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\begin{center}
\vskip 1cm{\LARGE\bf Some Observations on the Lah and Laguerre \\
\vskip .1in
Transforms of Integer Sequences}
\vskip 1cm 
\large
Paul Barry\\
School of Science\\
Waterford Institute of Technology\\
Ireland\\
\href{mailto:pbarry@wit.ie}{\tt pbarry@wit.ie} \\
\end{center}
\vskip .2 in

\begin{abstract} We study the Lah and related Laguerre transforms
within the context of exponential Riordan arrays. Links to the
Stirling numbers are explored. Results for finite matrices are
generalized, leading to a number of useful matrix factorizations.
\end{abstract}

\section{Integer sequences and transforms on them}
In this note, we shall consider integer sequences
$$a:\mathbf{N}_0 \to \mathbf{Z} $$ with general term $a_n=a(n)$. Normally, sequences will be
described by either by their ordinary generating function (o.g.f.),
that is, the function $g(x)$ such that
$$g(x)=\sum_{n=0}^{\infty}a_nx^n,$$ or by their exponential
generating function (e.g.f.) $f(x)$, where
$$f(x)=\sum_{n=0}^{\infty}a_n\frac{x^n}{n!}.$$

We shall encounter transformations that operate on integer sequences
during the course of this note. An example of such a transformation
that is widely used in the study of integer sequences is the
so-called Binomial transform \cite{Bin}, which associates to the
sequence with general term $a_n$ the sequence with general term
$b_n$ where
\begin{equation}b_n=\sum_{k=0}^n{{n}\choose{k}}a_k.\end{equation}
If we consider the sequence with general term $a_n$ to be the column vector
$\mathbf{a}=(a_0,a_1,\ldots)'$
 then we obtain the binomial transform of the sequence by multiplying this (infinite)
 vector by the lower-triangle matrix $\mathbf{B}$ whose $(n,k)$-th element is equal to
$\binom{n}{k}$:
\begin{displaymath}\mathbf{B}=\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 2 &amp; 1 &amp;
0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 3 &amp; 3 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 4 &amp; 6 &amp;
4 &amp; 1 &amp; 0 &amp; \ldots
\\1 &amp; 5 &amp; 10 &amp; 10 &amp; 5 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath} This
transformation is invertible, with
\begin{equation}a_n=\sum_{k=0}^n
{{n}\choose{k}}(-1)^{n-k}b_k.\end{equation} We note that
$\mathbf{B}$ corresponds to Pascal's triangle. Its row sums are
$2^n$, while its diagonal sums are the Fibonacci numbers $F(n+1)$.
If $\mathbf{B}^m$ denotes the $m-$th power of $\mathbf{B}$, then the
$n-$th term of $\mathbf{B}^m\mathbf{a}$ where $\mathbf{a}=\{a_n\}$
is given by $\sum_{k=0}^n m^{n-k}{{n}\choose{k}}a_k$.

If $\mathcal{A}(x)$ is the ordinary generating function of the
sequence $a_n$, then the ordinary generating function of the
transformed sequence $b_n$ is
$\frac{1}{1-x}\mathcal{A}(\frac{x}{1-x})$. Similarly, if
$\mathcal{G}(x)$ is the exponential generating function (e.g.f.) of
the sequence $a_n$, then the exponential generating function of the
binomial transform of $a_n$ is $\exp(x)\mathcal{G}(x)$.

 The binomial transform is an element of the exponential Riordan
group, which can be defined as follows.

The \emph{exponential Riordan group} \cite {PasTri}, \cite{ProdMat},
\cite{DeutschShap}, is a set of infinite lower-triangular integer
matrices, where each matrix is defined by a pair of generating
functions $g(x)=1+g_1x+g_2x^2+\ldots$ and $f(x)=f_1x+f_2x^2+\ldots$
where $f_1\ne 0$. The associated matrix is the matrix whose $k$-th
column has exponential generating function $g(x)f(x)^k/k!$ (the
first column being indexed by 0). The matrix corresponding to the
pair $f, g$ is denoted by $[g, f]$. The group law is then given by
\begin{displaymath} [g, f]*[h, l]=[g(h\circ f), l\circ
f].\end{displaymath} The identity for this law is $I=[1,x]$ and the
inverse of $[g, f]$ is $[g, f]^{-1}=[1/(g\circ \bar{f}), \bar{f}]$
where $\bar{f}$ is the compositional inverse of $f$.

If $\mathbf{M}$ is the matrix $[g,f]$, and $\mathbf{a}=\{a_n\}$ is
an integer sequence with exponential generating function
$\mathcal{A}$ $(x)$, then the sequence $\mathbf{M}\mathbf{a}$ has
exponential generating function $g(x)\mathcal{A}(f(x))$.

We shall use the notation $(g(x),f(x))$ to denote an element of the
(ordinary) Riordan group \cite{GSWW}. Riordan group techniques have
been used to provide alternative proofs of many binomial identities
that originally appeared in works such as \cite{Riordan},
\cite{Riordan2}. See for instance \cite{SpruId}, \cite{Spru}.

\begin{example}
The Binomial matrix $\mathbf{B}$ is the element $[e^x,x]$ of the
exponential Riordan group. More generally, $\mathbf{B}^k$ is the
element $[e^{kx},x]$ of the exponential Riordan group. It is easy to
show that the inverse $\mathbf{B}^{-k}$ of $\mathbf{B}^k$ is given
by $[e^{-kx},x]$. Note that as an element of the (ordinary) Riordan
group, $\mathbf{B}$ is given by $(\frac{1}{1-x},\frac{x}{1-x})$.
Similarly $\mathbf{B}^k$ is the element
$(\frac{1}{1-kx},\frac{x}{1-kx})$ of the Riordan group.
\end{example}
\begin{example}
The exponential generating function of the row sums of the matrix
$[g, f]$ is obtained by applying $[g, f]$ to $e^x$, the e.g.f. of
the sequence $1,1,1,\ldots$. Hence the row sums of $[g,f]$ have
e.g.f. $g(x)e^{f(x)}$.
\end{example}

We shall frequently refer to sequences by their sequence number in
the On-Line Encylopedia of Integer Sequences \cite{SL1}, \cite{SL2}.
For instance, Pascal's triangle is \seqnum{A007318} while the
Fibonacci numbers $F(n)$ are
\seqnum{A000045}.

\begin{example} \textbf{The Permutation matrix $\mathbf{P}$ and its inverse}. We
consider the matrix $$\mathbf{P}=[\frac{1}{1-x},x].$$ The general
term $P(n,k)$ of this matrix is easily found: \begin{eqnarray*}
P(n,k)&amp;=&amp;\frac{n!}{k!}[x^n]\frac{x^k}{1-x}\\
&amp;=&amp;\frac{n!}{k!}[x^{n-k}]\frac{1}{1-x}\\
&amp;=&amp;\frac{n!}{k!}[x^{n-k}]\sum_{j=0}^{\infty}x^j\\
&amp;=&amp;\frac{n!}{k!}\qquad \mathrm{if} \qquad n-k\ge 0, \qquad =0,
\qquad
\mathrm{otherwise,}\\
&amp;=&amp; [k\le n]\frac{n!}{k!}.\end{eqnarray*}
Here, we have used the Iverson bracket notation \cite{Concrete}, defined by $[\mathcal{P}]=1$ if the proposition $\mathcal{P}$ is true, and
$[\mathcal{P}]=0$ if $\mathcal{P}$ is false. For instance, $\delta_{ij}=[i=j]$, while
$\delta_n=[n=0]$.

Clearly, the inverse of
this matrix is $\mathbf{P}^{-1}=[(1-x),x]$. The general term of this
matrix is given by
\begin{eqnarray*} P^{-1}(n,k)&amp;=&amp;\frac{n!}{k!}[x^n](1-x)x^k\\
&amp;=&amp;\frac{n!}{k!}[x^{n-k}](1-x)\\
&amp;=&amp;\frac{n!}{k!}(\delta_{n-k}-\delta_{n-k-1})\\
&amp;=&amp;\delta_{n-k}-(k+1)\delta_{n-k-1}\end{eqnarray*} Thus
\begin{displaymath}\mathbf{P}=\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 2 &amp; 2 &amp; 1 &amp;
0 &amp; 0 &amp; 0 &amp; \ldots \\ 6 &amp; 6 &amp; 3 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 24 &amp; 24 &amp;
12&amp; 4 &amp; 1 &amp; 0 &amp; \ldots
\\120 &amp; 120 &amp; 60 &amp; 20 &amp; 5 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath} while

\begin{displaymath}\mathbf{P}^{-1}=\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\-1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; -2 &amp; 1
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; -3 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp;
0&amp; -4 &amp; 1 &amp; 0 &amp; \ldots
\\0 &amp; 0 &amp; 0 &amp; 0 &amp; -5 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath}
\end{example}
\section{The Lah transform}
Introduced by Jovovic (see, for instance, \seqnum{A103194}), the Lah
transform is the transformation on integer sequences whose matrix is
given by $$\mathbf{Lah}=[1,\frac{x}{1-x}].$$ Properties of the
matrix obtained from the $n \times n$ principal sub-matrix of
$\mathbf{Lah}$, and related matrices have been studied in
\cite{LahMatrix}. From the above definition, we see that the matrix
$\mathbf{Lah}$ has general term $\mathrm{Lah}(n,k)$ given by
\begin{eqnarray*}\mathrm{Lah}(n,k)&amp;=&amp;\frac{n!}{k!}[x^n]\frac{x^k}{(1-x)^k}\\
&amp;=&amp;\frac{n!}{k!}[x^{n-k}]\sum_{j=0}^{\infty}\binom{-k}{j}(-1)^jx^j\\
&amp;=&amp;\frac{n!}{k!}[x^{n-k}]\sum{j}\binom{k+j-1}{j}x^j\\
&amp;=&amp;\frac{n!}{k!}\binom{n-1}{n-k}\end{eqnarray*} Thus if $b_n$ is the
Lah transform of the sequence $a_n$, we have $$b_n=\sum_{k=0}^n
\frac{n!}{k!}\binom{n-1}{n-k}a_k.$$ It is clear that the inverse of
this matrix $\mathbf{Lah}^{-1}$ is given by $[1,\frac{x}{1+x}]$ with
general term $\mathrm{Lah}(n,k)(-1)^{n-k}$. Thus
$$a_n=\sum_{k=0}^n(-1)^{n-k}\frac{n!}{k!}\binom{n-1}{n-k}b_k.$$
Numerically, we have
\begin{displaymath}\mathbf{Lah}=\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\0 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 2 &amp; 1 &amp;
0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 6 &amp; 6 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 24 &amp; 36
&amp; 12 &amp; 1 &amp; 0 &amp; \ldots
\\0 &amp; 120 &amp; 240 &amp; 120 &amp; 20 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath} Operating on
the sequence with e.g.f. $f(x)$, it produces the sequence with
e.g.f. $f(\frac{x}{1-x})$. \begin{example} The row sums of the
matrix $\mathbf{Lah}$, obtained by operating on the sequence
$1,1,1\ldots$ with e.g.f. $e^x$, is the sequence $1, 1, 3, 13, 73,
501, \ldots$ (\seqnum{A000262}) with e.g.f. $e^\frac{x}{1-x}$.
We observe that this is
$n!L(n,-1,-1)=n!L_n^{(-1)}(-1)$ (see Appendix for notation). This sequence counts the number of partitions of $\{1,..,n\}$ into any number of lists, where a list means an ordered subset.

\end{example}
\section{The generalized Lah transform}
Extending the definition in \cite{LahMatrix}, we can define, for the
parameter $t$, the generalized Lah matrix
$$\mathbf{Lah}[t]=[1,\frac{x}{1-tx}].$$ It is immediate that
$\mathbf{Lah}[0]=[1,x]=\mathbf{I}$, and
$\mathbf{Lah}[1]=\mathbf{Lah}$. The general term of the matrix
$\mathbf{Lah}[t]$ is easily computed:
\begin{eqnarray*}\mathrm{Lah}[t](n,k)&amp;=&amp;\frac{n!}{k!}[x^n]\frac{x^k}{(1-tx)^k}\\
&amp;=&amp;\frac{n!}{k!}[x^{n-k}]\sum_{j=0}^{\infty}\binom{-k}{j}(-1)^jt^jx^j\\
&amp;=&amp;\frac{n!}{k!}[x^{n-k}]\sum{j}\binom{k+j-1}{j}t^jx^j\\
&amp;=&amp;\frac{n!}{k!}\binom{n-1}{n-k}t^{n-k}.\end{eqnarray*} We can
easily establish that $\mathbf{Lah}[t]^{-1}=[1,\frac{x}{1+tx}]=\mathbf{Lah}[-t]$ with
general term $\frac{n!}{k!}\binom{n-1}{n-k}t^{n-k}(-1)^{n-k}.$ We
also have $$\mathbf{Lah}[u+v]=\mathbf{Lah}[u]\cdot\mathbf{Lah}[v].$$
This follows since
\begin{eqnarray*}\mathbf{Lah}[u]\cdot\mathbf{Lah}[v]&amp;=&amp;[1,\frac{x}{1-ux}][1,\frac{x}{1-vx}]\\
&amp;=&amp;[1,\frac{\frac{x}{1-vx}}{1-\frac{ux}{1-vx}}]\\
&amp;=&amp;[1,\frac{\frac{x}{1-vx}}{\frac{1-vx-ux}{1-vx}}]\\
&amp;=&amp;[1,\frac{x}{1-(u+v)x}]\\
&amp;=&amp;\mathbf{Lah}[u+v].\end{eqnarray*} For integer $m$, it follows
that $$\mathbf{Lah}[mt]=(\mathbf{Lah}[t])^m.$$
\section{Laguerre related transforms}
In this section, we will define the Laguerre transform on integer
sequences to be the transform with matrix given by
$$\mathbf{Lag}=[\frac{1}{1-x},\frac{x}{1-x}].$$ We favour this denomination
through analogy with the Binomial transform, whose matrix is given
by $$(\frac{1}{1-x},\frac{x}{1-x}).$$ Numerically, we have
\begin{displaymath}\mathbf{Lag}=\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 2 &amp; 4 &amp; 1 &amp;
0 &amp; 0 &amp; 0 &amp; \ldots \\ 6 &amp; 18 &amp; 9 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 24 &amp; 96 &amp;
72 &amp; 16 &amp; 1 &amp; 0 &amp; \ldots
\\120 &amp; 600 &amp; 600 &amp; 200 &amp; 25 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath} The inverse of
the Laguerre transform, as we understand it in this section, is
given by
$$\mathbf{Lag}^{-1}=[\frac{1}{1+x},\frac{x}{1+x}].$$ Clearly, the
general term $\mathrm{Lag}(n,k)$ of the matrix $\mathbf{Lag}$ is
given by
$$\mathrm{Lag}(n,k)=\frac{n!}{k!}\binom{n}{k}.$$ Thus if $b_n$ is the Laguerre transform of the sequence
$a_n$, we have
$$b_n=\sum_{k=0}^n\frac{n!}{k!}\binom{n}{k}a_k.$$ The e.g.f. of $b_n$ is given by $\frac{1}{1-x}f(\frac{x}{1-x})$ where $f(x)$ is the e.g.f. of $a_n$. The
inverse matrix of $\mathbf{Lag}$ has general term given by
$(-1)^{n-k}\frac{n!}{k!}\binom{n}{k}$. Thus
$$a_n=\sum_{k=0}^n(-1)^{n-k}\frac{n!}{k!}\binom{n}{k}b_k.$$
The relationship
between the Lah transform with matrix $\mathbf{Lah}$ and the
Laguerre transform with matrix $\mathbf{Lag}$ is now clear:
\begin{eqnarray*} \mathbf{Lag}&amp;=&amp;[\frac{1}{1-x},\frac{x}{1-x}]\\
&amp;=&amp;[\frac{1}{1-x},x][1,\frac{x}{1-x}]\\
&amp;=&amp;\mathbf{P}\cdot \mathbf{Lah}\end{eqnarray*} We note that this
implies that \begin{eqnarray*}
\mathrm{Lag}(n,k)&amp;=&amp;\frac{n!}{k!}\binom{n}{k}\\
&amp;=&amp;\sum_{i=0}^n [i\le
n]\frac{n!}{i!}\frac{i!}{k!}\binom{i-1}{i-k}\\
&amp;=&amp;\sum_{i=0}^n[k \le n]\frac{n!}{k!}\binom{i-1}{i-k}\\
&amp;=&amp;\frac{n!}{k!}\sum_{i=0}^n \binom{i-1}{i-k}\end{eqnarray*} which
indeed is true since $$\binom{n}{k}=\sum_{i=0}^n \binom{i-1}{i-k}.$$
Numerically, we have
\begin{eqnarray*}\mathbf{P}\cdot \mathbf{Lah}&amp;=&amp;\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 2 &amp; 2 &amp; 1 &amp;
0 &amp; 0 &amp; 0 &amp; \ldots \\ 6 &amp; 6 &amp; 3 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 24 &amp; 24 &amp; 12
&amp; 4 &amp; 1 &amp; 0 &amp; \ldots
\\120 &amp; 120 &amp; 60 &amp; 20 &amp; 5 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right) \left(\begin{array}{ccccccc} 1 &amp;
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\0 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp;
2&amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 6 &amp; 6 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 0&amp; 24
&amp; 36 &amp; 12 &amp; 1 &amp; 0 &amp; \ldots
\\0 &amp; 120 &amp; 240 &amp; 120 &amp; 20 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\\
&amp;=&amp; \left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1
&amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 2 &amp; 4 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 6&amp;
18 &amp; 9 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 24 &amp; 96 &amp; 72 &amp; 16 &amp; 1 &amp; 0 &amp; \ldots
\\120 &amp; 600 &amp; 600 &amp; 200 &amp; 25 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\\&amp;=&amp;\mathbf{Lag}\end{eqnarray*}
Similarly we have
$$\mathbf{Lag}^{-1}=\mathbf{Lah}^{-1}\cdot\mathbf{P}^{-1},$$ which
implies that
\begin{eqnarray*}\mathrm{Lag}^{-1}(n,k)&amp;=&amp;(-1)^{n-k}\frac{n!}{k!}\binom{n}{k}\\
&amp;=&amp;\sum_{i=0}^n
(-1)^{n-i}\frac{n!}{i!}\binom{n-1}{n-i}(\delta_{i-k}-(k+1)\delta_{i-k-1}).\end{eqnarray*}
It is of course possible to pass from $\mathbf{Lag}$ to
$\mathbf{Lah}$ by:
$$\mathbf{Lah}=\mathbf{P}^{-1}\cdot\mathbf{Lag}.$$ Thus
\begin{eqnarray*}\mathrm{Lah}(n,k)&amp;=&amp;\frac{n!}{k!}\binom{n-1}{n-k}\\
&amp;=&amp;\sum_{i=0}^n
(\delta_{n-i}-(i+1)\delta_{n-i-1})\frac{i!}{k!}\binom{i}{k}
\end{eqnarray*}
We note in passing that this gives us the identity
$$n!\binom{n-1}{n-k}\\
=\sum_{i=0}^n (\delta_{n-i}-(i+1)\delta_{n-i-1})i!\binom{i}{k}.$$
Numerically, we have
\begin{eqnarray*}\mathbf{P}^{-1}\cdot \mathbf{Lag}&amp;=&amp;\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\-1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; -2 &amp; 1
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; -3 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; 0
&amp; -4 &amp; 1 &amp; 0 &amp; \ldots
\\0 &amp; 0 &amp; 0 &amp; 0 &amp; -5 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right) \left(\begin{array}{ccccccc} 1 &amp;
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 2 &amp;
4&amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 6 &amp; 18 &amp; 9 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 24 &amp;
96 &amp; 72 &amp; 16 &amp; 1 &amp; 0 &amp; \ldots
\\120 &amp; 600 &amp; 600 &amp; 200 &amp; 25 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\\
&amp;=&amp; \left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\0
&amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 2 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0
&amp; 6 &amp; 6 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 24 &amp; 36 &amp; 12 &amp; 1 &amp; 0 &amp; \ldots
\\0 &amp; 120 &amp; 240 &amp; 120 &amp; 20 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\\&amp;=&amp;\mathbf{Lah}\end{eqnarray*}
Thus if the Laguerre transform of $a_n$ has general term $b_n$, then the
general term of the Lah transform of $a_n$ will be given by
$$b_n-nb_{n-1}$$ (for $n &gt; 0)$.
\begin{example}
The row sums of the matrix $\mathbf{Lag}$, that is, the transform of
the sequence $1,1,1,\ldots$ with e.g.f. $e^x$, is the sequence $1,
2, 7, 34, 209, 1546, 13327, \ldots$ with e.g.f.
$\frac{1}{1-x}e^{\frac{x}{1-x}}$. This is \seqnum{A002720}. Among
other things, it counts the number of matchings in the bipartite
graph $K(n,n)$. It general term is $\sum_{k=0}^n
\frac{n!}{k!}\binom{n}{k}$. This is equal to $L_n(-1)$ where
$L_n(x)$ is the $n$-th Laguerre polynomial.
\end{example}
\begin{example} The row sums of the matrix $\mathbf{Lag}^{-1}$ yield
the sequence $1, 0, -1, 4, -15, 56, -185, 204, \ldots$ with e.g.f.
 $\frac{1}{1+x}e^{\frac{x}{1+x}}$. It has general term $\sum_{k=0}^n
 (-1)^{n-k}
\frac{n!}{k!}\binom{n}{k}$ which is equal to $(-1)^n L_n(1)$.
\end{example}
\section{The Associated Laguerre transforms}
The Lah and Laguerre transforms, as defined above, are elements of a one-parameter family
of transforms, whose general element is given by
$$\mathbf{Lag}^{(\alpha)}=[\frac{1}{(1-x)^{\alpha+1}},\frac{x}{1-x}].$$
We can calculate the general term of this matrix in the usual way:
\begin{eqnarray*} \mathrm{Lag}^{(\alpha)}(n,k)&amp;=&amp;\frac{n!}{k!}[x^n](1-x)^{-(\alpha+1)}x^k(1-x)^{-k}\\
&amp;=&amp;\frac{n!}{k!}[x^{n-k}](1-x)^{-(\alpha+k+1)}\\
&amp;=&amp;\frac{n!}{k!}\sum_{j=0}^{\infty}\binom{\alpha+k+j}{j}x^j\\
&amp;=&amp;\frac{n!}{k!}\binom{n+\alpha}{n-k}.\end{eqnarray*}
We note that $\mathbf{Lah}=\mathbf{Lag}^{(-1)}$ while $\mathbf{Lag}=\mathbf{Lag}^{(0)}$.
We can factorize $\mathbf{Lag}^{(\alpha)}$ as follows:
\begin{eqnarray*}\mathbf{Lag}^{(\alpha)}&amp;=&amp;[\frac{1}{(1-x)^{\alpha}},x][\frac{1}{1-x},\frac{x}{1-x}]\\
&amp;=&amp;\mathbf{P}^{(\alpha)}\cdot\mathbf{Lag}\end{eqnarray*}
where $\mathbf{P}^{(\alpha)}$ has general term $\frac{n!}{k!}\binom{n+\alpha-k-1}{n-k}$. Clearly,
$\mathbf{P}^{(1)}=\mathbf{P}$. In fact, we have $\mathbf{P}^{(\alpha)}=\mathbf{P}^\alpha$.
The transform of the sequence $a_n$ by the associated Laguerre transform for $\alpha$ is the sequence $b_n$ with general term
$b_n=\sum_{k=0}^n \frac{n!}{k!}\binom{n+\alpha}{n-k}a_k,$ which has e.g.f. $\frac{1}{(1-x)^{\alpha+1}}f(\frac{x}{1-x})$.

We note that in the literature of Riordan arrays, the subset of
matrices of the form $(1,f(x))$ forms a sub-group, called \emph{the
associated group}. We trust that this double use of the term
``associated'' does not cause confusion.
\section{The Generalized Laguerre transform}
We define, for the parameter $t$, the generalized Laguerre matrix
$\mathbf{Lag}[t]$ to be
$$\mathbf{Lag}[t]=[\frac{1}{1-tx},\frac{x}{1-tx}].$$
We immediately have
\begin{eqnarray*}\mathbf{Lag}[t]&amp;=&amp;[\frac{1}{1-tx},\frac{x}{1-tx}]\\
&amp;=&amp;[\frac{1}{1-tx},x][1,\frac{x}{1-tx}]\\
&amp;=&amp;\mathbf{P}[t]\cdot\mathbf{Lah}[t].\end{eqnarray*} where the
generalized permutation matrix $\mathbf{P}[t]$ has general term $[k
\le n]\frac{n!}{k!}t^{n-k}$. It is clear that
$$\mathbf{Lag}[t]^{-1}=\mathbf{Lag}[-t]=[\frac{1}{1+tx},\frac{x}{1+tx}].$$
It is possible to generalize the associated Laguerre transform
matrices in similar fashion.
\section{Transforming the expansion of $\frac{x}{1-\mu x-\nu x^2}$}
The e.g.f. of the expansion of $\frac{x}{1-\mu x-\nu x^2}$ takes the form
$$f(x)=A(\mu,\nu)e^{r_1 x}+B(\mu,\nu)e^{r_2 x}$$ which follows immediately from the Binet form of the general term.
 Thus the transform of this sequence by $\mathbf{Lag}^{(\alpha)}$ will have e.g.f.
 $$ \frac{A(\mu,\nu)}{(1-x)^{\alpha+1}}e^{\frac{r_1 x}{1-x}}+\frac{B(\mu,\nu)}{(1-x)^{\alpha+1}}e^{\frac{r_2 x}{1-x}}.$$
In the case of the Lah transform ($\alpha=-1$), we get the simple form
$$Ae^{\frac{r_1 x}{1-x}}+Be^{\frac{r_2 x}{1-x}}$$ while in the Laguerre case ($\alpha=0$) we get
$$A\frac{e^{\frac{r_1 x}{1-x}}}{1-x}+B\frac{e^{\frac{r_2 x}{1-x}}}{1-x}$$
But $\frac{e^{\frac{r x}{1-x}}}{1-x}$ is the e.g.f. of the sequence $n!L_n(-r)$. Thus is this case, the transformed
sequence has general term $A n!L_n(-r_1)+B n!L_n(-r_2)$.
\begin{example} The Laguerre transform of the Fibonacci numbers
$$F(n)=\frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2})^n-\frac{1}{\sqrt{5}}(\frac{1-\sqrt{5}}{2})^n$$
is given by $$\frac{1}{\sqrt{5}}n!L_n(-\frac{1+\sqrt{5}}{2})-\frac{1}{\sqrt{5}}n!L_n(-\frac{1-\sqrt{5}}{2}).$$
This is \seqnum{A105277}. It begins $0, 1, 5, 29, 203, 1680, 16058, \ldots$. \end{example}
\begin{example} The Laguerre transform of the Jacobsthal numbers \cite{BarryJac}, \cite{Jac} (expansions of $\frac{x}{1-x-2x^2}$)
$$J(n)=\frac{2^n}{3}-\frac{(-1)^n}{3}$$ is given by
$$\frac{1}{3}n!L_n(-2)-\frac{1}{3}n!L_n(1).$$ This is \seqnum{A129695}. It begins $0,1,5,30,221,1936,19587,\ldots$.
We can use this result to express the Lah transform of the Jacobsthal numbers, since this is equal to $b_n-nb_{n-1}$ where $b_n$ is
the Laguerre transform of $J(n)$. We find
$$\frac{n!}{3}(L_n(-2)-L_{n-1}(-2)-(L_n(1)-L_{n-1}(1))).$$
\end{example}
\begin{example}
We calculate the $\mathbf{Lag}^{(1)}$ transform of the Jacobsthal
numbers $J(n)$. Since
$\mathbf{Lag}^{(1)}=\mathbf{P}\cdot\mathbf{Lag}$, we apply
$\mathbf{P}$ to the Laguerre transform of $J(n)$. This gives us
$$\sum_{k=0}^n
\frac{n!}{k!}(k!L_k(-2)-k!L_k(1))/3=\frac{n!}{3}\sum_{k=0}^n
(L_k(-2)-L_k(1)).$$ This sequence has e.g.f.
$\frac{1}{(1-x)^2}\frac{e^{\frac{2x}{(1-x)}}-e^{\frac{-x}{1-x}}}{3}$.
\end{example}
\section{The Lah and Laguerre transforms and Stirling numbers}
In this section, we follow the notation of \cite{Concrete}. Thus the
Stirling numbers of the first kind, denoted by ${n \brack k}$, are
the elements of the matrix
$$\mathbf{s}=[1, \ln(\frac{1}{1-x})].$$ ${n \brack k}$ counts the number of ways
to arrange $n$ objects into $k$ cycles.

 The Stirling numbers of the
second kind, denoted by ${n \brace k}$, count the number of ways to
partition a set of $n$ things into $k$ nonempty subsets. ${n \brace
k}$ are the elements of the matrix
$$\mathbf{S}=[1, e^x-1].$$ We have
$${n \brace k}=\sum_{i=0}^k \binom{k}{i}\frac{i^n}{k!}.$$
We note that the matrix $[1, e^x-1]$ of Stirling numbers of the
second kind is the inverse of the matrix with elements $(-1)^{n-k}{n
\brack k}$, which is the matrix $[1, \ln(1+x)]$.

Related matrices include $$[\frac{1}{1-x},\ln(\frac{1}{1-x})]$$
whose elements are given by ${n+1 \brack k+1}$
 and its signed version, $$[\frac{1}{1+x},\ln(\frac{1}{1+x})]$$ whose elements are given by $(-1)^{n-k}{n+1 \brack
 k+1}$,
 along with their inverses, given respectively by $[e^{-x},
 1-e^{-x}]$, with general element $(-1)^{n-k}{n+1 \brace k+1}$, and
 $[e^x, e^x-1]$ with general element ${n+1 \brace k+1}$.

We can generalize a result linking the Lah matrix to the Stirling
numbers \cite{LahMatrix} to the infinite matrix case as follows:
$$\mathbf{Lah}=\mathbf{s}\cdot\mathbf{S}.$$
This is because we have
\begin{eqnarray*}
\mathbf{s}\cdot\mathbf{S}&amp;=&amp;[1,\ln(\frac{1}{1-x})][1,e^x-1]\\
&amp;=&amp;[1, e^{\ln(\frac{1}{1-x})}-1]\\
&amp;=&amp;[1,\frac{1}{1-x}-1]\\
&amp;=&amp;[1,\frac{x}{1-x}]=\mathbf{Lah}.\end{eqnarray*} Thus we have
$$\mathbf{S}=\mathbf{s}^{-1}\cdot\mathbf{Lah}, \qquad\qquad
\mathbf{s}=\mathbf{Lah}\cdot\mathbf{S}^{-1}.$$
 We now observe that
 \begin{eqnarray*} [\frac{1}{1-x},\frac{x}{1-x}][1,
 \ln(1+x)]&amp;=&amp;[\frac{1}{1-x},\ln(1+\frac{x}{1-x})]\\
 &amp;=&amp;[\frac{1}{1-x},\ln(\frac{1}{1-x})]\end{eqnarray*} or
 $$\mathbf{Lag}\cdot\left((-1)^{n-k}{n\brack k}\right)=\left({n+1
 \brack k+1}\right).$$
 We deduce the identity
 $$ {n+1 \brack k+1}=\sum_{j=0}^n
 \frac{n!}{j!}\binom{n}{j}(-1)^{j-k}{j \brack k}.$$

 Taking the inverse of the matrix identity above, we obtain
 $$\left({n+1
 \brack k+1}\right)^{-1}=\left((-1)^{n-k}{n\brack
 k}\right)^{-1}\cdot \mathbf{Lag}^{-1}$$
 which can be established alternatively by noting that
 \begin{eqnarray*}
 [1,e^x-1][\frac{1}{1+x},\frac{x}{1+x}]&amp;=&amp;[1.\frac{1}{1+e^x-1},
 \frac{e^x-1}{1+e^x-1}]\\
 &amp;=&amp;[e^{-x},1-e^{-x}].\end{eqnarray*}
 This establishes the identity
 $$(-1)^{n-k}{n+1 \brace k+1}=\sum_{j=0}^n
 (-1)^{j-k}\binom{j}{k}\frac{j!}{k!}{n \brace j}.$$

Finally, from the matrix identity
$$\mathbf{Lag}\cdot\left((-1)^{n-k}{n\brack k}\right)=\left({n+1
 \brack k+1}\right).$$ we deduce that
\begin{eqnarray*}\mathbf{Lag}&amp;=&amp;\left({n+1
 \brack k+1}\right)\cdot \left((-1)^{n-k}{n\brack k}\right)^{-1}\\
&amp;=&amp;\left({n+1
 \brack k+1}\right)\cdot \left({n \brace k}\right).\end{eqnarray*}
 Thus
 $$\mathrm{Lag}(n,k)=\sum_{j=0}^n {n+1 \brack j+1} {j \brace k}.$$
This is equivalent to the factorization
$$\mathbf{Lag}=[\frac{1}{1-x},\frac{x}{1-x}]=[\frac{1}{1-x},\ln(\frac{1}{1-x})][1,e^x-1].$$
This implies (see Appendix A) that
$$L_n(x)=\frac{1}{n!}\sum_{k=0}^n\sum_{j=0}^n {n+1 \brack j+1}{j \brace k}(-x)^k.$$

It is natural in this context to define as \emph{associated Stirling
numbers of the first kind} the elements ${n \brack k }_{\alpha}$ of
the matrices
$$[\frac{1}{(1-x)^{\alpha}},\ln(\frac{1}{1-x})].$$
For instance, ${n \brack k }_{0}={n \brack k }$ and ${n \brack k
}_{1}={n+1 \brack k+1 }$. We note that signed versions of these
numbers have been documented by Lang (see for instance
\seqnum{A049444} and \seqnum{A049458}). To calculate ${n \brack k
}_{2}$, we proceed as follows:
\begin{eqnarray*} \left({n \brack k }_{2}\right)&amp;=&amp;[\frac{1}{(1-x)^{2}},\ln(\frac{1}{1-x})]\\
&amp;=&amp;\mathbf{P} \cdot [\frac{1}{1-x},\ln(\frac{1}{1-x})]\\
&amp;=&amp;\left([k\le n]\frac{n!}{k!}\right) \left( {n+1 \brack
k+1}\right).\end{eqnarray*} Thus
$$ {n \brack k}_2=\sum_{j=0}^n \frac{n!}{j!}{j+1 \brack k+1}=\sum_{j=0}^n \frac{n!}{j!}{n \brack k}_1.$$

 More generally, since
 $[\frac{1}{(1-x)^{\alpha}},\ln(\frac{1}{1-x})]=\mathbf{P}[\frac{1}{(1-x)^{\alpha-1}},\ln(\frac{1}{1-x})]$,
 we have
 $${n \brack k}_{\alpha}=\sum_{j=0}^n \frac{n!}{j!}{n \brack
 k}_{\alpha-1}.$$

 $$\mathrm{Lag}^{(\alpha)}(n,k)=\sum_{j=0}^n {n \brack j}_{\alpha+1}{n \brace
 k}.$$

For example,
\begin{eqnarray*}\mathbf{Lag}^{(1)}&amp;=&amp;[\frac{1}{(1-x)^2},\ln(\frac{1}{1-x})][1,e^x-1]\\
&amp;=&amp;\left(\sum_{j=0}^n {n \brack j}_2 {j \brace k}\right)\\
&amp;=&amp;\left(\sum_{j=0}^n\sum_{i=0}^n\frac{n!}{i!}{n+1 \brack j+1}{j
\brace k}\right).\end{eqnarray*}

In general, we have
$$\mathbf{Lag}^{(\alpha)}=[\frac{1}{(1-x)^{\alpha+1}},\frac{x}{1-x}]=[\frac{1}{(1-x)^{\alpha+1}},\ln(\frac{1}{1-x})][1,e^x-1].$$
This implies that
$$\mathrm{Lag}^{(\alpha)}(n,k)=\sum_{j=0}^n {n \brack j}_{\alpha+1}{n \brace
 k}.$$

\section{The generalized Lah, Laguerre and Stirling matrices}
Given a parameter $t$ define the generalized Stirling numbers of the
first kind to be the elements of the matrix
$$\mathbf{s}[t]=[1,\frac{1}{t}\ln(\frac{1}{1-tx})].$$
Similarly, we define the generalized Stirling numbers of the second
kind to be the elements of the matrix
$$\mathbf{S}[t]=[1,\frac{e^{tx}-1}{t}].$$
Then $$\mathbf{S}[t]^{-1}=\mathbf{s}[-t].$$ For instance,
\begin{eqnarray*}\mathbf{s}[-t]\cdot\mathbf{S}[t]&amp;=&amp;
[1,-\frac{1}{t}\ln(\frac{1}{1+tx})][1,\frac{e^{tx}-1}{t}]\\
&amp;=&amp;[1,-\frac{1}{t}\ln(\frac{1}{1+t\frac{e^{tx}-1}{t}})]\\
&amp;=&amp;[1,-\frac{1}{t}\ln(\frac{1}{e^{tx}})]\\
&amp;=&amp;[1,\frac{1}{t}\ln(e^{tx})]\\
&amp;=&amp;[1,x]=\mathbf{I}.\end{eqnarray*} The general term of
$\mathbf{s}[t]$ is given by $t^{n-k}{n \brack j}$ and that of
$\mathbf{S}[t]$ is given by $t^{n-k}{n \brace j}$. An easy
calculation establishes that
$$\mathbf{Lah}[t]=\mathbf{s}[t]\mathbf{S}[t].$$ From this we immediately deduce that
$$\mathbf{Lag}[t]=\mathbf{P}[t]\mathbf{s}[t]\mathbf{S}[t].$$
Similarly results for the generalized associated Laguerre transform
matrices can be derived.


\section{Appendix A - the Laguerre and associated Laguerre functions}
The associated Laguerre polynomials \cite{Laguerre} are defined by
$$L_n^{(\alpha)}(x)=\frac{1}{n!}\sum_{k=0}^n \frac{n!}{k!}\binom{n+\alpha}{n-k}(-x)^k.$$
Their exponential generating function is
$$\frac{e^{\frac{xz}{1-z}}}{(1-z)^{\alpha+1}}$$
The Laguerre polynomials are given by $L_n(x)=L_n^{(0)}(x).$
The associated Laguerre polynomials are orthogonal on the interval $[0,\infty)$ for the weight
$e^{-x}x^{\alpha}$.

Using the notation developed above, we have
\begin{eqnarray*}L_n^{(\alpha)}(x)&amp;=&amp;\frac{1}{n!}\sum_{k=0}^n
\mathrm{Lag}^{(\alpha)}(n,k)(-x)^k\\
&amp;=&amp;\frac{1}{n!}\sum_{k=0}^n\sum_{i=0}^n {n \brack j}_{\alpha+1}{j
\brace k}(-x)^k.\end{eqnarray*}

In particular,
$$L_n(x)=\frac{1}{n!}\sum_{k=0}^n\sum_{i=0}^n {n+1 \brack j+1}{j
\brace k}(-x)^k.$$

\section{Acknowledgements}
The author would like to thank an anonymous reviewer for their
careful reading of this manuscript and their constructive remarks.

\section{Appendix B - Lah and Laguerre transforms in the OEIS}

\begin{center}
\textbf{Table 1. \ Table of Lah transforms }
\end{center}
\begin{center}
\begin{tabular}{|c|c|}\hline
$a_n$&amp;Lah transform $b_n$\\\hline
$A000290$&amp;$A103194$\\\hline
$A104600$&amp;$A121020$\\\hline
$A000262$&amp;$A025168$\\\hline
$A000079$&amp;$A052897$\\\hline
$A000110$&amp;$A084357$\\\hline
$A000085$&amp;$A049376$\\\hline
$A000670$&amp;$A084358$\\\hline
\end{tabular}\end{center}

\begin{center}
\textbf{Table 2. \ Table of Laguerre transforms }
\end{center}
\begin{center}
\begin{tabular}{|c|c|}\hline
$a_n$&amp;Laguerre transform $b_n$\\\hline
$A000007$&amp;$A000142$\\\hline
$A000012$&amp;$A002720$\\\hline
$A000045$&amp;$A105277$\\\hline
$A000079$&amp;$A087912$\\\hline
$A001045$&amp;$A129695$\\\hline
\end{tabular}\end{center}


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 P. Barry, Triangle Geometry and Jacobsthal Numbers,
\emph{Irish Math. Soc. Bulletin} \textbf{51} (2003), 45--57.

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defined by exponential Riordan arrays}, \emph{J. Integer Sequences}
{\bf 10}, Article 07.3.5.

\bibitem{ProdMat} E. Deutsch, L. Ferrari, and S. Rinaldi,
\href{http://arxiv.org/abs/math/0702638v1}{Production matrices and Riordan
arrays}, arXiv:math.CO/0702638v1, 2007.

\bibitem{DeutschShap} E. Deutsch, L. Shapiro, Exponential
Riordan Arrays, Lecture Notes, Nankai University, 2004, available
electronically at \href{http://www.combinatorics.net/ppt2004/Louis{\%}20W.{\%}20Shapiro/shapiro.pdf}{\tt http://www.combinatorics.net/ppt2004/Louis{\%}20W.{\%}20 Shapiro/shapiro.pdf}, 2007.

\bibitem{GSWW} S. Getu, L. Shapiro, W. Woan, and  L. Woodson,
The Riordan group, {\it Discr. Appl. Math.} {\bf 34} (1991), 229--239.

\bibitem{Concrete} R. L. Graham, D. E. Knuth, and O. Patashnik,
\emph{Concrete Mathematics}, Addison-Wesley, 1994.

\bibitem{LahMatrix} Tan Mingshu, Wang Tianming, Lah Matrix and its
algebraic properties, {\it Ars Combinatoria} {\bf 70} (2004), 97--108.

\bibitem{Riordan} J. Riordan, \emph{An Introduction to Combinatorial Analysis}, Dover, 2002.

\bibitem{Riordan2} J. Riordan, \emph{Combinatorial Identities}, John Wiley \&amp; Sons, 1968.

\bibitem{SL1} N. J. A.~Sloane, \emph{The
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sequences, \emph{Notices of the AMS}, \textbf{50} (2003), 912--915.

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\bibitem{Bin} E. W. Weisstein,
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\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}: Primary
11B83; Secondary 05A19, 15A30, 33C45.

\noindent \emph{Keywords:} Lah transform, Laguerre polynomials,
Stirling numbers, exponential Riordan array.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000079},
\seqnum{A000085},
\seqnum{A000110},
\seqnum{A000142},
\seqnum{A000262},
\seqnum{A000290},
\seqnum{A000670},
\seqnum{A002720},
\seqnum{A007318},
\seqnum{A025168},
\seqnum{A049376},
\seqnum{A052897},
\seqnum{A084357},
\seqnum{A084358},
\seqnum{A087912},
\seqnum{A103194},
\seqnum{A104600},
\seqnum{A105277},
\seqnum{A121020}, and
\seqnum{A129695}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received April 18 2007;
revised version received  May 1 2007.
Published in {\it Journal of Integer Sequences}, May 2 2007.

\bigskip
\hrule
\bigskip

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



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