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\begin{document}

\centerline{\psfig{file=logo0018.ps,width=4.5in}}
\vskip 1cm
\centerline{\LARGE\bf Two Analogues of a Classical Sequence}
\vskip 1.5cm

\centerline{\large Ruedi Suter}

\bigskip

\centerline{Mathematikdepartement}
\centerline{ETH Z\"urich}
\centerline{8092 Z\"urich, Switzerland}
\medskip
\centerline{Email address: \href{mailto:suter@math.ethz.ch}{suter@math.ethz.ch}}
%\urladdr{http://www.math.ethz.ch/$\tilde{\phantom{a}}$suter}
%\thanks{}

%\date{ }

%\dedicatory{}

\vskip2.5cm
\centerline{\bf {Abstract}}
\textit{We compute exponential generating functions for the numbers of edges in the Hasse
diagrams for the $\mathsf B$- and $\mathsf D$-analogues of the partition lattices.}

\vspace*{+.1in}
{\footnotesize
1991 {\em Mathematics Subject Classification.}
Primary 05A15, 52B30; Secondary 05A18, 05B35, 06A07, 11B73, 11B83, 15A15, 20F55}

\section*{Introduction}
When one looks up the sequence $1$, $6$, $31$, $160$, $856$, $4802$, $28\,337$,
$175\,896$, \ldots in one of Sloane's integer sequence identifiers \cite{HIS,EIS,OIS},
one learns that these numbers are the numbers of driving-point impedances of an
$n$-terminal network for $n=2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $\ldots$ as
described in an old article by Riordan~\cite{Ri}.

In combinatorics there are two common ways of generalizing classical enumerative facts.
One such generalization arises by replacing the set $[n]=\{1,\dots,n\}$ by an
$n$-dimensional vector space over the finite field $\mathbb F_q$ to
get a $q$-analogue.
The other generalization or extension is by considering
``$\mathsf B$- and $\mathsf D$-analogues'' of an ``$\mathsf A$-case''. This terminology
stems from Lie theory.
(There is no ``$\mathsf C$-case'' here since it
coincides with the ``$\mathsf B$-case''.) Of course one may try to combine the two
approaches and supply $q$-$\mathsf B$- and $q$-$\mathsf D$-analogues.

In this note I shall describe $\mathsf B$- and $\mathsf D$-analogues of the numbers of
driving-point impedances of an $n$-terminal network. To assuage any possible
curiosity about how these sequences look, here are their first few terms:
\begin{align*}
\mathsf B\mbox{-analogue}\qquad
&\mbox{$1$, $8$, $58$, $432$, $3396$, $28\,384$, $252\,456$, $2\,385\,280$, $\ldots$}\\
\mathsf D\mbox{-analogue}\qquad
&\mbox{$0$, $4$, $31$, $240$, $1931$, $16\,396$, $147\,589$, $1\,408\,224$, $\ldots$}
\end{align*}

I should probably emphasize that I will only give mathematical arguments and will not attempt to provide a
physical realization of $\mathsf B$-~and $\mathsf D$-networks.

We start from certain classical hyperplane arrangements.
A hyperplane
arrangement defines a family of subspaces, namely those subspaces which can be written
as intersections of some of the hyperplanes in the arrangement. For each such subspace
we will choose a normal form that represents the subspace. Such a normal form consists
of an equivalence class of partial $\{\pm1\}$-partitions in the terminology of Dowling
\cite{Do}. Dowling actually constructed $G$-analogues of the partition lattices for any
finite group $G$. Using the concept of voltage graphs (or signed graphs for $|G|=2$) or
more generally biased graphs, Zaslavsky gave a far-reaching generalization of Dowling's work. It is
amusing to see that not only the network but also the mathematical treatment of
hyperplane arrangements carries a graph-theoretical flavour. Here we will stick to the
normal form and not translate things into the framework of graph theory, despite the
success this approach has had for example in \cite{BjSa}. In some sense the normal form
approach pursues a strategy opposite to that of Zaslavsky's graphs.

Whitney numbers and characteristic polynomials for hyperplane arrangements or more
generally for subspace arrangements, that is, the numbers of vertices with fixed rank
in the Hasse diagrams and the M\"obius functions, have been studied by many authors.
Apparently little attention has been paid so far to the numbers of edges in the Hasse
diagrams.

There is another point worth mentioning. It concerns a dichotomy among the $\mathsf A$-,
$\mathsf B$-, and $\mathsf D$-series. We will see that everything is very easy for the
first two series whereas for the $\mathsf D$-series we must work a little harder.
Such a dichotomy between the $\mathsf A$- and $\mathsf B$-series on the one hand and the
$\mathsf D$-series on the other also occurs in other contexts, e.\,g., in the problem
of counting reduced decompositions of the longest element in the
corresponding Coxeter groups (see \cite{St} for the initial paper).
In contrast, in the Lie theory one has a different dichotomy, namely, between the simply
laced (like $\mathsf A$ and $\mathsf D$) and the non-simply-laced (like $\mathsf B$
and $\mathsf C$) types.

Finally, an obvious generalization, which, however, we do not go into, concerns
hyperplane arrangements for the infinite families of unitary reflection groups.

\section*{Hyperplane arrangements and their intersection lattices}
Let $\mathcal A=\{H_1,\dots,H_N\}$ be a collection of subspaces of codimension~$1$ in
the vector space $\mathbb R^n$.
We let $L(\mathcal A)$ denote the poset of all
intersections $H_{i_1}\cap\dots\cap H_{i_r}$, ordered by reverse inclusion. This poset
$L(\mathcal A)$ is actually a geometric lattice. Its bottom element $\widehat0$ is the
intersection over the empty index set, i.\,e., $\mathbb R^n$. The atoms are the
hyperplanes $H_1,\dots,H_N$, and the top element $\widehat1$ is $H_1\cap\dots\cap H_N$.
For many further details the reader is referred to Cartier's Bourbaki talk \cite{Ca},
Bj\"orner's exposition \cite{Bj} for more general subspace arrangements, and
the monograph by Orlik and Terao~\cite{OT}
for a
thorough exposition of the theory.

A theorem due to Orlik and Solomon states that for a finite irreducible Coxeter group
$W$ with Coxeter arrangement $\mathcal A=\mathcal A(W)$ we have the equality
\begin{equation}\label{AH}
|\mathcal A^H|=|\mathcal A|+1-h
\end{equation}
where $H\in\mathcal A$ is any hyperplane of the arrangement, $h$ is the Coxeter
number of $W$, and $\mathcal A^H$ is the hyperplane arrangement in $H$ with
the hyperplanes $H\cap H'$ for $H'\in\mathcal A-\{H\}$. In other words,
(\ref{AH}) says that each atom in the intersection lattice $L(\mathcal A)$ is
covered by $|\mathcal A|+1-h$ elements. One may wonder what can be said about the
number of elements that cover an arbitrary element in $L(\mathcal A)$.

The intersection lattices that concern us here come from the following hyperplanes
in $\mathbb R^n$.
\renewcommand{\arraystretch}{1.4}
$$\begin{array}{|l|l|}\hline
\multicolumn{1}{|c|}{\mbox{type of $\mathcal A$}}&
\multicolumn{1}{c|}{\mbox{elements of $\mathcal A$}}\\\hline
\quad(\mathsf A_1)^n&\{x_a=0\}_{a=1,\dots,n}\\
\quad\mathsf A_{n-1}&\phantom{\{x_a=0\}_{a=1,\dots,n},\ }\,
\{x_b=x_c\}_{1\leqslant b<c\leqslant n}\\
\quad\mathsf B_n&\{x_a=0\}_{a=1,\dots,n},\ \{x_b=x_c\}_{1\leqslant b<c\leqslant n},\
\{x_b=-x_c\}_{1\leqslant b<c\leqslant n}\\
\quad\mathsf D_n&\phantom{\{x_a=0\}_{a=1,\dots,n},\ }\,
\{x_b=x_c\}_{1\leqslant b<c\leqslant n},\
\{x_b=-x_c\}_{1\leqslant b<c\leqslant n}\\\hline
\end{array}$$
\renewcommand{\arraystretch}{1}%
Note that $\bigcap\limits_{H\in\mathcal A}H$ is the line $x_1=\dots=x_n$ for type
$\mathsf A_{n-1}$ (so the rank is $n-1$ in this case if $n>0$) whereas for the other
types the hyperplanes only meet in the zero vector. We agree to let $\mathsf A_{-1}$
denote the empty hyperplane arrangement in $0$. So the intersection lattices for
$\mathsf A_{-1}$ and $\mathsf A_0$ are isomorphic. Also there is a slight abuse of
notation for type $\mathsf A_1$ because it can be considered as $(\mathsf A_1)^1$ or as
$\mathsf A_{2-1}$. But this will not cause trouble.

For each subspace $E\in L(\mathcal A)$ we define the subset $B_E\subseteq[n]=
\{1,\dots,n\}$ by the property that
$$C_E:=\bigcap_{a\in[n]-B_E}\{x_a=0\}$$
is the smallest intersection of coordinate hyperplanes that contains $E$. For
instance if $\mathcal A$ is of type $\mathsf A_{n-1}$, we have $B_E=[n]$ for all
$E\in L(\mathcal A)$. For the hyperplane
$E=\{x_1\!=\!x_2\}\cap\{x_2\!=\!x_3\}\cap\{x_1\!=\!-x_3\}\cap\{x_4\!=\!x_7\}\cap
\{x_5\!=\!x_8\}\cap\{x_8\!=\!0\}\subseteq\mathbb R^8$
we get $B_E=\{4,6,7\}$.

Regarded as a subspace of $C_E$, $E$ is described
by a partition of $B_E$
together with
a function $\zeta:B_E\to\{\pm1\}$. If $\{B_1,\dots,B_k\}$ is a partition of $B_E$
into $k$ blocks, then $E$ is the $k$-dimensional subspace
$$E=\bigl\{(x_1,\dots,x_n)\in C_E\bigm|{b,c}\in B_j\mbox{ for some $j$}
\Longrightarrow
\zeta(b)\,x_b=\zeta(c)\,x_c\bigr\}.$$
Clearly, the correspondence between $E$ and
$\bigl(\{B_1,\dots,B_k\},\zeta\bigr)$ is $1$ to $2^k$ because for each block there is
a choice of sign.

This correspondence gives us a convenient notation for the subspaces in $L(\mathcal
A)$. We write down a partition of some $B\subseteq[n]$ and decorate the numbers $a\in B$
with $\zeta(a)=-1$ with an overbar. Having the possibility of choosing an overall sign
for each block, we agree that the smallest number in each block does not have an
overbar. As an example take
the Coxeter arrangement of type $\mathsf B_3$. There are $24$ subspaces to be
considered. Their representations as ``signed permutations'' are shown in the vertices
(boxes) of the following Hasse diagram.

\begin{center}
\setlength{\unitlength}{1.25mm}
\begin{picture}(90,47)
\put(42,42){\framebox(6,4){}}
\put(0,28){\framebox(6,4){$3$}}
\put(7,28){\framebox(6,4){$123$}}
\put(14,28){\framebox(6,4){$1\overline23$}}
\put(21,28){\framebox(6,4){$13$}}
\put(28,28){\framebox(6,4){$12$}}
\put(35,28){\framebox(6,4){$1\overline2$}}
\put(42,28){\framebox(6,4){$2$}}
\put(49,28){\framebox(6,4){$2\overline3$}}
\put(56,28){\framebox(6,4){$23$}}
\put(63,28){\framebox(6,4){$1\overline3$}}
\put(70,28){\framebox(6,4){$12\overline3$}}
\put(77,28){\framebox(6,4){$1\overline2\overline3$}}
\put(84,28){\framebox(6,4){$1$}}
\put(6,14){\framebox(6,4){$2|3$}}
\put(15,14){\framebox(6,4){$12|3$}}
\put(24,14){\framebox(6,4){$1\overline2|3$}}
\put(33,14){\framebox(6,4){$13|2$}}
\put(42,14){\framebox(6,4){$1|3$}}
\put(51,14){\framebox(6,4){$1\overline3|2$}}
\put(60,14){\framebox(6,4){$1|2\overline3$}}
\put(69,14){\framebox(6,4){$1|23$}}
\put(78,14){\framebox(6,4){$1|2$}}
\put(42,0){\framebox(6.8,4){$1|2|3$}}
\put(45,4){\line(-36,10){36}}\put(45,4){\line(-27,10){27}}\put(45,4){\line(-18,10){18}}
\put(45,4){\line(-9,10){9}}\put(45,4){\line(0,10){10}}\put(45,4){\line(9,10){9}}
\put(45,4){\line(18,10){18}}\put(45,4){\line(27,10){27}}\put(45,4){\line(36,10){36}}
\put(9,18){\line(-6,10){6}}\put(9,18){\line(36,10){36}}
\put(9,18){\line(43,10){43}}\put(9,18){\line(50,10){50}}
\put(18,18){\line(-15,10){15}}\put(18,18){\line(-8,10){8}}
\put(18,18){\line(13,10){13}}\put(18,18){\line(55,10){55}}
\put(27,18){\line(-24,10){24}}\put(27,18){\line(-10,10){10}}
\put(27,18){\line(11,10){11}}\put(27,18){\line(53,10){53}}
\put(36,18){\line(-26,10){26}}\put(36,18){\line(-19,10){19}}
\put(36,18){\line(-12,10){12}}\put(36,18){\line(9,10){9}}
\put(45,18){\line(-42,10){42}}\put(45,18){\line(-21,10){21}}
\put(45,18){\line(21,10){21}}\put(45,18){\line(42,10){42}}
\put(54,18){\line(-9,10){9}}\put(54,18){\line(12,10){12}}
\put(54,18){\line(19,10){19}}\put(54,18){\line(26,10){26}}
\put(63,18){\line(-46,10){46}}\put(63,18){\line(-11,10){11}}
\put(63,18){\line(10,10){10}}\put(63,18){\line(24,10){24}}
\put(72,18){\line(-62,10){62}}\put(72,18){\line(-13,10){13}}
\put(72,18){\line(8,10){8}}\put(72,18){\line(15,10){15}}
\put(81,18){\line(-50,10){50}}\put(81,18){\line(-43,10){43}}
\put(81,18){\line(-36,10){36}}\put(81,18){\line(6,10){6}}
\put(45,42){\line(-42,-10){42}}\put(45,42){\line(-35,-10){35}}
\put(45,42){\line(-28,-10){28}}\put(45,42){\line(-21,-10){21}}
\put(45,42){\line(-14,-10){14}}\put(45,42){\line(-7,-10){7}}
\put(45,42){\line(42,-10){42}}\put(45,42){\line(35,-10){35}}
\put(45,42){\line(28,-10){28}}\put(45,42){\line(21,-10){21}}
\put(45,42){\line(14,-10){14}}\put(45,42){\line(7,-10){7}}
\put(45,42){\line(0,-10){10}}
\end{picture}\par
{\small\textsc{Figure} 1. Hasse diagram of the $\mathsf B_3$ lattice}
\end{center}

\noindent
For instance $3$ stands for the line $x_1=x_2=0$, $1\overline23$ is for $x_1=-x_2=x_3$,
$1|2\overline3$ denotes the plane $x_2=-x_3$, $1|2$ means $x_3=0$ etc.

\section*{Vertices in the Hasse diagrams}
\begin{lemma}\label{conditions}
For a partition $\{B_1,\dots,B_k\}$ of a subset $B\subseteq[n]$ and a function
$\zeta:B\to\{\pm1\}$ the $k$-dimensional subspace
$$\left\{(x_1,\dots,x_n)\in\mathbb R^n\,\left|\
\begin{array}{@{}l@{}}
a\in[n]-B\,\Longrightarrow\,x_a=0\\
{b,c}\in B_j\mbox{\textup{ for some $j$}}\,\Longrightarrow
\,\zeta(b)\,x_b=\zeta(c)\,x_c\end{array}\right.\right\}$$
belongs to $L(\mathcal A)$ according to the following table.
\renewcommand{\arraystretch}{1.4}
$$\begin{array}{|l|l|}\hline
\multicolumn{1}{|c|}{\mbox{\textup{type of $\mathcal A$}}}&
\multicolumn{1}{c|}{\mbox{\textup{condition}}}\\\hline
\quad(\mathsf A_1)^n&|B|=k,\,\zeta=1\\
\quad\mathsf A_{n-1}&B=[n],\,\zeta=1\\
\quad\mathsf B_n&\mbox{---}\\
\quad\mathsf D_n&\bigl|[n]-B\bigr|\neq1\\\hline
\end{array}$$
\renewcommand{\arraystretch}{1}%
\end{lemma}
\begin{proof}
The conditions in the table above should be clear. For the types $(\mathsf A_1)^n$ and
$\mathsf A_{n-1}$ we put $\zeta=1$ for simplicity (literally, $\zeta$ must only be
constant on each block $B_j$). The condition for $\mathsf D_n$
simply takes into account that the hyperplanes $x_a=0$ do not belong to $L(\mathcal
A)$. But for instance $x_1=\dots=x_r=0$ for $r\geqslant2$ can be written as $x_1=-x_2$,
$x_1=\dots=x_r$ and hence this subspace is an element of~$L(\mathcal A)$.
\end{proof}

For integers ${n,k}\geqslant0$ and $b>0$ let $S_b(n,k)$ denote the number of partitions
of $[n]$ into $k$ blocks each containing at least $b$ elements. So $S_1(n,k)=S(n,k)$ is
a Stirling number of the second kind. Besides $b=1$ we shall only need the case
where $b=2$, which one knows from P\'olya-Szeg\H{o} \cite[Part I, Chap.~4, \S\,3;
Part VIII, Chap.~1, No.~22.3]{PS}. Nevertheless we state the following more general
proposition.

\begin{proposition}\label{Sb}
For every integer $b>0$ the generating function for the numbers $S_b(n,k)$ of partitions
of $[n]$ into $k$ blocks of length at least $b$ is
$$\sum_{{n,k}\geqslant0}S_b(n,k)\frac{x^n}{n!}\,y^k=\exp\left(y\cdot\Bigl(e^x-1-x-
\frac{x^2}{2!}-\dots-\frac{x^{b-1}}{(b-1)!}\Bigr)\right).$$
\end{proposition}
\begin{proof}
For $k\geqslant1$ we have the recurrence relation
\begin{equation}\label{rec}
S_b(n,k)=k\,S_b(n-1,k)+\tbinom{n-1}{b-1}\,S_b(n-b,k-1).
\end{equation}
In fact, to obtain a partition of $[n]$ into $k$ blocks of lengths at least $b$, we
can either take a partition of $[n-1]$ into $k$ blocks of lengths at least $b$ and
append the element $n$ to any one of the $k$ blocks, or we can take $b-1$ elements from
$[n-1]$ which together with $n$ constitute a block with $b$ elements and partition the
remaining $n-b$ elements into $k-1$ blocks of lengths at least~$b$.\par
To prove the proposition we must show that for every integer $k\geqslant0$
\begin{equation}\label{ind}
f_k(x):=\sum_{n\geqslant0}S_b(n,k)\frac{x^n}{n!}=\frac{1}{k!}\Bigl(e^x-1-x-
\frac{x^2}{2!}-\dots-\frac{x^{b-1}}{(b-1)!}\Bigr)^k.
\end{equation}
This follows by induction on $k$. The case $k=0$ is clear: 
$S_b(n,0)=\delta_{n,0}$. For $k\geqslant1$ we get a
differential equation for $f_k(x)$, namely
\begin{align*}
f_k'(x)&=\sum_nS_b(n,k)\frac{x^{n-1}}{(n-1)!}\\
&\stackrel{\Nbx{(\ref{rec})}}{=}
\sum_nk\,S_b(n-1,k)\frac{x^{n-1}}{(n-1)!}+\sum_n\tbinom{n-1}{b-1}\,S_b(n-b,k-1)
\frac{x^{n-1}}{(n-1)!}\\
&=k\,f_k(x)+\frac{x^{b-1}}{(b-1)!}\,f_{k-1}(x)\\
&=k\,f_k(x)+\frac{x^{b-1}}{(b-1)!}\frac{1}{(k-1)!}\Bigl(e^x-1-x-
\frac{x^2}{2!}-\dots-\frac{x^{b-1}}{(b-1)!}\Bigr)^{k-1}
\end{align*}
whose unique solution satisfying $f_k(0)=0$ is in fact given by the right hand side in
equation~(\ref{ind}).
\end{proof}

The lattices $L(\mathcal A)$ are graded posets with rank function the codimension. The
$r$th Whitney number of the second kind of a graded poset is by definition the number
of elements of rank $r$.
We begin by making
the Whitney numbers quite explicit.
We fix one of our hyperplane arrangements $\mathcal A$ in $\mathbb R^n$ and let
$W(n,r)$ be the $r$th Whitney number (of the second kind) of the intersection lattice
$L(\mathcal A)$. The Whitney numbers $W(n,n-k)$ when written in an array can be seen as
a generalization of Pascal's triangle. In fact, Pascal's triangle arises for the
Boolean lattices of type~$(\mathsf A_1)^n$.

\vspace{.1in}

Let us digress for a moment to consider such generalized Pascal triangles or arrays.
The (upper left) corner in the arrays that follow carry the Whitney number $W(0,0)$,
and the entries $(p,q)$ for the other Whitney numbers $W(p,q)$ are in accordance with
the following diagram.

{\footnotesize
$$\begin{CD}
W(n,n-k)@>>>W(n+1,n-k+1)\\
@VVV\\W(n+1,n-k)
\end{CD}$$}

\begin{itemize}
\item \textbf{Pascal arrangements} = Coxeter arrangements of type $(\mathsf A_1)^n$.\\
$W(n,n-k)=\binom nk$.\par
{\footnotesize
\newlength{\twofigures}\settowidth{\twofigures}{$00$}
$$\begin{CD}
1@>>>\makebox[\twofigures][c]{$1$}@>>>\phantom{0}1\phantom{0}@>>>
\makebox[2\twofigures][c]{$1$}@>>>\cdots\\
@VVV@VVV@VVV@VVV\\
1@>>>2@>>>3@>>>4@>>>\cdots\\
@VVV@VVV@VVV@VVV\\
1@>>>3@>>>6@>>>10@>>>\cdots\\
@VVV@VVV@VVV@VVV\\
1@>>>4@>>>10@>>>20@>>>\cdots\\
@VVV@VVV@VVV@VVV\\
\vdots&&\vdots&&\vdots&&\vdots
\end{CD}$$}
%\newpage
\item \textbf{Stirling arrangements} = Coxeter arrangements of type $\mathsf
A_{n-1}$.\\
$W(n,n-k)=S(n,k)$.
For the $\mathsf A_{n-1}$ lattices the analogue of the equation
$\tbinom
nk=\tbinom{n-1}{k}+\tbinom{n-1}{k-1}$ reads $S(n,k)=k\,S(n-1,k)+S(n-1,k-1)$, the case
$b=1$ of (\ref{rec}).\\[-3mm]
{\footnotesize
$$\begin{CD}
1@>{{}\cdot0}>>\makebox[\twofigures][c]{$0$}@>>>\phantom{0}0\phantom{0}
@>>>\makebox[2\twofigures][c]{$0$}@>>>\cdots\\
@VVV@VVV@VVV@VVV\\
1@>>>1@>>>1@>>>1@>>>\cdots\\
@VVV@VVV@VVV@VVV\\
1@>{{}\cdot2}>>3@>{{}\cdot2}>>7@>{{}\cdot2}>>15@>{{}\cdot2}>>\cdots\\
@VVV@VVV@VVV@VVV\\
1@>{{}\cdot3}>>6@>{{}\cdot3}>>25@>{{}\cdot3}>>90@>{{}\cdot3}>>\cdots\\
@VVV@VVV@VVV@VVV\\
1@>{{}\cdot4}>>10@>{{}\cdot4}>>65@>{{}\cdot4}>>350@>{{}\cdot4}>>\cdots\\
@VVV@VVV@VVV@VVV\\
\vdots&&\vdots&&\vdots&&\vdots
\end{CD}$$}
\pagebreak
\item \textbf{$2$-Dowling arrangements} = Coxeter arrangements of type $\mathsf B_n$.\\
$W(n,n-k)=T(n,k)$.
For the $\mathsf B_n$ lattices the Whitney numbers
satisfy the relation $T(n,k)=(2k+1)\,T(n-1,k)+T(n-1,k-1)$ (see
Corollary~\ref{corT}).
{\footnotesize
$$\begin{CD}
1@>>>\makebox[\twofigures][c]{$1$}@>>>\phantom{0}1\phantom{0}@>>>
\makebox[2\twofigures][c]{$1$}@>>>\cdots\\
@VVV@VVV@VVV@VVV\\
1@>{{}\cdot3}>>4@>{{}\cdot3}>>13@>{{}\cdot3}>>40@>{{}\cdot3}>>\cdots\\
@VVV@VVV@VVV@VVV\\
1@>{{}\cdot5}>>9@>{{}\cdot5}>>58@>{{}\cdot5}>>330@>{{}\cdot5}>>\cdots\\
@VVV@VVV@VVV@VVV\\
1@>{{}\cdot7}>>16@>{{}\cdot7}>>170@>{{}\cdot7}>>1520@>{{}\cdot7}>>\cdots\\
@VVV@VVV@VVV@VVV\\
\vdots&&\vdots&&\vdots&&\vdots
\end{CD}$$}
\end{itemize}

Continuing in the obvious way, one gets Whitney numbers of Dowling lattices
corresponding to the complete monomial groups $(\mathbb Z/m\mathbb Z)\wr\mathfrak S_n$,
the wreath product of the symmetric group of degree $n$ acting on
$(\mathbb Z/m\mathbb Z)^n$.
This is straightforward, and calculations can be found in \cite{Be1,Be2}.
For the $\mathsf D_n$ lattices the situation is more subtle.
The following table
suggests why this is so.
\renewcommand{\arraystretch}{1.4}
$$\begin{array}{|c|l|}\hline
\mbox{type}&\multicolumn{1}{c|}{\mbox{exponents}}\\\hline
(\mathsf A_1)^n&1,1,\dots,1\\
\mathsf A_n&1,2,\dots,n\\
\mathsf B_n&1,3,\dots,2n-1\\
\mathsf D_n&1,3,\dots,2n-3,n-1\\\hline
\end{array}$$
\renewcommand{\arraystretch}{1}%
The maverick exponent $n-1$ for type $\mathsf D_n$ reveals the fact that the
determinant of a $2n$\/$\times$\/$2n$ skew-symmetric matrix is the square of a
polynomial in the matrix entries.

This ends our digression. Also from now on we will neglect the nearly trivial
case of type $(\mathsf A_1)^n$.

\begin{proposition}\label{Whitney}
The Whitney numbers $W(n,n-k)$ are given by the following formulae.
\renewcommand{\arraystretch}{1.6}
$$\begin{array}{|l|l|}\hline
\multicolumn{1}{|c|}{\mbox{\textup{type}}}&\multicolumn{1}{c|}{W(n,n-k)}\\\hline
\mathsf A_{n-1}&S(n,k)\phantom{\sum\limits_{j=k}^n}\\
\mathsf B_n&\sum\limits_{j=k}^n2^{j-k}\tbinom nj\,S(j,k)\\
\mathsf D_n&\sum\limits_{j=k}^n2^{j-k}\tbinom nj\,S(j,k)-2^{n-1-k}n\,S(n-1,k)\\\hline
\end{array}$$
\renewcommand{\arraystretch}{1}%
\end{proposition}
\begin{proof}
The proof follows by elementary combinatorial reasoning from the table in
Lemma~\ref{conditions}. (Recall that $S(n,k)$ is a Stirling number of the second kind.)
\end{proof}
The table in Proposition~\ref{Whitney}
can also be found in the last corollary of \cite{Za}.

\begin{theorem}\label{vertices}
The generating functions for the Whitney numbers are as given in the following table.
\renewcommand{\arraystretch}{1.6}
$$\begin{array}{|c|l|}\hline
\mbox{\textup{type}}&\multicolumn{1}{c|}{\phantom{\biggl(}\sum\limits_{{n,k}\geqslant0}
W(n,n-k)\dfrac{x^n\mathstrut}{n!}\,y^k\phantom{\biggr(}}\\\hline
\mathsf A&\exp\bigl(y\cdot(e^x-1)\bigr)\phantom{\Bigl(}\\
\mathsf B&e^x\exp\Bigl(\dfrac y2\cdot\bigl(e^{2x}-1\bigr)\Bigr)\\
\mathsf D&(e^x-x)\exp\Bigl(\dfrac{y}{2\mathstrut}\cdot\bigl(e^{2x}-1\bigr)
\Bigr)\\\hline
\end{array}$$
\renewcommand{\arraystretch}{1}%
\end{theorem}
\begin{proof}
For type $\mathsf A$ this is Proposition~\ref{Sb} with $b=1$. For type $\mathsf B$ the
coefficients
$$a_n(y)=\sum_{k\geqslant0}\sum_{j=k}^n2^{j-k}\tbinom nj\,S(j,k)\,y^k\in\mathbb Z[y]$$
are the binomial transforms of
$$b_j(y)=\sum_{k\geqslant0}2^{j-k}S(j,k)\,y^k\in\mathbb Z[y].$$
Hence
\begin{align*}
\sum_{n\geqslant0}a_n(y)\frac{x^n}{n!}&=e^x\sum_{j\geqslant0}b_j(y)\frac{x^j}{j!}\\
&=e^x\sum_{j,k}S(j,k)\frac{(2x)^j}{j!}\Bigl(\frac y2\Bigr)^k=
e^x\exp\Bigl(\frac y2\cdot\bigl(e^{2x}-1\bigr)\Bigr).
\end{align*}
Finally, for type $\mathsf D$ we need to subtract
\begin{align*}
\sum_{n,k}2^{n-1-k}n\,S(n-1,k)\frac{x^n}{n!}\,y^k&=x\sum_{n,k}
S(n-1,k)\frac{(2x)^{n-1}}{(n-1)!}\Bigl(\frac y2\Bigr)^k\\
&=x\exp\Bigl(\frac y2\cdot\bigl(e^{2x}-1\bigr)\Bigr)
\end{align*}
from the generating function for type $\mathsf B$.
\end{proof}

Setting $y=1$ in Theorem~\ref{vertices} we get the exponential generating
function for the numbers of vertices in the Hasse diagrams. The coefficients in this
exponential generating function are the Bell numbers for type $\mathsf A$ and the
Dowling numbers for type $\mathsf B$.
For type $\mathsf D$ these numbers are apparently unnamed.

\begin{corollary}\label{corT}
The Whitney numbers $T(n,k)=W(n,n-k)$ for the
$2$-Dowling arrangements satisfy the recurrence relation
$$T(n,k)=(2k+1)\,T(n-1,k)+T(n-1,k-1).$$
\end{corollary}
\begin{proof}
$\Bigl(\dfrac{\partial}{\partial x}-2\,y\,\dfrac{\partial}{\partial y}-1-y\Bigr)\,
e^x\exp\Bigl(\dfrac y2\cdot\bigl(e^{2x}-1\bigr)\Bigr)=0$.
\end{proof}

\section*{Edges in the Hasse diagrams}
There are two obvious ways to count edges in a Hasse diagram. Namely, go through
all vertices and add up the numbers of edges that go upwards, or, dually, that go
downwards. As the result of Orlik and Solomon for the elements of rank $1$ suggests,
it is easier here
to count edges corresponding to vertices that cover a given vertex 
than to count those
edges corresponding to vertices that are covered by a given vertex.

An edge in the Hasse diagram for $L(\mathcal A)$ emanating in an upward direction from
$E\in L(\mathcal A)$ corresponds to a subspace $E'\in L(\mathcal A)$ of codimension
$1$ in $E$.
We shall count how many such subspaces are contained in $E$.

Schematically, we have
$$\bigl(\{B_1,\dots,B_k\},\zeta\bigr)\rightsquigarrow
\bigl(\{B_1',\dots,B_{k-1}'\},\zeta'\bigr)$$
with
$$E=\left\{(x_1,\dots,x_n)\in\mathbb R^n\,\left|\
\begin{array}{@{}l@{}}
a\in[n]-B\,\Longrightarrow\,x_a=0\\
{b,c}\in B_j\mbox{ for some $j$}\,\Longrightarrow
\,\zeta(b)\,x_b=\zeta(c)\,x_c\end{array}\right.\right\}$$
where $B=B_1\cup\dots\cup B_k$, and $E'$ is obtained by imposing a further equation,
$$E'=\left\{(x_1,\dots,x_n)\in\mathbb R^n\,\left|\
\begin{array}{@{}l@{}}
a\in[n]-B'\,\Longrightarrow\,x_a=0\\
{b,c}\in B_j'\mbox{ for some $j$}\,\Longrightarrow
\,\zeta'(b)\,x_b=\zeta'(c)\,x_c\end{array}\right.\right\}$$
where $B'=B_1'\cup\dots\cup B_{k-1}'$.

Imposing a further equation may have two different types of incarnations in terms of
normal forms. (As usual, $\widehat{B_k}$ means that $B_k$ is omitted.)

\begin{itemize}
\item \textbf{Fusing two blocks}.
Choose $1\leqslant i<j\leqslant k$, $\varepsilon\in\{\pm1\}$.
$$\left|\begin{array}{l}
\makebox[10cm][l]{$\{B_1',\dots,B_{k-1}'\}:=\{B_1,\dots,\widehat{B_i},\dots,
\widehat{B_j},\dots,B_k,B_i\cup B_j\}$}\\[1mm]
\zeta'(a):=\begin{cases}
\zeta(a)&\mbox{if $a\in B-B_j$}\\
\varepsilon\cdot\zeta(a)&\mbox{if $a\in B_j$}
\end{cases}
\end{array}\right.$$
\item \textbf{Dropping one block}.
Choose $1\leqslant i\leqslant k$.
$$\left|\begin{array}{l}
\makebox[10cm][l]{$\{B_1',\dots,B_{k-1}'\}:=\{B_1,\dots,
\widehat{B_i},\dots,B_k\}$}\\[1mm]
\zeta'(a):=\zeta(a)\mbox{ for all $a\in B-B_i$}
\end{array}\right.$$
\end{itemize}

\begin{lemma} For 
$$\bigl(\{B_1,\dots,B_k\},\zeta\bigr)\rightsquigarrow
\bigl(\{B_1',\dots,B_{k-1}'\},\zeta'\bigr)$$
with fixed $\bigl(\{B_1,\dots,B_k\},\zeta\bigr)$ there are the following numbers
of possibilities for fusing two blocks or dropping one block.
\renewcommand{\arraystretch}{1.7}
$$\begin{array}{|l|l|l|l|}\hline
\multicolumn{1}{|c|}{\mbox{\textup{type}}}&
\multicolumn{1}{c|}{\mbox{\textup{conditions}}}
&\multicolumn{1}{c|}{\mbox{\textup{fusing}}}
&\multicolumn{1}{c|}{\mbox{\textup{dropping}}}\\\hline
\mathsf A_{n-1}&\begin{array}{@{}l@{}}B=[n],\,\zeta=1\\[-0.5em]B'=[n],\,\zeta'=1
\end{array}&\binom k2&\ \ 0\\[2mm]
\mathsf B_n&\mbox{---}&\binom k2\cdot2&\ \ k\\[2mm]
\mathsf D_n&\begin{array}{@{}l@{}}\bigl|[n]-B\bigr|\neq1\\[-0.4em]
\bigl|[n]-B'\bigr|\neq1\end{array}&\binom k2\cdot2&
\begin{cases}k&\mbox{\textup{if} $B\neq[n]$}\\
\#\bigl\{i\bigm||B_i|\geqslant2\bigr\}&\mbox{\textup{if} $B=[n]$}\end{cases}\\\hline
\end{array}$$
\renewcommand{\arraystretch}{1}%
\end{lemma}

The total number of subspaces of dimension $k-1$ in $L(\mathcal A)$ lying in some
fixed subspace $E\in L(\mathcal A)$ of dimension $k$ is thus $\binom k2$ for type
$\mathsf A$ and $k^2$ for type $\mathsf B$, while for type $\mathsf D$ this number is
not specified by the dimension alone and can vary between $k^2-k$ and $k^2$.

The following diagrams give a rough idea of how the Hasse diagrams look for the
first few lattices in the $\mathsf D$-series. The first diagram abbreviates the
relevant piece of information for the Hasse diagram of the $\mathsf B_3$ lattice, whose
full form was given earlier. For instance the Hasse diagram for $\mathsf D_4$ contains
$$1\cdot12+12\cdot7+16\cdot3+18\cdot4+24\cdot1+1\cdot0=240$$
edges.\\

\setlength{\unitlength}{0.4mm}
\begin{center}
\begin{tabular}[t]{c}\\
\FFL{\mathsf B_3}\\
\setcounter{numx}{0}\EdgesEven{0}{1}\\[8mm]
\EdgesOdd{1}{13}\\[8mm]
\setcounter{numx}{2}\EdgesEven{4}{9}\\[8mm]
\setcounter{numx}{4}\EdgesOdd{9}{1}
\end{tabular}
\qquad\quad
\begin{tabular}[t]{c}\\
\FFL{\ \mathsf D_2=(\mathsf A_1)^2}\\
\setcounter{numx}{0}\EdgesEven{0}{1}\\[8mm]
\EdgesOdd{1}{2}\\[8mm]
\setcounter{numx}{1}\EdgesEven{2}{1}
\end{tabular}
\qquad
\begin{tabular}[t]{c}\\
\FFL{\mathsf D_3=\mathsf A_3}\\
\setcounter{numx}{0}\EdgesEven{0}{1}\\[8mm]
\EdgesOdd{1}{7}\\[8mm]
\setcounter{numx}{1}\EdgesOdd{3}{6}\\[8mm]
\setcounter{numx}{3}\EdgesEven{6}{1}
\end{tabular}
\qquad
\begin{tabular}[t]{c}\\
\FFL{\mathsf D_4}\\
\setcounter{numx}{0}\EdgesEven{0}{1}\\[8mm]
\EdgesOdd{1}{24}\\[8mm]
\setcounter{numx}{1}\EdgesOdd{3}{16}\setcounter{numx}{2}\EdgesEven{4}{18}\\[8mm]
\setcounter{numx}{3}\EdgesOdd{7}{12}\\[8mm]
\setcounter{numx}{6}\EdgesEven{12}{1}
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}[t]{c}\\
\FFL{\mathsf D_5}\\
\setcounter{numx}{0}\EdgesEven{0}{1}\\[8mm]
\EdgesOdd{1}{81}\\[8mm]
\setcounter{numx}{1}\EdgesOdd{3}{40}\setcounter{numx}{2}\EdgesEven{4}{150}\\[8mm]
\setcounter{numx}{3}\EdgesOdd{7}{40}\setcounter{numx}{4}\EdgesEven{8}{60}
\EdgesOdd{9}{10}\\[8mm]
\setcounter{numx}{6}\EdgesOdd{13}{20}\\[8mm]
\setcounter{numx}{10}\EdgesEven{20}{1}
\end{tabular}
\qquad
\begin{tabular}[t]{c}\\
\FFL{\mathsf D_6}\\
\setcounter{numx}{0}\EdgesEven{0}{1}\\[8mm]
\EdgesOdd{1}{268}\\[8mm]
\setcounter{numx}{1}\EdgesOdd{3}{96}\setcounter{numx}{2}\EdgesEven{4}{955}\\[8mm]
\setcounter{numx}{3}\EdgesOdd{7}{120}\setcounter{numx}{4}\EdgesEven{8}{480}
\EdgesOdd{9}{320}\\[8mm]
\setcounter{numx}{6}\EdgesOdd{13}{80}\setcounter{numx}{7}\EdgesEven{14}{180}
\setcounter{numx}{8}\EdgesEven{16}{15}\\[8mm]
\setcounter{numx}{10}\EdgesOdd{21}{30}\\[8mm]
\setcounter{numx}{15}\EdgesEven{30}{1}\\[2mm]
\end{tabular}\par
{\small\textsc{Figure} 2. Abbreviated Hasse diagrams}
\end{center}


\begin{theorem}
The exponential generating functions for the numbers of edges in the Hasse diagrams for
types $(\mathsf A_{n-1})_{n\geqslant0}$, $(\mathsf B_n)_{n\geqslant0}$, and
$(\mathsf D_n)_{n\geqslant0}$ are as given in the following table.
\renewcommand{\arraystretch}{1.7}
$$\begin{array}{|c|l|}\hline
\mbox{\textup{type}}&\multicolumn{1}{c|}{\textup{exponential generating
function}}\\\hline
\mathsf A&\dfrac{1\mathstrut}{2}(e^x-1)^2\exp(e^x-1)\\[1em]
\mathsf B&e^x\dfrac14\bigl(e^{4x}-1\bigr)
\exp\Bigl(\dfrac12\bigl(e^{2x}-1\bigr)\Bigr)\\[1em]
\mathsf D&(e^x-1-x)\dfrac14\bigl(e^{4x}-1\bigr)
\exp\Bigl(\dfrac12\bigl(e^{2x}-1\bigr)\Bigr)\\[0.3em]
&\quad{}+e^x\dfrac{1}{4\mathstrut}\,\bigl(e^{4x}-1-4x\bigr)
\exp\Bigl(\dfrac12\bigl(e^{2x}-1-2x\bigr)\Bigr)\\\hline
\end{array}$$
\renewcommand{\arraystretch}{1}%
\end{theorem}
\begin{proof}
For type $\mathsf A_{n-1}$ there are $S(n,k)$ $k$-dimensional subspaces each
containing $\binom k2$ subspaces in $L(\mathcal A)$ of codimension~$1$. Thus we get the
generating function
\begin{align*}
\makebox[1cm][l]{$\displaystyle\sum_{n,k}S(n,k)\tbinom k2\frac{x^n}{n!}=
\frac12\frac{\partial^2}{\partial y^2}\sum_{n,k}S(n,k)
\frac{x^n}{n!}\,y^k\Bigr|_{y=1}$}&\\
&=\frac12\frac{\partial^2}{\partial y^2}\exp\bigl(y\cdot(e^x-1)\bigr)\Bigr|_{y=1}=
\frac12(e^x-1)^2\exp(e^x-1).
\end{align*}

For type $\mathsf B_n$ there are $\sum\limits_{j=k}^n
2^{j-k}\tbinom nj\,S(j,k)$ $k$-dimensional subspaces
each containing $k^2$ subspaces in $L(\mathcal A)$ of codimension~$1$. Thus we get the
generating function
\begin{align*}
\makebox[1cm][l]{$\displaystyle\sum_{n,k}\sum_{j=k}^n2^{j-k}\tbinom nj\,
S(j,k)\,k^2\frac{x^n}{n!}$}&\\
&=\sum_{j,n}\frac{x^{n-j}}{(n-j)!}\frac{\partial}{\partial y}\,y
\frac{\partial}{\partial y}\sum_k2^{j-k}S(j,k)\frac{x^j}{j!}\,y^k\Bigr|_{y=1}\\
&=\sum_{m\geqslant0}\frac{x^m}{m!}\frac{\partial}{\partial y}\,y
\frac{\partial}{\partial y}\sum_{j,k}2^{j-k}S(j,k)\frac{x^j}{j!}\,y^k\Bigr|_{y=1}\\
&=\sum_{m\geqslant0}\frac{x^m}{m!}\frac{\partial}{\partial y}\,y
\frac{\partial}{\partial y}\exp\Bigl(\frac
y2\cdot\bigl(e^{2x}-1\bigr)\Bigr)\Bigr|_{y=1}\\
&=\sum_{m\geqslant0}\frac{x^m}{m!}\frac14\bigl(e^{4x}-1\bigr)
\exp\Bigl(\frac12\bigl(e^{2x}-1\bigr)\Bigr)\\
&=e^x\frac14\bigl(e^{4x}-1\bigr)\exp\Bigl(\frac12\bigl(e^{2x}-1\bigr)\Bigr).
\end{align*}
The reader may wonder why we did not insert the formula for
$$\sum\limits_{n,k}\sum\limits_j2^{j-k}\tbinom nj\,S(j,k)\frac{x^n}{n!}\,y^k$$
directly. The reason for going through the seemingly arcane substitution $m=n-j$ is
that we can then use this calculation for type $\mathsf D$. Namely, for type
$\mathsf D$ we must subtract the terms for $j=n-1$ and $j=n$, that is, for $m=1$ and
$m=0$ in the generating function for $\mathsf B$ and then add the modified term
corresponding to $j=n$. 

Let us direct our attention to the case $B=[n]$ for $\mathsf D_n$. We get a partition
of $[n]$ into $k$ blocks with exactly $h$ blocks of length $1$ by choosing $h$ elements
from $[n]$ and partitioning the remaining set of $n-h$ elements into $k-h$ blocks of
lengths at least $2$. Taking into account also the choice of $\zeta:[n]\to\{\pm1\}$,
we have
$$2^{n-k}\tbinom nh\,S_2(n-h,k-h)$$
elements of rank $n-k$ in the Hasse diagram for $\mathsf D_n$ which are covered by
$k^2-h$ elements. The modified term corresponding to $j=n$ is thus
{\allowdisplaybreaks
\begin{align*}
\makebox[2mm][l]{$\displaystyle\sum_{n,k}\sum_h2^{n-k}\tbinom
nh\,S_2(n-h,k-h)\,(k^2-h)\frac{x^n}{n!}$}&\\
&=\sum_{n,k}\sum_h\frac{x^h}{h!}\,2^{n-k}S_2(n-h,k-h)\,(k^2-h)\frac{x^{n-h}}{(n-h)!}
\,y^k\Bigr|_{y=1}\\
&=\sum_h\frac{x^h}{h!}\Bigl(\frac{\partial}{\partial y}\,y\frac{\partial}{\partial y}
-h\Bigr)y^h\sum_{n,k}2^{n-k}S_2(n-h,k-h)\frac{x^{n-h}}{(n-h)!}\,y^{k-h}\Bigr|_{y=1}\\
&=\sum_h\frac{x^h}{h!}\Bigl(\frac{\partial}{\partial y}\,y\frac{\partial}{\partial y}
-h\Bigr)y^h\exp\Bigl(\frac y2\cdot\bigl(e^{2x}-1-2x\bigr)\Bigr)\Bigr|_{y=1}\\
&=\sum_h\frac{x^h}{h!}\Bigl(h^2-h+(2h+1)\frac12\bigl(e^{2x}-1-2x\bigr)+\frac14
\bigl(e^{2x}-1-2x\bigr)^2\Bigr)\\[-0.4em]
&\hspace{19em}
\times\exp\Bigl(\frac12\bigl(e^{2x}-1-2x\bigr)\Bigr)\\
&=e^x\Bigl(x^2+(2x+1)\frac12\bigl(e^{2x}-1-2x\bigr)+\frac14
\bigl(e^{2x}-1-2x\bigr)^2\Bigr)\\
&\hspace{19em}
\times\exp\Bigl(\frac12\bigl(e^{2x}-1-2x\bigr)\Bigr)\\
&=e^x\frac14\bigl(e^{4x}-1-4x\bigr)\exp\Bigl(\frac12\bigl(e^{2x}-1-2x\bigr)
\Bigr).
\end{align*}
}

The exponential generating function for the numbers of edges for the $\mathsf D$-series
therefore takes the form
$$(e^x-1-x)\frac14\bigl(e^{4x}-1\bigr)\exp\Bigl(\frac12\bigl(e^{2x}-1\bigr)\Bigr)
+e^x\frac14\bigl(e^{4x}-1-4x\bigr)
\exp\Bigl(\frac12\bigl(e^{2x}-1-2x\bigr)\Bigr).$$
\end{proof}

\section*{A curious determinant}
Apparently it was A.~Lenard who discovered that the Hankel determinant with the Bell
numbers as entries is a superfactorial (see the reference in \cite{We}).  Let
us compute its $\mathsf B$-analogue. So let the Dowling numbers $D_n$ be given by
$$\sum_{n\geqslant0}D_n\frac{x^n}{n!}=e^x
\exp\Bigl(\frac12\bigl(e^{2x}-1\bigr)\Bigr).$$

\begin{proposition}
\renewcommand{\arraystretch}{1.4}
$$\left|\begin{array}{cccc}
D_0&D_1&\dots&D_n\\D_1&D_2&\dots&D_{n+1}\\
\vdots&\vdots&&\vdots\\D_n&D_{n+1}&\dots&D_{2n}\end{array}\right|=
2^{n(n+1)/2}\prod_{k=1}^nk!$$
\end{proposition}

We shall prove the following generalization which involves the numbers $G_n$ (for
$l=0$) that occurred in Kerber's note \cite[(7)]{Ke} in connexion with multiply
transitive groups and also in M.~Bernstein's and Sloane's ``eigen-sequence paper''
\cite[Table~1(a)]{BeSl} in a new setting.

\begin{proposition}
Define the sequence of generalized Bell numbers $(G_n)_{n\geqslant0}$ depending on
$l$ and $m$ by
\begin{equation}\label{GBell}
\sum_{n\geqslant0}G_n\frac{x^n}{n!}=e^{lx}
\exp\Bigl(\frac1m\bigl(e^{mx}-1\bigr)\Bigr).
\end{equation}
Then
\renewcommand{\arraystretch}{1.4}
\begin{equation}\label{DetG}
\left|\begin{array}{cccc}
G_0&G_1&\dots&G_n\\G_1&G_2&\dots&G_{n+1}\\
\vdots&\vdots&&\vdots\\G_n&G_{n+1}&\dots&G_{2n}\end{array}\right|=
m^{n(n+1)/2}\prod_{k=1}^nk!
\end{equation}
\end{proposition}
\begin{proof}
The statement in \cite[p.~113/114]{Ko} can be rephrased by saying that a Hankel
determinant does not change its value when the matrix entries are subject to a
binomial transform. Hence the
determinant in $(\ref{DetG})$ is independent of $l\in\mathbb Z$
and consequently also independent of $l$ when $l$ is considered as an indeterminate.
Therefore we will assume that $l=0$ in the
definition $(\ref{GBell})$ of the numbers~$G_n$.

As an aside let us mention that
the invariance under binomial transform gives the following identity between Hankel
determinants with Bell numbers as entries.
\renewcommand{\arraystretch}{1.4}
$$\left|\begin{array}{cccc}
B_0&B_1&\dots&B_n\\B_1&B_2&\dots&B_{n+1}\\
\vdots&\vdots&&\vdots\\B_n&B_{n+1}&\dots&B_{2n}\end{array}\right|=
\left|\begin{array}{cccc}
B_1&B_2&\dots&B_{n+1}\\B_2&B_3&\dots&B_{n+2}\\
\vdots&\vdots&&\vdots\\B_{n+1}&B_{n+2}&\dots&B_{2n+1}\end{array}\right|$$

To compute the determinant (\ref{DetG}) we proceed by induction. Let us first define
$H_{n,k}\in\mathbb Q[m]$ by
\begin{equation}\label{defH}
\sum_{n\geqslant0}H_{n,k}\frac{y^n}{n!}=\frac{1}{k!}\,e^{-y}\frac{1}{m^k}\,\bigl(
\log(1+my)\bigr)^k\qquad(k={0,1,2,\dots}).
\end{equation}
Note that $H_{n,n}=1$. Hence with
\begin{equation}\label{defI}
I_{h,n}=\sum_{k=0}^nG_{h+k}\,H_{n,k}
\end{equation}
we have
\renewcommand{\arraystretch}{1.4}
\begin{equation}\label{Ind}
\left|\begin{array}{cccc}
G_0&G_1&\dots&G_n\\G_1&G_2&\dots&G_{n+1}\\
\vdots&\vdots&&\vdots\\G_n&G_{n+1}&\dots&G_{2n}\end{array}\right|=
\left|\begin{array}{cccc}
G_0&\dots&G_{n-1}&I_{0,n}\\
\vdots&&\vdots&\vdots\\G_{n-1}&\dots&G_{2n-2}&I_{n-1,n}\\
G_n&\dots&G_{2n-1}&I_{n,n}\end{array}\right|.
\end{equation}
From
\begin{equation}\label{Identity}
\sum_{h,n}I_{h,n}\frac{x^h}{h!}\frac{y^n}{n!}=
\exp\Bigl(\frac1m\bigl(e^{mx}-1\bigr)\Bigr)\exp\Bigl(y\cdot\bigl(e^{mx}-1\bigr)\Bigr)
\end{equation}
we see that $I_{0,n}=\dots=I_{n-1,n}=0$ and $I_{n,n}=m^n\cdot n!$. Hence
$(\ref{DetG})$ follows from $(\ref{Ind})$ by induction.\par
We must finally prove $(\ref{Identity})$.
So let us compute:
{\allowdisplaybreaks
\begin{align*}
\sum_{h,n}I_{h,n}\frac{x^h}{h!}\frac{y^n}{n!}&\stackrel{\Nbx{(\ref{defI})}}{=}
\sum_{h,k,n}G_{h+k}\,H_{n,k}\frac{x^h}{h!}\frac{y^n}{n!}\\
&\stackrel{\Nbx{(\ref{defH})}}{=}
\sum_{h,k}G_{h+k}\frac{x^h}{h!}\frac{1}{k!}\,e^{-y}\frac{1}{m^k}\,\bigl(\log(1+my)
\bigr)^k\\
&=e^{-y}\sum_{h,k}G_{h+k}\frac{1}{(h+k)!}\,\tbinom{h+k}{h}\,x^h\frac{1}{m^k}\,\bigl(
\log(1+my)\bigr)^k\\
&=e^{-y}\sum_nG_n\frac{1}{n!}\Bigl(x+\frac{1}{m}\log(1+my)\Bigr)^n\\
&\stackrel{\Nbx{(\ref{GBell})}}{=}
e^{-y}\exp\biggl(\frac{1}{m}\,\Bigl(e^{m\bigl(x+\frac{1}{m}\log(1+my)\bigr)}-1
\Bigr)\biggr)\\
&=\exp\Bigl(\frac1m\bigl(e^{mx}-1\bigr)\Bigr)\exp\Bigl(y\cdot\bigl(e^{mx}-1\bigr)\Bigr).
\end{align*}
}
We have thus verified equation~(\ref{Identity}).
\end{proof}

{\footnotesize
\noindent\textit{Acknowledgements.}
Without Sloane's integer sequence database I would probably never have come across the
reference \cite{Ri}. Also at one instance the \texttt{gfun}
Maple package by Salvy and Zimmermann \cite{SZ} was helpful.}

%\bibliographystyle{amsplain}
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\end{thebibliography}

\vspace*{+.5in}
\centerline{\rule{5in}{.01in}}

\vspace*{+.1in}
\noindent
{\small
(Concerned with sequences
\htmladdnormallink{A003128}{http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=003128},
\htmladdnormallink{A039755}{http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=039755},
\htmladdnormallink{A039756}{http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=039756},
\htmladdnormallink{A039757}{http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=039757},
\htmladdnormallink{A039758}{http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=039758},
\htmladdnormallink{A039759}{http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=039759},
\htmladdnormallink{A039760}{http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=039760},
\htmladdnormallink{A039761}{http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=039761},
\htmladdnormallink{A039762}{http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=039762},
\htmladdnormallink{A039763}{http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=039763},
\htmladdnormallink{A039764}{http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=039764},
\htmladdnormallink{A039765}{http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=039765}.)
}

\centerline{\rule{5in}{.01in}}

\vspace*{+.1in}
\noindent
Received Jan.\ 13, 2000;
published in Journal of Integer Sequences March 10, 2000.

\centerline{\rule{5in}{.01in}}

\vspace*{+.1in}
\noindent
Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.research.att.com/~njas/sequences/JIS/}.

\centerline{\rule{5in}{.01in}}


\end{document}

