Riordan Arrays, Orthogonal Polynomials as Moments, and Hankel Transforms

Paul Barry School of Science Waterford Institute of Technology Ireland

                         pbarry@wit.ie

Abstract

Taking the examples of Legendre and Hermite orthogonal polynomials, we show how to interpret the fact that these orthogonal polynomials are moments of other orthogonal polynomials in terms of their associated Riordan arrays. We use these means to calculate the Hankel transforms of the associated polynomial sequences.

1 Introduction

In this note, we shall re-interpret some of the results of Ismail and Stanton [15, 16] in terms of Riordan arrays. These authors give functionals [15] whose moments are the Hermite, Laguerre, and various Meixner families of polynomials. In this note, we shall confine ourselves to Legendre and Hermite polynomials. Indeed, the types of orthogonal polynomials representable with Riordan arrays is very limited (see below), but it is nevertheless instructive to show that a number of them can be exhibited as moments, again using (parameterized) Riordan arrays.

The essence of the paper is to show that a Riordan array L (either ordinary or exponential) defines a family of orthogonal polynomials (via its inverse L^{−1}) if and only if its production matrix [8, 9, 10] is tri-diagonal. The sequence of moments μ_{n} associated to the family of orthogonal polynomials then appears as the elements of the first column of L. In terms of generating functions, this means that if L = (g, f) (or L = [g, f]), then g(x) is the generating function of the moment sequence. By defining suitable parameterized Riordan arrays, we can exhibit the Legendre and Hermite polynomials as such moment sequences.

While partly expository in nature, the note assumes a certain familiarity with integer sequences, generating functions, orthogonal polynomials [5, 12, 30], Riordan arrays [25, 29], production matrices [10, 23], and the Hankel transform of sequences [2, 7, 20]. We provide background material in this note to give a hopefully coherent narrative. Many interesting examples of sequences and Riordan arrays can be found in Neil Sloane’s On-Line Encyclopedia of Integer Sequences (OEIS), [27, 28]. 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. Preliminaries on integer sequences and (ordinary) Riordan arrays

3. Orthogonal polynomials and Riordan arrays

4. Exponential Riordan arrays and orthogonal polynomials

5. The Hankel transform of an integer sequence

6. Legendre polynomials

7. Legendre polynomials as moments

8. Hermite polynomials

9. Hermite polynomials as moments

10. Acknowledgements

11. Appendix - The Stieltjes transform of a measure

2 Preliminaries on integer sequences and Riordan arrays

For an integer sequence a_{n}, that is, an element of ℤ^{ℕ}, the power series f(x) = ∑_{n=0}^{∞} a_{n}x^{n} is called the ordinary generating function or g.f. of the sequence. a_{n} is thus the coefficient of x^{n} in this series. We denote this by a_{n} = [x^{n}]f(x). For instance, F_{n} = [x^{n}]\frac{x}{1−x−x^{2}} is the n-th Fibonacci number \underline{A000045}, while C_{n} = [x^{n}]\frac{1−\sqrt{1−4x}}{2x} is the n-th Catalan number \underline{A000108}. The article [21] gives examples of the use of the operator [x^{n}]. We use the notation 0^{n} = [x^{n}]1 for the sequence 1, 0, 0, 0, …, \underline{A000007}. Thus 0^{n} = [n = 0] = δ_{n,0} = \binom{0}{n}. Here, we have used the Iverson bracket notation [13], defined by [P] = 1 if the proposition P is true, and [P] = 0 if P is false.

For a power series f(x) = ∑_{n=0}^{∞} a_{n}x^{n} with f(0) = 0 we define the reversion or compositional inverse of f to be the power series f̄(x) such that f(f̄(x)) = x. We shall sometimes write this as f̄ = Revf.

For a lower triangular matrix (a_{n,k})_{n,k≥0} the row sums give the sequence with general term ∑_{k=0}^{n} a_{n,k} while the diagonal sums form the sequence with general term

∑_{k=0}^{⌊ \frac{n}{2} ⌋} a_{n−k,k}.

The Riordan group [25, 29], is a set of infinite lower-triangular integer matrices, where each matrix is defined by a pair of generating functions g(x) = 1 + g_{1}x + g_{2}x^{2} + ⋯ and f(x) = f_{1}x + f_{2}x^{2} + ⋯ where f_{1} ≠ 0 [29]. We assume in addition that f_{1} = 1 in what follows. The associated matrix is the matrix whose i-th column is generated by g(x)f(x)^{i} (the first column being indexed by 0). The matrix corresponding to the pair g, f is denoted by (g, f) or R(g, f). The group law is then given by

(g, f) · (h, l) = (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.

A Riordan array of the form (g(x), x), where g(x) is the generating function of the sequence a_{n}, is called the sequence array of the sequence a_{n}. Its general term is a_{n−k} (or more precisely, [k ≤ n]a_{n−k}). Such arrays are also called Appell arrays as they form the elements of the so-called Appell subgroup.

If M is the matrix (g, f), and a = (a_{0}, a_{1}, …)′ is an integer sequence with ordinary generating function A (x), then the sequence Ma has ordinary generating function g(x)A(f(x)). The (infinite) matrix (g, f) can thus be considered to act on the ring of integer sequences ℤ^{ℕ} by multiplication, where a sequence is regarded as a (infinite) column vector. We can extend this action to the ring of power series ℤ[[x]] by

(g, f) : A(x) ↦ (g, f) · A(x) = g(x)A(f(x)).

Example 1. The so-called binomial matrix B is the element (\frac{1}{1−x}, \frac{x}{1−x}) of the Riordan group. It has general element \binom{n}{k}, and hence as an array coincides with Pascal’s triangle. More generally, B^{m} is the element (\frac{1}{1−mx}, \frac{x}{1−mx}) of the Riordan group, with general term \binom{n}{k}m^{n−k}. It is easy to show that the inverse B^{−m} of B^{m} is given by (\frac{1}{1+mx}, \frac{x}{1+mx}).

Example 2. If a_{n} has generating function g(x), then the generating function of the sequence

b_{n} = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} a_{n−2k}

is equal to

\frac{g(x)}{1 − x^{2}} = (\frac{1}{1 − x^{2}} , x) · g(x),

while the generating function of the sequence

d_{n} = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n − k}{k}a_{n−2k}

is equal to

\frac{1}{1 − x^{2}} g (\frac{x}{1 − x^{2}} ) = (\frac{1}{1 − x^{2}} , \frac{x}{1 − x^{2}} ) · g(x).

The row sums of the matrix (g, f) have generating function

(g, f) · \frac{1}{1 − x} = \frac{g(x)}{1 − f(x)}

while the diagonal sums of (g, f) (sums of left-to-right diagonals in the North East direction) have generating function g(x)/(1 − xf(x)). These coincide with the row sums of the “generalized” Riordan array (g, xf):

(g, xf) · \frac{1}{1 − x} = \frac{g(x)}{1 − xf(x)}.

For instance the Fibonacci numbers F_{n+1} are the diagonal sums of the binomial matrix B given by (\frac{1}{1−x}, \frac{x}{1−x}):

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 2 & 1 & 0 & 0 & 0 & … \\ 1 & 3 & 3 & 1 & 0 & 0 & … \\ 1 & 4 & 6 & 4 & 1 & 0 & … \\ 1 & 5 & 10 & 10 & 5 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

while they are the row sums of the “generalized” or “stretched” [6] Riordan array (\frac{1}{1−x}, \frac{x^{2}}{1−x}):

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

Each Riordan array (g(x), f(x)) has bi-variate generating function given by

\frac{g(x)}{1 − yf(x)}.

For instance, the binomial matrix B has generating function

\frac{\frac{1}{1−x}}{1 − y \frac{x}{1−x}} = \frac{1}{1 − x(1 + y)}.

For a sequence a_{0}, a_{1}, a_{2}, … with g.f. g(x), the “aeration” of the sequence is the sequence a_{0}, 0, a_{1}, 0, a_{2}, … with interpolated zeros. Its g.f. is g(x^{2}).

The aeration of a (lower-triangular) matrix M with general term m_{i,j} is the matrix whose general term is given by

m_{\frac{i+j}{2} , \frac{i−j}{2}}^{r} \frac{1 + (−1)^{i−j}}{2} ,

where m_{i,j}^{r} is the i, j-th element of the reversal of M:

m_{i,j}^{r} = m_{i,i−j}.

In the case of a Riordan array (or indeed any lower triangular array), the row sums of the aeration are equal to the diagonal sums of the reversal of the original matrix.

Example 3. The Riordan array (c(x^{2}), xc(x^{2})) is the aeration of (c(x), xc(x)) \underline{A033184}. Here

c(x) = \frac{1 − \sqrt{1 − 4x}}{2x}

is the g.f. of the Catalan numbers. Indeed, the reversal of (c(x), xc(x)) is the matrix with general element

[k ≤ n + 1]\binom{n + k}{k}\frac{n − k + 1}{n + 1},

which begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 2 & 2 & 0 & 0 & 0 & … \\ 1 & 3 & 5 & 5 & 0 & 0 & … \\ 1 & 4 & 9 & 14 & 14 & 0 & … \\ 1 & 5 & 14 & 28 & 42 & 42 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

This is \underline{A009766}. Then (c(x^{2}), xc(x^{2})) has general element

\binom{n + 1}{\frac{n−k}{2}} \frac{k + 1}{n + 1} \frac{1 + (−1)^{n−k}}{2} ,

and begins

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

This is \underline{A053121}. Note that

(c(x^{2}), xc(x^{2})) = (\frac{1}{1 + x^{2}} , \frac{x}{1 + x^{2}} )^{−1} .

We observe that the diagonal sums of the reverse of (c(x), xc(x)) coincide with the row sums of (c(x^{2}), xc(x^{2})), and are equal to the central binomial coefficients \binom{n}{⌊ \frac{n}{2} ⌋} \underline{A001405}.

An important feature of Riordan arrays is that they have a number of sequence characterizations [4, 14]. The simplest of these is as follows.

Proposition 4. [14, Theorem 2.1, Theorem 2.2] Let D = [d_{n,k}] be an infinite triangular matrix. Then D is a Riordan array if and only if there exist two sequences A = [a_{0}, a_{1}, a_{2}, …] and Z = [z_{0}, z_{1}, z_{2}, …] with a_{0} ≠ 0, z_{0} ≠ 0 such that

• d_{n+1,k+1} = ∑_{j=0}^{∞} a_{j}d_{n,k+j}, (k, n = 0, 1, …)

• d_{n+1,0} = ∑_{j=0}^{∞} z_{j}d_{n,j}, (n = 0, 1, …).

The coefficients a_{0}, a_{1}, a_{2}, … and z_{0}, z_{1}, z_{2}, … are called the A-sequence and the Zsequence of the Riordan array D = (g(x), f(x)), respectively. Letting A(x) be the generating function of the A-sequence and Z(x) be the generating function of the Z-sequence, we have

A(x) = \frac{x}{f̄(x)}, Z(x) = \frac{1}{f̄(x)} (1 − \frac{1}{g(f̄(x))}) .  (1)

3 Orthogonal polynomials and Riordan arrays

By an orthogonal polynomial sequence (p_{n}(x))_{n≥0} we shall understand [5, 12] an infinite sequence of polynomials p_{n}(x), n ≥ 0, of degree n, with real coefficients (often integer coefficients) that are mutually orthogonal on an interval [x_{0}, x_{1}] (where x_{0} = −∞ is allowed, as well as x_{1} = ∞), with respect to a weight function w : [x_{0}, x_{1}] → ℝ :

∫_{x_{0}}^{x_{1}} p_{n}(x)p_{m}(x)w(x)dx = δ_{nm}\sqrt{h_{n}h_{m}},

where

∫_{x_{0}}^{x_{1}} p_{n}^{2}(x)w(x)dx = h_{n}.

We assume that w is strictly positive on the interval (x_{0}, x_{1}). Every such sequence obeys a so-called “three-term recurrence” :

p_{n+1}(x) = (a_{n}x + b_{n})p_{n}(x) − c_{n}p_{n−1}(x)

for coefficients a_{n}, b_{n} and c_{n} that depend on n but not x. We note that if

p_{j}(x) = k_{j}x^{j} + k′_{j}x^{j−1} + … j = 0, 1, …

then

a_{n} = \frac{k_{n+1}}{k_{n}} ,  b_{n} = a_{n} (\frac{k′_{n+1}}{k_{n+1}} − \frac{k′_{n}}{k_{n}} ) ,  c_{n} = a_{n} (\frac{k_{n−1}h_{n}}{k_{n}h_{n−1}} ) .

Since the degree of p_{n}(x) is n, the coefficient array of the polynomials is a lower triangular (infinite) matrix. In the case of monic orthogonal polynomials (where k_{n} = 1 for all n) the diagonal elements of this array will all be 1. In this case, we can write the three-term recurrence as

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 moments associated to the orthogonal polynomial sequence are the numbers

μ_{n} = ∫_{x_{0}}^{x_{1}} x^{n}w(x)dx.

We can find p_{n}(x), α_{n} and β_{n} (and in the right circumstances, w(x) - see the Appendix) from a knowledge of these moments. To do this, we let Δ_{n} be the Hankel determinant |μ_{i+j}|_{i,j≥0}^{n} and Δ_{n,x} be the same determinant, but with the last row equal to 1, x, x^{2}, . . .. Then

p_{n}(x) = \frac{Δ_{n,x}}{Δ_{n−1}} .

More generally, we let H ( \begin{matrix}u_{1} & … & u_{k} \\ v_{1} & … & v_{k} \end{matrix} ) be the determinant of Hankel type with (i, j)-th

term μ_{u_{i}+v_{j}}. Let

Δ_{n} = H ( \begin{matrix}0 & 1 & … & n \\ 0 & 1 & … & n \end{matrix} ) , Δ′_{n} = H_{n} ( \begin{matrix}0 & 1 & … & n − 1 & n \\ 0 & 1 & … & n − 1 & n + 1 \end{matrix} ) .

Then we have

α_{n} = \frac{Δ′_{n}}{Δ_{n}} − \frac{Δ′_{n−1}}{Δ_{n−1}} , β_{n} = \frac{Δ_{n−2}Δ_{n}}{Δ_{n−1}^{2}} .

We shall say that a family of polynomials {p_{n}(x)}_{n≥0} is [31] formally orthogonal, if there exists a linear functional L on polynomials such that

1. p_{n}(x) is a polynomial of degree n,

2. L(p_{n}(x)p_{m}(x)) = 0 for m ≠ n,

3. L(p_{n}^{2}(x)) ≠ 0.

Consequences of this definition include [31] that

L(x^{m}p_{n}(x)) = κ_{n}δ_{mn}, 0 ≤ m ≤ n, κ_{n} ≠ 0,

and if q(x) = ∑_{k=0}^{n} a_{k}p_{k}(x), then a_{k} = L(qp_{k})/L(p_{k}^{2}).

The sequence of numbers μ_{n} = L(x^{n}) is called the sequence of moments of the family of orthogonal polynomials defined by L. Note that where a suitable weight function w(x) exists, then we can realize the functional L as

L(p(x)) = ∫_{ℝ} p(x)w(x) dx.

If the family p_{n}(x) is an orthogonal family for the functional L, then it is also orthogonal for cL, where c ≠ 0. In the sequel, we shall always assume that μ_{0} = L(p_{0}(x)) = 1.

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, three term recurrences, and the recurrence coefficients and the g.f. of the moment sequence of the orthogonal polynomials.

Theorem 5. [18] (Cf. [31], Théorème 9 on p.I-4, or [32], 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 6. [18] (Cf. [31], Proposition 1, (7), on p. V-5, or [32], 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 − ⋯}}}}.

Given a family of monic orthogonal polynomials

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},

we can write

p_{n}(x) = ∑_{k=0}^{n} a_{n,k}x^{k}.

Then we have

∑_{k=0}^{n+1} a_{n+1,k}x^{k} = (x − α_{n}) ∑_{k=0}^{n} a_{n,k}x^{k} − β_{n} ∑_{k=0}^{n−1} a_{n−1,k}x^{k}

from which we deduce

a_{n+1,0} = −α_{n}a_{n,0} − β_{n}a_{n−1,0}  (2)

and

a_{n+1,k} = a_{n,k−1} − α_{n}a_{n,k} − β_{n}a_{n−1,k}  (3)

We note that if α_{n} and β_{n} are constant, equal to α and β, respectively, then the sequence (1, −α, −β, 0, 0, …) forms an A-sequence for the coefficient array. The question immediately arises as to the conditions under which a Riordan array (g, f) can be the coefficient array of a family of orthogonal polynomials. A partial answer is given by the following proposition.

Proposition 7. Every Riordan array of the form

(\frac{1}{1 + rx + sx^{2}} , \frac{x}{1 + rx + sx^{2}} )

is the coefficient array of a family of monic orthogonal polynomials.

Proof. The array (\frac{1}{1+rx+sx^{2}} , \frac{x}{1+rx+sx^{2}} ) [17] has a C-sequence C(x) = ∑_{n≥0} c_{n}x^{n} given by

\frac{x}{1 + rx + sx^{2}} = \frac{x}{1 − xC(x)},

and thus

C(x) = −r − sx.

This means that the Riordan array (\frac{1}{1+rx+sx^{2}} , \frac{x}{1+rx+sx^{2}} ) is determined by the fact that

a_{n+1,k} = a_{n,k−1} + ∑_{i≥0} c_{i}a_{n−i,k} for n, k = 0, 1, 2, …

where a_{n,−1} = 0. In the case of (\frac{1}{1+rx+sx^{2}} , \frac{x}{1+rx+sx^{2}} ) we have

a_{n+1,k} = a_{n,k−1} − ra_{n,k} − sa_{n−1,k}.

Working backwards, this now ensures that

p_{n+1}(x) = (x − r)p_{n}(x) − sp_{n−1}(x),

where p_{n}(x) = ∑_{k=0}^{n} a_{n,k}x^{n}. The result now follows from Theorem 5.

We note that in this case the three-term recurrence coefficients α_{n} and β_{n} are constants. We have in fact the following proposition (see the next section for information on the Chebyshev polynomials).

Proposition 8. The Riordan array (\frac{1}{1+rx+sx^{2}} , \frac{x}{1+rx+sx^{2}} ) is the coefficient array of the modified Chebyshev polynomials of the second kind given by

P_{n}(x) = (\sqrt{s})^{n}U_{n} (\frac{x − r}{2\sqrt{s}}) , n = 0, 1, 2, …

Proof. The production array of (\frac{1}{1+rx+sx^{2}} , \frac{x}{1+rx+sx^{2}} )^{−1} is given by

( \begin{matrix}r & 1 & 0 & 0 & 0 & 0 & … \\ s & r & 1 & 0 & 0 & 0 & … \\ 0 & s & r & 1 & 0 & 0 & … \\ 0 & 0 & s & r & 1 & 0 & … \\ 0 & 0 & 0 & s & r & 1 & … \\ 0 & 0 & 0 & 0 & s & r & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

The result is now a consequence of the article [11] by Elouafi, for instance.

The complete answer can be found by considering the associated production matrix of a Riordan arrray, in the following sense.

The concept of a production matrix [8, 9, 10] is a general one, but for this work 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

ĪA_{P} = A_{P} P

where Ī = (δ_{i+1,j})_{i,j≥0} (where δ is the usual Kronecker symbol):

Ī = ( \begin{matrix}0 & 1 & 0 & 0 & 0 & 0 & … \\ 0 & 0 & 1 & 0 & 0 & 0 & … \\ 0 & 0 & 0 & 1 & 0 & 0 & … \\ 0 & 0 & 0 & 0 & 1 & 0 & … \\ 0 & 0 & 0 & 0 & 0 & 1 & … \\ 0 & 0 & 0 & 0 & 0 & 0 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

We have

P = A_{P}^{−1}ĪA_{P} .  (4)

Writing \overline{A_{P}} = ĪA_{P}, we can write this equation as

P = A_{P}^{−1}\overline{A_{P}} .  (5)

Note that \overline{A_{P}} is a “beheaded” version of A_{P}; that is, it is A_{P} with the first row removed.

The production matrix P is sometimes [23, 26] called the Stieltjes matrix S_{A_{P}} associated to A_{P}. Other examples of the use of production matrices can be found in [1], for instance. The sequence formed by the row sums of A_{P} often has combinatorial significance and is called the sequence associated to P. Its general term a_{n} is given by a_{n} = u^{T} P^{n}e where

e = ( 1 1 1 ⋮ )

In the context of Riordan arrays, the production matrix associated to a proper Riordan array takes on a special form :

Proposition 9. [10, Proposition 3.1] Let P be an infinite production matrix and let A_{P} be the matrix induced by P. Then A_{P} is an (ordinary) Riordan matrix if and only if P is of the form

P = ( \begin{matrix}ξ_{0} & α_{0} & 0 & 0 & 0 & 0 & … \\ ξ_{1} & α_{1} & α_{0} & 0 & 0 & 0 & … \\ ξ_{2} & α_{2} & α_{1} & α_{0} & 0 & 0 & … \\ ξ_{3} & α_{3} & α_{2} & α_{1} & α_{0} & 0 & … \\ ξ_{4} & α_{4} & α_{3} & α_{2} & α_{1} & α_{0} & … \\ ξ_{5} & α_{5} & α_{4} & α_{3} & α_{2} & α_{1} & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

where ξ_{0} ≠ 0, α_{0} ≠ 0. Moreover, columns 0 and 1 of the matrix P are the Z- and Asequences, respectively, of the Riordan array A_{P}.

We recall that we have

A(x) = \frac{x}{f̄(x)}, Z(x) = \frac{1}{f̄(x)} (1 − \frac{1}{g(f̄(x))}) .

Example 10. We consider the Riordan array L where

L^{−1} = (\frac{1 − λx − μx^{2}}{1 + ax + bx^{2}} , \frac{x}{1 + ax + bx^{2}} ) .

The production matrix (Stieltjes matrix) of

L = (\frac{1 − λx − μx^{2}}{1 + ax + bx^{2}} , \frac{x}{1 + ax + bx^{2}} )^{−1}

is given by

P = S_{L} = ( \begin{matrix}a + λ & 1 & 0 & 0 & 0 & 0 & … \\ b + μ & a & 1 & 0 & 0 & 0 & … \\ 0 & b & a & 1 & 0 & 0 & … \\ 0 & 0 & b & a & 1 & 0 & … \\ 0 & 0 & 0 & b & a & 1 & … \\ 0 & 0 & 0 & 0 & b & a & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

We note that since

L^{−1} = (\frac{1 − λx − μx^{2}}{1 + ax + bx^{2}} , \frac{x}{1 + ax + bx^{2}} )
= (1 − λx − μx^{2}, x) · (\frac{1}{1 + ax + bx^{2}} , \frac{x}{1 + ax + bx^{2}} ) ,

we have

L = (\frac{1 − λx − μx^{2}}{1 + ax + bx^{2}} , \frac{x}{1 + ax + bx^{2}} )^{−1} = (\frac{1}{1 + ax + bx^{2}} , \frac{x}{1 + ax + bx^{2}} )^{−1}·(\frac{1}{1 − λx − μx^{2}} , x) .

If we now let

L_{1} = (\frac{1}{1 + ax}, \frac{x}{1 + ax}) · L,

then [22] we obtain that the Stieltjes matrix for L_{1} is given by

S_{L_{1}} = ( \begin{matrix}λ & 1 & 0 & 0 & 0 & 0 & … \\ b + μ & 0 & 1 & 0 & 0 & 0 & … \\ 0 & b & 0 & 1 & 0 & 0 & … \\ 0 & 0 & b & 0 & 1 & 0 & … \\ 0 & 0 & 0 & b & 0 & 1 & … \\ 0 & 0 & 0 & 0 & b & 0 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

We have in fact the following general result [22] :

Proposition 11. If L = (g(x), f(x)) is a Riordan array and P = S_{L} is tridiagonal, then necessarily

P = S_{L} = ( \begin{matrix}a_{1} & 1 & 0 & 0 & 0 & 0 & … \\ b_{1} & a & 1 & 0 & 0 & 0 & … \\ 0 & b & a & 1 & 0 & 0 & … \\ 0 & 0 & b & a & 1 & 0 & … \\ 0 & 0 & 0 & b & a & 1 & … \\ 0 & 0 & 0 & 0 & b & a & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

where

f(x) = Rev\frac{x}{1 + ax + bx^{2}}  and  g(x) = \frac{1}{1 − a_{1}x − b_{1}xf} ,

and vice-versa.

This leads to the important corollary

Corollary 12. If L = (g(x), f(x)) is a Riordan array and P = S_{L} is tridiagonal, with

P = S_{L} = ( \begin{matrix}a_{1} & 1 & 0 & 0 & 0 & 0 & … \\ b_{1} & a & 1 & 0 & 0 & 0 & … \\ 0 & b & a & 1 & 0 & 0 & … \\ 0 & 0 & b & a & 1 & 0 & … \\ 0 & 0 & 0 & b & a & 1 & … \\ 0 & 0 & 0 & 0 & b & a & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,  (6)

then L^{−1} is the coefficient array of the family of orthogonal polynomials p_{n}(x) where p_{0}(x) = 1, p_{1}(x) = x − a_{1}, and

p_{n+1}(x) = (x − a)p_{n}(x) − b_{n}p_{n−1}(x), n ≥ 2,

where b_{n} is the sequence 0, b_{1}, b, b, b, . . ..

Proof. By Favard’s theorem, it suffices to show that L^{−1} defines a family of polynomials {p_{n}(x)} that obey the above three-term recurrence. Now L is lower-triangular and so L^{−1} is the coefficient array of a family of polynomials p_{n}(x) (with the degree of p_{n}(x) being n), where

L^{−1} ( 1 ; x ; x^{2} ; x^{3} ; ⋮ ) = ( p_{0}(x) ; p_{1}(x) ; p_{2}(x) ; p_{3}(x) ; ⋮ ) .

We have

S_{L} · L^{−1} = L^{−1} · L̄ · L^{−1} = L^{−1} · Ī· L · L^{−1} = L^{−1} · Ī.

Thus

S_{L} · L^{−1} · (1, x, x^{2}, …)^{T} = L^{−1} · Ī· (1, x, x^{2}, …)^{T} = L^{−1} · (x, x^{2}, x^{3}, …)^{T} .

We therefore obtain

( \begin{matrix}a_{1} & 1 & 0 & 0 & 0 & 0 & … \\ b_{1} & a & 1 & 0 & 0 & 0 & … \\ 0 & b & a & 1 & 0 & 0 & … \\ 0 & 0 & b & a & 1 & 0 & … \\ 0 & 0 & 0 & b & a & 1 & … \\ 0 & 0 & 0 & 0 & b & a & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ( p_{0}(x) ; p_{1}(x) ; p_{2}(x) ; p_{3}(x) ; ⋮ ) = ( xp_{0}(x) ; xp_{1}(x) ; xp_{2}(x) ; xp_{3}(x) ; ⋮ ) ,

from which we infer that

p_{1}(x) = x − a_{1},

and

p_{n+1}(x) + ap_{n}(x) + b_{n}p_{n−1}(x) = xp_{n}(x), n ≥ 1,

or

p_{n+1}(x) = (x − a)p_{n}(x) − b_{n}p_{n−1}(x), n ≥ 1.

If we now start with a family of orthogonal polynomials {p_{n}(x)}, p_{0}(x) = 1, p_{1}(x) = x − a_{1}, that for n ≥ 1 obey a three-term recurrence

p_{n+1}(x) = (x − a)p_{n}(x) − b_{n}p_{n−1}(x),

where b_{n} is the sequence 0, b_{1}, b, b, b, …, then we can define [23] an associated Riordan array L = (g(x), f(x)) by

f(x) = Rev\frac{x}{1 + ax + bx^{2}} and g(x) = \frac{1}{1 − a_{1}x − b_{1}xf} .

Clearly, L^{−1} is then the coefficient array of the family of polynomials {p_{n}(x)}. Combining these results, we have

Theorem 13. A Riordan array L = (g(x), f(x)) is the inverse of the coefficient array of a family of orthogonal polynomials if and only if its production matrix P = S_{L} is tri-diagonal.

Proof. If L has a tri-diagonal production matrix, then by Corollary (12), L^{−1} is the coefficient array of orthogonal polynomials. It conversely L^{−1} is the coefficient array of a family of orthogonal polynomials, then using the fact they these polynomials obey a three-term recurrence and the uniqueness of the Z- and A-sequences, we see using equations (2) and (3) along with Proposition 9, that the production matrix is tri-diagonal.

Proposition 14. Let L = (g(x), f(x)) be a Riordan array with tri-diagonal production matrix S_{L}. Then

[x^{n}]g(x) = L(x^{n}),

where L is the linear functional that defines the associated family of orthogonal polynomials.

Proof. Let L = (l_{i,j})_{i,j≥0}. We have [31]

x^{n} = ∑_{i=0}^{n} l_{n,i}p_{i}(x).

Applying L, we get

L(x^{n}) = L (∑_{i=0}^{n} l_{n,i}p_{i}(x)) = ∑_{i=0}^{n} l_{n,i}L(p_{i}(x)) = ∑_{i=0}^{n} l_{n,i}δ_{i,0} = l_{n,0} = [x^{n}]g(x).

Thus under the conditions of the proposition, by Theorem 6, g(x) is the g.f. of the moment sequence μ_{n} = L(x^{n}). Hence g(x) has the continued fraction expansion

g(x) = \frac{1}{1 − a_{1}x − \frac{b_{1}x^{2}}{1 − ax − \frac{bx^{2}}{1 − ax − \frac{bx^{2}}{1 − ax − ⋯}}}}.

This can also be established directly. To see this, we use the

Lemma 15. Let

f(x) = Rev\frac{x}{1 + ax + bx^{2}} .

Then

\frac{f}{x} = \frac{1}{1 − ax − bx^{2}(f/x)}.

Proof. By definition, f(x) is the solution u(x), with u(0) = 0, of

\frac{u}{1 + au + bu^{2}} = x.

We find

u(x) = \frac{1 − ax − \sqrt{1 − 2ax + (a^{2} − 4b)x^{2}}}{2bx} .

Solving the equation

v(x) = \frac{1}{1 − av − bx^{2}v} ,

we obtain

v(x) = \frac{1 − ax − \sqrt{1 − 2ax + (a^{2} − 4b)x^{2}}}{2bx^{2}} = \frac{f(x)}{x}.

Thus we have

\frac{f(x)}{x} = \frac{1}{x}Rev\frac{x}{1 + ax + bx^{2}} = \frac{1}{1 − ax − \frac{bx^{2}}{1 − ax − \frac{bx^{2}}{1 − ⋯}}}.

Now

g(x) = \frac{1}{1 − a_{1}x − b_{1}xf} = \frac{1}{1 − a_{1}x − b_{1}x^{2}(f/x)}

immediately implies by the above lemma that

g(x) = \frac{1}{1 − a_{1}x − \frac{b_{1}x^{2}}{1 − ax − \frac{bx^{2}}{1 − ax − \frac{bx^{2}}{1 − …}}}}.

We note that the elements of the rows of L^{−1} can be identified with the coefficients of the characteristic polynomials of the successive principal sub-matrices of P.

Example 16. We consider the Riordan array

(\frac{1}{1 + ax + bx^{2}} , \frac{x}{1 + ax + bx^{2}} ) .

Then the production matrix (Stieltjes matrix) of the inverse Riordan array (\frac{1}{1+ax+bx^{2}} , \frac{x}{1+ax+bx^{2}} )^{−1} left-multiplied by the k-th binomial array

(\frac{1}{1 − kx}, \frac{x}{1 − kx}) = (\frac{1}{1 − x}, \frac{x}{1 − x})^{k}

is given by

P = ( \begin{matrix}a + k & 1 & 0 & 0 & 0 & 0 & … \\ b & a + k & 1 & 0 & 0 & 0 & … \\ 0 & b & a + k & 1 & 0 & 0 & … \\ 0 & 0 & b & a + k & 1 & 0 & … \\ 0 & 0 & 0 & b & a + k & 1 & … \\ 0 & 0 & 0 & 0 & b & a + k & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

and vice-versa. This follows since

(\frac{1}{1 + ax + bx^{2}} , \frac{x}{1 + ax + bx^{2}} )·(\frac{1}{1 + kx}, \frac{x}{1 + kx}) = (\frac{1}{1 + (a + k)x + bx^{2}} , \frac{x}{1 + (a + k)x + bx^{2}} ) .

In fact we have the more general result :

\begin{matrix}(\frac{1 + λx + μx^{2}}{1 + ax + bx^{2}} , \frac{x}{1 + ax + bx^{2}} ) · (\frac{1}{1 + kx}, \frac{x}{1 + kx}) = \\ (\frac{1 + λx + μx^{2}}{1 + (a + k)x + bx^{2}} , \frac{x}{1 + (a + k)x + bx^{2}} ) . \end{matrix}

The inverse of this last matrix therefore has production array

( \begin{matrix}a + k − λ & 1 & 0 & 0 & 0 & 0 & … \\ b − μ & a + k & 1 & 0 & 0 & 0 & … \\ 0 & b & a + k & 1 & 0 & 0 & … \\ 0 & 0 & b & a + k & 1 & 0 & … \\ 0 & 0 & 0 & b & a + k & 1 & … \\ 0 & 0 & 0 & 0 & b & a + k & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

4 Exponential Riordan arrays

The exponential Riordan group [3, 8, 10], 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. In what follows, we shall assume

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. 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] 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 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}).

Example 17. We consider the exponential Riordan array [\frac{1}{1−x}, x], \underline{A094587}. This array 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!}, and 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, …).

Example 18. We consider the exponential Riordan array L = [1, \frac{x}{1−x}]. The general term of this matrix 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).

We will use the following [8, 10],important result concerning matrices that are production matrices for exponential Riordan arrays.

Proposition 19. 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} + …  (7)

be two formal power series that that

r(f(x)) = f′(x)  (8)
c(f(x)) = \frac{g′(x)}{g(x)}.  (9)

Then

(i)  a_{n+1,0} = ∑_{i} i!c_{i}a_{n,i}  (10)
(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}  (11)

or, assuming c_{k} = 0 for k < 0 and r_{k} = 0 for k < 0,

a_{n+1,k} = \frac{1}{k!} ∑_{i≥k−1} i!(c_{i−k} + kr_{i−k+1})a_{n,i}.  (12)

Conversely, starting from the sequences defined by (7), the infinite array (a_{n,k})_{n,k≥0} defined by (12) is an exponential Riordan array.

A consequence of this proposition is that the production matrix P = (p_{i,j})_{i,j≥0} for an exponential Riordan array obtained as in the proposition satisfies [10]

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

Furthermore, the bivariate exponential 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)),  (13)

and

c(x) = \frac{g′(f̄(x))}{g(f̄(x))}.  (14)

Example 20. The production matrix of [1, \frac{x}{1+x}] \underline{A111596} is given by

( \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} ) .

The row sums of L^{−1} have e.g.f. exp (\frac{x}{1+x}), and start 1, 1, −1, 1, 1, −19, 151, . . .. This is \underline{A111884}. This follows since we have g(x) = 1 and so g′(x) = 0, implying that c(x) = 0, and f(x) = \frac{x}{1+x} which gives us f̄(x) = \frac{x}{1−x} and f′(x) = \frac{1}{(1+x)^{2}} . Thus f′(f̄(x)) = c(x) = (1 − x)^{2}. Hence the bivariate generating function of P is e^{xy}(1 − x)^{2}y, as required.

Example 21. 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}.

It is closely related to the Laguerre polynomials. Its inverse A^{−1} is the exponential Riordan array [\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 of the matrix A^{−1} = [\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} ) .

This follows since we have g(x) = \frac{1}{1−x}, and so g′(x) = \frac{1}{(1−x)^{2}} , and f(x) = \frac{x}{1−x} which yields f̄(x) = \frac{x}{1+x} and f′(x) = \frac{1}{(1−x)^{2}} . Then

c(x) = \frac{g′(f̄(x))}{g(f̄(x))} = 1 + x

while

r(x) = f′(f̄(x)) = (1 + x)^{2}.

Thus the bivariate generating function of P is given by

e^{xy}(1 + x + (1 + x)^{2}y).

We note that

A = exp(S),

where

S = ( \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 0 & 0 & 0 & 0 & 0 & … \\ 0 & 4 & 0 & 0 & 0 & 0 & … \\ 0 & 0 & 9 & 0 & 0 & 0 & … \\ 0 & 0 & 0 & 16 & 0 & 0 & … \\ 0 & 0 & 0 & 0 & 25 & 0 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

Example 22. 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) [12, 24, 30]. 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 bi-variate generating function for the production matrix of the inverse array is

e^{tz}(z − 1 + t(1 − z)),

which is what is required.

We can infer the following result from the article [23] by Peart and Woan.

Proposition 23. If L = [g(x), f(x)] is an exponential Riordan array and P = S_{L} is tridiagonal, then necessarily

P = S_{L} = ( \begin{matrix}α_{0} & 1 & 0 & 0 & 0 & 0 & … \\ β_{1} & α_{1} & 1 & 0 & 0 & 0 & … \\ 0 & β_{2} & α_{2} & 1 & 0 & 0 & … \\ 0 & 0 & β_{3} & α_{3} & 1 & 0 & … \\ 0 & 0 & 0 & β_{4} & α_{4} & 1 & … \\ 0 & 0 & 0 & 0 & β_{5} & α_{5} & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

where \{α_{i}\}_{i≥0} is an arithmetic sequence with common difference α, \{\frac{β_{i}}{i} \}_{i≥1} is an arithmetic sequence with common difference β, and

ln(g) = ∫ (α_{0} + β_{1}f) dx, g(0) = 1,

where f is given by

f′ = 1 + αf + βf^{2}, f(0) = 0,

and vice-versa.

In the above, we note that α_{0} = a_{1} where g(x) is the g.f. of a_{0} = 1, a_{1}, a_{2}, . . .. We have the important

Corollary 24. If L = [g(x), f(x)] is an exponential Riordan array and P = S_{L} is tridiagonal, with

P = S_{L} = ( \begin{matrix}α_{0} & 1 & 0 & 0 & 0 & 0 & … \\ β_{1} & α_{1} & 1 & 0 & 0 & 0 & … \\ 0 & β_{2} & α_{2} & 1 & 0 & 0 & … \\ 0 & 0 & β_{3} & α_{3} & 1 & 0 & … \\ 0 & 0 & 0 & β_{4} & α_{4} & 1 & … \\ 0 & 0 & 0 & 0 & β_{5} & α_{5} & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

then L^{−1} is the coefficient array of the family of monic orthogonal polynomials p_{n}(x) where p_{0}(x) = 1, p_{1}(x) = x − a_{1} = x − α_{0}, and

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

Proof. By Favard’s theorem, it suffices to show that L^{−1} defines a family of polynomials {p_{n}(x)} that obey the above three-term recurrence. Now L is lower-triangular and so L^{−1} is the coefficient array of a family of polynomials p_{n}(x), where

L^{−1} ( 1 ; x ; x^{2} ; x^{3} ; ⋮ ) = ( p_{0}(x) ; p_{1}(x) ; p_{2}(x) ; p_{3}(x) ; ⋮ ) .

We have

S_{L} · L^{−1} = L^{−1} · L̄ · L^{−1} = L^{−1} · Ī· L · L^{−1} = L^{−1} · Ī.

Thus

S_{L} · L^{−1} · (1, x, x^{2}, …)^{T} = L^{−1} · Ī· (1, x, x^{2}, …)^{T} = L^{−1} · (x, x^{2}, x^{3}, …)^{T} .

We therefore obtain

( \begin{matrix}α_{0} & 1 & 0 & 0 & 0 & 0 & … \\ β_{1} & α_{1} & 1 & 0 & 0 & 0 & … \\ 0 & β_{2} & α_{2} & 1 & 0 & 0 & … \\ 0 & 0 & β_{3} & α_{3} & 1 & 0 & … \\ 0 & 0 & 0 & β_{4} & α_{4} & 1 & … \\ 0 & 0 & 0 & 0 & β_{5} & α_{5} & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ( p_{0}(x) ; p_{1}(x) ; p_{2}(x) ; p_{3}(x) ; ⋮ ) = ( xp_{0}(x) ; xp_{1}(x) ; xp_{2}(x) ; xp_{3}(x) ; ⋮ ) ,

from which we infer

p_{1}(x) = x − α_{0},

and

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

or

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

If we now start with a family of orthogonal polynomials {p_{n}(x)} that obeys a three-term recurrence

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

with p_{0}(x) = 1, p_{1}(x) = x − α_{0}, where {α_{i}}_{i≥0} is an arithmetic sequence with common difference α and \{\frac{β_{i}}{i} \}_{i≥1} is an arithmetic sequence with common difference β, then we can define [23] an associated exponential Riordan array L = [g(x), f(x)] by

f′ = 1 + αf + βf^{2}, f(0) = 0,

and

ln(g) = ∫ (α_{0} + β_{1}f) dx, g(0) = 1.

Clearly, L^{−1} is then the coefficient array of the family of polynomials {p_{n}(x)}. Gathering these results, we have

Theorem 25. An exponential Riordan array L = [g(x), f(x)] is the inverse of the coefficient array of a family of orthogonal polynomials if and only if its production matrix P = S_{L} is tri-diagonal.

Proposition 26. Let L = [g(x), f(x)] be an exponential Riordan array with tri-diagonal production matrix S_{L}. Then

n![x^{n}]g(x) = L(x^{n}) = μ_{n},

where L is the linear functional that defines the associated family of orthogonal polynomials.

Proof. Let L = (l_{i,j})_{i,j≥0}. We have

x^{n} = ∑_{i=0}^{n} l_{n,i}p_{i}(x).

Applying L, we get

L(x^{n}) = L (∑_{i=0}^{n} l_{n,i}p_{i}(x)) = ∑_{i=0}^{n} l_{n,i}L(p_{i}(x)) = ∑_{i=0}^{n} l_{n,i}δ_{i,0} = l_{n,0} = n![x^{n}]g(x).

Corollary 27. Let L = [g(x), f(x)] be an exponential Riordan array with tri-diagonal production matrix S_{L}. Then the moments μ_{n} of the associated family of orthogonal polynomials are given by the terms of the first column of L.

Thus under the conditions of the proposition, g(x) is the g.f. of the moment sequence L(x^{n}). Hence g(x) has the continued fraction expansion

g(x) = \frac{1}{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 − ⋯}}}}.

When we come to study Hermite polynomials, we shall be working with elements of the exponential Appell subgroup. By the exponential Appell subgroup AeR of eR we understand the set of arrays of the form [f(x), x].

Let A ∈ AeR correspond to the sequence (a_{n})_{n≥0}, with e.g.f. f(x). Let B ∈ AeR correspond to the sequence (b_{n}), with e.g.f. g(x). Then we have 1. The row sums of A are the binomial transform of (a_{n}).

2. The inverse of A is the sequence array for the sequence with e.g.f. \frac{1}{f(x)}.

3. The product AB is the sequence array for the exponential convolution a ∗ b(n) = ∑_{k=0}^{n} \binom{n}{k}a_{k}b_{n−k} with e.g.f. f(x)g(x).

For instance, the row sums of A = [f(x), x] will have e.g.f. given by

[f(x), x] · e^{x} = f(x)e^{x} = e^{x}f(x) = [e^{x}, x] · f(x),

which is the e.g.f. of the binomial transform of (a_{n}).

Example 28. We consider the matrix [cosh(x), x], \underline{A119467}, with elements

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 0 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 0 & 1 & 0 & 0 & 0 & … \\ 0 & 3 & 0 & 1 & 0 & 0 & … \\ 1 & 0 & 6 & 0 & 1 & 0 & … \\ 0 & 5 & 0 & 10 & 0 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

The row sums of this matrix have e.g.f. cosh(x) exp(x), which is the e.g.f. of the sequence 1, 1, 2, 4, 8, 16, . . .. The inverse matrix is [sech(x), x], \underline{A119879}, with entries

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 0 & 1 & 0 & 0 & 0 & 0 & … \\ −1 & 0 & 1 & 0 & 0 & 0 & … \\ 0 & −3 & 0 & 1 & 0 & 0 & … \\ 5 & 0 & −6 & 0 & 1 & 0 & … \\ 0 & 25 & 0 & −10 & 0 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

The row sums of this matrix have e.g.f. sech(x) exp(x). This is \underline{A155585}.

5 The Hankel transform of an integer sequence

The Hankel transform 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} | .  (15)

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 − ⋯}}}}

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}.  (16)

Note that this independent from α_{n}.

We note that α_{n} and β_{n} are in general not integers. Now let H ( \begin{matrix}u_{1} & … & u_{k} \\ v_{1} & … & v_{k} \end{matrix} ) be the

determinant of Hankel type with (i, j)-th term μ_{u_{i}+v_{j}}. Let

Δ_{n} = H ( \begin{matrix}0 & 1 & … & n \\ 0 & 1 & … & n \end{matrix} ) , Δ′_{n} = H_{n} ( \begin{matrix}0 & 1 & … & n − 1 & n \\ 0 & 1 & … & n − 1 & n + 1 \end{matrix} ) .

Then we have

α_{n} = \frac{Δ′_{n}}{Δ_{n}} − \frac{Δ′_{n−1}}{Δ_{n−1}} ,  β_{n} = \frac{Δ_{n−2}Δ_{n}}{Δ_{n−1}^{2}} .  (17)

6 Legendre polynomials

We recall that the Legendre polynomials P_{n}(x) can be defined by

P_{n}(x) = ∑_{k=0}^{n}(−1)^{k}\binom{n}{k}^{2} (\frac{1 + x}{2})^{n−k} (\frac{1 − x}{2})^{k} .

Their generating function is given by

\frac{1}{\sqrt{1 − 2xt + t^{2}}} = ∑_{n=0}^{∞} P_{n}(x)t^{n}.

We note that the production matrix of the inverse of the coefficient array of these polynomials is given by

( \begin{matrix}0 & 1 & 0 & 0 & 0 & 0 & … \\ \frac{1}{3} & 0 & \frac{2}{3} & 0 & 0 & 0 & … \\ 0 & \frac{2}{5} & 0 & \frac{3}{5} & 0 & 0 & … \\ 0 & 0 & \frac{3}{7} & 0 & \frac{4}{7} & 0 & … \\ 0 & 0 & 0 & \frac{4}{9} & 0 & \frac{5}{9} & … \\ 0 & 0 & 0 & 0 & \frac{5}{11} & 0 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

which corresponds to the fact that the P_{n}(x) satisfy the following three-term recurrence

(n + 1)P_{n+1}(x) = (2n + 1)xP_{n}(x) − nP_{n−1}(x).

The shifted Legendre polynomials P̃_{n}(x) are defined by

P̃_{n}(x) = P_{n}(2x − 1).

They satisfy

P̃_{n}(x) = (−1)^{n} ∑_{k=0}^{n} \binom{n}{k}\binom{n + k}{k}(−x)^{k} = ∑_{k=0}^{n}(−1)^{n−k}\binom{n + k}{2k}\binom{2k}{k}x^{k}.

Their coefficient array begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ −1 & 2 & 0 & 0 & 0 & 0 & … \\ 1 & −6 & 6 & 0 & 0 & 0 & … \\ −1 & 12 & −30 & 20 & 0 & 0 & … \\ 1 & −20 & 90 & −140 & 70 & 0 & … \\ −1 & 30 & −210 & 560 & −630 & 252 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

and so the first few terms begin

1, 2x − 1, 6x^{2} − 6x + 1, 20x^{3} − 30x^{2} + 12x − 1, …

We clearly have

\frac{1}{\sqrt{1 − 2(2x − 1)t + t^{2}}} = ∑_{n=0}^{∞} P̃_{n}(x)t^{n}.

7 Legendre polynomials as moments

Our goal in this section is to represent the Legendre polynomials as the first column of a Riordan array whose production matrix is tri-diagonal. We first of all consider the so-called shifted Legendre polynomials. We have

Proposition 29. The inverse L of the Riordan array

(\frac{1 + r(1 − r)x^{2}}{1 + (2r − 1)x + r(r − 1)x^{2}} , \frac{x}{1 + (2r − 1)x + r(r − 1)x^{2}} )

has as its first column the shifted Legendre polynomials P̃_{n}(r). The production matrix of L is tri-diagonal.

Proof. Indeed, standard Riordan array techniques show that we have

L = (\frac{1 + r(1 − r)x^{2}}{1 + (2r − 1)x + r(r − 1)x^{2}} , \frac{x}{1 + (2r − 1)x + r(r − 1)x^{2}} )^{−1}
= (\frac{1}{\sqrt{1 − 2(2r − 1)x + x^{2}}} , \frac{1 − (2r − 1)x − \sqrt{1 − 2(2r − 1)x + x^{2}}}{2r(r − 1)x} ) .

This establishes the first part. Now using equations (1), with

f(x) = \frac{1 − (2r − 1)x − \sqrt{1 − 2(2r − 1)x + x^{2}}}{2r(r − 1)x} , f̄(x) = \frac{x}{1 + (2r − 1)x + r(r − 1)x^{2}} ,

and

g(x) = \frac{1}{\sqrt{1 − 2(2r − 1)x + x^{2}}} ,

we find that

Z(x) = (2r − 1) + 2rx(r − 1), A(x) = 1 + x(2r − 1) + x^{2}r(r − 1).

Hence by Proposition 9 the production matrix P = S_{L} of L is given by

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

Corollary 30. The shifted Legendre polynomials are moments of the family of orthogonal polynomials whose coefficient array is given by

L^{−1} = (\frac{1 + r(1 − r)x^{2}}{1 + (2r − 1)x + r(r − 1)x^{2}} , \frac{x}{1 + (2r − 1)x + r(r − 1)x^{2}} ) .

Proof. This follows from the above result and Proposition 14. Proposition 31. The Hankel transform of the sequence P̃_{n}(r) is given by 2^{n}(r(r − 1))^{\binom{n+1}{2}}. Proof. From the above, the g.f. of P̃_{n}(r) is given by

\frac{1}{1 − (2r − 1)x − \frac{2r(r − 1)x^{2}}{1 − (2r − 1)x − \frac{r(r − 1)x^{2}}{1 − (2r − 1)x − \frac{r(r − 1)x^{2}}{1 − ⋯}}}}.

The result now follows from Equation (16).

We note that

P̃_{n}(r) = \frac{1}{π} ∫_{−2\sqrt{r(r−1)}+2r−1}^{2\sqrt{r(r−1)}+2r−1} \frac{x^{n}}{\sqrt{−x^{2} + 2(2r − 1)x − 1}} dx

gives an explicit moment representation for P̃_{n}(r).

Turning now to the Legendre polynomials P_{n}(x), we have the following result.

Proposition 32. The inverse L of the Riordan array

(\frac{1 + \frac{1−r^{2}}{4} x^{2}}{1 + rx + \frac{r^{2}−1}{2}x^{2}} , \frac{x}{1 + rx + \frac{r^{2}−1}{2}x^{2}} )

has as its first column the Legendre polynomials P_{n}(r). The production matrix of L is tridiagonal.

Proof. We have

L = (\frac{1 + \frac{1−r^{2}}{4} x^{2}}{1 + rx + \frac{r^{2}−1}{2}x^{2}} , \frac{x}{1 + rx + \frac{r^{2}−1}{2}x^{2}} )^{−1}

= (\frac{1}{\sqrt{1 − 2rx + x^{2}}} , \frac{2(1 − rx − \sqrt{1 − 2rx + x^{2}})}{x(r^{2} − 1)}) .

This proves the first assertion. Using equations (9) again, we obtain the following tri-diagonal matrix as the production matrix of L:

( \begin{matrix}r & 1 & 0 & 0 & 0 & 0 & … \\ \frac{r^{2}−1}{2} & r & 1 & 0 & 0 & 0 & … \\ 0 & \frac{r^{2}−1}{4} & r & 1 & 0 & 0 & … \\ 0 & 0 & \frac{r^{2}−1}{4} & r & 1 & 0 & … \\ 0 & 0 & 0 & \frac{r^{2}−1}{4} & r & 1 & … \\ 0 & 0 & 0 & 0 & \frac{r^{2}−1}{4} & r & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

Corollary 33.

L^{−1} = (\frac{1 + \frac{1−r^{2}}{4} x^{2}}{1 + rx + \frac{r^{2}−1}{2}x^{2}} , \frac{x}{1 + rx + \frac{r^{2}−1}{2}x^{2}} )

is the coefficient array of a set of orthogonal polynomials for which the Legendre polynomials are moments.

Proposition 34. The Hankel transform of P_{n}(r) is given by

\frac{(r^{2} − 1)^{\binom{n+1}{2}}}{2^{n^{2}}} .

Proof. From the above, we obtain that the g.f. of P_{n}(r) can be expressed as

\frac{1}{1 − rx − \frac{\frac{r^{2}−1}{2}x^{2}}{1 − rx − \frac{\frac{r^{2}−1}{4}x^{2}}{1 − rx − \frac{\frac{r^{2}−1}{4}x^{2}}{1 − ⋯}}}}.

The result now follows from Equation (16). We end this section by noting that

P_{n}(r) = \frac{1}{π} ∫_{r−\sqrt{r^{2}−1}}^{r+\sqrt{r^{2}−1}} \frac{x^{n}}{\sqrt{−x^{2} + 2rx − 1}} dx

gives an explicit moment representation for P_{n}(r).

8 Hermite polynomials

The Hermite polynomials may be defined as

H_{n}(x) = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \frac{(−1)^{k}(2x)^{n−2k}}{k!(n − 2k)!} .

The generating function for H_{n}(x) is given by

e^{2xt−t^{2}} = ∑_{n=0}^{∞} H_{n}(x)\frac{t^{n}}{n!}.

The unitary Hermite polynomials (also called normalized Hermite polynomials) are given by

He_{n}(x) = 2^{− \frac{n}{2}} H_{n}(\sqrt{2}x) = ∑_{k=0}^{n} \frac{n!}{(−2)^{\frac{n−k}{2}} k! (\frac{n−k}{2})!} \frac{1 + (−1)^{n−k}}{2} x^{k}.

Their generating function is given by

e^{xt− \frac{t^{2}}{2}} = ∑_{n=0}^{∞} He_{n}(x)\frac{t^{n}}{n!}.

We note that the coefficient array of He_{n} is a proper exponential Riordan array, equal to [e^{− \frac{x^{2}}{2}} , x] .

This array \underline{A066325} begins

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

It is the aeration of the alternating sign version of the Bessel coefficient array \underline{A001497}. The inverse of this matrix has production matrix

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

which corresponds to the fact that we have the following three-term recurrence for He_{n}:

He_{n+1}(x) = xHe_{n}(x) − nHe_{n−1}(x).

9 Hermite polynomials as moments

Proposition 35. The proper exponential Riordan array

L = [e^{rx− \frac{x^{2}}{2}} , x]

has as first column the unitary Hermite polynomials He_{n}(r). This array has a tri-diagonal production array.

Proof. The first column of L has generating function e^{rx− \frac{x^{2}}{2}} , from which the first assertion follows. We now use equations (13) and (14) to calculate the production matrix P = S_{L}. We have f(x) = x, so that f̄(x) = x and f′(f̄(x)) = 1 = r(x). Also, g(x) = e^{rx− \frac{x^{2}}{2}} implies that g′(x) = g(x)(r − x), and hence

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

Thus the production array of L is indeed tri-diagonal, beginning

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

We note that L starts

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ r & 1 & 0 & 0 & 0 & 0 & … \\ r^{2} − 1 & 2r & 1 & 0 & 0 & 0 & … \\ r(r^{2} − 3) & 3(r^{2} − 1) & 3r & 1 & 0 & 0 & … \\ r^{4} − 6r^{2} + 3 & 4r(r^{2} − 3) & 6(r^{2} − 1) & 4r & 1 & 0 & … \\ r(r^{4} − 10r^{2} + 15) & 5(r^{4} − 6r^{2} + 3) & 10r(r^{2} − 3) & 10(r^{2} − 1) & 5r & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

Thus

L^{−1} = [e^{−rx+ \frac{x^{2}}{2}} , x]

is the coefficient array of a set of orthogonal polynomials which have as moments the unitary Hermite polynomials. These new orthogonal polynomials satisfy the three-term recurrence

ℌ_{n+1}(x) = (x − r)ℌ_{n}(x) + nℌ_{n−1}(x),

with ℌ_{0} = 1, ℌ_{1} = x − r. Proposition 36. The Hankel transform of the sequence He_{n}(r) is given by (−1)^{\binom{n+1}{2}} ∏_{k=0}^{n} k! Proof. By the above, the g.f. of H_{n}(r) is given by

\frac{1}{1 − rx + \frac{2x^{2}}{1 − rx + \frac{3x^{2}}{1 − rx + \frac{4x^{2}}{1 − …}}}}.

The result now follows from Equation (16).

Turning now to the Hermite polynomials H_{n}(x), we have the following result.

Proposition 37. The proper exponential Riordan array

L = [e^{2rx−x^{2}}, x]

has as first column the Hermite polynomials H_{n}(r). This array has a tri-diagonal production array.

Proof. The first column of L has generating function e^{2rx−x^{2}}, from which the first assertion follows. Using equations (13) and (14) we find that the production array of L is indeed tri-diagonal, beginning

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

We note that L starts

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 2r & 1 & 0 & 0 & 0 & 0 & … \\ 2(2r^{2} − 1) & 4r & 1 & 0 & 0 & 0 & … \\ 4r(2r^{2} − 3) & 6(2r^{2} − 1) & 6r & 1 & 0 & 0 & … \\ 4(4r^{3} − 12r^{2} + 3) & 16r(2r^{2} − 3) & 12(2r^{2} − 1) & 8r & 1 & 0 & … \\ 8r(4r^{4} − 20r^{2} + 15) & 20(4r^{4} − 12r^{2} + 3) & 40r(2r^{2} − 3) & 20(2r^{2} − 1) & 10r & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

Thus

L^{−1} = [e^{−2rx+x^{2}}, x]

is the coefficient array of a set of orthogonal polynomials which have as moments the Hermite polynomials. These new orthogonal polynomials satisfy the three-term recurrence

ℌ_{n+1}(x) = (x − 2r)ℌ_{n}(x) + 2nℌ_{n−1}(x),

with ℌ_{0} = 1, ℌ_{1} = x − 2r.

Proposition 38. The Hankel transform of the sequence H_{n}(r) is given by (−1)^{\binom{n+1}{2}} ∏_{k=0}^{n} 2^{k}k!

Proof. By the above, the g.f. of H_{n}(r) is given by

\frac{1}{1 − 2rx + \frac{2x^{2}}{1 − 2rx + \frac{4x^{2}}{1 − 2rx + \frac{6x^{2}}{1 − …}}}}.

The result now follows from Equation (16).

10 Acknowledgements

The author would like to thank an anonymous referee for their careful reading and cogent suggestions which have hopefully led to a clearer paper.

11 Appendix - The Stieltjes transform of a measure

The Stieltjes transform of a measure μ on ℝ is a function G_{μ} defined on ℂ \ ℝ by

G_{μ}(z) = ∫_{ℝ} \frac{1}{z − t}μ(t).

If f is a bounded continuous function on ℝ, we have

∫_{ℝ} f(x)μ(x) = − lim_{y→0^{+}} ∫_{ℝ} f(x)ℑG_{μ}(x + iy)dx.

If μ has compact support, then G_{μ} is holomorphic at infinity and for large z,

G_{μ}(z) = ∑_{n=0}^{∞} \frac{a_{n}}{z^{n+1}} ,

where a_{n} = ∫_{ℝ} t^{n}μ(t) are the moments of the measure. If μ(t) = dψ(t) = ψ′(t)dt then (Stieltjes-Perron)

ψ(t) − ψ(t_{0}) = −\frac{1}{π} lim_{y→0^{+}} ∫_{t_{0}}^{t} ℑG_{μ}(x + iy)dx.

If now g(x) is the generating function of a sequence a_{n}, with g(x) = ∑_{n=0}^{∞} a_{n}x^{n}, then we can define

G(z) = \frac{1}{z}g (\frac{1}{z}) = ∑_{n=0}^{∞} \frac{a_{n}}{z^{n+1}} .

By this means, under the right circumstances we can retrieve the density function for the measure that defines the elements a_{n} as moments.

Example 39. We let g(z) = \frac{1−\sqrt{1−4z}}{2z} be the g.f. of the Catalan numbers. Then

G(z) = \frac{1}{z}g (\frac{1}{z}) = \frac{1}{2} (1 − \sqrt{\frac{x − 4}{x}}) .

Then

ℑG_{μ}(x + iy) = −\frac{\sqrt{2}\sqrt{\sqrt{x^{2} + y^{2}}\sqrt{x^{2} − 8x + y^{2} + 16} − x^{2} + 4x − y^{2}}}{4\sqrt{x^{2} + y^{2}}} ,

and so we obtain

ψ′(x) = −\frac{1}{π} lim_{y→0^{+}} \{−\frac{\sqrt{2}\sqrt{\sqrt{x^{2} + y^{2}}\sqrt{x^{2} − 8x + y^{2} + 16} − x^{2} + 4x − y^{2}}}{4\sqrt{x^{2} + y^{2}}} \}

= \frac{1}{2π} \frac{\sqrt{x(4 − x)}}{x}.

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2010 Mathematics Subject Classification: Primary 42C05; Secondary 11B83, 11C20, 15B05, 15B36, 33C45. Keywords: Legendre polynomials, Hermite polynomials, integer sequence, orthogonal polynomials, moments, Riordan array, Hankel determinant, Hankel transform.

(Concerned with sequences \underline{A000007}, \underline{A000045}, \underline{A000108}, \underline{A000262}, \underline{A001405}, \underline{A001497}, \underline{A007318}, \underline{A009766}, \underline{A021009}, \underline{A033184}, \underline{A053121}, \underline{A066325}, \underline{A094587}, \underline{A094816}, \underline{A111596}, \underline{A111884}, \underline{A119467}, \underline{A119879}, and \underline{A155585}.)

Received May 14 2010; revised version received December 6 2010; January 31 2011. Published in Journal of Integer Sequences, February 19 2011.

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