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\begin{center}
\vskip 1cm{\LARGE\bf A Note on the Hankel
Transform of the  \\
\vskip .12in
Central  Binomial Coefficients
}
\vskip 1cm
\large
Mario Garcia Armas\\
Facultad de Matem\a'{a}tica y Computaci\a'{o}n \\
Universidad de La Habana \\
Havana, Cuba \\
\href{marioga@matcom.uh.cu}{\tt marioga@matcom.uh.cu} \\
\ \\
B. A. Sethuraman \\
Dept. of Mathematics\\
California State University, Northridge\\
Northridge, CA 91330 \\
USA \\
\href{al.sethuraman@csun.edu}{\tt al.sethuraman@csun.edu} \\
\end{center}

\vskip .2 in
\begin{abstract}
We show that the $n\times n$ Hankel matrix formed from the successive
even central binomial coefficients ${2l \choose l}$, $l=0, 1,\dots$
arises naturally when considering the trace form in the number ring of
the maximal real subfield of suitable cyclotomic fields. By considering
the trace form in two different integral bases of the number ring we
get a factorization of this matrix which immediately yields the
well-known zeroth and first Hankel transforms of the sequence.
\end{abstract}

\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{defin}[theorem]{Definition}
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%*********************************%
% Al's notation

\def\QQ{\mathbb{Q}}
\def\ZZ{\mathbb{Z}}
\def\Qi{\mathbb{Q}(\imath)}
\def\root2{\sqrt{2}}
\def\om{\omega}
\def\ominv{\omega^{-1}}
\def\th{\theta}
\def\ds{\displaystyle}
\def\o{\om}
\def\O{\mathcal{O}}
\def\Tr{{\rm Tr}_{K/\QQ}}
\def\thet{\theta}
%\def\v{\mathbf{v}}
\def\v{V}
%\def\w{\mathbf{w}}
\def\w{W}
\def\bb{\mathbf{b}}
\def\aa{\mathbf{a}}
\def\BB{\mathbf{B}}
\def\AA{\mathbf{A}}



\section{Introduction}
\label{SecnIntro}

Given a sequence ${a_l}$, $l=0,1,\dots$, the
$n\times n$ Hankel matrix $H_{n}^{(k)}$, $k=0,1,\dots$, formed
from this sequence is the matrix
$$
 \left( \begin{array} {ccccc} a_k&a_{k+1}&a_{k+2}&\ldots & a_{k+n-1}\\
a_{k+1}&a_{k+2}&a_{k+3}&\ldots&a_{k+n}\\
a_{k+2}&a_{k+3}&a_{k+4}&\ldots&a_{k+n+1}\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
a_{k+n-1}&a_{k+n}&a_{k+n+1}&\ldots&a_{k+2n-2}
\end{array} \right)
$$

The  \textsl{$k$-th Hankel Transform} of the sequence ${a_l}$,
$l=0,1,\dots$ is the sequence of determinants $d_n^{(k)}$ of the
matrices $H_{n}^{(k)}$ above, $n=1,2,\dots$. It is worth mentioning
that some authors refer to the Hankel transform only as the sequence
$d_n = \det H_n^{(0)}$ (see, for example, \cite{Lay}).

Hankel matrices have been studied extensively, and connections
between Hankel matrices and other areas of mathematics are well
known (see \cite{Tamm} for a very nice survey of Hankel matrices
especially in relation to combinatorics and coding theory). The term
Hankel transform was introduced in Sloane's sequence
\seqnum{A055878} and first studied
in \cite{Lay}. Since then, there have been numerous further studies
of Hankel transforms of sequences, for instance, \cite{Aig,CF,CRI,ERR,SpSt}.

Consider the particular sequence $a_l = {2l \choose l}$,
$l=0,1,\dots$, and denote $H_{n}^{(0)}$ by simply $H_n$.  The main purpose of this paper is to show that the matrices $H_n := H_{n}^{(0)}$ and $H_{n}^{(1)}$ that define the zeroth and first Hankel transforms of this particular sequence arise very naturally when considering the trace form ${\rm Tr} (x,y) = \Tr(xy)$ on the number ring
$\O_K$, where $K$ is the maximal real subfield of the $2^N$-th
cyclotomic field, for any $N$ such that $2^N \ge 8n$.  By
considering the same trace form with respect to two different
integral bases of $\O_K$, we obtain in a very natural way a factorization of $H_n$ as
\begin{equation} \label{basicfactorization}
H_n = B_n D_n B_n^T
\end{equation}
where
\begin{equation} \label{BnDef}
B_n  = \left( \begin{array} {ccccc}
1  & 0& 0 & \ldots& 0 \\[0.05 in]
%\\
2 \choose 1& 1 & 0&\ldots& 0 \\ [0.05 in]
% \\
 4 \choose 2& 4\choose 1 & 1&\ldots& 0 \\ [0.05 in]
% \\
% 6 \choose 3& 6\choose 2 & 6\choose 1&1&\ldots& 0\\
% \\
\vdots & \vdots &\vdots  &\ddots & \vdots \\[0.05 in]
%\\
2(n-1) \choose n-1 & 2(n-1) \choose n-2 & 2(n-1)\choose n-3
&\ldots & 1
\end{array} \right)
\end{equation} and

\begin{equation} \label{DnDef}
D_n = \left( \begin{array} {ccccc} 1&0&0&\ldots & 0\\
0&2&0&\ldots&0\\
0&0&2&\ldots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
0&0&0&\ldots&2
\end{array} \right)
\end{equation}

Similarly, we get the factorization 
\begin{equation} \label{basic_Hn1_factorization}
H_n^{(1)}=2 C_n C_n^T
\end{equation}
where
\begin{equation} \label{CnDef}
C_n  = \left( \begin{array} {ccccc}
1  & 0& 0 & \ldots& 0 \\[0.05 in]
%\\
3 \choose 1& 1 & 0&\ldots& 0 \\[0.05 in]
% \\
 5 \choose 2& 5\choose 1 & 1&\ldots& 0 \\[0.05 in]
% \\
% 6 \choose 3& 6\choose 2 & 6\choose 1&1&\ldots& 0\\
% \\
\vdots & \vdots &\vdots  &\ddots & \vdots \\[0.05 in]
%\\
2n-1 \choose n-1 & 2n-1 \choose n-2 & 2n-1\choose n-3 &\ldots & 1
\end{array} \right)
\end{equation}


These factorizations yield at once the  following well-known result 
(see, for example, \cite{Aig,Pea,Rad}):
\begin{corollary} \label{cor:zeroNone}
The zeroth Hankel transform $d_n^{(0)}$ of the sequence ${2l
\choose l}$, $l=0,1,\dots$ is the sequence $2^{n-1}$,
$n=1,2,\dots$, and the first Hankel transform $d_n^{(1)}$ is the
sequence $2^n$, $n=1,2\dots$.
\end{corollary}

Of course, once the factorization has been guessed at, Eq.~(\ref{basicfactorization})  and Eq.~ (\ref{basic_Hn1_factorization})
can  be proved by elementary means: our point here is only to
show that the Hankel matrices $H_n$ and $H_n^{(1)}$ and the factorizations above arise completely naturally
in number theory.





\section{Trace Calculations in $\O_K$} \label{SecnTraceCalculns}

Given a positive integer $N$ we write $\om$ for the primitive
$2^N$-th root of unity $e^{2\pi \imath/2^N}$.  We write $\thet$
for the element $\om+\ominv$. We write $\th_j$ ($j=0,1,\dots,$)
for the element $\om^j + \om^{-j}$, so that $\th_1 = \th$ and
$\th_0 = 2$. We write $L$ for the field $\QQ(\om)$, and $K$ for
the real subfield $\QQ(\thet)$. Note that $[L:\QQ] = 2^{N-1}$ and
$[K:\QQ] = 2^{N-2}$.  We write $m$ for $2^{N-2}$.

We begin by computing traces of the elements $\th^i$ as well as of products $\th_i\th_j$.  (Lemmas \ref{trace_thet_powers} and
\ref{trace_thet_subs} also appear in \cite{OS}, and are implicit in \cite[Prop. 4.3]{BFN}.)

%Our first result is the following:

\begin{lemma} \label{trace_thet_powers}
For $1 \le s < 2 \cdot m$,
\begin{equation*}
{\rm Tr}_{K/\QQ}(\thet^s) = \begin{cases} 0, & \text{if $s$ is odd;}
\\
m{s \choose {s/2}}, &\text{if $s$ is even.}
\end{cases}
\end{equation*}
\end{lemma}

\begin{proof}
Observe that ${\rm Tr}_{K/\QQ} (\thet^s) = {\rm Tr}_{L/\Qi}(\thet^s)$. Now
expanding $\thet^s = (\om+\ominv)^s$ binomially, we find
$$\thet^s=\sum_{j=0}^s {s\choose j} \om^j \om^{-(s-j)} = \sum_{j=0}^s {s\choose j}
\om^{2j-s}$$  Notice that when $s$ is odd, only odd powers of
$\om$ appear in this expansion. Since $\om$ raised to any odd
power is also a primitive $2^N$-th root of unity, it has minimal
polynomial $x^{m} \pm\imath$ over $\Qi$, and consequently, any odd
power of $\om$ has trace zero from $L$ to $\Qi$. It follows that
${\rm Tr}_{L/\Qi}(\thet^s) = 0$ when $s$ is odd. (Notice that this is
true for all odd $s$ and not just those in the range of the statement of the lemma.)

When $s$ is even, we first assume that $s < m$.  Then, the terms
in the expansion of $\thet^s$ above have even powers of $\om$ that
run through $s, s-2, \dots, 2, 0, -2, \dots, -(s-2), -s$. Given
any nonzero even integer $2l$ in this set, we write it as $2^{e} a$ for some
$e$ and odd integer $a$.  Then $\om^{2l}$ is a primitive
$2^{N-e}$-root of unity, and $[L:\QQ(\om^{2l})] = 2^{e}$. Since,
by assumption, $e < N-2$,   $\QQ(\om^{2l})$ strictly contains
$\Qi$. Now, ${\rm Tr}_{L/\Qi}(\om^{2l}) = {\rm Tr}_{\QQ(\om^{2l})/\Qi}
{\rm Tr}_{L/\QQ(\om^{2l})}(\om^{2l}) =
2^{e}{\rm Tr}_{\QQ(\om^{2l})/\Qi}(\om^{2l}) $.  Just as in the previous
paragraph, ${\rm Tr}_{\QQ(\om^{2l})/\Qi}(\om^{2l})$ is zero since the
minimal polynomial of $\om^{2l}$ is  $x^{2^{N-e-2}} \pm\imath$.
Hence, all nonzero powers of $\om$ contribute nothing to the
trace, so  ${\rm Tr}_{L/\Qi}(\thet^s)$ is $m$ times the coefficient of
the term $\om^0$ which is ${s \choose s/2}$.

When $2\cdot m > s \ge m$, we need a small modification. The
expansion of $\thet^s$ will have only even powers of $\om$ as
before, but continuing to write these powers as $2l$,
% = 2^{e} a$,
we will now also have powers where $2\cdot m > 2l \ge m$.  We
first consider the powers  $2l > m$: we factor $\om^m$ out to find
$\om^{2l} = \imath \om^{2l-m}$. Thus, ${\rm Tr}_{L/\Qi}(\om^{2l}) =
\imath {\rm Tr}_{L/\Qi}(\om^{2l-m})$. From our assumptions we find that
$2l-m$ is a positive even integer and that $m
> 2l-m$, so
the arguments of the previous paragraph show that this
trace is zero. Thus, we are left with the terms $\om^{m}$,
$\om^0$, and $\om^{-{m}}$. But $ {\rm Tr}_{L/\Qi}(\om^{m}) =
{\rm Tr}_{L/\Qi}(\imath) = m \imath$, while $ {\rm Tr}_{L/\Qi}(\om^{-{m}}) =
{\rm Tr}_{L/\Qi}(-\imath) = -m \imath$, so these two terms cancel each
other out.  Once again, we are left with the term $\om^0$ whose
trace is $m {s \choose s/2}$.
\end{proof}

\begin{lemma} \label{trace_thet_subs}
For $1 \le j < 2  m$,
\begin{equation}
{\rm Tr}_{K/\QQ}(\th_j) = 0
\end{equation}
and for $1 \le i,j <   m$
\begin{equation}
{\rm Tr}_{K/\QQ}(\th_{i}\th_{j}) = \begin{cases} 0, & \text{if $i\neq j$;}\\
2m, & \text{if $i=j$}. \end{cases}
\end{equation}
\end{lemma}

\begin{proof} The proof of the first part is essentially contained
in the proof of Lemma \ref{trace_thet_powers} above.  We have
${\rm Tr}_{K/\QQ}(\th_j) = {\rm Tr}_{L/\Qi}(\th_j) = {\rm Tr}_{L/\Qi}(\om^j +
\om^{-j})$.  We saw in that proof that ${\rm Tr}_{L/\Qi}(\om^j)=0$ for
all $1 \le j <2m$ except when $j=m$, so ${\rm Tr}_{L/\Qi}(\th_j)=0$ for
all such $j$.  When $j=m$, we have ${\rm Tr}_{L/\Qi}(\om^m)=m\imath$ and
${\rm Tr}_{L/\Qi}(\om^{-m})=-m\imath$.  Hence ${\rm Tr}_{L/\Qi}(\th_m)=0$ as
well.


For the second assertion, note that $\th_{i}\th_{j} = \th_{i+j} +
\th_{j-i}$ where we can assume without loss of generality that
$j-i \ge 0$. The result immediately follows from the calculations
of ${\rm Tr}_{K/\QQ}(\th_{j})$ above, noting that $i+j < 2m$, and $\th_0
= 2$.

\end{proof}

Note that $\O_K = \ZZ[\th]$ (see \cite[Exer.\ 35, Chap.\ 2]{Mar} for
instance). Expanding each power $\th^s$ binomially and collecting
terms we find
\begin{equation} \label{transform_formulas}
\th^s = \begin{cases} \sum\limits_{j=0}^{\lfloor s/2 \rfloor} {s
\choose j} \th_{s-2j}, & \text{if $s$ is odd;}
\\  \sum\limits_{j=0}^{  (s/2) -1} {s
\choose j} \th_{s-2j} + {s \choose s/2}, &\text{if $s$ is even.}
\end{cases}
\end{equation}

For any positive integer $n$, let $B_n$ be as in (\ref{BnDef}),
and   $C_n$ as in (\ref{CnDef}). Let
\begin{eqnarray*}
\v_e&=& (1,\th^2,\th^4,\dots,\th^{m-2})^T\\
\v_o&=& (\th,\th^3,\th^5,\dots,\th^{m-1})^T\\
\w_e&=& (1,\th_2,\th_4,\dots,\th_{m-2})^T\\
\w_o&=& (\th_1 = \th,\th_3,\th_5,\dots,\th_{m-1})^T
\end{eqnarray*}

Then Eq.~(\ref{transform_formulas}) splits as two matrix
relations:

\begin{eqnarray}
\v_e &=& B_{m/2} \w_e \label{vw_e_connection};\\
\v_o &=& C_{m/2} \w_o \label{vw_o_connection}.
\end{eqnarray}

Since $1$, $\th$, $\th^2$, $\dots,$ $\th^{m-1}$ is a $\ZZ$-basis for $\O_K$, and since $B_m$ and $C_m$ are integer matrices with determinant $1$, these relations
show that $1$, $\th_1 =\th$, $\th_2$, $\dots,$ $\th_{m-1}$ is also a $\ZZ$-basis for $\O_K$. But these relations show us even
more: if we define
\begin{eqnarray}
M_e &=& \ZZ \oplus \ZZ \th^2 \oplus \ZZ \th^4\oplus\cdots\oplus \ZZ \th^{m-2}\quad \text{and} \\
M_o &=& \ZZ  \oplus \ZZ  \th \oplus \ZZ \th^3\oplus\cdots\oplus \ZZ \th^{m-1}
\end{eqnarray}
then $1$, $\th_2$, $\th_4$, $\dots$, $\th_{m-2}$ is also a $\ZZ $ basis for $M_e$, and $\th$, $\th_3$, $\dots$, $\th_{m-1}$ is
also a $\ZZ $ basis for $M_o$.

Hence, since $\Tr$ is $\ZZ$ linear, for any $x\in M_e$, $x =  b_0 + b_2 \th^2 + \cdots + b_{m-2} \th^{m-2}$ and any $y\in M_e$,
$y = \sum c_0 + c_2 \th^2 + \cdots + c_{m-2} \th^{m-2}$,
the value of ${\rm Tr}(x,y)=\Tr(xy)$ is determined by the values of
$\Tr(\th^i\th^j)$, $i,j=0,2\dots,m-2$.  By a similar reasoning, writing $x$ and $y$ in terms of the basis $1$, $\th_2$, $\th_4$,
$\dots$, $\th_{m-2}$, the values of ${\rm Tr}(x,y)$ on $M_e$ is also determined by the values of $\Tr(\th_i\th_j)$,
$i,j=0,2\dots,m-2$. Lemmas \ref{trace_thet_powers} and \ref{trace_thet_subs} immediately give us the following result which
connects our Hankel matrix to the trace form:

\begin{corollary} \label{H_n is tracematrix}
The matrix $(\Tr(\theta^i\theta^j))$ ($i,j=0,2\dots,m-2$) equals $m$ times the Hankel matrix $H_{m/2}$
and %the matrix
$(\Tr(\theta_i\theta_j)$ ($i,j=0,2\dots,m-2$) equals $m$ times the matrix $D_{m/2}$ defined in Equation~(\ref{DnDef}).
\end{corollary}

Similarly, by considering the values of ${\rm Tr}(x,y)$ on the $\ZZ$ module $M_o$ in the two bases $\th$, $\th^3$, $\dots$,
$\th^{m-1}$ and $\th$, $\th_3$, $\dots,$ $\th_{m-1}$, we have the following:
\begin{corollary}\label{H_n_Sup_1_is_tracematrix}
The matrix $(\Tr(\theta^i\theta^j))$ ($i,j=1,3\dots,m-1$) equals $m$ times the Hankel matrix $H_{m/2}^{(1)}$
and %the matrix
$(\Tr(\theta_i\theta_j)$ ($i,j=1,3\dots,m-1$) equals $2m$ times the identity matrix.
\end{corollary}

Now observe that the matrix $(\theta^i\theta^j)$
($i,j=0,2\dots,m-2$) is just $\v_e \cdot {\v_e}^T$ (a product of
$n\times 1$ and $1\times n$ matrices), and that $(\theta_i\theta_j)$
($i,j=0,2\dots,m-2$) equals $\w_e \cdot {\w_e}^T$. Similarly,
$(\theta^i\theta^j)$ ($i,j=1,3\dots,m-1$) equals $\v_o \cdot
{\v_o}^T$ and  $(\theta_i\theta_j)$ ($i,j=1,3\dots,m-1$) equals
$\w_e \cdot {\w_e}^T$.

Equations~(\ref{vw_e_connection}) and (\ref{vw_o_connection}),
the $\ZZ$ bilinearity of $\Tr$, and Corollaries \ref{H_n is
tracematrix} and \ref{H_n_Sup_1_is_tracematrix} now give us the following (here, given a matrix $M$, $\Tr (M)$ stands for the
matrix whose entries are the traces of the entries of $M$):
\begin{eqnarray*} \label{EvenCasePowerOf2Factorization}
H_{m/2} &=& \frac{1}{m} (\Tr(\theta^i\theta^j))_{i,j=0,2\dots,m-2} = \frac{1}{m} \Tr (\v_e \cdot {\v_e}^T) \\
        &=& \frac{1}{m} B_{m/2} \Tr (\w_e \cdot {\w_e}^T) B_{m/2}^T \\
        &=& \frac{1}{m} B_{m/2}(\Tr(\theta_i\theta_j)_{i,j=0,2\dots,m-2}B_{m/2}^T \\
        &=& B_{m/2}D_{m/2}B_{m/2}^T .\\
H_{m/2}^{(1)} &=& \frac{1}{m} (\Tr(\theta^i\theta^j))_{i,j=1,3\dots,m-1} = \frac{1}{m} \Tr (\v_o \cdot {\v_o}^T) \\
        &=& \frac{1}{m} C_{m/2} \Tr (\w_o \cdot {\w_o}^T)C_{m/2}^T \\
        &=& \frac{1}{m} C_{m/2}(\Tr(\theta_i\theta_j)_{i,j=1,3\dots,m-1}C_{m/2}^T\\
         &=& 2C_{m/2}C_{m/2}^T .
\end{eqnarray*}

Let $G_n^{(k)}$ denote  the
Hankel matrices  formed from the sequence of
\textit{odd} central binomial coefficients ${2l+1\choose l}$,
$l=0,1,\dots$. 
Note that since ${2n \choose n} = 2 {2n-1 \choose n-1}$, we have the relation
 $H_n^{(k+1)} = 2 G_n^{(k)}$.

The discussions
above now immediately yield the following theorem:

\begin{theorem} \label{factorization_theorem} For any $n\ge 1$ and any $k\ge 1$, we have the
factorizations:

\renewcommand\theenumi{\roman{enumi}}
\begin{enumerate}
\item  \label{HnFactor} $H_n = B_n D_n B_n^T$, and
\item \label{Hn1Factor} $H_n^{(1)}=2 C_n C_n^T$.
\item  \label{GnFactor} $G_n :=G_n^{(0)}  = C_n C_n^T$.
\item  \label{HnKFactorWithB} $H_{n}^{(k)} = B_{n+k,n} D_n B_n^T$, where  $B_{n+k,n}$
denotes the lower left $n\times n$ block of $B_{n+k}$.
\item \label{GnKFactorWithB} $G_{n}^{(k)} = {1 \over 2} B_{n+k+1,n} D_n B_n^T$,
\item \label{HnKFactorWithC}$H_n^{(k)} = 2 C_{n+k-1,n} C_n^T$.
(Of course, when $k=1$, this is the same as Part (\ref{Hn1Factor})
above.)
\item \label{GnKFactorWithC}$G_n^{(k)} = C_{n+k,n} C_n^T$ .
\end {enumerate}


\end{theorem}

\begin{proof}
We pick an $N$ such that $2^N \ge 8n$, and work in the maximal real subfield $K$ of the $2^N$-th cyclotomic extension of $\QQ$.
The equations preceding the statement of the theorem yield the factorizations $H_{m/2} = B_{m/2} D_{m/2} B_{m/2}^T$ and
$H_{m/2}^{(1)}=2 C_{m/2} C_{m/2}^T$.  By the choice of $N$, we have $n \le m/2$. Note that $H_n$, $H_n^{(1)}$, $B_n$, $C_n$ and
$D_n$ are all just the upper left $n\times n$ blocks of the corresponding matrices $H_{m/2}$, $H_{m/2}^{(1)}$, $B_{m/2}$,
$C_{m/2}$ and $D_{m/2}$. Studying the upper left $n\times n$ blocks of the products $B_{m/2} D_{m/2} B_{m/2}^T$ and $C_{m/2}
C_{m/2}^T$, and noting the lower triangular nature of $B_{m/2}$ and $C_{m/2}$ and the diagonal nature of $D_{m/2}$, we get the
first two factorizations of the theorem.

We now substitute $n+k$ for $n$ throughout in the first two
factorizations, and observe that $H_n^{(k)}$ is the lower left
$n\times n$ block of $H_{n+k}$, as also the lower left $n\times n$
block of $H_{n+k-1}^{(1)}$.  Studying the products $B_{n+k} D_{n+k} B_{n+k}^T$ and
$C_{n+k-1} C_{n+k-1}^T$ yields the two factorizations in
(\ref{HnKFactorWithB}) and (\ref{HnKFactorWithC}) as well.

The factorizations in (\ref{GnFactor}), (\ref{GnKFactorWithB}),
and (\ref{GnKFactorWithC}) are a direct consequence of the
relation $H_n^{(k+1)} = 2 G_n^{(k)}$.
\end{proof}



Note that Factorization (\ref{GnFactor}) of $G_n$ was described in
\cite[Prop.\ 6]{Aig} by showing that odd binomial coefficients could be
regarded as \textsl{Catalan-like numbers}.  (In the notation of
\cite[Prop.\ 6]{Aig}, the odd binomial coefficients are $C_n^{(3,2)}$,
the matrix $\tilde{A_n}$ is our $G_{n+1}$ and the matrix $A_n$ is our
$C_{n+1}$.)

Taking the
determinants on both sides of Parts (\ref{HnFactor}) and (\ref{Hn1Factor}) of
Theorem \ref{factorization_theorem} above yeilds Corollary \ref{cor:zeroNone}. 

Let $c_n^{(k)}$ denote the determinant of $G_n^{(k)}$,
and note that the relation $H_n^{(k+1)} = 2 G_n^{(k)}$ shows that
$d_n^{(k+1)} = 2^n c_n^{(k)}$.  We therefore also have the following:

\begin{corollary} (See \cite[Eq.\ 1.5]{Tamm}, also \cite[Prop.\ 6]{Aig}.) $c_n^{(0)} = 1$.
\end{corollary}

\begin{remark} Parts (\ref{HnKFactorWithB}) or (\ref{HnKFactorWithC}) of Theorem
\ref{factorization_theorem} show that the computation of
$d_n^{(k)}$  for $k\ge 2$ can be accomplished by computing the determinants of  $B_{n+k,n}$ or
$C_{n+k,n}$. Since the primary goal of this note is to establish the connection between the $H_n^{(k)}$ and number theory we will not do this here, but we note that
these can be computed using, for instance, the very general techniques described in \cite[Thm.\ 26]{Kratt} (as can the determinants of the original $H_n^{(k)}$ themselves!).  Computing these determinants shows that for $k\ge 2$
\begin{equation} \label{dnk_formula}
d_n^{(k)} = 2^n \prod_{1 \leq i \leq j \leq k-1}{i+j-1+2n \over i+j-1}
\end{equation}
(recovering, for example, \cite[Eq.\ 1.5]{Tamm}).
\end{remark}


\section{Acknowledgements} The second author would like to thank B.M. \'{A}brego for introducing him to the literature on Hankel transforms, and to
Fr\'{e}d\'{e}rique Oggier for some preliminary discussions and computations.  The second author was supported in part by an NSF grant.



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\end{thebibliography}

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\noindent 2000 {\it Mathematics Subject Classification}: Primary
11B83; Secondary 11R04.

\noindent \emph{Keywords:} Hankel transform, Hankel matrix, 
central binomial coefficients, trace forms, ring of algebraic integers.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A055878}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received November 10 2008;
revised version received  November 25 2008.
Published in {\it Journal of Integer Sequences}, December 14 2008.

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\noindent
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