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\begin{center}
\vskip 1cm{\LARGE\bf Converting Between Generalized Bell, Lah, \\
\vskip .1in
Stirling, and Tanh Numbers
}
\vskip 1cm
\large
Giacomo Della Riccia \\
Department of Mathematics and Computer Science \\
University of Udine \\
Via delle Scienze 206 \\
33100 Udine \\
Italy \\
\href{mailto:dlrca@uniud.it}{\tt dlrca@uniud.it}
\end{center}

\vskip .2 in
\begin{abstract}
We define a Riordan triangle for generalized Bell numbers and we
establish general identities connecting Lah, Stirling, Tanh and the
generalized Bell numbers. Several familiar inverse relations
converting between special sequences are shown to be particular
cases of the general identities.
\end{abstract}


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\section{Introduction}
We consider, in the context of the (exponential) Riordan group
$\mathcal{(R;\ \star)}$, inverse relations
\begin{equation}\label{array}
c(n,m)=\sum_ia(n,i)b(i,m)\Leftrightarrow b(n,m)=
\sum_iA(n,i)c(i,m),\quad n,m\geq0
\end{equation}
converting between arrays of numbers. The concept of Riordan group
was introduced by Shapiro et al.\ \cite{Shapiro1991} as a
generalization of a study of Rogers \cite{Rogers1978} on
\textit{renewal arrays} and used in several applications, including
\cite{Sprugnoli1994,Sprugnoli1995,Shapiro2003,ZhaoWang2003,CamNkw2005,MerSprVer2005}.
The last reference, pointed out to us by the referee, contains an
application of the concept of a Riordan array to the
Akiyama-Tanigawa transformation  analogous to the formalism
developed in the present paper. An element $R\in (\mathcal{R};\
\star)$ is denoted by $R=(q(u),\ R(u))$ where $q(u)$ is an
exponential generating function such that $q(0)=1$ and $R(u)$ is an
exponential generating function such that $R(0)=0$ and $R'(0)=1$.
The numbers $R(n,m)$, defined recursively by
$\frac{1}{m!}q(u)R^m(u)=\sum_{n\geq0}R(n,m)u^n/n!,\ m=0,1,2,\cdots$,
are the entries of an \textit{invertible} infinite lower triangular
matrix $(R(n,m)_{n,m\geq0})$, called a \textit{Riordan array}. The
Riordan group product $\star$ is defined by
\begin{equation}\label{Gproduct}
c=a\star b=(g(u),a(u))\star (h(u),b(u))=(g(u)h(a(u)),\ b(a(u))).
\end{equation}
This notation and the matrix notation $ c(n,m)= \sum_ia(n,i)b(i,m)$
are isomorphic. The unit element is $I=(1,\ u)$ and the inverse $r$
of $R$ is $r=(g(u),\ R(u))^{-1}=(1/g(\overline{R}(u)),\
\overline{R}(u))$ where $\overline{R}(u)$ is the compositional
inverse of $R(u): R(\overline{R}(u))=\overline{R}(R(u))=u$.

In Section 2, we consider pairs of Riordan arrays $\{a,A\}$,
$\{b,B\}$ and $\{c,C\}$ corresponding to numbers that are inverse of
each other and from an identity $c=a\star b$, we derive new
identities $C=B\star A$, $A=b\star C$, etc.. If numbers $R(n,m)$ can
be extended to all integers $n$ and $m$, then we will use
\textit{dual} elements $\widetilde{R}(n,m)=R(-m,-n)$ and dual
identities defined by
\begin{eqnarray*}
\widetilde{c}(n,m)&amp;=&amp;\sum_ia(-m,i)b(i,-n)=\sum_ia(-m,-i)b(-i,-n)\\
&amp;=&amp;\sum_i\widetilde{b}(n,i)\widetilde{a}(i,m),
\end{eqnarray*}
that is, $\widetilde{c}=\widetilde{b}\star\widetilde{a}$. The change
of $i$ into $-i$ is legal because, conventionally, all numbers are
null when $n&lt;0\leq m$ and the summation ranges are unrestricted. We
will also consider inverse relations that are duals (``$\sim$'') of
each other:
$$c=a\star b\Leftrightarrow b=A\star c\quad\sim\quad\widetilde{c}=
\widetilde{b}\star\widetilde{a}\Leftrightarrow
\widetilde{b}=\widetilde{c}\star\widetilde{A}.$$

In Section 3, we will discuss inverse relations converting between
\textit{sequences} that are special cases $m=1,\ 2$ of inverse
relations (\ref{array}) converting between \textit{arrays}.
\section{Converting between Stirling, Lah, Tanh and Bell numbers}
The Riordan group elements considered in this paper belong to the
so-called \textit{associated} subgroup of $\mathcal{(R;\ \star)}$,
i.e., they are of the form $R=(1,\ R(u))$. In this case the
expression (\ref{Gproduct}) of the product of two Riordan arrays
$c=a\star b$ reduces to
\begin{eqnarray*}
c_m(u)&amp;=&amp;\sum_{n\geq0}c(n,m)\frac{u^n}{n!}=
\sum_{n\geq0}\sum_ia(n,i)b(i,m)\frac{u^n}{n!}=
\sum_ib(i,m)\frac{a^i(u)}{i!}\\
&amp;=&amp;b_m(a(u)).
\end{eqnarray*}
The referee suggested the use of the \textit{negation} rule\ ``-''\
: for any $R=(1,\ R(u))$, $-R=(1,\ -R(-u))$ and
$-(\Psi\star\Omega)=-\Psi\star -\Omega$; the rule replaces $R(n,m)$
in the formulas by $(-1)^{n-m}R$ and this allows a more elegant
formulation of our results. We shall also use a \textit{scaling}
rule\ ``$\mu$''\ : for any $R=(1,\ R(u))$, $\mu R=(1,\ \mu
R(u/\mu))$ and $\mu(\Psi\star\Omega)=\mu\Psi\star\mu\Omega$ for all
real or complex numbers $\mu$; the rule replaces $R(n,m)$ in the
formulas by $R(n,m)/\mu^{n-m}$. Negation is a scaling with $\mu=-1$
and $\lim_{\mu\rightarrow\pm\infty}\mu(1,\ R(u))=(1,\ u)=I$.

\underline{Stirling numbers}. The Stirling numbers of the first and
second kind are denoted $s(n,m)$ and $S(n,m)$, respectively,
\cite{Comtet1974}. The \emph{unsigned} numbers are ${n\brack
m}=(-1)^{n-m}s(n,m)$ and ${n\brace m}=S(n,m)$. Formal properties of
the Stirling numbers \cite{Graham1994} include
\begin{eqnarray*}
&amp;&amp;s=(1,\ s(u)=\ln(1+u)),\quad S=(1,\ S(u)=e^u-1),\\
&amp;&amp;s\star S=S\star s=I,\quad\widetilde{S}=-s\quad\widetilde{s}=-S.
\end{eqnarray*}

\underline{Lah numbers} We define Lah numbers of the first and
second kind $\lambda(n,m)$ and $\Lambda(n,m)$ as follows:
$$\lambda(n,m)=\frac{(-1)^m}{2^{n-m}}L(n,m),\quad
\Lambda(n,m)=\frac{(-1)^n}{2^{n-m}}L(n,m),$$where instead of the
familiar explicit expression $$L(n,m)=(-1)^n
\frac{n!}{m!}{n-1\choose m-1},\quad n\geq m&gt;0,$$ we will use
$$L(n,m)=(-1)^n(n-m)!{n\choose n-m}{n-1\choose n-m},\quad\mbox{all integers}\
n,m,$$that has the advantage of extending $L(n,m)$ to all integers
$n$ and $m$, and providing a definition for Lah dual numbers:
\begin{eqnarray}
&amp;&amp;\widetilde{L}(n,m)=L(-m,-n)=(-1)^m(n-m)!{-m\choose
n-m}{-m-1\choose n-m}\label{lahdual}=\\
&amp;&amp;(-1)^m(n-m)!{n-1\choose n-m}{n\choose
n-m}=(-1)^{n-m}L(n,m).\nonumber
\end{eqnarray}

     From the exponential generating function of $L(n,m)$ (see \cite[Problem 16 (b), p.\ 44]{Riordan1980})
$$L_m(u)=\sum_nL(n,m)\frac{u^n}{n!}=\frac{1}{m!}\left(\frac{-u}{1+u}\right)^m,$$
we derive the exponential generating functions
$$\lambda_m(u)=\frac{1}{m!}\left(\frac{u}{1+u/2}\right)^m,\quad
\Lambda_m(u)=\frac{1}{m!}\left(\frac{u}{1-u/2}\right)^m.$$ These
exponential generating functions, the fact that
$\lambda(u)=-\Lambda(-u)$ and (\ref{lahdual}) yield the formal
properties:
\begin{eqnarray*}
&amp;&amp;\lambda=\left(1,\
\lambda(u)=\frac{u}{1+u/2}\right),\quad\Lambda=\left(1,\
\Lambda(u)=\frac{u}{1+u/2}\right),\\
&amp;&amp;\Lambda(u)=\overline{\lambda}(u)\quad\lambda\star\Lambda=\Lambda\star\lambda=
I,\quad\widetilde{\Lambda}=\Lambda=-\lambda\quad
\widetilde{\lambda}=\lambda=-\Lambda.
\end{eqnarray*}

\underline{Tanh numbers}. Let $t=(1,\ t(u)=\arctan(u))$ and $T=(1,\
T(u)=\tan(u))$ be the Riordan arrays of the arctan $t(n,m)$ and
tangent $T(n,m)$ numbers, respectively. The recurrence relations
(see \cite[pp.\ 259--260]{Comtet1974})
\begin{eqnarray*}
&amp;&amp;t(n+1,m)=t(n,m-1)-n(n-1)t(n-1,m),\\
&amp;&amp;T(n+1,m)=T(n,m-1)+m(m+1)T(n,m+1),\\
&amp;&amp;t(0,m)=T(0,m)=[m=0],\quad t(n,0)=T(n=0)=[n=0],
\end{eqnarray*}
imply $\widetilde{T}(n,m)=T(-m,-n)=t(n,m)$ and
$\widetilde{t}(n,m)=t(-m,-n)=T(n,m)$.

Our first and second kind Tanh numbers $\theta(n,m)$ and
$\Theta(n,m)$ are
$$\theta(n,m)=\frac{(-1)^{(n-m)/2}}{2^{n-m}}t(n,m),\quad
\Theta(n,m)=\frac{(-1)^{(n-m)/2}}{2^{n-m}}T(n,m),$$ that is,
$\theta=\mu t$ and $\Theta=\mu T$ with $\mu=2/\imath,\ \imath^2=-1$.
Summarizing:
\begin{eqnarray*}
&amp;&amp;\theta= \left(1,\
\theta(u)=\ln\frac{1+u/2}{1-u/2}\right),\quad\Theta=\left(1,\
\Theta(u)=2 \frac{e^{u}-1}{e^{u}+1}\right),\\
&amp;&amp;\Theta(u)=\overline{\theta}(u)\quad\theta\star\Theta=
\Theta\star\theta=I,\quad\widetilde{\Theta}=
\theta=-\theta\quad\widetilde{\theta}=\Theta=-\Theta,
\end{eqnarray*}
where $\theta=-\theta$ and $\Theta=-\Theta$ because $\theta(u)$ and
$\Theta(u)$ are odd functions of $u$.
\begin{thm}
Numbers in each pair $\{s,S\}$, $\{\lambda,\Lambda\}$ and
$\{\theta,\Theta\}$ convert between numbers in the other two pairs.
\end{thm}
\begin{proof}With $S(u)=e^u-1$, $\lambda(u)=u/(1+u/2)$
and $\Theta(u)=2(e^{u}-1)/(e^{u}+1)$,
\begin{eqnarray*}
&amp;&amp;\lambda(S(u))=\lambda(e^u-1)=\frac{e^u-1}{1+(e^u-1)/2}=
\Theta(u),\quad\mbox{therefore}\\
&amp;&amp;\Theta=S\star\lambda\quad\mbox{and in matrix
notation}\quad\Theta(n,m)=\sum_iS(n,i)\lambda(i,m).
\end{eqnarray*}
The identities that can be derived from $\Theta=S\star\lambda$
are\\
\bigskip
\begin{tabular}{ll}
$\theta=\Lambda\star s$ &amp; Riordan group inversion\\
$S=\Theta\star\Lambda$ &amp; right multiplication by $\Lambda$\\
$s=\lambda\star\theta$ &amp; left multiplication by $s$/right multiplication by $\theta$\\
$\Lambda=\theta\star S$ &amp; left
multiplication by $\theta$/right multiplication by $\Lambda$\\
$\lambda=s\star\Theta$ &amp; left multiplication by s
\end{tabular}\\
and the inverse relations\\
\bigskip
\begin{tabular}{lll}
$\Theta=S\star\lambda\Leftrightarrow\lambda=s\star\Theta$ &amp;
$\sim$ &amp; $\theta=\Lambda\star s\Leftrightarrow \Lambda=\theta\star S$\\
$\theta=\Lambda\star s\Leftrightarrow s=\lambda\star\theta$ &amp; $\sim$
&amp; $\Theta= S\star\lambda\Leftrightarrow S=\Theta\star \Lambda$\\
$S=\Theta\star\Lambda\Leftrightarrow\Lambda=\theta\star S$ &amp; $\sim$
&amp; $s=\lambda\star\theta\Leftrightarrow \lambda=s\star\Theta$.
\end{tabular}\\
The above identities prove the theorem. In deriving dual inverse
relations, we tacitly used duality and negation rules.
\end{proof}
\begin{cor}\label{cor}
Tanh numbers are represented by linear combinations of Stirling
numbers, and conversely:
\begin{eqnarray*}
&amp;&amp;\theta(n,m)=n!\sum_{i=m}^n{n-1\choose
i-1}\frac{1}{2^{n-i}i!}s(i,m),\\
&amp;&amp;\Theta(n,m)=\frac{1}{m!}\sum_{i=m}^n{i-1\choose
m-1}\frac{(-1)^{i-m}i!}{2^{i-m}}S(n,i);\\
&amp;&amp;s(n,m)=n!\sum_{i=m}^n{n-1\choose
i-1}\frac{(-1^{n-i}}{2^{n-i}}\theta(i,m),\\
&amp;&amp;S(n,m)=\frac{1}{m!}\sum_{i=m}^n{i-1\choose
m-1}\frac{i!}{2^{i-m}}\Theta(n,i).
\end{eqnarray*}
\end{cor}
\begin{proof}
Write explicitly $\lambda$ and $\Lambda$ in $\theta=\Lambda\star
s$,\ $\Theta=S\star\lambda$,\ $s=\lambda\star\theta$ and
$S=\Theta\star\Lambda$.
\end{proof}
\underline{Bell numbers}. Bell numbers $A_n,\ n\geq1,$ can be
defined by $\mathcal{B}(u)=\exp(e^u-1)-1=\sum_{n\geq1}A_nu^n/n!$,
\cite{Carlitz1980}. We define generalized Bell numbers of the second
kind $\mathcal{B}(n,m)$ by
$$\mathcal{B}_m(u)=\frac{1}{m!}(\exp(e^u-1)-1)^m=
\sum_{n\geq1}\mathcal{B}(n,m)\frac{u^n}{n!}.$$ With $S(u)= e^u-1$
and $s(u)=\overline{S}(u)=\ln(1+u)$, we have
$$\mathcal{B}(u)=S(S(u))\quad\mbox{and}\quad\beta(u)=\overline{\mathcal{B}}(u)=
s(s(u))=\ln(1+\ln(1+u)).$$
 Thus, generalized Bell numbers of the
first kind $\beta(n,m)$ can be defined by
$$\beta_m(u)=\frac{1}{m!}(\ln(1+\ln(1+u)))^m=\sum_{n\geq1}\beta(n,m)\frac{u^n}{n!}.$$
\begin{thm}Bell numbers $\left\{\beta,\mathcal{B}\right\}$ can be represented
as squares (group multiplication) of Stirling numbers: $\beta=s\star
s\ \mbox{and}\ \mathcal{B}=S\star S$.
\end{thm}
\begin{proof}The Riordan array products $\beta=s\star s$ and $\mathcal{B}=S\star S$
hold in view of the functional relations $\beta(u)=s(s(u))$ and
$\mathcal{B}(u)=S(S(u))$.
\end{proof}
Summarizing:
\begin{eqnarray*}
&amp;&amp;\beta=(1,\ \beta(u)=\ln(1+\ln(1+u))),\\
&amp;&amp;\mathcal{B}=(1,\ \mathcal{B}(u)=\exp(e^u-1)-1)\\
&amp;&amp;\beta\star\mathcal{B}=\mathcal{B}\star\beta=I,\quad\widetilde{\mathcal{B}}=
 -\beta,\quad\widetilde{\beta}=-\mathcal{B}
\end{eqnarray*}
where the dual relations follow from Stirling number dual relations.
\begin{thm}
Bell numbers in $\{\beta,\mathcal{B}\}$ and Stirling numbers in
$\{s,S\}$ convert between each other.
\end{thm}
\begin{proof}
The identities that can be derived from $\mathcal{B}=S\star
S$ are\\
\bigskip
\begin{tabular}{ll}
$\beta=s\star
s$ &amp; Riordan group inversion\\
$S=s\star\mathcal{B}$ &amp; left multiplication by\ $s$\\
$s=S\star\beta$ &amp; left multiplication by $s$ and right
multiplication by $\beta$,
\end{tabular}\\
and the inverse relations\\
\bigskip
\begin{tabular}{lll}
$\mathcal{B}=S\star S\Leftrightarrow S=s\star\mathcal{B}$ &amp;
$\sim$ &amp; $\beta=s\star s\Leftrightarrow s=\beta\star S$\\
$\beta=s\star s\Leftrightarrow s=S\star\beta$ &amp; $\sim$ &amp;
$\mathcal{B}=S\star S\Leftrightarrow S=\mathcal{B}\star s$.
\end{tabular}\\
The above identities prove the theorem. In deriving dual inverse
relations, we tacitly used duality and negation rules.
\end{proof}
\begin{cor}
Bell numbers $\{\beta,\mathcal{B}\}$ and Stirling numbers $\{s,S\}$
satisfy the commutation relations $S\star\beta=\beta\star S=s$ and
$s\star\mathcal{B}=\mathcal{B}\star s=S$.
\end{cor}
\begin{proof}
Left multiplication by $S$ and right multiplication by $S$ of
$\beta=s\star s$ give $S\star\beta=s$ and $\beta\star S=s$,
respectively. By inversion, we get $\mathcal{B}\star
s=s\star\mathcal{B}=S$.
\end{proof}
The Riordan triangle $(\mathcal{B}(n,m)_{n,m\geq0})$ is different
from the Pascal type triangle proposed by Shallit
\cite{Shallit1980}. This is obtained by flipping and reorienting a
\emph{rectangular} array given by Cohn et al.\ \cite{Cohn1962} for
an efficient calculation of the Bell numbers. The reader may find it
quite interesting to see the nice properties of Shallit's triangle.
\section{Special cases}
The following examples are mainly special cases $m=1, 2$ of the
above relations and they will be compared with identities commonly
found in the literature.

1) When $m=1$,
$\Theta=S\star\lambda\Leftrightarrow\lambda=s\star\Theta$ gives
\begin{eqnarray*}
\frac{C_n}{2^n}&amp;=&amp;(-1)^{(n+1)/2}\frac{T_n}{2^n}+[n=0]=2(1-2^{n+1})\frac{B_{n+1}}{n+1}
\\
&amp;=&amp;\sum_{i=0}^n{n\brace i}\frac{(-1)^ii!}{2^i}\\
&amp;\Leftrightarrow&amp;\ \frac{n!}{2^n}=\sum_{i=0}^n{n\brack
i}(-1)^i\frac{C_i}{2^i}= \sum_{i=0}^n{n\brack
i}2(1-2^{i+1})\frac{B_{i+1}}{i+1}.
\end{eqnarray*}
The first identity appears in (\cite[p.\ 585]{Knuth1968}) and
(\cite[Exercise 6.76]{Graham1994}).  It can also be found in
Sprugnoli (\cite[p.\ 288]{Sprugnoli1994}) as a result of the theory
of Riordan arrays and in Chen \cite{Wu2001} by application of the
Akiyama-Tanigawa algorithm \cite{Kaneko2000}.

2) Since $\theta(n,m)=\Theta(n,m)=0$ when $n-m$ is odd, Corollary
\ref{cor} yields
$$\sum_{i=m}^n{n-1\choose
i-1}\frac{2^i}{i!}s(i,m)=\sum_{i=m}^n{i-1\choose
m-1}\frac{(-1)^ii!}{2^i}S(n,i)=0,\quad n-m\ \mbox{odd}.$$ For $m=1$,
the first identity leads to a trivial result, whereas the second
gives the identity $\sum_{i=1}^{n}{{2n}\brace i}(-1)^ii!/2^i=0,\
n&gt;0$ found by Lengyel (\cite[p.\ 7]{Lengyel1996}) in his study of
the series $\sum_{k=0}^\infty
k^nx^k$.\\
For $m=2$ with $s(n,2)=(-1)^n(n-1)!H_{n-1}$ where $H_n$ are harmonic
numbers, we find
$$\sum_{i=2}^{2n+1}{2n\choose i-1}\frac{2^i}{i!}s(i,2)=
\sum_{i=1}^{2n}{2n+1\choose i+1}(-2)^iH_i=0,\ n&gt;0.$$

3) For $m=1$, $S=\Theta\star\Lambda\Leftrightarrow
\Lambda=\theta\star S$ gives
$$1+[n=0]=\sum_{i=0}^n\Theta(n,i)\frac{i!}{2^{i-1}}\
\Leftrightarrow\
\frac{n!}{2^{n-1}}=\sum_{i=0}^n\theta(n,i)\left(1+[n=0]\right).$$

4) For $m=1$, two identities in Lah's original paper (\cite[Eqs.\
(34) and (43)]{Lah1955}):
$L(n,m)=\sum_{i=0}^ns(n,i)(-1)^iS(i,m)\Leftrightarrow(-1)^nS(n,m)=
\sum_{i=0}^nS(n,i)L(i,m)$, yield the simple inverse relation
$$\sum_{i=0}^n{n \brack i}=n!\Leftrightarrow\sum_{i=0}^n{n \brace
i}(-1)^ii!=(-1)^n$$.

5) The inverse of $$\sum_i{n \brace i}{i+1\brack
m}\frac{(-1)^{i+1-m}}{i+1}={n+1\choose m}\frac{B_{n+1-m}}{n+1}$$
that appears in (\cite[Section 6.5, Eq.\ (6.99)]{Graham1994}) is
$$\sum_i{n\brack i}(-1)^{i+1-m}{i+1\choose
m}\frac{B_{i+1-m}}{i+1}=\frac{{n+1\brack m}}{n+1}.$$ For $m=1$, we
find the well-known inverse relations (\cite[pp.\
220--221]{Comtet1974})
$$\sum_i{n\brace{i}}\frac{(-1)^ii!}{i+1}=B_n\Leftrightarrow\sum_i{n\brack
i}(-1)^iB_i=\frac{n!}{n+1}.$$

6) From $\mathcal{B}=S\star S$ and $S=\Theta\star\lambda$ we obtain
$\mathcal{B}= S\star\Theta\star\Lambda$, which in matrix notation is
$\mathcal{B}(n,m)= \sum_i\sum_jS(n,i)\Theta(i,j)\Lambda(j,m)$. To
hold down the size of this paper, we give only one example of triple
factorization.
\section{Main results and Conclusion}
Group theory provides a rapid and systematic way to derive from a
given identity connecting Riordan arrays, further identities and
inverse relations. Applying this approach to
$\Theta(n,m)=\sum_iS(n,i)\lambda(i,m)$, we found that special cases
$m=1,\ 2$ of the general identities are results obtained in the
literature by (often more complex) procedures which seem at first to
be unrelated to each other, whereas here they have a common origin
and are obtained by a unifying procedure.

The identities $\beta=s\star s$ and $\mathcal{B}=S\star S$ connect
Bell, Stirling, Lah, and Tanh numbers through, for instance, the identity
$\mathcal{B}=S\star\Theta\star\Lambda$. Triple factorizations
$d=a\star b\star c$ may lead in special cases $m=1,\ 2$ to
interesting identities connecting three Riordan arrays; we hope to
investigate in this direction in a future publication.

\section{Acknowledgement}
The author wishes to thank the referee for valuable comments and
suggestions which have greatly improved the clarity of this paper.
\bibliographystyle{amsplain}
\providecommand{\bysame}{\leavevmode\hbox
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\noindent 2000 {\it Mathematics Subject Classification}:
Primary 05A19; Secondary 05A15.

\noindent \emph{Keywords: } Riordan group, conversions, Bell triangle.

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\hrule
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\vspace*{+.1in}
\noindent
Received February 2 2006;
revised version received July 21 2006. 
Published in {\it Journal of Integer Sequences}, July 21 2006.

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