Eulerian Polynomials as Moments, via Exponential Riordan Arrays

Paul Barry School of Science Waterford Institute of Technology Ireland

                     pbarry@wit.ie

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.

1 Introduction

The Eulerian polynomials [9, 14, 17, 21]

P_{n}(x) = ∑_{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}, …, with the well-known triangle of Eulerian numbers [16]

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 4 & 1 & 0 & 0 & 0 & … \\ 1 & 11 & 11 & 1 & 0 & 0 & … \\ 1 & 26 & 66 & 26 & 1 & 0 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

as coefficient array. These coefficients W_{n,k} obey the recurrence [17]

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} = ∑_{i=0}^{n−k}(−1)^{i}\binom{n + 1}{i}(n − k − i)^{n}

holds. The polynomials P_{n}(x) were introduced by Euler [13] in the form

∑_{k=0}^{∞}(k + 1)^{n}t^{k} = \frac{P_{n}(t)}{(1 − t)^{n+1}} .

They have exponential generating function

∑_{n=0}^{∞} P_{n}(x)\frac{t^{n}}{n!} = \frac{(1 − x)e^{(1−x)t}}{1 − xe^{(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 [20] 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 [5, 6]

1. 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;

2. If the production matrix [10, 11, 12] 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; 3. 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′(f̄(x)),

and

Z(x) = \frac{g′(f̄(x))}{g(f̄(x))},

where f̄(x) is the compositional inverse (series reversion) of f(x).

We recall that for an exponential Riordan array

L = [g, f]

the production matrix P_{L} of L [10, 11, 12] is the matrix

P_{L} = L^{−1}L̄,

where L̄ 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 [8, 15, 28]. Continued fractions will be referred to in the sequel; [30] is a general reference, while [18, 19] discuss the connection between continued fractions and orthogonal polynomials, moments and Hankel transforms [20, 25]. We recall that for a given sequence a_{n} its Hankel transform is the sequence of determinants h_{n} = |a_{i+j}|_{0≤i,j≤n}. Many interesting examples of number triangles, including exponential Riordan arrays, can be found in Neil Sloane’s On-Line Encyclopedia of Integer Sequences [26, 27]. Sequences are frequently referred to by their OEIS number. For instance, the binomial matrix (Pascal’s triangle) B with (n, k)-th element \binom{n}{k} is \underline{A007318}.

The following well-known results (the first is the well-known “Favard’s Theorem”), which we essentially reproduce from [18], 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.

Theorem 1. [18] (Cf. [29, Théorème 9 on p.I-4], or [30, Theorem 50.1]). Let (p_{n}(x))_{n≥0} be a sequence of monic polynomials, the polynomial p_{n}(x) having degree n = 0, 1, … Then the sequence (p_{n}(x)) is (formally) orthogonal if and only if there exist sequences (α_{n})_{n≥0} and (β_{n})_{n≥1} with β_{n} ≠ 0 for all n ≥ 1, such that the three-term recurrence

p_{n+1} = (x − α_{n})p_{n}(x) − β_{n}p_{n−1}(x), for n ≥ 1,

holds, with initial conditions p_{0}(x) = 1 and p_{1}(x) = x − α_{0}.

Theorem 2. [18] (Cf. [29, Proposition 1, (7), on p. V-5], or [30, Theorem 51.1]). Let (p_{n}(x))_{n≥0} be a sequence of monic polynomials, which is orthogonal with respect to some functional L. Let

p_{n+1} = (x − α_{n})p_{n}(x) − β_{n}p_{n−1}(x), for n ≥ 1,

be the corresponding three-term recurrence which is guaranteed by Favard’s theorem. Then the generating function

g(x) = ∑_{k=0}^{∞} μ_{k}x^{k}

for the moments μ_{k} = L(x^{k}) satisfies

g(x) = \frac{μ_{0}}{1 − α_{0}x − \frac{β_{1}x^{2}}{1 − α_{1}x − \frac{β_{2}x^{2}}{1 − α_{2}x − \frac{β_{3}x^{2}}{1 − α_{3}x − ⋯}}}}.

The Hankel transform [20] of a given sequence A = {a_{0}, a_{1}, a_{2}, ...} is the sequence of Hankel determinants {h_{0}, h_{1}, h_{2}, … } where h_{n} = |a_{i+j}|_{i,j=0}^{n}, i.e

A = \{a_{n}\}_{n∈ℕ_{0}} → h = \{h_{n}\}_{n∈ℕ_{0}} : h_{n} = | \begin{matrix}a_{0} & a_{1} & ⋯ & a_{n} \\ a_{1} & a_{2} & & a_{n+1} \\ ⋮ & & ... \\ a_{n} & a_{n+1} & & a_{2n} \end{matrix} | .  (1)

The Hankel transform of a sequence a_{n} and that of its binomial transform are equal. In the case that a_{n} has g.f. g(x) expressible in the form

g(x) = \frac{a_{0}}{1 − α_{0}x − \frac{β_{1}x^{2}}{1 − α_{1}x − \frac{β_{2}x^{2}}{1 − α_{2}x − \frac{β_{3}x^{2}}{1 − α_{3}x − ⋯}}}}

then we have [18]

h_{n} = a_{0}^{n+1}β_{1}^{n}β_{2}^{n−1} ⋯ β_{n−1}^{2}β_{n} = a_{0}^{n+1} ∏_{k=1}^{n} β_{k}^{n+1−k}.  (2)

Note that this is independent of α_{n}.

2 The Eulerian polynomials P_{n}(x)

We consider the sequence with e.g.f.

\frac{(α − β)e^{(α−β)t}}{α − βe^{(α−β)t}} .

This is the sequence that begins

1, α, α(α + β), α(α^{2} + 4αβ + β^{2}), α(α^{3} + 11α^{2}β + 11αβ^{2} + β^{3}), ….

Setting α = 1 and β = x gives us the Eulerian polynomials P_{n}(x). We have the Proposition 3. The production matrix of the exponential Riordan array

[\frac{(α − β)e^{(α−β)t}}{α − βe^{(α−β)t}} , \frac{e^{(α−β)t} − 1}{α − βe^{(α−β)t}} ]

is tri-diagonal. Proof. Writing the above exponential Riordan array as [g, f], we have

f(t) = \frac{e^{(α−β)t} − 1}{α − βe^{(α−β)t}}

and hence

f′(t) = \frac{e^{(α+β)t}(α − β)^{2}}{βe^{αt} − αe^{βt}} ,

and

f̄(t) = \frac{1}{α − β} ln (\frac{αt + 1}{βt + 1}) .

Then

A(t) = f′(f̄(t)) = (αt + 1)(βt + 1) = 1 + (α + β)t + αβt^{2}.

We have

g(t) = \frac{(α − β)e^{(α−β)t}}{α − βe^{(α−β)t}}

and hence

g′(t) = \frac{αe^{(α+β)t}(α − β)^{2}}{(βe^{αt} − αe^{βt})^{2}} ,

and so

Z(t) = \frac{g′(f̄(t))}{g(f̄(t))} = α(βt + 1) = α + αβt.

Thus the production matrix, which has bivariate g.f. given by

e^{ty}(α + αβt + (1 + (α + β)t + αβt^{2})y),

is tri-diagonal.

The production matrix takes the form

( \begin{matrix}α & 1 & 0 & 0 & 0 & 0 & … \\ αβ & 2α + β & 1 & 0 & 0 & 0 & … \\ 0 & 4αβ & 3α + 2β & 1 & 0 & 0 & … \\ 0 & 0 & 9αβ & 4α + 3β & 1 & 0 & … \\ 0 & 0 & 0 & 16αβ & 5α + 4β & 1 & … \\ 0 & 0 & 0 & 0 & 25αβ & 6α + 5β & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

For completeness, we observe that while in the special case α = β the Riordan array is not obviously well-defined, the production matrix is, and it leads in this special case to the exponential Riordan array

[\frac{1}{1 − αt}, \frac{t}{1 − αt}]

which has general element \binom{n}{k} \frac{n!}{k!}α^{n−k}. In the case α = β = 1, we get the exponential Riordan array

[\frac{1}{1 − t}, \frac{t}{1 − t}]

whose inverse is the coefficient array of the Laguerre polynomials [3].

Returning now to the Eulerian polynomials, we set α = 1 and β = x, to get Theorem 4. 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}xQ_{n−2}(t).

Proof. The initial polynomial terms of the sequence Q_{n}(t) can be read from the elements of

[\frac{(1 − x)e^{(1−x)t}}{1 − xe^{(1−x)t}} , \frac{e^{(1−x)t} − 1}{1 − xe^{(1−x)t}} ]^{−1} = [\frac{1}{1 + t}, \frac{1}{1 − x} ln (\frac{1 + t}{1 + xt})] ,

which begins

( \begin{matrix}1 & 0 & 0 & 0 & … \\ −1 & 1 & 0 & 0 & … \\ 2 & −x − 3 & 1 & 0 & … \\ −6 & 2x^{2} + 5x + 11 & −3(x + 2) & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

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{matrix}1 & 1 & 0 & 0 & 0 & 0 & … \\ x & 2 + x & 1 & 0 & 0 & 0 & … \\ 0 & 4x & 3 + 2x & 1 & 0 & 0 & … \\ 0 & 0 & 9x & 4 + 3x & 1 & 0 & … \\ 0 & 0 & 0 & 16x & 5 + 4x & 1 & … \\ 0 & 0 & 0 & 0 & 25x & 6 + 5x & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

Corollary 5. The sequence of Eulerian polynomials P_{n}(x) has ordinary generating function given by the continued fraction

\frac{1}{1 − t − \frac{xt^{2}}{1 − (2 + x)t − \frac{4xt^{2}}{1 − (3 + 2x)t − \frac{9xt^{2}}{1 − ⋯}}}}.

Corollary 6. The Hankel transform of the sequence of Eulerian polynomials P_{n}(x) is given by

h_{n} = x^{\binom{n+1}{2}} ∏_{k=1}^{n} k!^{2}.

3 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{(α − β)^{2}e^{(α+β)t}}{βe^{αt} − αe^{βt}} ,

where we retain the use of g(t) = \frac{(α−β)e^{(α−β)t}}{α−βe^{(α−β)t}} from the previous section.

When α = 1 and β = x, g′(t) generates the shifted sequence P_{n+1}(x). We then have

Proposition 7. The production matrix of the exponential Riordan array

[\frac{(α − β)^{2}e^{(α+β)t}}{βe^{αt} − αe^{βt}} , \frac{e^{(α−β)t} − 1}{α − βe^{(α−β)t}} ]

is tri-diagonal.

Proof. As in the previous proposition, we obtain

A(t) = f′(f̄(t)) = (αt + 1)(βt + 1) = 1 + (α + β)t + αβt^{2},

where

f̄(t) = \frac{1}{α − β} ln (\frac{αt + 1}{βt + 1}) .

Then

Z(t) = \frac{g′′(f̄(t))}{g′(f̄(t))} = (α + β) + 2αβt.

The bivariate generating function of the production matrix is then

e^{ty}((α + β) + 2αβt + (1 + (α + β)t + αβt^{2})y),

and hence the production matrix is tri-diagonal.

The production matrix in this case begins

( \begin{matrix}α + β & 1 & 0 & 0 & … \\ 2αβ & 2(α + β) & 1 & 0 & … \\ 0 & 6αβ & 3(α + β) & 1 & … \\ 0 & 0 & 12αβ & 4(α + β) & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

In the case α = β, we obtain the exponential Riordan array

[\frac{1}{(1 − αt)^{2}} , \frac{t}{1 − αt}] ,

with (n, k)-th element \binom{n+1}{k+1} \frac{n!}{k!}α^{n−k}. For α = β = 1 this gives us

[\frac{1}{(1 − t)^{2}} , \frac{t}{1 − t}] ,

which is \underline{A105278}.

Specializing to the values α = 1, β = x, we get the

Theorem 8. 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 > 1,

R_{n}(t) = (t − n(1 + x))R_{n−1}(t) − n(n − 1)xR_{n−2}(t).

Proof. The initial terms of the polynomial sequence R_{n}(t) can be read from the elements of the inverse matrix

[\frac{(α − β)^{2}e^{(α+β)t}}{βe^{αt} − αe^{βt}} , \frac{e^{(1−x)t} − 1}{1 − xe^{(1−x)t}} ]^{−1} = [\frac{1}{(1 + t)(1 + tx)}, \frac{1}{1 − x} ln (\frac{1 + t}{1 + xt})] ,

which begins

( \begin{matrix}1 & 0 & 0 & 0 & … \\ −x − 1 & 1 & 0 & 0 & … \\ 2x^{2} + 2x + 2 & −3(x + 1) & 1 & 0 & … \\ −6(x^{3} + x^{2} + x + 1) & 11x^{2} + 14x + 11 & −6(x + 1) & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

The three-term recurrence is derived from the production matrix, which in this case is

( \begin{matrix}1 + x & 1 & 0 & 0 & 0 & 0 & … \\ 2x & 2(1 + x) & 1 & 0 & 0 & 0 & … \\ 0 & 6x & 3(1 + x) & 1 & 0 & 0 & … \\ 0 & 0 & 12x & 4(1 + x) & 1 & 0 & … \\ 0 & 0 & 0 & 20x & 5(1 + x) & 1 & … \\ 0 & 0 & 0 & 0 & 30x & 6(1 + x) & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

Corollary 9. The sequence of shifted Eulerian polynomials P_{n+1}(x) has ordinary generating function given by the continued fraction

\frac{1}{1 − (1 + x)t − \frac{2xt^{2}}{1 − 2(1 + x)t − \frac{6xt^{2}}{1 − 3(1 + x)t − \frac{12xt^{2}}{1 − ⋯}}}}.

Corollary 10. The Hankel transform of the shifted Eulerian polynomials P_{n+1}(x) is given by

h_{n} = (2x)^{\binom{n+1}{2}} ∏_{k=1}^{n} \binom{k + 2}{2}^{n−k}.

4 The Eulerian number triangles

As with the Narayana numbers and their associated number triangles [7], we can distinguish between three distinct but related triangles of Eulerian numbers. Thus we have the triangle \underline{A173018} [16, 17]

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 4 & 1 & 0 & 0 & 0 & … \\ 1 & 11 & 11 & 1 & 0 & 0 & … \\ 1 & 26 & 66 & 26 & 1 & 0 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

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} = ∑_{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 \underline{A123125} of the coefficients A_{n,k} [1] where

A_{n,k} = ∑_{i=0}^{k}(−1)^{i}\binom{n + 1}{i}(k − i)^{n},

which begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 0 & 1 & 0 & 0 & 0 & 0 & … \\ 0 & 1 & 1 & 0 & 0 & 0 & … \\ 0 & 1 & 4 & 1 & 0 & 0 & … \\ 0 & 1 & 11 & 11 & 1 & 0 & … \\ 0 & 1 & 26 & 66 & 26 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

and finally we have the Pascal-like triangle of coefficients

Ã_{n,k} = A_{n+1,k+1} = ∑_{i=0}^{k+1}(−1)^{i}\binom{n + 2}{i}(k − i)^{n+1},

which begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 4 & 1 & 0 & 0 & 0 & … \\ 1 & 11 & 11 & 1 & 0 & 0 & … \\ 1 & 26 & 66 & 26 & 1 & 0 & … \\ 1 & 57 & 302 & 302 & 57 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

This is \underline{A008292}. We have

Ã_{n,k} = (n − k + 1)Ã_{n−1,k−1} + (k + 1)Ã_{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.

Example 11. The sequence a_{n} = ∑_{k=0}^{n} W_{n,k}2^{k} is the sequence \underline{A000670} of preferential arrangements, or rankings of competitors in a race, with ties [23]. The sequence

b_{n} = ∑_{k=0}^{n} A_{n,k}2^{k} = ∑_{k=0}^{n} W_{n,n−k}2^{k}

or \underline{A000629} is the sequence of rankings of competitors in a race, with ties and dropouts [22]. Note that from our results above, the sequence a_{n} has generating function given by

\frac{1}{1 − x − \frac{2x^{2}}{1 − 4x − \frac{8x^{2}}{1 − 7x − \frac{18x^{2}}{1 − ⋯}}}}.

The g.f. of the sequence a_{n+1} is given by

\frac{1}{1 − 3x − \frac{4x^{2}}{1 − 6x − \frac{12x^{2}}{1 − 9x − \frac{24x^{2}}{1 − ⋯}}}}.

In this case it happens that b_{n} is the binomial transform of a_{n}, and hence [4] its g.f. has continued fraction expression

\frac{1}{1 − 2x − \frac{2x^{2}}{1 − 5x − \frac{8x^{2}}{1 − 8x − \frac{18x^{2}}{1 − ⋯}}}}.

We deduce from this that the Hankel transform of b_{n} is

h_{n} = 2^{\binom{n+1}{2}} ∏_{k=0}^{n} k!^{2},

which is the sequence \underline{A091804} that begins

1, 2, 32, 9216, 84934656, 39137889484800, …

5 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′(f̄(z)) = 1 + μz + νz^{2}.

Now substituting z = f(t) we obtain

f′(f̄(f(t))) = 1 + μf(t) + νf(t)^{2}

or

f′(t) = 1 + μf(t) + νf(t)^{2}

or

\frac{dy}{dt} = 1 + μy + νy^{2},

where y = f(t). In the Eulerian case above, we have

\frac{dy}{dt} = (1 + αy)(1 + βy),

with initial condition y(0) = 0. The form of y = f(t) follows from this variant of the logistic equation.

6 Appendix: exponential Riordan arrays

The exponential Riordan group [2, 10, 12], 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_{1}x + g_{2}x^{2} + ⋯ and f(x) = f_{1}x + f_{2}x^{2} + ⋯ where g_{0} ≠ 0 and f_{1} ≠ 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

[g, f] · [h, l] = [g(h ∘ f), l ∘ f].

The identity for this law is I = [1, x] and the inverse of [g, f] is [g, f]^{−1} = [1/(g ∘ f̄), f̄] where f̄ is the compositional inverse of f.

If M is the matrix [g, f], and u = (u_{n})_{n≥0} is an integer sequence with exponential generating function U (x), then the sequence Mu has exponential generating function g(x)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, … has exponential generating function e^{x}.

As an element of the group of exponential Riordan arrays, the binomial matrix B with (n, k)-th element \binom{n}{k} is given by 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 [11, 12] a matrix P called its production matrix, which has bivariate g.f. given by

e^{xy}(Z(x) + A(x)y)

where

A(x) = f′(f̄(x)), Z(x) = \frac{g′(f̄(x))}{g(f̄(x))}.

We have

P = L^{−1}L̄

where L̄ [24, 30] is the matrix L with its top row removed.

References

[1] M. Aigner, A Course in Enumeration, Springer, 2007.

[2] P. Barry, On a family of generalized Pascal triangles defined by exponential Riordan array, J. Integer Seq., 10 (2007), Article 07.3.5.

[3] P. Barry, Some observations on the Lah and Laguerre transforms of integer sequences, J. Integer Seq., 10 (2007), Article 07.4.6.

[4] P. Barry, Continued Fractions and Transformations of Integer Sequences, J. Integer Seq., 12, (2009), Article 09.7.6.

[5] P. Barry and A. Hennessy, Meixner-type results for Riordan arrays and associated integer sequences, J. Integer Seq., 13 (2010), Article 10.9.4.

[6] P. Barry, Riordan arrays, orthogonal polynomials as moments, and Hankel transforms, J. Integer Seq., 14 (2011), Article 11.2.2.

[7] P. Barry and A. Hennessy, A note on Narayana triangles and related polynomials, Riordan arrays, and MIMO capacity calculations, J. Integer Seq., 14 (2011), Article 11.3.8.

[8] T. S. Chihara, An Introduction to Orthogonal Polynomials, Dover Publications, 2011.

[9] L. Comtet, Advanced Combinatorics, Springer.

[10] E. Deutsch and L. Shapiro, Exponential Riordan arrays, Lecture Notes, Nankai University, 2004, available electronically at http://www.combinatorics.net/ppt2004/Louis%20W.%20Shapiro/shapiro.htm.

[11] E. Deutsch, L. Ferrari, and S. Rinaldi, Production matrices, Adv. in Appl. Math., 34 (2005), 101–122.

[12] E. Deutsch, L. Ferrari, and S. Rinaldi, Production matrices and Riordan arrays, Ann. Comb., 13 (2009), 65–85.

[13] L. Euler, Remarques sur un beau rapport entre les séries des puissances tant directes que réciproques, 1768. E352 (Eneström Index).

[14] D. Foata, Eulerian Polynomials: from Euler’s Time to the Present, in K. Alladi, J. R. Klauder, and C. R. Rao, eds., The Legacy of Alladi Ramakrishnan in the Mathematical Sciences, Springer, 2010.

[15] W. Gautschi, Orthogonal Polynomials: Computation and Approximation, OUP, 2004.

[16] I. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics, 2nd ed., Addison– Wesley, 1994.

[17] F. Hirzebruch, Eulerian polynomials, Munster J. Math., 1 (2008), 9–14.

[18] C. Krattenthaler, Advanced determinant calculus, Séminaire Lotharingien Combin. 42 (1999), Article B42q, available electronically at

http://www.mat.univie.ac.at/ slc/wpapers/s42kratt.html.

[19] C. Krattenthaler, Advanced determinant calculus: A complement, Linear Algebra Appl., 411 (2005), 68–166.

[20] J. W. Layman, The Hankel transform and some of its properties, J. Integer Seq., 4 (2001), Article 01.1.5.

[21] P. Luschny, Eulerian polynomials, http://www.luschny.de/math/euler/EulerianPolynomials.html, 2011.

[22] D. MacHale, personal communication.

[23] E. Mendelson, Races with ties, Math. Mag., 55 (1982), 170–175.

[24] P. Peart and W.-J. Woan, Generating functions via Hankel and Stieltjes matrices, J. Integer Seq., 3(2000), Article 00.2.1.

[25] Ch. Radoux, Déterminants de Hankel et théorème de Sylvester, available electronically

    at www.mat.univie.ac.at/∼slc/opapers/s28radoux.pdf, 2011.

[26] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences. Published electronically at http://oeis.org, 2011.

[27] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, Notices Amer. Math. Soc., 50 (2003), 912–915.

[28] G. Szegö, Orthogonal Polynomials, 4th ed., Amer. Math. Soc., 1975.

[29] G. Viennot, Une théorie combinatoire des polynômes orthogonaux généraux, UQAM, 1983.

[30] H. S. Wall, Analytic Theory of Continued Fractions, AMS Chelsea Publishing, 2001.

2010 Mathematics Subject Classification: Primary 11B83; Secondary 33C45, 42C05, 15B36, 15B05, 11C20. Keywords: Eulerian number, Eulerian polynomial, Euler’s triangle, exponential Riordan array, orthogonal polynomials, moments, Hankel transform.

(Concerned with sequences \underline{A000629}, \underline{A000670}, \underline{A007318}, \underline{A008292}, \underline{A091804}, \underline{A105278}, \underline{A123125}, and \underline{A173018}.)

Received May 6 2011; revised versions received September 16 2011; October 17 2011. Published in Journal of Integer Sequences, October 17 2011.

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