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\begin{center}
\vskip 1cm{\LARGE\bf Eulerian Polynomials as Moments, via \\
\vskip .1in
Exponential Riordan Arrays} \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} 
Using the theory of exponential Riordan arrays and
orthogonal polynomials, we demonstrate that the Eulerian polynomials
and the shifted Eulerian polynomials are moment sequences for a simple
family of orthogonal polynomials. The coefficient arrays of these
families of orthogonal polynomials are shown to be exponential Riordan
arrays. Using the theory of orthogonal polynomials we are then able to
characterize the generating functions of the Eulerian and shifted
Eulerian polynomials in continued fraction form, and to calculate their
Hankel transforms.
\end{abstract}

\section{Introduction}
The Eulerian polynomials \cite{Comtet, Foata, Hirzebruch, Luschny}
$$P_n(x)=\sum_{k=0}^n W_{n,k} x^k$$ form the sequence ${P_n(x)}$ which begins
$$P_0(x)=1, P_1(x)=1, P_2(x)=1+x,P_3(x)=1+4x+x^2,\ldots,$$ with the well-known triangle of Eulerian numbers \cite{Concrete}
\begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\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; 4 &amp; 1 &amp; 0
&amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 11 &amp; 11 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots
\\1 &amp; 26 &amp; 66 &amp; 26 &amp; 1 &amp; 0 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right) \end{displaymath}
as coefficient array. These coefficients $W_{n,k}$ obey the recurrence \cite{Hirzebruch}
$$W_{n,k}=(k+1) W_{n-1,k}+(n-k) W_{n-1, k-1}$$ with appropriate boundary conditions. The closed form expression
$$W_{n,k}=\sum_{i=0}^{n-k} (-1)^i \binom{n+1}{i} (n-k-i)^n$$ holds. The polynomials ${P_n(x)}$ were introduced by Euler \cite{Euler} in the form
$$\sum_{k=0}^{\infty}(k+1)^n t^k = \frac{P_n(t)}{(1-t)^{n+1}}.$$
They have exponential generating function
$$\sum_{n=0}^{\infty} P_n(x) \frac{t^n}{n!} = \frac{(1-x)e^{(1-x)t}}{1-x e^{(1-x)t}}.$$
In this note we show that the sequence of Eulerian polynomials $P_n(x)$ is the moment sequence of a family of orthogonal polynomials. In addition, we show that the sequence of shifted Eulerian polynomials $P_{n+1}(x)$ is similarly the moment sequence of a family of orthogonal polynomials. This will enable us to find the form of the continued fraction expressions for the generating functions of these polynomials, as well as giving us a method for calculating the Hankel transform \cite{Layman} of the Eulerian and shifted Eulerian polynomials.
For this, we will require three results from the theory of exponential Riordan arrays (see Appendix for an introduction to exponential Riordan arrays). These are \cite{Barry_Meixner, Barry_Moment}
\begin{enumerate}
\item The inverse of an exponential Riordan array $[g,f]$ is the coefficient array of a family of orthogonal polynomials if and only if the production matrix of $[g,f]$ is tri-diagonal;
\item If the production matrix \cite{DeutschShap, ProdMat_0, ProdMat} of $[g,f]$ is tri-diagonal, then the elements of the first column of $[g,f]$ are the moments of the corresponding family of orthogonal polynomials;
\item The bivariate generating function of the production matrix of $[g,f]$ is given by
$$e^{xy}(Z(x)+A(x)y)$$ where
$$A(x)=f'(\bar{f}(x)),$$ and
$$Z(x)=\frac{g'(\bar{f}(x))}{g(\bar{f}(x))},$$ where $\bar{f}(x)$ is the compositional inverse (series reversion) of
$f(x)$.
\end{enumerate}

We recall that for an exponential Riordan array $$L=[g,f]$$ the production matrix $P_L$ of $L$ \cite{DeutschShap, ProdMat_0, ProdMat} is the matrix
$$P_L= L^{-1} \bar{L},$$ where $\bar{L}$ is the matrix $L$ with the first row removed.

A quick introduction to exponential Riordan arrays can be found in the Appendix to this note. For general information on orthogonal polynomials and moments, see \cite{Chihara, Gautschi, Szego}. Continued fractions will be referred to in the sequel; \cite{Wall} is a general reference, while \cite{Kratt, Kratt1} discuss the connection between continued fractions and orthogonal polynomials, moments and Hankel transforms \cite{Layman, Radoux}. We recall that for a given sequence $a_n$ its Hankel transform is the sequence of determinants $h_n=|a_{i+j}|_{0 \le i,j \le n}$. Many interesting examples of number triangles, including exponential Riordan arrays, can be found in Neil Sloane's On-Line
Encyclopedia of Integer Sequences \cite{SL1, SL2}. Sequences are frequently referred to by their
OEIS number. For instance, the binomial matrix (Pascal's triangle) $\mathbf{B}$ with $(n,k)$-th element $\binom{n}{k}$ is \seqnum{A007318}.

The following well-known results (the first is the well-known ``Favard's Theorem''), which we essentially reproduce from
\cite{Kratt}, specify the links between orthogonal polynomials, the three-term recurrences that define them, the recurrence coefficients of those three-term recurrences, and the g.f. of
the moment sequence of the orthogonal polynomials.
\begin{theorem} \label{ThreeT}\cite{Kratt} (Cf. \cite[Th\'eor\`eme 9 on p.I-4]{Viennot}, or \cite[Theorem 50.1]{Wall}). Let $(p_n(x))_{n\ge 0}$
be a sequence of monic polynomials, the polynomial $p_n(x)$ having degree $n=0,1,\ldots$ Then the sequence $(p_n(x))$ is (formally)
orthogonal if and only if there exist sequences $(\alpha_n)_{n\ge 0}$ and $(\beta_n)_{n\ge 1}$ with $\beta_n \neq 0$ for all $n\ge 1$,
such that the three-term recurrence
$$p_{n+1}=(x-\alpha_n)p_n(x)-\beta_n p_{n-1}(x), \quad \text{for}\quad n\ge 1, $$
holds, with initial conditions $p_0(x)=1$ and $p_1(x)=x-\alpha_0$.
\end{theorem}

\begin{theorem} \label{CF} \cite{Kratt} (Cf. \cite[Proposition 1, (7), on p. V-5]{Viennot}, or \cite[Theorem 51.1]{Wall}). Let $(p_n(x))_{n\ge
0}$ be a sequence of monic polynomials, which is orthogonal with respect to some functional $\mathcal{L}$. Let
$$p_{n+1}=(x-\alpha_n)p_n(x)-\beta_n p_{n-1}(x), \quad \text{for}\quad n\ge 1, $$ be the corresponding three-term recurrence which is
guaranteed by Favard's theorem. Then the generating function
$$g(x)=\sum_{k=0}^{\infty} \mu_k x^k $$ for the moments $\mu_k=\mathcal{L}(x^k)$ satisfies
$$g(x)=\cfrac{\mu_0}{1-\alpha_0 x-
\cfrac{\beta_1 x^2}{1-\alpha_1 x -
\cfrac{\beta_2 x^2}{1-\alpha_2 x -
\cfrac{\beta_3 x^2}{1-\alpha_3 x -\cdots}}}}.$$
\end{theorem}
\noindent
The {\it
Hankel
transform} \cite{Layman} of a given sequence
$A=\{a_0,a_1,a_2,...\}$ is the
sequence of Hankel determinants $\{h_0, h_1, h_2,\dots \}$
where
$h_{n}=|a_{i+j}|_{i,j=0}^{n}$, i.e

\begin{center} \begin{equation}
 \label{gen1}
 A=\{a_n\}_{n\in\mathbb N_0}\quad \rightarrow \quad
 h=\{h_n\}_{n\in\mathbb N_0}:\quad
h_n=\left| \begin{array}{ccccc}
 a_0\ &amp; a_1\  &amp; \cdots &amp; a_n  &amp;  \\
 a_1\ &amp; a_2\  &amp;        &amp; a_{n+1}  \\
\vdots &amp;      &amp; \ddots &amp;          \\
 a_n\ &amp; a_{n+1}\ &amp;    &amp; a_{2n}
\end{array} \right|. \end{equation} \end{center} The Hankel
transform of a sequence $a_n$ and that of its binomial transform are
equal.

\noindent In the case that $a_n$ has g.f. $g(x)$ expressible in the form
$$g(x)=\cfrac{a_0}{1-\alpha_0 x-
\cfrac{\beta_1 x^2}{1-\alpha_1 x-
\cfrac{\beta_2 x^2}{1-\alpha_2 x-
\cfrac{\beta_3 x^2}{1-\alpha_3 x-\cdots}}}}$$ then
we have \cite{Kratt}
\begin{equation}\label{Kratt} h_n = a_0^{n+1} \beta_1^n\beta_2^{n-1}\cdots \beta_{n-1}^2\beta_n=a_0^{n+1}\prod_{k=1}^n
\beta_k^{n+1-k}.\end{equation}
Note that this is independent of $\alpha_n$.
\section{The Eulerian polynomials $P_n(x)$}
We consider the sequence with e.g.f.
$$\frac{(\alpha-\beta)e^{(\alpha-\beta)t}}{\alpha-\beta e^{(\alpha-\beta)t}}.$$ This is the sequence that begins
$$1, \alpha, \alpha(\alpha+\beta), \alpha(\alpha^2+4\alpha \beta+\beta^2), \alpha(\alpha^3+11\alpha^2 \beta+11\alpha \beta^2+\beta^3),\ldots.$$ Setting $\alpha=1$ and $\beta=x$ gives us the Eulerian polynomials $P_n(x)$.
We have the
\begin{proposition} The production matrix of the exponential Riordan array
$$\left[\frac{(\alpha-\beta)e^{(\alpha-\beta)t}}{\alpha-\beta e^{(\alpha-\beta)t}}, \frac{e^{(\alpha-\beta)t}-1}{\alpha-\beta e^{(\alpha-\beta)t}}\right]$$ is tri-diagonal.
\end{proposition}
\begin{proof}
Writing the above exponential Riordan array as $[g,f]$, we have
$$f(t)=\frac{e^{(\alpha-\beta)t}-1}{\alpha-\beta e^{(\alpha-\beta)t}}$$ and hence
$$f'(t)=\frac{e^{(\alpha+\beta)t}(\alpha-\beta)^2}{\beta e^{\alpha t}-\alpha e^{\beta t}},$$ and
$$\bar{f}(t)=\frac{1}{\alpha-\beta}\ln\left(\frac{\alpha t+1}{\beta t+1}\right).$$
Then $$A(t)=f'(\bar{f}(t))=(\alpha t+1)(\beta t+1)=1+(\alpha+\beta)t+ \alpha \beta t^2.$$
We have $$ g(t)=\frac{(\alpha-\beta)e^{(\alpha-\beta)t}}{\alpha-\beta e^{(\alpha-\beta)t}}$$ and
hence $$g'(t)=\frac{\alpha e^{(\alpha+\beta)t}(\alpha-\beta)^2}{(\beta e^{\alpha t}-\alpha e^{\beta t})^2},$$
and so $$Z(t)=\frac{g'(\bar{f}(t))}{g(\bar{f}(t))}=\alpha (\beta t+1)=\alpha+\alpha \beta t.$$
Thus the production matrix, which has bivariate g.f. given by
$$ e^{t y}(\alpha+\alpha \beta t +(1+(\alpha+\beta)t+ \alpha \beta t^2)y),$$ is tri-diagonal.
\end{proof}
\noindent The production matrix takes the form
\begin{displaymath}\left(\begin{array}{ccccccc}
\alpha
&amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\\alpha \beta  &amp; 2\alpha+\beta &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp;
\ldots \\ 0 &amp; 4\alpha \beta &amp; 3\alpha+2 \beta &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; 9 \alpha \beta &amp; 4\alpha+3\beta  &amp; 1
&amp;
0 &amp; \ldots \\ 0 &amp; 0 &amp; 0 &amp; 16\alpha \beta &amp; 5\alpha+4 \beta &amp; 1 &amp; \ldots \\0 &amp; 0  &amp; 0 &amp; 0
&amp;
25\alpha \beta &amp; 6\alpha+5 \beta
&amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp;
\ddots\end{array}\right).\end{displaymath}
\noindent
For completeness, we observe that while in the special case $\alpha=\beta$ the Riordan array is not obviously well-defined, the production matrix is, and it leads in  this special case to the exponential Riordan array
$$\left[\frac{1}{1-\alpha t},\frac{t}{1-\alpha t}\right]$$ which has general element $\binom{n}{k}\frac{n!}{k!}\alpha^{n-k}$. In the case $\alpha=\beta=1$, we get the exponential Riordan array
 $$\left[\frac{1}{1-t}, \frac{t}{1-t}\right]$$ whose inverse is the coefficient array of the Laguerre polynomials \cite{Lah}.

Returning now to the Eulerian polynomials, we set $\alpha=1$ and $\beta=x$, to get
\begin{theorem}
The Eulerian polynomials $P_n(x)$ are the moments of the family of orthogonal polynomials $Q_n(t)$ defined by
$Q_0(t)=1$, $Q_1(t)=t-1$, and
$$Q_n(t)=(t-((n-1)x+n))Q_{n-1}(t)-(n-1)^2 x Q_{n-2}(t).$$
\end{theorem}
\begin{proof}
The initial polynomial terms of the sequence $Q_n(t)$ can be read from the elements of
$$\left[\frac{(1-x)e^{(1-x)t}}{1-xe^{(1-x)t}},\frac{e^{(1-x)t}-1}{1-xe^{(1-x)t}}\right]^{-1}=\left[\frac{1}{1+t},\frac{1}{1-x}\ln\left(\frac{1+t}{1+xt}\right)\right],$$ which begins
\begin{displaymath}\left(\begin{array}{ccccc} 1 &amp; 0 &amp;
0
&amp; 0  &amp; \ldots \\-1 &amp; 1 &amp; 0 &amp; 0  &amp; \ldots \\ 2 &amp; -x-3
&amp; 1 &amp; 0 &amp; \ldots \\ -6 &amp; 2x^2+5x+11 &amp; -3(x+2) &amp; 1  &amp; \ldots \\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots  &amp;
\ddots\end{array}\right).\end{displaymath}
Hence in particular $Q_0(t)=1$ and $Q_1(t)=t-1$. The three-term recurrence is derived from the production matrix, which in this case is \begin{displaymath}\left(\begin{array}{ccccccc}
1
&amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ x  &amp; 2+x &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp;
\ldots \\ 0 &amp; 4 x &amp; 3+2 x &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; 9  x &amp; 4+3x  &amp; 1
&amp;
0 &amp; \ldots \\ 0 &amp; 0 &amp; 0 &amp; 16 x &amp; 5+4 x &amp; 1 &amp; \ldots \\0 &amp; 0  &amp; 0 &amp; 0
&amp;
25 x &amp; 6+5 x
&amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp;
\ddots\end{array}\right).\end{displaymath}
\end{proof}
\begin{corollary} The sequence of Eulerian polynomials $P_n(x)$ has ordinary generating function given by the continued fraction
$$\cfrac{1}{1-t-
\cfrac{xt^2}{1-(2+x)t-
\cfrac{4xt^2}{1-(3+2x)t-
\cfrac{9xt^2}{1-\cdots}}}}.$$
\end{corollary}
\begin{corollary}
The Hankel transform of the sequence of Eulerian polynomials $P_n(x)$ is given by
$$h_n = x^{\binom{n+1}{2}}\prod_{k=1}^n k!^2.$$
\end{corollary}
\section{The shifted Eulerian polynomials $P_{n+1}(x)$}
For the shifted Eulerian polynomials $P_{n+1}(x)$, we consider the exponential Riordan array
$$[g'(t), f(t)],$$ where
$$g'(t)=\frac{(\alpha-\beta)^2 e^{(\alpha+\beta)t}}{\beta e^{\alpha t}-\alpha e^{\beta t}},$$ where we retain the use of
$g(t)=\frac{(\alpha-\beta)e^{(\alpha-\beta)t}}{\alpha-\beta e^{(\alpha-\beta)t}}$ from the previous section.

When $\alpha=1$ and $\beta=x$, $g'(t)$ generates the shifted sequence $P_{n+1}(x)$.
We then have
\begin{proposition} The production matrix of the exponential Riordan array
$$\left[\frac{(\alpha-\beta)^2 e^{(\alpha+\beta)t}}{\beta e^{\alpha t}-\alpha e^{\beta t}}, \frac{e^{(\alpha-\beta)t}-1}{\alpha-\beta e^{(\alpha-\beta)t}}\right]$$ is tri-diagonal.
\end{proposition}
\begin{proof} As in the previous proposition, we obtain
$$A(t)=f'(\bar{f}(t))=(\alpha t+1)(\beta t+1)=1+(\alpha+\beta)t+ \alpha \beta t^2,$$
where
$$\bar{f}(t)=\frac{1}{\alpha-\beta}\ln\left(\frac{\alpha t+1}{\beta t+1}\right).$$
Then
$$Z(t)=\frac{g''(\bar{f}(t))}{g'(\bar{f}(t))}=(\alpha+\beta)+2\alpha \beta t.$$ The bivariate generating function of the
production matrix is then
$$e^{ty} ((\alpha+\beta)+2\alpha \beta t+(1+(\alpha+\beta)t+ \alpha \beta t^2)y),$$ and hence the
production matrix is tri-diagonal.
\end{proof}
\noindent The production matrix in this case begins
\begin{displaymath}\left(\begin{array}{ccccc} \alpha+\beta &amp; 1 &amp;
0
&amp; 0  &amp; \ldots \\2 \alpha \beta &amp; 2(\alpha+\beta) &amp; 1 &amp; 0  &amp; \ldots \\ 0 &amp; 6\alpha \beta
&amp; 3(\alpha+\beta) &amp; 1 &amp; \ldots \\ 0 &amp; 0 &amp; 12\alpha \beta  &amp; 4(\alpha +\beta)  &amp; \ldots \\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots  &amp;
\ddots\end{array}\right).\end{displaymath}
In the case $\alpha=\beta$, we obtain the exponential Riordan array
$$\left[ \frac{1}{(1-\alpha t)^2},\frac{t}{1-\alpha t}\right],$$ with $(n,k)$-th element $\binom{n+1}{k+1}\frac{n!}{k!}\alpha^{n-k}$. For $\alpha=\beta=1$ this gives us
$$\left[\frac{1}{(1-t)^2}, \frac{t}{1-t}\right],$$ which is \seqnum{A105278}.

Specializing to the values $\alpha=1$, $\beta=x$, we get the
\begin{theorem} The shifted Eulerian polynomials $P_{n+1}(x)$ are the moments of the family of
orthogonal polynomials $R_n(t)$ given by $R_0(t)=1$, $R_1(t)=t-x-1$, and for $n &gt;1$,
$$R_n(t)=(t-n(1+x))R_{n-1}(t)-n(n-1)x R_{n-2}(t).$$
\end{theorem}
\begin{proof}
The initial terms of the polynomial sequence $R_n(t)$ can be read from the elements of the inverse matrix
$$\left[\frac{(\alpha-\beta)^2 e^{(\alpha+\beta)t}}{\beta e^{\alpha t}-\alpha e^{\beta t}},\frac{e^{(1-x)t}-1}{1-xe^{(1-x)t}}\right]^{-1}=\left[\frac{1}{(1+t)(1+tx)},\frac{1}{1-x}\ln\left(\frac{1+t}{1+xt}\right)\right],$$
which begins
\begin{displaymath}\left(\begin{array}{ccccc} 1 &amp; 0 &amp;
0
&amp; 0  &amp; \ldots \\-x-1 &amp; 1 &amp; 0 &amp; 0  &amp; \ldots \\ 2x^2+2x+2 &amp; -3(x+1)
&amp; 1 &amp; 0 &amp; \ldots \\ -6(x^3+x^2+x+1) &amp; 11x^2+14x+11 &amp; -6(x+1) &amp; 1  &amp; \ldots \\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots  &amp;
\ddots\end{array}\right).\end{displaymath}
The three-term recurrence is derived from the production matrix, which in this case is \begin{displaymath}\left(\begin{array}{ccccccc}
1+x
&amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 2x  &amp; 2(1+x) &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp;
\ldots \\ 0 &amp; 6 x &amp; 3(1+x) &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; 12x   &amp; 4(1+x)  &amp; 1
&amp;
0 &amp; \ldots \\ 0 &amp; 0 &amp; 0 &amp; 20 x &amp; 5(1+x) &amp; 1 &amp; \ldots \\0 &amp; 0  &amp; 0 &amp; 0
&amp;
30 x &amp; 6(1+x)
&amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp;
\ddots\end{array}\right).\end{displaymath}
\end{proof}
\begin{corollary} The sequence of shifted Eulerian polynomials $P_{n+1}(x)$ has ordinary generating function given by the continued fraction
$$\cfrac{1}{1-(1+x)t-
\cfrac{2xt^2}{1-2(1+x)t-
\cfrac{6xt^2}{1-3(1+x)t-
\cfrac{12xt^2}{1-\cdots}}}}.$$
\end{corollary}
\begin{corollary} The Hankel transform of the shifted Eulerian polynomials $P_{n+1}(x)$ is given by
$$h_n= (2x)^{\binom{n+1}{2}}\prod_{k=1}^n \binom{k+2}{2}^{n-k}.$$
\end{corollary}
\section{The Eulerian number triangles}
As with the Narayana numbers and their associated number triangles \cite{Narayana}, we can distinguish between three distinct but related triangles of Eulerian numbers.
Thus we have the triangle \seqnum{A173018} \cite{Concrete, Hirzebruch}
\begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\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; 4 &amp; 1 &amp; 0
&amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 11 &amp; 11 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots
\\1 &amp; 26 &amp; 66 &amp; 26 &amp; 1 &amp; 0 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right) \end{displaymath} of Eulerian numbers
$W_{n,k}$ that obey the recurrence
$$W_{n,k}=(k+1) W_{n-1,k}+(n-k) W_{n-1, k-1}$$ with appropriate boundary conditions, for which the  closed form expression
$$W_{n,k}=\sum_{i=0}^{n-k} (-1)^i \binom{n+1}{i} (n-k-i)^n$$ holds.
We have the reversal of this triangle, which is the triangle \seqnum{A123125} of the coefficients $A_{n,k}$ \cite{Aigner} where
$$A_{n,k}=\sum_{i=0}^k (-1)^i \binom{n+1}{i} (k-i)^n,$$ which begins
\begin{displaymath}\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; 1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 1 &amp; 4 &amp; 1
&amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 1 &amp; 11 &amp; 11 &amp; 1 &amp; 0 &amp; \ldots
\\0 &amp; 1 &amp; 26 &amp; 66 &amp; 26 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right), \end{displaymath} and finally we have the Pascal-like triangle of
coefficients $$\tilde{A}_{n,k}=A_{n+1,k+1}=\sum_{i=0}^{k+1} (-1)^i \binom{n+2}{i} (k-i)^{n+1},$$ which begins
\begin{displaymath}\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; 4 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 11 &amp; 11 &amp; 1
&amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 26 &amp; 66 &amp; 26 &amp; 1 &amp; 0 &amp; \ldots
\\1 &amp; 57 &amp; 302 &amp; 302 &amp; 57 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right). \end{displaymath} This is \seqnum{A008292}.

\noindent We have
$$\tilde{A}_{n,k}=(n-k+1)\tilde{A}_{n-1,k-1}+(k+1)\tilde{A}_{n-1,k},$$ with appropriate boundary conditions.
As with the Narayana numbers, each of these triangles has significant combinatorial applications and it is often important to distinguish one from the other.
\begin{example} The sequence $a_n=\sum_{k=0}^n W_{n,k} 2^k$ is the sequence \seqnum{A000670} of preferential arrangements, or rankings of competitors in a race, with ties \cite{Mendelson}. The sequence
$$b_n=\sum_{k=0}^n A_{n,k}2^k = \sum_{k=0}^n W_{n,n-k} 2^k$$ or \seqnum{A000629} is the sequence of rankings of competitors in a race, with ties and dropouts \cite{MacHale}. Note that from our results above, the sequence $a_n$ has generating function given by
$$\cfrac{1}{1-x-
\cfrac{2x^2}{1-4x-
\cfrac{8x^2}{1-7x-
\cfrac{18x^2}{1-\cdots}}}}.$$
The g.f. of the sequence $a_{n+1}$ is given by
$$\cfrac{1}{1-3x-
\cfrac{4x^2}{1-6x-
\cfrac{12x^2}{1-9x-
\cfrac{24x^2}{1-\cdots}}}}.$$
In this case it happens that $b_n$ is the binomial transform of $a_n$, and hence \cite{Barry_CF} its g.f. has continued fraction expression
$$\cfrac{1}{1-2x-
\cfrac{2x^2}{1-5x-
\cfrac{8x^2}{1-8x-
\cfrac{18x^2}{1-\cdots}}}}.$$
\noindent We deduce from this that the Hankel transform of $b_n$ is
$$h_n = 2^{\binom{n+1}{2}}\prod_{k=0}^n k!^2 ,$$ which is the sequence \seqnum{A091804} that begins
$$1, 2, 32, 9216, 84934656, 39137889484800, \ldots$$
\end{example}
\section{A related ODE}
The form of $f(t)$ above is related to a simple ODE. This arises as follows. In order to have a tri-diagonal production matrix, we need to have an expression of the form
$$A(z)=f'(\bar{f}(z))=1+\mu z+ \nu z^2.$$
Now substituting $z= f(t)$ we obtain
$$f'(\bar{f}(f(t)))=1 + \mu f(t)+ \nu f(t)^2$$ or
$$f'(t)=1+ \mu f(t)+ \nu f(t)^2$$ or
$$ \frac{dy}{dt} = 1+ \mu y +\nu y^2,$$ where $y=f(t)$.
In the Eulerian case above, we have
$$\frac{dy}{dt}=(1+\alpha y)(1+\beta y),$$ with initial condition $y(0)=0$. The form of $y=f(t)$  follows from this variant of the logistic equation.
\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 $g(x)=g_0+g_1x+g_2x^2+\cdots$ and
$f(x)=f_1x+f_2x^2+\cdots$ 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
$g(x)f(x)^i/i!$ (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 $I=[1,x]$ 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
$\mathbf{u}=(u_n)_{n \ge 0}$
is an integer sequence with exponential generating function
$\mathcal{U}$
$(x)$, then the sequence $\mathbf{M}\mathbf{u}$ has
exponential
generating function $g(x)\mathcal{U}(f(x))$. Thus the row sums
of the array
$[g,f]$ have exponential generating function given by $g(x)e^{f(x)}$ since the sequence
$1,1,1,\ldots$ has exponential generating function $e^x$.

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
 $\mathbf{B}=[e^x,x]$. By the above, the exponential
generating function of
its row sums is given by $e^x e^x = e^{2x}$, as expected
($e^{2x}$ 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
$$e^{xy}(Z(x)+A(x)y)$$ where
$$A(x)=f'(\bar{f}(x)), \quad Z(x)=\frac{g'(\bar{f}(x))}{g(\bar{f}(x))}.$$
We have $$P=L^{-1}\bar{L}$$ where $\bar{L}$ \cite{PPWW, Wall} is the matrix $L$ with its top row removed.

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\bigskip
\hrule
\bigskip
\noindent 2010 {\it Mathematics Subject Classification}: Primary
11B83; Secondary 33C45, 42C05, 15B36, 15B05, 11C20.


\noindent \emph{Keywords:} Eulerian number, Eulerian polynomial, Euler's triangle, exponential Riordan array, orthogonal polynomials, moments, Hankel transform.

\bigskip
\hrule
\bigskip
\noindent (Concerned with sequences
\seqnum{A000629},
\seqnum{A000670},
\seqnum{A007318},
\seqnum{A008292},
\seqnum{A091804},
\seqnum{A105278},
\seqnum{A123125}, and
\seqnum{A173018}.)

\vspace*{+.1in}
\noindent
Received May 6 2011;
revised versions received  September 16 2011; October 17 2011.
Published in {\it Journal of Integer Sequences}, October 17 2011.

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