Combinatorial Polynomials as Moments, Hankel Transforms, and Exponential Riordan Arrays

Paul Barry School of Science Waterford Institute of Technology Ireland

                         pbarry@wit.ie

Abstract In the case of two combinatorial polynomials, we show that they can exhibited as moments of paramaterized families of orthogonal polynomials, and hence derive their Hankel transforms. Exponential Riordan arrays are the main vehicles used for this.

1 Introduction

Let [n] = 1, 2, …, n, and let SP_{n} be the set of set-partitions of [n]. For a set-partition π ∈ SP_{n}, let |π| be the number of parts in π. Then the n-th exponential polynomial, also known as the n-th Touchard polynomial (and sometimes called the n-th Bell polynomial [23]), is given by

e_{n}(z) = ∑_{π} z^{|π|} = ∑_{k=0}^{n} S(n, k)z^{k},

where

S(n, k) = \frac{1}{k!} ∑_{j=0}^{k}(−1)^{k−j}\binom{k}{j}j^{n}

is the general element of the exponential Riordan array

[1, e^{x} − 1] .

This is the matrix of Stirling numbers of the second kind \underline{A008277}, which begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 0 & 1 & 0 & 0 & 0 & 0 & … \\ 0 & 1 & 1 & 0 & 0 & 0 & … \\ 0 & 1 & 3 & 1 & 0 & 0 & … \\ 0 & 1 & 7 & 6 & 1 & 0 & … \\ 0 & 1 & 15 & 25 & 10 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

It is well known [11, 15, 18] that the Hankel transform of these polynomials is given by

z^{\binom{n+1}{2}} ∏_{k=1}^{n} k!.

Now let

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

be the general term of the triangle of Eulerian numbers. The matrix of these numbers \underline{A008292} 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} ) .

A(n, k) is the number of permutations in 𝔖_{n} with k excedances. The Eulerian polynomials EU_{n}(z) are defined by

EU_{n}(z) = ∑_{k=0}^{n} A(n, k)z^{k}.

It is shown in [18] that the Hankel transform of these polynomials is given by

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

These two results are consequences of the following two theorems.

Theorem 1. The polynomials e_{n}(z) are moments of the family of orthogonal polynomials whose coefficient array is given by the inverse of the exponential Riordan array

[e^{z(e^{x}−1)}, e^{x} − 1] .

Theorem 2. The polynomials EU_{n}(z) are moments of the family of orthogonal polynomials whose coefficient array is given by the inverse of the exponential Riordan array

[\frac{e^{zx}(1 − z)}{e^{zx} − ze^{x}}, \frac{e^{x} − e^{zx}}{e^{zx} − ze^{x}} ] .

Note that in the case z = 1, the above matrix is taken to be [\frac{1}{1−x}, \frac{x}{1−x}], whose inverse is the coefficient array of the Laguerre polynomials [2].

While partly expository in nature, this note assumes a certain familiarity with integer sequences, generating functions, orthogonal polynomials [5, 10, 22], Riordan arrays [17, 21], production matrices [9, 14], and the integer Hankel transform [4, 6, 13]. Many interesting examples of sequences and Riordan arrays can be found in Neil Sloane’s On-Line Encyclopedia of Integer Sequences (OEIS), [19, 20]. Sequences are frequently referred to by their OEIS number. For instance, the binomial matrix B (“Pascal’s triangle”) is \underline{A007318}. The plan of the paper is as follows:

1. This Introduction

2. Integer sequences, Hankel transforms, exponential Riordan arrays, orthogonal polynomials

3. Proof of Theorem 1

4. Proof of Theorem 2

2 Integer sequences, Hankel transforms, exponential Riordan arrays, orthogonal polynomials

In this section, we recall known results on integer sequences, Hankel transforms, exponential Riordan arrays and orthogonal polynomials that will be useful for the sequel.

For an integer sequence a_{n}, that is, an element of ℤ^{ℕ}, the power series f_{o}(x) = ∑_{k=0}^{∞} a_{k}x^{k} is called the ordinary generating function or g.f. of the sequence, while f_{e}(x) = ∑_{k=0}^{∞} \frac{a_{k}}{k!} x^{k} is called the exponential generating function or e.g.f. of the sequence. a_{n} is thus the coefficient of x^{n} in f_{o}(x). We denote this by a_{n} = [x^{n}]f_{o}(x). Similarly, a_{n} = n![x^{n}]f_{e}(x). For instance, F_{n} = [x^{n}]\frac{x}{1−x−x^{2}} is the n-th Fibonacci number \underline{A000045}, while n! = n![x^{n}]\frac{1}{1−x}, which says that \frac{1}{1−x} is the e.g.f. of n! \underline{A000142}. For a power series f(x) = ∑_{n=0}^{∞} a_{n}x^{n} with f(0) = 0 and f′(0) ≠ 0 we define the reversion or compositional inverse of f to be the power series f̄(x) = f^{[−1]}(x) such that f(f̄(x)) = x. We sometimes write f̄ = Revf.

The Hankel transform [13] 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 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 − ⋯}}}}

(with β_{i} ≠ 0 for all i) then we have [11, 12, 24]

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

Note that this is independent from α_{n}. In general α_{n} and β_{n} are not integers. Such a continued fraction is associated to a monic family of orthogonal polynomials which obey the three term recurrence

p_{n+1}(x) = (x − α_{n})p_{n}(x) − β_{n}p_{n−1}(x),  p_{0}(x) = 1,  p_{1}(x) = x − α_{0}.

The terms appearing in the first column of the inverse of the coefficient array of these polynomials are the moments of the family.

The exponential Riordan group [1, 7, 9], 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 f_{1} ≠ 0. 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]. It is monic if g_{0} = 1. 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. We use the notation eR to denote this group. 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] are given by g(x)e^{f(x)} since the sequence 1, 1, 1, … has exponential generating function e^{x}.

Example 3. The binomial matrix is the matrix with general term \binom{n}{k}. It is realized by Pascal’s triangle. As an exponential Riordan array, it is given by [e^{x}, x]. We further have

([e^{x}, x])^{m} = [e^{mx}, x].

Example 4. We have

[e^{z(e^{x}−1)}, e^{x} − 1] = [e^{z(e^{x}−1)}, x] · [1, e^{x} − 1] .

A more interesting factorization is given by Proposition 5. The general term of the matrix L = [e^{z(e^{x}−1)}, e^{x} − 1] is given by

L_{n,k} = ∑_{j=0}^{n} S(n, j)\binom{j}{k}z^{j−k}.

Proof. A straight-forward calculation shows that

[e^{z(e^{x}−1)}, e^{x} − 1] = [1, e^{x} − 1] · [e^{zx}, x] .

The assertion now follows since the general term of [1, e^{x} − 1] is S(n, k) and that of [e^{zx}, x] is \binom{n}{k}z^{n−k}.

As an example of the calculation of an inverse, we have the following proposition.

Proposition 6.

[e^{z(e^{x}−1)}, e^{x} − 1]^{−1} = [e^{−zx}, ln(1 + x)] .

Proof. This follows since with

f(x) = e^{x} − 1

we have

f̄(x) = ln(1 + x).

Proposition 7.

[\frac{e^{zx}(1 − z)}{e^{zx} − ze^{x}}, \frac{e^{x} − e^{zx}}{e^{zx} − ze^{x}} ]^{−1} = [1 + zx, \frac{1}{z − 1} ln (\frac{1 + zx}{1 + x})] .

Note that in the case z = 1, we have

[\frac{1}{1 − x}, \frac{x}{1 − x}]^{−1} = [\frac{1}{1 + x}, \frac{x}{1 + x}] .

Proof. This follows since with

f(x) = \frac{e^{zx}(1 − z)}{e^{zx} − ze^{x}}

we have

f̄(x) = \frac{1}{z − 1} ln (\frac{1 + zx}{1 + x}) .

An important concept for the sequel is that of production matrix. The concept of a production matrix [8, 9] is a general one, but for this note we find it convenient to review it in the context of Riordan arrays. Thus let P be an infinite matrix (most often it will have integer entries). Letting r_{0} be the row vector

r_{0} = (1, 0, 0, 0, …),

we define r_{i} = r_{i−1}P, i ≥ 1. Stacking these rows leads to another infinite matrix which we denote by A_{P}. Then P is said to be the production matrix for A_{P}. If we let

u^{T} = (1, 0, 0, 0, …, 0, …)

then we have

A_{P} = ( u^{T} ; u^{T} P ; u^{T} P^{2} ; ⋮ )

and

DA_{P} = A_{P} P

where D = (δ_{i,j+1})_{i,j≥0} (where δ is the usual Kronecker symbol). In [14] P is called the Stieltjes matrix associated to A_{P}. In [9], we find the following result concerning matrices that are production matrices for exponential Riordan arrays.

Proposition 8. Let A = (a_{n,k})_{n,k≥0} = [g(x), f(x)] be an exponential Riordan array and let

c(y) = c_{0} + c_{1}y + c_{2}y^{2} + …,  r(y) = r_{0} + r_{1}y + r_{2}y^{2} + …  (3)

be two formal power series that that

r(f(x)) = f′(x)  (4)
c(f(x)) = \frac{g′(x)}{g(x)}.  (5)

Then

(i)  a_{n+1,0} = ∑_{i} i!c_{i}a_{n,i}  (6)
(ii)  a_{n+1,k} = r_{0}a_{n,k−1} + \frac{1}{k!} ∑_{i≥k} i!(c_{i−k} + kr_{i−k+1})a_{n,i}  (7)

or, defining c_{−1} = 0,

a_{n+1,k} = \frac{1}{k!} ∑_{i≥k−1} i!(c_{i−k} + kr_{i−k+1})a_{n,i}.  (8)

Conversely, starting from the sequences defined by (3), the infinite array (a_{n,k})_{n,k≥0} defined by (8) is an exponential Riordan array.

A consequence of this proposition is that P = (p_{i,j})_{i,j≥0} where

p_{i,j} = \frac{i!}{j!}(c_{i−j} + jr_{r−j+1}) (c_{−1} = 0).

Furthermore, the bivariate exponential generating function

ϕ_{P} (t, z) = ∑_{n,k} p_{n,k}t^{k} \frac{z^{n}}{n!}

of the matrix P is given by

ϕ_{P} (t, z) = e^{tz}(c(z) + tr(z)).

Note in particular that we have

r(x) = f′(f̄(x))

and

c(x) = \frac{g′(f̄(x))}{g(f̄(x))}.

Example 9. We consider the exponential Riordan array [\frac{1}{1−x}, x], \underline{A094587}. This array [2] has elements

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 1 & 0 & 0 & 0 & 0 & … \\ 2 & 2 & 1 & 0 & 0 & 0 & … \\ 6 & 6 & 3 & 1 & 0 & 0 & … \\ 24 & 24 & 12 & 4 & 1 & 0 & … \\ 120 & 120 & 60 & 20 & 5 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

and general term [k ≤ n]\frac{n!}{k!} with inverse

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ −1 & 1 & 0 & 0 & 0 & 0 & … \\ 0 & −2 & 1 & 0 & 0 & 0 & … \\ 0 & 0 & −3 & 1 & 0 & 0 & … \\ 0 & 0 & 0 & −4 & 1 & 0 & … \\ 0 & 0 & 0 & 0 & −5 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

which is the array [1 − x, x]. In particular, we note that the row sums of the inverse, which begin 1, 0, −1, −2, −3, … (that is, 1 − n), have e.g.f. (1 − x) exp(x). This sequence is thus the binomial transform of the sequence with e.g.f. (1 − x) (which is the sequence starting 1, −1, 0, 0, 0, …). In order to calculate the production matrix P of [\frac{1}{1−x}, x] we note that f(x) = x, and hence we have f′(x) = 1 so f′(f̄(x)) = 1. Also g(x) = \frac{1}{1−x} leads to g′(x) = \frac{1}{(1−x)^{2}} , and so, since f̄(x) = x, we get

\frac{g′(f̄(x))}{g(f̄(x))} = \frac{1}{1 − x}.

Thus the generating function for P is

e^{tz} (\frac{1}{1 − z} + t) .

Thus P is the matrix [\frac{1}{1−x}, x] with its first row removed.

Example 10. We consider the exponential Riordan array [1, \frac{x}{1−x}]. The general term of this matrix [2] may be calculated as follows:

T_{n,k} = \frac{n!}{k!}[x^{n}]\frac{x^{k}}{(1 − x)^{k}}
= \frac{n!}{k!}[x^{n−k}](1 − x)^{−k}
= \frac{n!}{k!}[x^{n−k}] ∑_{j=0}^{∞} \binom{−k}{j}(−1)^{j}x^{j}

= \frac{n!}{k!}[x^{n−k}] ∑_{j=0}^{∞} \binom{k + j − 1}{j}x^{j}

= \frac{n!}{k!}\binom{k + n − k − 1}{n − k}
= \frac{n!}{k!}\binom{n − 1}{n − k}.

Thus its row sums, which have e.g.f. exp (\frac{x}{1−x}), have general term ∑_{k=0}^{n} \frac{n!}{k!} \binom{n−1}{n−k}. This is \underline{A000262}, the ‘number of “sets of lists”: the number of partitions of \{1, .., n\} into any number of lists, where a list means an ordered subset’. Its general term is equal to (n−1)!L_{n−1}(1, −1). The inverse of [1, \frac{x}{1−x}] is the exponential Riordan array [1, \frac{x}{1+x}], \underline{A111596}. The row sums of this sequence have e.g.f. exp (\frac{x}{1+x}), and start 1, 1, −1, 1, 1, −19, 151, . . .. This is \underline{A111884}. To calculate the production matrix of [1, \frac{x}{1+x}] we note that g′(x) = 0, while f̄(x) = \frac{x}{1+x} with f′(x) = \frac{1}{(1+x)^{2}} . Thus

f′(f̄(x)) = (1 + x)^{2},

and so the generating function of the production matrix is given by

e^{tz}t(1 + z)^{2}.

The production matrix of the inverse begins

( \begin{matrix}0 & 1 & 0 & 0 & 0 & 0 & … \\ 0 & 2 & 1 & 0 & 0 & 0 & … \\ 0 & 2 & 4 & 1 & 0 & 0 & … \\ 0 & 0 & 6 & 6 & 1 & 0 & … \\ 0 & 0 & 0 & 12 & 8 & 1 & … \\ 0 & 0 & 0 & 0 & 20 & 10 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

Example 11. The exponential Riordan array A = [\frac{1}{1−x}, \frac{x}{1−x}], or

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 1 & 0 & 0 & 0 & 0 & … \\ 2 & 4 & 1 & 0 & 0 & 0 & … \\ 6 & 18 & 9 & 1 & 0 & 0 & … \\ 24 & 96 & 72 & 16 & 1 & 0 & … \\ 120 & 600 & 600 & 200 & 25 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

has general term

T_{n,k} = \frac{n!}{k!}\binom{n}{k}.

Its inverse is [\frac{1}{1+x}, \frac{x}{1+x}] with general term (−1)^{n−k} \frac{n!}{k!} \binom{n}{k}. This is \underline{A021009}, the triangle of coefficients of the Laguerre polynomials L_{n}(x). The production matrix [\frac{1}{1−x}, \frac{x}{1−x}] is given by

( \begin{matrix}1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 3 & 1 & 0 & 0 & 0 & … \\ 0 & 4 & 5 & 1 & 0 & 0 & … \\ 0 & 0 & 9 & 7 & 1 & 0 & … \\ 0 & 0 & 0 & 16 & 9 & 1 & … \\ 0 & 0 & 0 & 0 & 25 & 11 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

Example 12. The exponential Riordan array [e^{x}, ln (\frac{1}{1−x})], or

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 3 & 1 & 0 & 0 & 0 & … \\ 1 & 8 & 6 & 1 & 0 & 0 & … \\ 1 & 24 & 29 & 10 & 1 & 0 & … \\ 1 & 89 & 145 & 75 & 15 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

is the coefficient array for the polynomials

2F_{0}(−n, x; −1)

which are an unsigned version of the Charlier polynomials (of order 0) [10, 16, 22]. This is \underline{A094816}. It is equal to

[e^{x}, x] [1, ln (\frac{1}{1 − x})] ,

or the product of the binomial array B and the array of (unsigned) Stirling numbers of the first kind. The production matrix of the inverse of this matrix is given by

( \begin{matrix}−1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & −2 & 1 & 0 & 0 & 0 & … \\ 0 & 2 & −3 & 1 & 0 & 0 & … \\ 0 & 0 & 3 & −4 & 1 & 0 & … \\ 0 & 0 & 0 & 4 & −5 & 1 & … \\ 0 & 0 & 0 & 0 & 5 & −6 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

which indicates the orthogonal nature of these polynomials. We can prove this as follows. We have

[e^{x}, ln (\frac{1}{1 − x})]^{−1} = [e^{−(1−e^{−x})}, 1 − e^{−x}] .

Hence g(x) = e^{−(1−e^{−x})} and f(x) = 1 − e^{−x}. We are thus led to the equations

r(1 − e^{−x}) = e^{−x},
c(1 − e^{−x}) = −e^{−x},

with solutions r(x) = 1 − x, c(x) = x − 1. Thus the bivariate generating function for the production matrix of the inverse array is

e^{tz}(z − 1 + t(1 − z)),

which is what is required.

3 Proof of Theorem 1

Proof. We show first that with L = [e^{z(e^{x}−1)}, e^{x} − 1], the matrix L^{−1} which is given by

L^{−1} = [e^{z(e^{x}−1)}, e^{x} − 1]^{−1} = [e^{−zx}, ln(1 + x)] ,

is the coefficient array of a family of orthogonal polynomials. To this end, we calculate the production array of [e^{z(e^{x}−1)}, e^{x} − 1]. We have f(x) = e^{x}−1, f′(x) = e^{x} and f̄(x) = ln(1+x). Thus

c(x) = f′(f̄(x)) = 1 + x.

Similarly, for g(x) = e^{z(e^{x}−1)}, we have g′(x) = ze^{z(e^{x}−1)+x} and so

r(x) = \frac{g′(f̄(x))}{g(f̄(x))} = \frac{ze^{zx}(1 + x)}{e^{zx}} = z(1 + x).

Thus the production matrix sought has generating function

e^{tw}(c(w) + tr(w)) = e^{tw}(1 + w + t(z(1 + w))).

Thus the production array P_{L} is tri-diagonal, beginning

( \begin{matrix}z & 1 & 0 & 0 & 0 & 0 & … \\ z & z + 1 & 1 & 0 & 0 & 0 & … \\ 0 & 2z & z + 2 & 1 & 0 & 0 & … \\ 0 & 0 & 3z & z + 3 & 1 & 0 & … \\ 0 & 0 & 0 & 4z & z + 4 & 1 & … \\ 0 & 0 & 0 & 0 & 5z & z + 5 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

Now it is well known that

∑_{k=0}^{n} \frac{e_{k}(z)}{k!}x^{k} = e^{z(e^{x}−1)},

and hence the polynomials e_{n}(z) are the moments of the family of orthogonal polynomials whose coefficient array is L^{−1}.

Corollary 13. The Hankel transform of e_{n}(z) is z^{\binom{n+1}{2}} ∏_{k=1}^{n} k!.

Proof. From the above, we have that the generating function of e_{n}(z) is given by the continued fraction

\frac{1}{1 − zx − \frac{zx^{2}}{1 − (z + 1)x − \frac{2zx^{2}}{1 − (z + 2)x − \frac{3zx^{2}}{1 − ⋯}}}}.

In other words, β_{n} = nz. Thus the Hankel transform of e_{n}(z) is given by

∏_{k=1}^{n} β_{k}^{n−k+1} = ∏_{k=1}^{n}(kz)^{n−k+1} = z^{\binom{n+1}{2}} ∏_{k=1}^{n} k!.

We note that the Hankel transform of the row sums of L = [e^{z(e^{x}−1)}, e^{x} − 1] is equal to

(z + 1)^{\binom{n+1}{2}} ∏_{k=1}^{n} k!.

Note also that if we take z = e^{t}, we obtain a solution to the restricted Toda chain [3].

4 Proof of Theorem 2

Proof. We show first that with L = [\frac{e^{zx}(1−z)}{e^{zx}−ze^{x}}, \frac{e^{x}−e^{zx}}{e^{zx}−ze^{x}} ], the matrix L^{−1} which is given by

L^{−1} = [\frac{e^{zx}(1 − z)}{e^{zx} − ze^{x}}, \frac{e^{x} − e^{zx}}{e^{zx} − ze^{x}} ]^{−1} = [1 + zx, \frac{1}{z − 1} ln (\frac{1 + zx}{1 + x})] ,

is the coefficient array of a family of orthogonal polynomials. To this end, we calculate the production array of L = [\frac{e^{zx}(1−z)}{e^{zx}−ze^{x}}, \frac{e^{x}−e^{zx}}{e^{zx}−ze^{x}} ]. We have f(x) = \frac{e^{x}−e^{zx}}{e^{zx}−ze^{x}}, f̄(x) = \frac{1}{z−1} ln (\frac{1+zx}{1+x}) and

f′(x) = \frac{(1 − z)^{2}e^{x(1+z)}}{(e^{zx} − ze^{x})^{2}} .

Thus

c(x) = f′(f̄(x)) = (1 + x)(1 + zx).

Also g(x) = \frac{e^{zx}(1−z)}{e^{zx}−ze^{x}}, which implies that g′(x) = xf′(x) and so

r(x) = \frac{g′(f̄(x))}{g(f̄(x))} = z(1 + x).

Thus the generating function of P_{L} is given by

e^{tw}(c(w) + tr(w)) = e^{tw}((1 + w)(1 + zw) + t(z(1 + w))).

Thus the production array P_{L} is tri-diagonal, beginning

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

Now it is known that

∑_{k=0}^{n} \frac{EU_{k}(z)}{k!}x^{k} = \frac{e^{zx}(1 − z)}{e^{zx} − ze^{x}},

and hence the polynomials e_{n}(z) are the moments of the family of orthogonal polynomials whose coefficient array is L^{−1}. Corollary 14. The Hankel transform of EU_{n}(z) is z^{\binom{n+1}{2}} ∏_{k=1}^{n} k!^{2}. Proof. From the above, we have that the generating function of EU_{n}(z) is given by the continued fraction

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

In other words, β_{n} = n^{2}z. Thus the Hankel transform of EU_{n}(z) is given by

∏_{k=1}^{n} β_{k}^{n−k+1} = ∏_{k=1}^{n}(k^{2}z)^{n−k+1} = z^{\binom{n+1}{2}} ∏_{k=1}^{n} k!^{2}.

References

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2010 Mathematics Subject Classification: Primary 11B83; Secondary 05A15, 11C20, 15B05, 15B36, 42C05. Keywords: Integer sequence, exponential Riordan array, Touchard polynomial, exponential polynomial, moments, orthogonal polynomials, Hankel determinant, Hankel transform.

(Concerned with sequences \underline{A000045}, \underline{A000142}, \underline{A000262}, \underline{A007318}, \underline{A008277}, \underline{A008292}, \underline{A021009}, \underline{A094587}, \underline{A094816}, \underline{A111596}, and \underline{A111884}.)

Received July 1 2010; revised version received May 31 2011. Published in Journal of Integer Sequences, June 10 2011.

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