{notn} Let $d(M,\vec r)_n=\det(M(n)_{\vec r})$.
In general, there are ${2k\choose k}$ vectors $\vec r$ with $-k&lt;r_1&lt;r_2&lt;\cdots&lt; r_k\leq k$, and we
will order these lexicographically, starting from $\langle -k+1, -k+2, \cdots, 0\rangle$ and ending at
$\langle 1, 2, 3, \cdots, k\rangle$.  
For a fixed value of $n$, the set of numbers $d(M,\vec r)_n$ thus forms a column vector of length
${2k\choose k}$, which we will denote $d(M)_n$.
\end{notn}

\begin{corollary}\label{cor:det}  Suppose $n\geq k$, and suppose $\vec r=\langle r_1, r_2, \cdots, r_k\rangle$
where \[-k&lt;r_1&lt;r_2&lt;\cdots&lt;r_k&lt; k.\]  Let $r_0=-k$.  Then
\[d(M,\vec r)_n=\sum_{l=0}^k (-1)^{k-l} m_{r_l} d(M,\delta_l\vec r)_{n-1}.\]
If $r_k=k$, then $d(M,\vec r)_n=m_{r_k} d(M,\delta_k\vec r)_{n-1}$.
\end{corollary}

\begin{proof} This follows immediately from Lemma \ref{lem:remove} using cofactor expansion along
the bottom row.
\end{proof}

Using Corollary \ref{cor:det}, we may find a recurrence relation for the sequence $d(M)_n$ in $\mathbb R^{2k\choose k}$.

\begin{example}\label{ex:k=1} Suppose $k=1$.  Then there are $2$ permissible vectors $\vec r$ of length $k=1$,
namely $\langle 0\rangle$ and $\langle 1\rangle$. By Corollary \ref{cor:det}, we have for $n\geq 1$
\[d(M,\langle 0\rangle)_n=-m_1d(M,\langle 1\rangle)_{n-1}+m_0 d(M,\langle 0\rangle)_{n-1},\]
and
\[d(M,\langle 1\rangle)_n=m_1 d(M,\langle 0\rangle)_{n-1}.\]
We will write $z, y$ for $m_0, m_1$ respectively.  Then, the system of
equations above can be written in matrix form as
\[d(M)_n
=\left(\begin{matrix}
z &amp; -y \\
y &amp; 0\end{matrix}\right)
d(M)_{n-1}\]
for all $n\geq 1$.  Shifting indices, we have for all $n\geq 0$:
\[d(M)_{n+1}=\left(\begin{matrix}
z &amp; -y \\
y &amp; 0\end{matrix}\right)
d(M)_n.\]
By Lemma \ref{lem:characteristic}, for each permissible vector $\vec r$,
the sequence $d(M, \vec r)_n$ satisfies a recurrence relation with characteristic
polynomial equal to the characteristic polynomial of the matrix
$\left(\begin{matrix} z &amp; -y \\ y &amp; 0\end{matrix}\right)$, which
is $\lambda^2-z\lambda +y^2$.  By Remark \ref{rem:case}, the sequence of
determinants $\det(M(n))$ coincides with the sequence $d(M,\langle 0\rangle)_n$.
Thus, the sequence of determinants $\det(M(n))$ satisfies the recurrence relation
with characteristic polynomial equal to $\lambda^2-z\lambda +y^2$.
\end{example}

\begin{example}\label{ex:k=2}
Now suppose $k=2$.  Then there are $6$ permissible vectors $\vec r$ of length $k=2$,
namely $\langle -1,0\rangle,
\langle -1, 1\rangle, \langle -1, 2\rangle, \langle 0,1\rangle, \langle 0,2\rangle,
\langle 1,2\rangle$.  Now we will write $z, y, x$ for $m_0, m_1, m_2$ respectively.
Using Corollary \ref{cor:det} again (and again shifting indices), we can compute that for $n\geq 1$
\[
d(M)_{n+1}=\left(\begin{matrix}
z &amp; -y &amp; 0 &amp; x &amp; 0 &amp; 0\\
y &amp; 0 &amp; -y &amp; 0 &amp; x &amp; 0\\
x &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0\\
0 &amp; y &amp; -z &amp; 0 &amp; 0 &amp; x\\
0 &amp; x &amp; 0 &amp; 0 &amp; 0 &amp; 0\\
0 &amp; 0 &amp; x &amp; 0 &amp; 0 &amp; 0\end{matrix}\right)
d(M)_n.\]

\end{example}

Here, the characteristic polynomial of the matrix in the above equation is
\[\lambda^6-z\lambda^5+(y^2-x^2)\lambda^4+(2x^2z-2xy^2)\lambda^3+(x^2y^2-x^4)\lambda^2-x^4z\lambda+x^6,\]
which factors as
\[(\lambda-x)^2(\lambda^4+(2x-z)\lambda^3+(2x^2+y^2-2x z)\lambda^2+(2x^3-x^2z)\lambda+x^4).\]

Again, it follows from Lemma \ref{lem:characteristic} and Remark \ref{rem:case}
that the sequence of determinants $\det(M(n))$ satisfies a recurrence relation
with the above characteristic polynomial for $n\geq 1$.

\begin{example}\label{ex:k=3}
Now suppose $k=3$.  In this case, there are ${6\choose 3}=20$ permissible vectors $\vec r$ of length $3$.
We then use Corollary \ref{cor:det} as in Example \ref{ex:k=2} above, and we get a $20\times 20$ matrix $A$
such that \[d(M)_{n+1}=A \cdot d(M)_{n}\] for all $n\geq 2$.  The matrix $A$ can be expressed as follows, where we
write $z, y, x, w$ for $m_0, m_1, m_2, m_3$ respectively.
{\scriptsize\[\left(\begin{array}{cccccccccccccccccccc}
z &amp; -y &amp; 0 &amp; 0 &amp; x &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; -w &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
y &amp; 0 &amp; -y &amp; 0 &amp; 0 &amp; x &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; -w &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
x &amp; 0 &amp; 0 &amp; -y &amp; 0 &amp; 0 &amp; x &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; -w &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
w &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; y &amp; -z &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; x &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; -w &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; x &amp; 0 &amp; -z &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; x &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; -w &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; w &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; x &amp; -y &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; x &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; -w &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; w &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; w &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; y &amp; -z &amp; 0 &amp; y &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; -w &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; x &amp; 0 &amp; -z &amp; 0 &amp; y &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; -w &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; w &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; x &amp; -y &amp; 0 &amp; 0 &amp; y &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; -w &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; w &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; w &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; x &amp; -y &amp; z &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; -w \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; w &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; w &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; w &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
\end{array}\right)\]}

\end{example}

The characteristic polynomial of the above matrix factors as the product
\begin{eqnarray*}
\lefteqn{(\lambda^6-x\lambda^5+wy\lambda^4-w^2z\lambda^3+w^3y\lambda^2-w^4x\lambda+w^6)^2 \cdot}  \\
&amp;&amp; (\lambda^8+(2x-z)\lambda^7+(w^2+2x^2-2wy+y^2-2xz)\lambda^6\\
&amp;&amp;+(2w^2x+2x^3-4wxy-w^2z-x^2z+2wyz)\lambda^5\\
&amp;&amp;+(4w^2x^2+x^4-4w^3y-4wx^2y+2w^2y^2+w^2z^2)\lambda^4\\
&amp;&amp;+(2w^4x+2w^2x^3-4w^3xy-w^4z-w^2x^2z+2w^3yz)\lambda^3\\
&amp;&amp;+(w^6+2w^4x^2-2w^5y+w^4y^2-2w^4xz)\lambda^2
+(2w^6x-w^6z)\lambda+
w^8
\end{eqnarray*}

Again, it follows from Lemma \ref{lem:characteristic} and Remark \ref{rem:case}
that the sequence of determinants $\det(M(n))$ satisfies a recurrence relation
with the above characteristic polynomial for $n\geq 2$.

\section{Proofs of Theorems \ref{thm:2} and \ref{thm:3}}\label{sec:end}

Before turning to the proofs of Theorems \ref{thm:2} and \ref{thm:3}, we
give another proof of Theorem \ref{thm:1}.  As in our previous proof, we
let $\{h_n\}$ denote the Hankel transform of the sequence $\alpha c_n+\beta c_{n+1}$,
and we recall that \[h_n=\det((\alpha LCU+\beta LC\{1\}U)(n)).\]  Thus, if we let
$M=\alpha LCU+\beta LC\{1\}U$, then $h_n=\det(M(n))$.
Now, by Proposition \ref{prop:LCKU2}, the matrix $M$ is a $1$-hyperdiagonal matrix,
with $m_1=\beta$ and $m_0=2\beta+\alpha$.  By Example \ref{ex:k=1},
substituting $y=\beta$ and $z=2\beta+\alpha$, the sequence $\det(M(n))$
satisfies a recurrence relation with characteristic polynomial
$\lambda^2-z\lambda +y^2$ for all $n\geq 0$.  That is, for all $n\geq 0$, we have 
\[h_{n+2}-(2\beta+\alpha)h_{n+1}+\beta^2 h_n=0.\]

Our proofs of Theorems \ref{thm:2} and \ref{thm:3} follow a similar pattern.

\begin{proof} (Theorem \ref{thm:2}).  We
let $\{h_n\}$ denote the Hankel transform of the sequence $\alpha c_n+\beta c_{n+1}+\gamma c_{n+2}$.
We then have \[h_n=\det((\alpha LCU+\beta LC\{1\}U+\gamma LC\{2\}U)(n)).\]  Thus, if we let
\[M=\alpha LCU+\beta LC\{1\}U+\gamma LC\{2\}U,\] then $h_n=\det(M(n))$.
Now, by Proposition \ref{prop:LCKU2}, the matrix $M$ is a $2$-hyperdiagonal matrix,
with \[m_2=\gamma, m_1=4\gamma+\beta, m_0=6\gamma+2\beta+\alpha.\]  By Example \ref{ex:k=2},
substituting $x=\gamma$, $y=4\gamma+\beta$ and $z=6\gamma+2\beta+\alpha$,
we find that for $n\geq 1$, the sequence $h_n=\det(M(n))$ satisfies the recurrence relation whose characteristic polynomial
is the product of $(\lambda-\gamma)^2$ and
\[\lambda^4+(-\alpha-2\beta-4\gamma)\lambda^3+(\beta^2-2\alpha\gamma+4\beta\gamma
+6\gamma^2)\lambda^2+(-\alpha-2\beta-4\gamma)\gamma^2\lambda+\gamma^4.\]
We compute directly that for $0\leq n\leq 2$, $h_n$ satisfies the recurrence relation with characteristic polynomial
\[\lambda^4+(-\alpha-2\beta-4\gamma)\lambda^3+(\beta^2-2\alpha\gamma+4\beta\gamma
+6\gamma^2)\lambda^2+(-\alpha-2\beta-4\gamma)\gamma^2\lambda+\gamma^4.\]
Our result now follows from Lemma \ref{lem:shorter} with $m=1$, $k_1=2$, and $k_2=4$.
\end{proof}


\begin{proof} (Theorem \ref{thm:3}).  We
let $\{h_n\}$ denote the Hankel transform of the sequence
\[\alpha c_n+\beta c_{n+1}+\gamma c_{n+2}+\delta c_{n+3}.\]
We then have \[h_n=\det((\alpha LCU+\beta LC\{1\}U+\gamma LC\{2\}U+\delta LC\{3\}U)(n)).\]
Thus, if we let
\[M=\alpha LCU+\beta LC\{1\}U+\gamma LC\{2\}U+\delta LC\{3\}U,\] then $h_n=\det(M(n))$.
Now, by Proposition \ref{prop:LCKU2}, the matrix $M$ is a $3$-hyperdiagonal matrix,
with \[m_3=\delta, m_2=6\delta+\gamma, m_1=15\delta+4\gamma+\beta, m_0=
20\delta+6\gamma+2\beta+\alpha.\]  By Example \ref{ex:k=3},
substituting $w=\delta$, $x=6\delta+\gamma$, $y=15\delta+4\gamma+\beta$
and $z=20\delta+6\gamma+2\beta+\alpha$, we find that the sequence
$h_n=\det(M(n))$ satisfies the recurrence relation whose characteristic polynomial factors
as the product of the square of

{\scriptsize
\begin{eqnarray*}
(\lambda^6+(-\gamma-6\delta)\lambda^5+(\beta+4\gamma+15\delta)\delta\lambda^4
+(-\alpha-2\beta-6\gamma-20\delta)\delta^2\lambda^3
+(\beta+4\gamma+15\delta)\delta^3\lambda^2
+(-\gamma-6\delta)\delta^4+\delta^6)
\end{eqnarray*}
}
and

{\scriptsize
\begin{eqnarray*}
\lambda^8\\
+
(-\alpha-2\beta-4\gamma-8\delta)\lambda^7\\
+(\beta^2-2\alpha\gamma+4\beta\gamma+6\gamma^2-12\alpha\delta+4\beta\delta+24\gamma\delta+28\delta^2)\lambda^6\\
+(-\alpha\gamma^2-2\beta\gamma^2-4\gamma^3+2\alpha\beta\delta+4\beta^2\delta-4\alpha\gamma\delta
-24\gamma^2\delta-7\alpha\delta^2+2\beta\delta^2-60\gamma\delta^2-56\delta^3)\lambda^5\\
+(\gamma^4-4\beta\gamma^2\delta+8\gamma^3\delta+\alpha^2\delta^2+4\alpha\beta\delta^2
+6\beta^2\delta^2+12\alpha\gamma\delta^2-8\beta\gamma\delta^2+36\gamma^2\delta^2+
40\alpha\delta^3-8\beta\delta^3
+80\gamma\delta^3+70\delta^4)\lambda^4\\
+(-\alpha\gamma^2-2\beta\gamma^2-4\gamma^3+2\alpha\beta\delta+4\beta^2\delta-4\alpha\gamma\delta
-24\gamma^2\delta-7\alpha\delta^2+2\beta\delta^2-60\gamma\delta^2-56\delta^3)\delta^2\lambda^3\\
+(\beta^2-2\alpha\gamma+4\beta\gamma+6\gamma^2-12\alpha\delta+4\beta\delta+24\gamma\delta+28\delta^2)
\delta^4\lambda^2\\
+(-\alpha-2\beta-4\gamma-8\delta)\delta^6\lambda^1\\
+\delta^8\lambda^0
\end{eqnarray*}

}

We compute directly that for $0\leq n\leq 13$, $h_n$ satisfies the recurrence relation with characteristic polynomial
equal to the second polynomial above.
Our result now follows from Lemma \ref{lem:shorter} with $m=2$, $k_1=12$, and $k_2=8$.
\end{proof}

\begin{thebibliography}{99}


\bibitem{BCCS} I.~Bajunaid,
J.~M.~Cohen, F.~Colonna, and D.~Signman, Function series, Catalan
numbers, and Random walks on trees, {\it Amer. Math. Monthly} {\bf 112}
(2005), 765--785.

\bibitem{BCQY} A.~T.~Benjamin, N.~T.~Cameron, J.~J.~Quinn, and C.~R.~Yerger,
Catalan determinants --- a combinatorial approach, {\it Congr. Numer.} {\bf 200}
(2010), 27--34.

\bibitem{CF} M.~Chamberland and
C.~French, Generalized Catalan numbers and generalized Hankel transformations,
{\it J. Integer Seq.} {\bf 10} (2007), 
\href{http://www.cs.uwaterloo.ca/journals/JIS/VOL10/Chamberland/chamberland12.html}{Article 07.1.1}.

\bibitem{CK} K.~L.~Collins and L.~B.~Krompart, The number of 
Hamiltonian paths in a rectangular grid, {\it Discrete Math.} {\bf 169}
(1997), 29--38.

\bibitem{CRI} A. Cvetkovi\'c, P. Rajkovi\'c, and
M. Ivkovi\'c, Catalan numbers, and Hankel transform,
and Fibonacci numbers, {\it J. Integer Seq.} {\bf 5} (2002), 
\href{http://www.cs.uwaterloo.ca/journals/JIS/VOL5/Ivkovic/ivkovic3.html}{Article 02.1.3}.

\bibitem{MW} M.~E.~Mays and J.~Wojciechowski, A determinant property of Catalan numbers.
{\it Discrete Math.} {\bf 211} (2000), 125--133.

\bibitem{OEIS} N.~J.~A.~Sloane, {\it The On-Line Encyclopedia of Integer Sequences.} Published electronically at \href{http://oeis.org}{\tt http://oeis.org}.

\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent {\it 2010 Mathematics Subject Classification:} Primary:
11C20; Secondary 11B37, 11B75, 15A15.

\noindent {\it Keywords:} Hankel transform, Catalan Numbers, recurrence
relations, integer sequences.


\bigskip
\hrule
\bigskip
 
\noindent (Concerned with sequences
\seqnum{A000159},
\seqnum{A014523}, and
\seqnum{A103433}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received January 28 2011;
revised version received April 5 2011. 
Published in {\it Journal of Integer Sequences}, April 20 2011.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}

                                                                                


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