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\begin{center}
\vskip 1cm{\LARGE\bf  Constructing Exponential Riordan Arrays \\
\vskip .1in
from Their $A$ and $Z$ 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 show how to construct an exponential Riordan array from a knowledge
of its $A$ and Z sequences. The effect of pre- and post-multiplication by
the binomial matrix on the $A$ and $Z$ sequences is examined, as well as
the effect of scaling the $A$ and $Z$ sequences. Examples are given,
including a discussion of related Sheffer orthogonal polynomials.
\end{abstract}

\section{Introduction}
One of the most fundamental results concerning Riordan arrays is that they have a sequence characterization \cite{He_character, SGWW}. This normally involves two sequences, called the $A$-sequence and the $Z$-sequence. For exponential Riordan arrays \cite{DeutschShap} (see Appendix), this characterization is equivalent to the fact that the production matrix \cite{ProdMat} of an exponential array $[g, f]$, with $A$-sequence $A(t)$ and $Z$-sequence $Z(t)$ has bivariate generating function
$$e^{zt}(Z(t)+A(t)z).$$
In this case we have
$$A(t)=f'(\bar{f}(t)), \quad Z(t)=\frac{g'(\bar{f}(t))}{g(\bar{f}(t))}.$$
Examples of exponential Riordan arrays and their production
matrices may be found in the {\it On-Line Encyclopedia of Integer Sequences}
\cite{SL1, SL2}. In that database, sequences are referred to by their
A-numbers. For known sequences, we shall adopt this convention in this
note.

A natural question to ask is the following. If we are given two suitable power series $A(t)$ and $Z(t)$, can we recover the corresponding exponential Riordan array $[g(t), f(t)]$ whose $A$ and $Z$ sequences correspond to the given power series $A(t)$ and $Z(t)$?

The next two simple results provide a means of doing this.
\begin{lemma} For an exponential Riordan array $[g(t), f(t)]$ with A-sequence $A(t)$, we have
$$ \frac{d}{dt}\bar{f}(t)=\frac{1}{A(t)}.$$
\end{lemma}
\begin{proof}
By definition of the compositional inverse, we have
$$ f(\bar{f}(t))=t.$$
Differentiating this with respect to $t$, we obtain
$$f'(\bar{f}(t)) \frac{d}{dt}\bar{f}(t)=1$$ or
$$ \frac{d}{dt}\bar{f}(t)=\frac{1}{f'(\bar{f}(t))}=\frac{1}{A(t)}.$$
\end{proof}
\begin{lemma} For an exponential Riordan array $[g(t), f(t)]$ with A-sequence $A(t)$ and Z-sequence $Z(t)$, we have
$$ \frac{d}{dt} \ln(g(\bar{f}(t)))=\frac{Z(t)}{A(t)}.$$
\end{lemma}
\begin{proof}
We have
$$\frac{d}{dt} \ln(g(\bar{f}(t)))=\frac{g'(\bar{f}(t))}{g(\bar{f}(t))} \frac{d}{dt}\bar{f}(t)=Z(t) \frac{1}{A(t)}=\frac{Z(t)}{A(t)}.$$ \end{proof}
Thus if we can easily carry out the reversion from $\bar{f}(t)$ to $f(t)$, a knowledge of $A(t)$ and $Z(t)$, along with the equations
\begin{equation} \frac{d}{dt}\bar{f}(t)=\frac{1}{A(t)}, \quad \quad \frac{d}{dt} \ln(g(\bar{f}(t)))=\frac{Z(t)}{A(t)} \end{equation} will allow us to find $f(t)$ and $g(t)$. The steps to achieve this are as follows.
\begin{itemize}
\item Using the equation $\frac{d}{dt}\bar{f}(t)=\frac{1}{A(t)}$, solve for $\bar{f}(t)$.
\item Revert $\bar{f}(t)$ to get $f(t)$.
\item Sove the equation $\frac{d}{dt} \ln(g(\bar{f}(t)))=\frac{Z(t)}{A(t)}$ and take the exponential to get $g(\bar{f}(t))$.
\item Solve for $g(t)$ by substituting $f(t)$ in place of $t$ in the last found expression.
\end{itemize}
Constants of integration may be determined using such conditions as $\bar{f}(0)=f(0)=0$, and $g(0)=1$.

\begin{example} We seek to find $[g(t), f(t)]$ where
$$A(t)=\frac{1}{1+t}, \quad \quad Z(t)=-\frac{1}{1+t}.$$
We start by solving the equation
$$\frac{d}{dt}\bar{f}(t)=1+t.$$
Since $\bar{f}(0)=0$, we find that
$$\bar{f}(t)=t+\frac{t^2}{2}=t\left(1+\frac{t}{2}\right).$$
We revert this to get
$$f(t)=\sqrt{1+2t}-1.$$
We now solve the equation
$$\frac{d}{dt}\ln(g(\bar{f}(t)))=\frac{Z(t)}{A(t)}=-1.$$
Thus we find that
$$\ln(g(\bar{f}(t)))=-t \Rightarrow g(\bar{f}(t))=e^{-t}.$$
Thus (since $\bar{f}(f(t))=t$) we get
$$g(t)=e^{-f(t)}=e^{1-\sqrt{1+2t}}.$$
Hence the exponential Riordan array with the given $A$ and $Z$ sequences is
$$[g, f]= \left[e^{1-\sqrt{1+2t}}, \sqrt{1+2t}-1\right].$$

We note that
$$[g,f]^{-1}=\left[e^t, t+\frac{t^2}{2}\right]$$ which is the Pascal-like matrix \seqnum{A100862} \cite{Barry_Pascal}.

In like manner, we can show that
$$ A(t)=\frac{1}{1+2t}, \quad \quad Z(t)=-\frac{1}{1+2t}$$ corresponds to the exponential Riordan array
$$[g, f]=\left[e^{\frac{1-\sqrt{1+4t}}{2}}, \frac{\sqrt{1+4t}-1}{2}\right],$$ whose inverse
$$[g, f]^{-1}=[e^t, t+t^2]$$ is Pascal-like \cite{Barry_Pascal}. In general, if $A(t)=-Z(t)=\frac{1}{1+rt}$, then
$$[g,f]=\left[e^{\frac{1}{r}(1-\sqrt{1+2rt})}, \frac{1}{r}(\sqrt{1+2rt}-1)\right].$$
Then $$[g,f]^{-1} = \left[e^t, t+r\frac{t^2}{2}\right]$$ is a Pascal-type matrix.

\end{example}
\section{Effect of the binomial transform}
The next proposition shows the effect of changing $Z(t)$ to $Z(t)+1$ and to $Z(t)+A(t)$, respectively. We recall that the binomial matrix $B=[e^t, t]$.
\begin{proposition}
Let $[g, f]$ be an exponential Riordan array with A and Z sequences $A(t)$ and $Z(t)$ respectively.
Then the exponential Riordan array  $ B \cdot [g,f]$ has A and Z sequences $A(t)$ and $Z(t)+1$ respectively, while
the exponential Riordan array $ [g, f]\cdot B$ has A and Z sequences $A(t)$ and $Z(t)+A(t)$ respectively.
\end{proposition}

\begin{proof} Firstly, we let the exponential Riordan array $[h, l]$ have A and Z sequences $A(t)$ and $Z(t)+1$ respectively.
Then we have $\frac{d}{dt}\bar{l}(t)=\frac{1}{A(t)}$, which implies that $l(t)=f(t)$ (since $l(0)=f(0)=0$).
Now
$$ \frac{d}{dt}\ln(h(\bar{l}(t)))=\frac{d}{dt}\ln(h(\bar{f}(t)))=\frac{Z(t)+1}{A(t)}=\frac{Z(t)}{A(t)}+\frac{1}{A(t)}.$$
Thus
$$\ln(h(\bar{f}(t)))=\ln(g(\bar{f}(t)))+\bar{f}(t) \Rightarrow h(\bar{f}(t))=g(\bar{f}(t))e^{\bar{f}(t)}.$$
We obtain that
$$h(t)=g(t)l^t$$ and so
$$[h(t), l(t)]= [e^t g(t), f(t)]=[e^t, t]\cdot [g(t), f(t)]=B\cdot [g(t), f(t)].$$
Secondly, we now assume that the exponential Riordan array $[h, l]$ have A and Z sequences $A(t)$ and $Z(t)+A(t)$ respectively.
As before, we see that $l(t)=f(t)$. Also,
$$ \frac{d}{dt}\ln(h(\bar{l}(t)))=\frac{d}{dt}\ln(h(\bar{f}(t)))=\frac{Z(t)+A(t)}{A(t)}=\frac{Z(t)}{A(t)}+1.$$
Thus
$$\ln(h(\bar{f}(t)))=\ln(g(\bar{f}(t)))+t \Rightarrow h(\bar{f}(t))=g(\bar{f}(t))e^t.$$
Now substituting $f(t)$ for $t$ gives us
$$ h(t)=e^{f(t)} g(t).$$
Thus
$$[h, l]=[e^{f(t)} g(t), f(t)]= [g(t), f(t)]\cdot [e^t, t]=[g(t), f(t)]\cdot B.$$
We shall see examples of these results in the next section.

\end{proof}

\section{Effect of Scaling}
In this section, we will assume that the exponential Riordan array with A and Z sequences $A(t)$ and $Z(t)$, respectively, is given by $[g(t), f(t)]$. We wish to characterize the exponential Riordan array $[g^*(t), f^*(t)]$ whose A and Z sequences are $A^*(t)=rA(t)$ and $Z^*(t)=sZ(t)$ respectively.
\begin{proposition}
We have
$$[g^*(t), f^*(t)] = \left[g(rt)^{\frac{s}{r}}, rf(t)\right].$$
\end{proposition}
\begin{proof}
We have
$$\frac{d}{dt} \bar{f^*}(t)=\frac{1}{rA}=\frac{1}{r}\frac{d}{dt}\bar{f}(t).$$ \noindent
Thus
$$\bar{f^*}(t)=\frac{1}{r}\bar{f}(t) \Rightarrow f^*(t)=r f(t).$$
Then
$$\frac{d}{dt} \ln(g^*(\bar{f^*}(t)))=\frac{sZ}{rA}=\frac{s}{r}\frac{d}{dt} \ln(g(\bar{f}(t))),$$ and so
$$\ln(g^*(\bar{f^*}(t)))=\frac{s}{r}\ln(g(\bar{f}(t)))=\ln\left(g(\bar{f}(t))^{\frac{s}{r}}\right).$$ Thus
$$g^*(\bar{f^*}(t))=g(\bar{f}(t))^{\frac{s}{r}} \Rightarrow g^*(\frac{1}{r}\bar{f}(t))=g(\bar{f}(t))^{\frac{s}{r}} \Rightarrow g^*(\frac{1}{r}t)=g(t)^{\frac{s}{r}}, $$
or
$$g^*(t)=g(rt)^{\frac{s}{r}}.$$
\begin{example}
We let $$A(t)=1+t,\quad \quad Z(t)=1+2t.$$
We find that the corresponding exponential array is
$$[g,f]=\left[e^{2e^t-t-2}, e^t-1\right],$$
which begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 & 0 & 0
&0 & 0 & \cdots \\1 & 1 & 0 & 0 & 0 & 0 & \cdots \\ 3 & 3 & 1 &
0 & 0 & 0 &
\cdots \\ 9 & 13 & 6 & 1 & 0 & 0 & \cdots \\ 35 & 59 & 37 & 10 &
1 & 0 & \cdots \\153 & 301  & 230 & 85 & 15 & 1 &\cdots\\ \vdots
& \vdots &
\vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right),\end{displaymath} with production matrix which begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 1 &
0
& 0 & 0 & 0 & \ldots \\2 & 2 & 1 & 0 & 0 & 0 & \ldots \\ 0 &
4
& 3  & 1 & 0 &
0 & \ldots \\ 0 & 0 & 6   & 4  & 1 & 0 & \ldots \\ 0 & 0 & 0
& 8   & 5  & 1 & \ldots \\0 & 0 & 0 & 0 & 10   & 6
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath}
We now take
$$A^*(t)=3(1+t), \quad \quad Z^*(t)=5(1+2t).$$
The corresponding exponential Riordan array is then given by
$$[g^*(t), f^*(t)]=\left[\left(e^{2e^{3t}-3t-2}\right)^{\frac{5}{3}}, 3(e^t-1)\right].$$
This array begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 & 0 & 0
&0 & 0 & \cdots \\5 & 3 & 0 & 0 & 0 & 0 & \cdots \\ 55 & 33 & 9 &
0 & 0 & 0 &
\cdots \\ 665 & 543 & 162 & 27 & 0 & 0 & \cdots \\ 9895 & 9033 & 3573 & 702 &
81 & 0 & \cdots \\165185 & 170103  & 76410 & 19575 & 2835 & 243 &\cdots\\ \vdots
& \vdots &
\vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right),\end{displaymath} with production matrix which begins
\begin{displaymath}\left(\begin{array}{ccccccc} 5 & 3 &
0
& 0 & 0 & 0 & \ldots \\10 & 8 & 3 & 0 & 0 & 0 & \ldots \\ 0 &
20
& 11  & 3 & 0 &
0 & \ldots \\ 0 & 0 & 30   & 14  & 3 & 0 & \ldots \\ 0 & 0 & 0
& 40  & 17  & 3 & \ldots \\0 & 0 & 0 & 0 & 50   & 20
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath}
\end{example}
\end{proof}
\section{Further examples}
\begin{example} We take the Stirling number related choice of
$$A(t)=1+t, \quad \quad Z(t)=1+t.$$
From
$$\frac{d}{dt}\bar{f}(t)=\frac{1}{1+t},$$ we obtain
$$\bar{f}(t)=\ln(1+t) \Rightarrow f(t)=e^t-1.$$
Then from
$$\frac{d}{dt}\ln(g(\bar{f}(t)))=\frac{Z(t)}{A(t)}=1$$ we obtain
$$\ln(g(\bar{f}(t)))=t \Rightarrow g(\bar{f}(t))=e^t,$$ and hence
$$g(t)=e^{e^t-1}.$$
Thus we obtain $$[g, f]=\left[e^{e^t-1}, e^t-1\right],$$ which is \seqnum{A049020}. We have
$$ [g,f] = S_2 \cdot B$$ where
$S_2$ is the matrix of Stirling numbers of the second kind (\seqnum{A048993}) and $B$ is the binomial matrix (\seqnum{A007318}). The production array of $[g, f]$ is given by
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 1 &
0
& 0 & 0 & 0 & \ldots \\1 & 2 & 1 & 0 & 0 & 0 & \ldots \\ 0 &
2
& 3  & 1 & 0 &
0 & \ldots \\ 0 & 0 & 3   & 4  & 1 & 0 & \ldots \\ 0 & 0 & 0
& 4   & 5  & 1 & \ldots \\0 & 0 & 0 & 0 & 5   & 6
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath}
Since this production matrix is tri-diagonal, the inverse matrix $[g, f]^{-1}$ is the coefficient array of a family of orthogonal polynomials \cite{Barry_Meixner, Barry_Moment}. The family in question is the family of Charlier polynomials, which has the Bell numbers (with e.g.f.\ $e^{e^t-1}$) as moments. The Charlier polynomials satisfy the three-term recurrence
$$P_n(t)=(t-n)P_{n-1}(t)-(n-1)P_{n-2}(t),$$ with
$P_0(t)=1$, $P_1(t)=t-1$.
\end{example}
\begin{example}
We take
$$A(t)=1+t \quad \quad Z(t)=1+t+t^2.$$
Again, we find that
$$f(t)=e^t-1.$$
Then
$$\frac{d}{dt}\ln(g(\bar{f}(t)))=\frac{Z(t)}{A(t)}=\frac{1+t+t^2}{1+t},$$ and hence
$$\ln(g(\bar{f}(t)))=\frac{t^2}{2}+\ln(1+t).$$
Thus
$$g(\bar{f}(t))=e^{\frac{t^2}{2}}(1+t),$$ and so
$$g(t)=e^{\frac{(e^t-1)^2}{2}}(1+e^t-1)=e^t e^{\frac{(e^t-1)^2}{2}}.$$
In this case, the production matrix is four-diagonal and begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 1 &
0
& 0 & 0 & 0 & \ldots \\1 & 2 & 1 & 0 & 0 & 0 & \ldots \\ 2 &
2
& 3  & 1 & 0 &
0 & \ldots \\ 0 & 6 & 3   & 4  & 1 & 0 & \ldots \\ 0 & 0 & 12
& 4   & 5  & 1 & \ldots \\0 & 0 & 0 & 20 & 5   & 6
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath}
The exponential Riordan array
$$ [g, f]=\left[e^t e^{\frac{(e^t-1)^2}{2}}, e^t-1\right]$$ begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 & 0 & 0
&0 & 0 & \cdots \\1 & 1 & 0 & 0 & 0 & 0 & \cdots \\ 2 & 3 & 1 &
0 & 0 & 0 &
\cdots \\ 7 & 10 & 6 & 1 & 0 & 0 & \cdots \\ 29 & 45 & 31 & 10 &
1 & 0 & \cdots \\136 & 241  & 180 & 75 & 15 & 1 &\cdots\\ \vdots
& \vdots &
\vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath}
The row sums of this array are the Dowling numbers \seqnum{A007405}.		

We note that the exponential Riordan array
$$ B^{-1}\cdot [g, f]=[e^{-t}, t]\cdot [g, f]=\left[ e^{\frac{(e^t-1)^2}{2}}, e^t-1\right]$$ has
$$A(t) = 1+t \quad \quad Z(t)=t+t^2.$$
This array begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 & 0 & 0
&0 & 0 & \cdots \\0 & 1 & 0 & 0 & 0 & 0 & \cdots \\ 1 & 1 & 1 &
0 & 0 & 0 &
\cdots \\ 3 & 4 & 3 & 1 & 0 & 0 & \cdots \\ 10 & 19 & 13 & 6 &
1 & 0 & \cdots \\45 & 91  & 75 & 35 & 10 & 1 &\cdots\\ \vdots
& \vdots &
\vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath}
The first column of this array is \seqnum{A060311}, while its row sums are given by \seqnum{A004211}. The production matrix of this array begins
\begin{displaymath}\left(\begin{array}{ccccccc} 0 & 1 &
0
& 0 & 0 & 0 & \ldots \\1 & 1 & 1 & 0 & 0 & 0 & \ldots \\ 2 &
2
& 2  & 1 & 0 &
0 & \ldots \\ 0 & 6 & 3   & 3  & 1 & 0 & \ldots \\ 0 & 0 & 12
& 4   & 4  & 1 & \ldots \\0 & 0 & 0 & 20 & 5   & 5
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right),\end{displaymath} where we see that the effect of the inverse binomial matrix is to subtract $1$ from the diagonal.

In this example, we have $Z(t)=1+t+t^2=A(t)+t^2$. Thus the exponential Riordan array $[g, f]$
is equal to the product
$$[ h, l] \cdot B$$ where the exponential Riordan array
$[h, l]$ has A and Z sequences of $1+t$ and $t^2$, respectively.
\end{example}
\begin{example} We take
$$A(t)=1+t^2, \quad \quad Z(t)=1+t+t^2.$$ 
Then
Thus $$f(t)=\tan(t).$$
Now
$$\frac{d}{dt}\ln(g(\bar{f}(t)))=\frac{Z(t)}{A(t)}=\frac{1+t+t^2}{1+t^2}=1+\frac{t}{1+t^2},$$ and so
$$\ln(g(\bar{f}(t)))=\ln\sqrt{1+t^2}+t.$$
Thus
$$g(\bar{f}(t))=e^t \sqrt{1+t^2} \Rightarrow g(t)=e^{\tan(t)}\sqrt{1+\tan^2(t)}=\frac{e^{\tan(t)}}{\cos(t)}.$$
Thus the sought-for exponential Riordan array is given by
$$[g, f]=\left[e^{\tan(t)} \sec(t), \tan(t)\right].$$
This matrix begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 & 0 & 0
&0 & 0 & \cdots \\1 & 1 & 0 & 0 & 0 & 0 & \cdots \\ 2 & 2 & 1 &
0 & 0 & 0 &
\cdots \\ 6 & 8 & 3 & 1 & 0 & 0 & \cdots \\ 20 & 32 & 20 & 4 &
1 & 0 & \cdots \\92 & 156  & 100 & 40 & 5 & 1 &\cdots\\ \vdots
& \vdots &
\vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right),\end{displaymath} with production matrix that begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 1 &
0
& 0 & 0 & 0 & \ldots \\1 & 1 & 1 & 0 & 0 & 0 & \ldots \\ 2 &
4
& 1  & 1 & 0 &
0 & \ldots \\ 0 & 6 & 9   & 1  & 1 & 0 & \ldots \\ 0 & 0 & 12
& 16   & 1  & 1 & \ldots \\0 & 0 & 0 & 20 & 25   & 1
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath} 
The first column is \seqnum{A009244}.
We note that we have the following factorization
$$[g,f]=\left[e^{\tan(t)} \sec(t), \tan(t)\right]=[\sec(t), \tan(t)]\cdot B.$$
Thus we can say that the exponential Riordan array $[\sec(t), \tan(t)]$, which begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 & 0 & 0
&0 & 0 & \cdots \\0 & 1 & 0 & 0 & 0 & 0 & \cdots \\ 1 & 0 & 1 &
0 & 0 & 0 &
\cdots \\ 0 & 5 & 0 & 1 & 0 & 0 & \cdots \\ 5 & 0 & 14 & 0 &
1 & 0 & \cdots \\0 & 61  & 0 & 30 & 0 & 1 &\cdots\\ \vdots
& \vdots &
\vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right),\end{displaymath} has
A sequence defined by $1+t^2$ and $Z$ sequence defined by $t$. Thus its production matrix is given by
\begin{displaymath}\left(\begin{array}{ccccccc} 0 & 1 &
0
& 0 & 0 & 0 & \ldots \\1 & 0 & 1 & 0 & 0 & 0 & \ldots \\ 0 &
4
& 0  & 1 & 0 &
0 & \ldots \\ 0 & 0 & 9 & 0  & 1 & 0 & \ldots \\ 0 & 0 & 0
& 16  & 0  & 1 & \ldots \\0 & 0 & 0 & 0 & 25  & 0
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath}
We can infer from this that the inverse array
$$[\sec(t), \tan(t)]^{-1}=\left[\frac{1}{\sqrt{1+t^2}}, \tan^{-1}(t)\right]$$ is the coefficient array of the
family of orthogonal polynomials
$$P_n(t)=t P_{n-1}(t)-(n-1)^2P_{n-2}(t),$$
with $P_0(t)=1$ and $P_1(t)=t$.
\end{example}
\begin{example} In this example, we let
$$A(t)=1+t, \quad \quad Z(t)=\frac{1}{1-t}.$$
As before, we get $f(t)=e^t-1$.
Now
$$\frac{d}{dt}\ln(g(\bar{f}(t)))=\frac{Z(t)}{A(t)}=\frac{1}{1-t^2},$$ and hence
$$\ln(g(\bar{f}(t)))=\frac{1}{2}\ln\left( \frac{1+t}{1-t}\right).$$
We infer that
$$g(t)=\sqrt{\frac{e^t}{2-e^t}}.$$
The function $g(t)$ generates the sequence \seqnum{A014307} which begins
$$1, 1, 2, 7, 35, 226, 1787, 16717, 180560, 2211181, 30273047, \ldots.$$ It has many combinatorial interpretations \cite{Callan, Klazar, Ren}.

The exponential Riordan array
$$[g,f] = \left[\sqrt{\frac{e^t}{2-e^t}}, e^t-1\right]$$ begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 & 0 & 0
&0 & 0 & \cdots \\1 & 1 & 0 & 0 & 0 & 0 & \cdots \\ 2 & 3 & 1 &
0 & 0 & 0 &
\cdots \\ 7 & 10 & 6 & 1 & 0 & 0 & \cdots \\ 35 & 45 & 31 & 10 &
1 & 0 & \cdots \\226 & 271  & 180 & 75 & 15 & 1 &\cdots\\ \vdots
& \vdots &
\vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right),\end{displaymath} with production matrix that begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 1 &
0
& 0 & 0 & 0 & \ldots \\1 & 2 & 1 & 0 & 0 & 0 & \ldots \\ 2 &
2
& 3  & 1 & 0 &
0 & \ldots \\ 6 & 6 & 3 & 4  & 1 & 0 & \ldots \\ 24 & 24 & 12
& 4  & 5  & 1 & \ldots \\120 & 120 & 60 & 20 & 5  & 6
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath}
In general, the exponential Riordan array with
$$A(t)=1+t, \quad \quad Z(t)=\frac{r}{1-t},$$ is given by
$$[g,f] = \left[\left(\frac{e^t}{2-e^t}\right)^{r/2}, e^t-1\right].$$
\end{example}
\begin{example} For this example, we take
$$A(t)=e^{-t}, \quad \quad Z(t)=e^t.$$
Then
$$\frac{d}{dt}\bar{f}(t)=\frac{1}{A(t)}=\frac{1}{e^{-t}}=e^t,$$ and so we get
$$\bar{f}(t)=e^t+C=e^t-1$$ since $\bar{f}(0)=0$.
Thus $$f(t)=\ln(1+t).$$
Now
$$\frac{d}{dt}\ln(g(\bar{f}(t)))=\frac{Z(t)}{A(t)}=\frac{e^t}{e^{-t}}=e^{2t},$$ and so
$$\ln(g(\bar{f}(t)))=\frac{e^{2t}}{2}-\frac{1}{2} \Rightarrow g(\bar{f}(t))=e^{\frac{1}{2}(e^{2t}-1)}.$$
Substituting $f(t)$ for $t$ we get
$$g(t)=e^{\frac{1}{2}(e^{2\ln(1+t)}-1)}=e^{t+\frac{t^2}{2}}.$$
Thus
$$[g, f]=\left[e^{t+\frac{t^2}{2}}, \ln(1+t)\right].$$
We note that if we have
$$A(t)=Z(t)=e^{-t},$$ then we obtain
$$[g, f]=[1+t, \ln(1+t)].$$
Interestingly, this last exponential Riordan array has a production matrix that is
equal the ordinary Riordan array
$$\left(\frac{1+2t}{1+t}, \frac{t}{1+t}\right)$$ with its first row removed.
\end{example}
\section{Orthogonal polynomials}
When $Z(t)=\alpha+\beta t$ and $A(t)=1+\gamma t+ \delta t^2$, the production matrix of the corresponding exponential Riordan array $[g, f]$ is tri-diagonal, beginning as follows.
\begin{displaymath}\left(\begin{array}{ccccccc} \alpha & 1 &
0
& 0 & 0 & 0 & \ldots \\\beta & \alpha+\gamma & 1 & 0 & 0 & 0 & \ldots \\ 0 &
2(\beta+\delta)
& \alpha+2 \gamma  & 1 & 0 &
0 & \ldots \\ 0 & 0 & 3(\beta+2\delta) & \alpha+3\gamma  & 1 & 0 & \ldots \\ 0 & 0 & 0
& 4(\beta+3\delta) & \alpha+4\gamma  & 1 & \ldots \\0 & 0 & 0 & 0 & 5(\beta+4\delta)  & \alpha+5\gamma
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath}
As a consequence, $[g, f]^{-1}$ is the coefficient array of the family of orthogonal polynomials $P_n(t)$ defined by the three-term recurrence \cite{Chihara, Gautschi, Szego}
$$P_n(t)=(t-(\alpha+(n-1)\gamma))P_{n-1}(t)-(n-1)(\beta+(n-2)\delta)P_{n-2}(t),$$ with $P_0(t)=1$ and $P_1(t)=x-\alpha$. These are precisely the Sheffer orthogonal polynomials \cite{Salam, He_character}.
\begin{example}
We take the case of
$$A(t)=1+t+t^2,\quad \quad Z(t)=1+t.$$
We have
$$\frac{d}{dt}\bar{f}(t)=\frac{1}{1+t+t^2}.$$
Choosing the constant of integration so that $\bar{f}(0)=0$, we get
$$\bar{f}(t)=\frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2t+1}{\sqrt{3}}\right)-\frac{\pi}{3\sqrt{3}}.$$
Thus
\begin{eqnarray*} f(t)&=&\frac{\sqrt{3}}{2}\tan\left(\frac{\sqrt{3}t}{2}+\frac{\pi}{6}\right)-\frac{1}{2}\\
&=&\frac{2 \sin\left(\frac{\sqrt{3}t}{2}\right)}{\sqrt{3}\cos\left(\frac{\sqrt{3}t}{2}\right)-\sin\left(\frac{\sqrt{3}t}{2}\right)}\\
&=& \frac{2 \tan\left(\frac{\sqrt{3}t}{2}\right)}{\sqrt{3}-\tan\left(\frac{\sqrt{3}t}{2}\right)}.\end{eqnarray*}
We now have
$$\frac{d}{dt}\ln(g(\bar{f}(t)))=\frac{Z(t)}{A(t)}=\frac{1+t}{1+t+t^2},$$ and hence
$$\ln(g(\bar{f}(t)))=\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2t+1}{\sqrt{3}}\right)+\frac{1}{2}\ln(1+t+t^2)-\frac{\pi}{6\sqrt{3}}.$$
From this we infer that
$$g(t)=\frac{\sqrt{3}e^{\frac{x}{2}}}{\sqrt{3}\cos\left(\frac{\sqrt{3}t}{2}\right)-\sin\left(\frac{\sqrt{3}t}{2}\right)}.$$
The function $g(t)$ generates the sequence \seqnum{A049774}, which counts the number of permutations of $n$ elements not containing the consecutive pattern $123$.

The sought-for matrix is thus
$$[g,f]=\left[\frac{\sqrt{3}e^{\frac{x}{2}}}{\sqrt{3}\cos\left(\frac{\sqrt{3}t}{2}\right)-\sin\left(\frac{\sqrt{3}t}{2}\right)},\frac{2 \sin\left(\frac{\sqrt{3}t}{2}\right)}{\sqrt{3}\cos\left(\frac{\sqrt{3}t}{2}\right)-\sin\left(\frac{\sqrt{3}t}{2}\right)}\right].$$
This exponential Riordan array is \seqnum{A182822}, which begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 & 0 & 0
&0 & 0 & \cdots \\1 & 1 & 0 & 0 & 0 & 0 & \cdots \\ 2 & 3 & 1 &
0 & 0 & 0 &
\cdots \\ 5 & 12 & 6 & 1 & 0 & 0 & \cdots \\ 17 & 53 & 39 & 10 &
1 & 0 & \cdots \\70 & 279  & 260 & 95 & 15 & 1 &\cdots\\ \vdots
& \vdots &
\vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right),\end{displaymath} with production matrix that begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 1 &
0
& 0 & 0 & 0 & \ldots \\1 & 2 & 1 & 0 & 0 & 0 & \ldots \\ 0 &
4
& 3  & 1 & 0 &
0 & \ldots \\ 0 & 0 & 9 & 4  & 1 & 0 & \ldots \\ 0 & 0 & 0
& 16  & 5  & 1 & \ldots \\0 & 0 & 0 & 0 & 25  & 6
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath}
\end{example}
\begin{example} We change the previous example slightly by taking
$$A(t)=1+2t+t^2=(1+t)^2,\quad \quad Z(t)=1+t.$$
Then we have
$$\frac{d}{dt}\bar{f}(t)=\frac{1}{(1+t)^2} \Rightarrow \bar{f}(t)=-\frac{1}{1+t}+1=\frac{t}{1+t}.$$
This means that
$$f(t)=\frac{t}{1-t}.$$
Now we have
$$\frac{d}{dt}\ln(g(\bar{f}(t)))=\frac{Z(t)}{A(t)}=\frac{1}{1+t},$$ and hence
$$\ln(g(\bar{f}(t)))=ln(1+t) \Rightarrow g(\bar{f}(t))=1+t.$$
This implies that
$$g(t)=1+f(t)=1+\frac{t}{1-t}=\frac{1}{1-t}.$$
Thus
$$[g,f]=\left[\frac{1}{1-t}, \frac{t}{1-t}\right].$$
Thus $[g,f]^{-1}$ is the coefficient array of the Laguerre polynomials \cite{Barry_Lah}.

We finish by noting that the simple addition of $t$ to $A(t)$ has allowed us to go from the relatively complicated exponential Riordan array
$$\left[\frac{\sqrt{3}e^{\frac{x}{2}}}{\sqrt{3}\cos\left(\frac{\sqrt{3}t}{2}\right)-\sin\left(\frac{\sqrt{3}t}{2}\right)},\frac{2 \sin\left(\frac{\sqrt{3}t}{2}\right)}{\sqrt{3}\cos\left(\frac{\sqrt{3}t}{2}\right)-\sin\left(\frac{\sqrt{3}t}{2}\right)}\right]$$ to the simple exponential Riordan array
$$\left[\frac{1}{1-t}, \frac{t}{1-t}\right].$$
\end{example}
\section{Appendix: exponential Riordan arrays}
 The \emph{exponential Riordan group} \cite
{Barry_Pascal, DeutschShap, ProdMat}, is a set of
infinite lower-triangular integer matrices, where each matrix
is defined by a pair
of generating functions $g(t)=g_0+g_1t+g_2t^2+\cdots$ and
$f(t)=f_1t+f_2t^2+\cdots$ where $g_0 \ne 0$ and $f_1\ne 0$. We usually assume that
$$g_0=f_1=1.$$
The associated
matrix is the matrix
whose $i$-th column has exponential generating function
$g(t)f(t)^i/i!$ (the first column being indexed by 0). The
matrix corresponding to
the pair $f, g$ is denoted by $[g, f]$.  The group law is given by \begin{displaymath}
[g,
f]\cdot [h,
l]=[g(h\circ f), l\circ f].\end{displaymath} The identity for
this law is $I=[1,t]$ 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
$\mathbf{u}=(u_n)_{n \ge 0}$
is an integer sequence with exponential generating function
$\mathcal{U}$
$(t)$, then the sequence $\mathbf{M}\mathbf{u}$ has
exponential
generating function $g(t)\mathcal{U}(f(t))$. Thus the row sums
of the array
$[g,f]$ have exponential generating function given by $g(t)e^{f(t)}$ since the sequence
$1,1,1,\ldots$ has exponential generating function $e^t$.

As an element of the group of exponential Riordan arrays, the binomial matrix $\mathbf{B}$ with $(n,k)$-th element $\binom{n}{k}$ is given by
 $\mathbf{B}=[e^t,t]$. By the above, the exponential
generating function of
its row sums is given by $e^t e^t = e^{2t}$, as expected
($e^{2t}$ is the e.g.f.\ of $2^n$).

To each exponential Riordan array $L=[g,f]$ is associated \cite{ProdMat_0, ProdMat} a matrix $P$ called its \emph{production} matrix, which has
bivariate g.f. given by
$$e^{zt}(Z(t)+A(t)z)$$ where
$$A(t)=f'(\bar{f}(t)), \quad Z(t)=\frac{g'(\bar{f}(t))}{g(\bar{f}(t))}.$$
We have $$P=L^{-1}\bar{L}$$ where $\bar{L}$ \cite{PPWW, Wall} is the matrix $L$ with its top row removed.

The ordinary Riordan group is described in \cite{SGWW}.


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\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}: Primary 11C20;
Secondary 11B83, 15B36, 33C45.  \\ 
\noindent \emph{Keywords:}
exponential Riordan array, $A$ sequence, $Z$ sequence, production
matrix, orthogonal polynomial.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A004211},
\seqnum{A007318},
\seqnum{A007405},
\seqnum{A009244},
\seqnum{A014307},
\seqnum{A048993},
\seqnum{A049020},
\seqnum{A049774},
\seqnum{A060311},
\seqnum{A100862}, and
\seqnum{A182822}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received October 8 2013;
revised version received   December 30 2013.
Published in {\it Journal of Integer Sequences}, January 6 2014.

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