A Note on Three Families of Orthogonal Polynomials defined by Circular Functions,

and Their Moment Sequences

Paul Barry School of Science Waterford Institute of Technology Ireland

                         pbarry@wit.ie

Abstract

Using the language of exponential Riordan arrays, we study three distinct families of orthogonal polynomials defined by trigonometric functions. We study the moment sequences of theses families, finding continued fraction expressions for their generating functions, and calculate the Hankel transforms of these moment sequences. Results related to the Euler or zigzag numbers, as well as the generalized Euler or Springer numbers, are found. In addition, we characterize the Dowling numbers as moments of a family of orthogonal polynomials.

1 Introduction

Riordan arrays [12] are known to have many applications in the area of combinatorics. They often allow us to express in concise form results about objects of combinatorial interest [14], and to infer analogues and generalizations, by virtue of their expressive form. In this note, we continue a study of the links between certain Riordan arrays and orthogonal polynomials [3, 7], and study three families of orthogonal polynomials each defined by trigonometric functions. Our results stem from the observation that if a Riordan array L has a tri-diagonal production matrix, then L^{−1} is the coefficient array of a family of orthogonal polynomials [1, 2]. We shall be concerned in this note with exponential Riordan arrays [4], that is, infinite lower-triangular matrices M (denoted by [g, f]) defined by a pair of power series f(x), where f(0) = 0, f′(0) ≠ 0, and g(x), where g(0) ≠ 0, such that the k-th column of M has exponential generating function g(x)(f(x))^{k}/k!. We recall that for an exponential Riordan array

L = [g, f]

the production matrix P_{L} of L [4, 5, 6] is the matrix

P_{L} = L^{−1}L̃,

where L̃ is the matrix L with the first row removed. The bivariate generating function of P_{L} is given by

e^{xy}(c(x) + yr(x)),

where we have the Deutsch equations

r(x) = f′(f̄(x)),  (1)

and

c(x) = \frac{g′(f̄(x))}{g(f̄(x))}.  (2)

The following well-known results (the first is the well-known “Favard’s Theorem”), which we essentially reproduce from [8], specify the links between orthogonal polynomials, the threeterm 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. [8] (Cf. [16, Théorème 9, p. I-4], or [17, 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. [8] (Cf. [16, Proposition 1, Eq. (7), p. V-5], or [17, 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 [10] 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} | .  (3)

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 [8]

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}.  (4)

Note that this is independent of α_{n}. The exponential Riordan arrays that we shall study in the sequel will have polynomial entries, where the polynomials have integer coefficients.

2 The exponential Riordan array L = [\frac{1}{cos^{r} x}, \frac{sin x}{cos x}]

The numbers with generating function \frac{1}{cos x} = sec x begin

1, 0, 1, 0, 5, 0, 61, 0, …

The “unaerated” sequence 1, 1, 5, 61, … is called the sequence of Euler, secant, or Zig numbers \underline{A000364}. Both sequences are of importance in combinatorics. For instance, they are closely associated to alternating permutations. We consider the related exponential Riordan array

L = [\frac{1}{cos^{r} x}, \frac{sin x}{cos x}] = [sec^{r} x, tan x],

which depends on the parameter r. The inverse of L is given by

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

We have the following proposition, relating these matrices to a family of orthogonal polynomials.

Proposition 3. The matrix

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

is the coefficient array of the family of orthogonal polynomials P_{n}^{(r)}(x) which satisfy the following three-term recurrence

P_{n}^{(r)}(x) = xP_{n−1}^{(r)}(x) − ((n − 1)r + (n − 1)(n − 2))P_{n−2}^{(r)}(x),

with P_{0}^{(r)}(x) = 1, P_{1}^{(r)}(x) = x.

Proof. We must show that the production matrix of L is tri-diagonal. We have f(x) = tan x and thus f̄(x) = tan^{−1}(x), and f′(x) = sec^{2} x. Similarly g(x) = sec^{r} x and hence g′(x) = r sin x sec^{r+1} x. Thus

r(x) = f′(f̄(x)) = 1 + x^{2},

while

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

This implies that the production matrix P_{L} of L is generated by

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

Thus the production matrix P_{L} has the form

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

which proves the assertion.

Note that we can write the above three-term recurrence as

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

Corollary 4. The generating function of the moment sequence associated to the family of orthogonal polynomials P_{n}^{(r)} is given by

sec^{r} x = \frac{1}{1 − \frac{rx^{2}}{1 − \frac{2(r + 1)x^{2}}{1 − \frac{3(r + 2)x^{2}}{1 − \frac{4(r + 3)x^{2}}{1 − ⋯}}}}}.

We note that this moment sequence begins

1, 0, r, 0, r(3r + 2), 0, r(15r^{2} + 30r + 16), 0, ….

This sequence appears as the first column of the matrix L.

Corollary 5. The Hankel transform of the moment sequence associated to the family of orthogonal polynomials P_{n}^{(r)} is given by

h_{n} = ∏_{k=0}^{n} k!(r + k)^{n−k}.

Proof. We have

h_{n} = ∏_{k=0}((k + 1)r + k(k + 1))^{n−k} = ∏_{k=0}^{n}(k + 1)^{n−k}(r + k)^{n−k} = ∏_{k=0}^{n} k!(r + k)^{n−k}.

Setting r = 1 shows that sec x is the generating function of the moments of the orthogonal polynomials

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

The Hankel transform of these moments (the aerated Euler numbers) is thus given by

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

This is \underline{A055209}.

3 The exponential Riordan array L = [\frac{1}{(cos x−sin x)^{r}}, \frac{sin x}{cos x−sin x}]

We now modify the denominator in the foregoing from cos x to cos x − sin x. We note that

\frac{1}{cos x − sin x}

is the generating function of the so-called Springer, or generalized Euler numbers \underline{A001586}. Thus we are led to consider the exponential Riordan array

L = [\frac{1}{(cos x − sin x)^{r}} , \frac{sin x}{cos x − sin x}] = [\frac{1}{(cos x − sin x)^{r}} , \frac{1}{1 − tan x} − 1] .

Note that

L = [(f′(x))^{r/2}, f(x)] where f(x) = \frac{sin x}{cos x − sin x}.

Again, we find that

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

is the coefficient array of a family of orthogonal polynomials. This is the content of

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

is the coefficient array of the family of orthogonal polynomials P_{n}^{(r)}(x) which satisfy the following three-term recurrence

P_{n}^{(r)}(x) = (x − (r + 2(n − 1)))P_{n−1}^{(r)}(x) − (2(n − 1)r + 2(n − 1)(n − 2))P_{n−2}^{(r)}(x),

with P_{0}^{(r)}(x) = 1, P_{1}^{(r)}(x) = x − r.

Proof. We have g(x) = \frac{1}{(cos x−sin x)^{r}}, and f(x) = \frac{sin x}{cos x−sin x}. Then using equations (1) and (2) we obtain that the generating function of the production array P_{L} of L is equal to

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

This implies that P_{L} has the form

( \begin{matrix}r & 1 & 0 & 0 & 0 & 0 & ⋯ \\ 2r & r + 2 & 1 & 0 & 0 & 0 & ⋯ \\ 0 & 4r + 4 & r + 4 & 1 & 0 & 0 & ⋯ \\ 0 & 0 & 6r + 12 & r + 6 & 1 & 0 & ⋯ \\ 0 & 0 & 0 & 8r + 24 & r + 8 & 1 & ⋯ \\ 0 & 0 & 0 & 0 & 10r + 40 & r + 10 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

which proves the assertion.

Corollary 7. The generating function of the moment sequence associated to the family of orthogonal polynomials P_{n}^{(r)} is given by

\frac{1}{(cos x − sin x)^{r}} = \frac{1}{1 − rx − \frac{2rx^{2}}{1 − (r + 2)x − \frac{4(r + 1)x^{2}}{1 − (r + 4)x − \frac{6(r + 2)x^{2}}{1 − (r + 6)x − \frac{8(r + 3)x^{2}}{1 − ⋯}}}}}.

We note that this moment sequence begins

1, r, r^{2} + 2r, r(r^{2} + 6r + 4), r(r^{3} + 12r^{2} + 28r + 16), ….

Corollary 8. The Hankel transform of the moment sequence associated to the family of orthogonal polynomials P_{n}^{(r)} is given by

h_{n} = 2^{\binom{n+1}{2}} ∏_{k=0}^{n} k!(r + k)^{n−k}.

Proof. We have

h_{n} = ∏_{k=0}^{n} 2^{k}k!(r + k)^{n−k} = 2^{\binom{n+1}{2}} ∏_{k=0}^{n} k!(r + k)^{n−k}.

Setting r = 1 shows that \frac{1}{cos x−sin x} is the generating function of the moments of the orthogonal polynomials

P_{n}^{(1)}(x) = (x − (2n − 1))P_{n−1}^{(1)}(x) − 2(n − 1)^{2}P_{n−2}^{(1)}(x).

The Hankel transform of these moments (the generalized Euler or Springer numbers) is thus given by

h_{n} = 2^{\binom{n+1}{2}} ∏_{k=0}^{n} k!(k + 1)^{n−k} = 2^{\binom{n+1}{2}} ∏_{k=0}^{n} k!^{2}.

This is \underline{A091804}. It is of interest to analyze the structure of the moment sequence

1, r, r^{2} + 2r, r(r^{2} + 6r + 4), r(r^{3} + 12r^{2} + 28r + 16)….

This is a sequence of polynomials in r, with coefficient array given by the exponential Riordan array

[1, ln (\frac{1}{cos x − sin x})] .

We can generalize the foregoing results as follows.

Proposition 9. The matrix

L^{−1} = [\frac{1}{(1 + 2x + 2sx^{2})^{\frac{r}{2}}} , \begin{matrix}tan & −1 & (\frac{\sqrt{2s−1}x}{1+x}) \\ √ & \overline{2s − 1} \end{matrix}]

is the coefficient array of the family of orthogonal polynomials P_{n}^{(r,s)}(x) which satisfy the following three-term recurrence

P_{n}^{(r,s)}(x) = (x − (r + 2(n − 1)))P_{n−1}^{(r,s)}(x) − (2(n − 1)r + 2s(n − 1)(n + r − 2))P_{n−2}^{(r,s)}(x),

with P_{0}^{(r,s)}(x) = 1, P_{1}^{(r,s)}(x) = x − r.

In fact, we find that the matrix L, where

L = [\frac{1}{(cos(\sqrt{2s − 1}x) − sin(\sqrt{2s − 1}x)/\sqrt{2s − 1})^{r}} , \frac{sin(\sqrt{2s − 1}x)/\sqrt{2s − 1}}{cos(\sqrt{2s − 1}x) − sin(\sqrt{2s − 1}x)/\sqrt{2s − 1}}] ,

has production matrix

( \begin{matrix}r & 1 & 0 & 0 & 0 & 0 & ⋯ \\ 2sr & r + 2 & 1 & 0 & 0 & 0 & ⋯ \\ 0 & 4s(r + 1) & r + 4 & 1 & 0 & 0 & ⋯ \\ 0 & 0 & 6s(r + 2) & r + 6 & 1 & 0 & ⋯ \\ 0 & 0 & 0 & 8s(r + 3) & r + 8 & 1 & ⋯ \\ 0 & 0 & 0 & 0 & 10s(r + 4) & r + 10 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

which shows that L^{−1} is indeed the coefficient array of the family of orthogonal polynomials P_{n}^{(r,s)}. In addition, we see that the generating function

\frac{1}{(cos(\sqrt{2s − 1}x) − sin(\sqrt{2s − 1}x)/\sqrt{2s − 1})^{r}}

has the following continued fraction expression:

\frac{1}{1 − rx − \frac{2srx^{2}}{1 − (r + 2)x − \frac{4s(r + 1)x^{2}}{1 − (r + 4)x − \frac{6s(r + 2)x^{2}}{1 − (r + 6)x − \frac{8s(r + 3)x^{2}}{1 − ⋯}}}}}.

Hence the Hankel transform of the moment sequence of the family of orthogonal polynomials P_{n}^{(r,s)}(x) is given by

h_{n}(r, s) = (2s)^{\binom{n+1}{2}} ∏_{k=0}^{n} k!(r + k)^{n−k}.

Writing u = 2s − 1, we have

Proposition 10. The matrix

L^{−1} = [\frac{1}{(1 + 2x + (u + 1)x^{2})^{\frac{r}{2}}} , \begin{matrix}tan & −1 & (\frac{\sqrt{u}x}{1+x}) \\ √ & \overline{u} \end{matrix}]

is the coefficient array of the family of orthogonal polynomials P_{n}^{(r,u)}(x) which satisfy the following three-term recurrence

P_{n}^{(r,s)}(x) = (x − (r + 2(n − 1)))P_{n−1}^{(r,s)}(x) − (n − 1)(n + r − 2)(u + 1)P_{n−2}^{(r,s)}(x),

with P_{0}^{(r,s)}(x) = 1, P_{1}^{(r,s)}(x) = x − r.

In this case, the production matrix of L is given by

( \begin{matrix}r & 1 & 0 & 0 & 0 & 0 & ⋯ \\ r(u + 1) & r + 2 & 1 & 0 & 0 & 0 & ⋯ \\ 0 & 2(r + 1)(u + 1) & r + 4 & 1 & 0 & 0 & ⋯ \\ 0 & 0 & 3(r + 2)(u + 1) & r + 6 & 1 & 0 & ⋯ \\ 0 & 0 & 0 & 4(r + 3)(u + 1) & r + 8 & 1 & ⋯ \\ 0 & 0 & 0 & 0 & 5(r + 4)(u + 1) & r + 10 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

We note that for orthogonality, we must have s ≠ 0, or equivalently, u ≠ −1. The case r = 1, s = 0 is interesting. In this case, we have

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

The matrix L^{−1}, which in this case is not the coefficient array of a family of orthogonal polynomials, begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & ⋯ \\ −1 & 1 & 0 & 0 & 0 & 0 & ⋯ \\ 3 & −4 & 1 & 0 & 0 & 0 & ⋯ \\ −15 & 23 & −9 & 1 & 0 & 0 & ⋯ \\ 105 & −176 & 86 & −16 & 1 & 0 & ⋯ \\ −945 & 1689 & −950 & 230 & −25 & 1 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

with L given by

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & ⋯ \\ 1 & 1 & 0 & 0 & 0 & 0 & ⋯ \\ 1 & 4 & 1 & 0 & 0 & 0 & ⋯ \\ 1 & 13 & 9 & 1 & 0 & 0 & ⋯ \\ 1 & 40 & 58 & 16 & 1 & 0 & ⋯ \\ 1 & 121 & 330 & 170 & 25 & 1 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

This is \underline{A039755}. It represents the 2-Dowling arrangements, corresponding to the Whitney numbers for the B_{n} lattices [15]. We have in this case

L = [e^{x}, e^{x} sinh x] ,

and the production matrix of L is given by

( \begin{matrix}1 & 1 & 0 & 0 & 0 & 0 & ⋯ \\ 0 & 3 & 1 & 0 & 0 & 0 & ⋯ \\ 0 & 0 & 5 & 1 & 0 & 0 & ⋯ \\ 0 & 0 & 0 & 7 & 1 & 0 & ⋯ \\ 0 & 0 & 0 & 0 & 9 & 1 & ⋯ \\ 0 & 0 & 0 & 0 & 0 & 11 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

The row sums of

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

give the sequence of Dowling numbers \underline{A007405} with e.g.f.

e^{x}e^{e^{x} sinh x} = e^{x} exp (\frac{e^{2x} − 1}{2}) .

This sequence begins

1, 2, 6, 24, 116, 648, 4088, 28640, …,

and has Hankel transform [15]

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

The sequence of Dowling numbers has g.f. given by the continued fraction

\frac{1}{1 − 2x − \frac{2x^{2}}{1 − 4x − \frac{4x^{2}}{1 − 6x − \frac{6x^{2}}{1 − 8x − ⋯}}}}.

This follows from the fact that the Dowling numbers coincide with the first column of the exponential Riordan array

L_{D} = [e^{x} exp(e^{x} sinh x), e^{x} sinh x]

which has production matrix

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

Thus the Dowling numbers are the moments of the family of orthogonal polynomials P_{n}^{D}(x) with coefficient array

L_{D}^{−1} = [\frac{e^{−x}}{\sqrt{1 + 2x}}, \frac{ln(1 + 2x)}{2}] = [\frac{e^{−x}}{\sqrt{1 + 2x}}, ln(\sqrt{1 + 2x})] ,

which satisfy the three-term recurrence

P_{n}^{D}(x) = (x − 2n)P_{n−1}^{D}(x) − 2(n − 1)P_{n−1}^{D}(x),

with P_{0}^{D}(x) = 1 and P_{1}^{D}(x) = x − 2.

4 The Riordan array L = [(\frac{1+sin x}{cos^{2} x})^{r} , \frac{sin x+1−cos x}{cos x}]

In this example, we start with f(x) = \frac{sin x}{cos x} and then modify the numerator expression by adding 1 − cos x. Our modified f(x) is now given by

f(x) = \frac{sin x + 1 − cos x}{cos x} = sec x + tan x − 1.

We note that the numbers with e.g.f. sec x + tan x, which begin

1, 1, 1, 2, 5, 16, 61, 272, 1385, 7936, …

are called “up-down” numbers (or Euler numbers) \underline{A000111}. We now let

g(x) = (f′(x))^{r} = (\frac{1 + sin x}{cos^{2} x})^{r} = ((1 + sin x) sec^{2} x)^{r}.

Thus we consider the exponential Riordan array

L = [(f′(x))^{r}, f(x)] = [((1 + sin x) sec^{2} x)^{r}, sec x + tan x − 1].

We have

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

Note that

(2 tan^{−1}(1 + x) − \frac{π}{2} )′ = \frac{2}{2 + 2x + x^{2}} .

Proposition 11. The matrix

L^{−1} = [(\frac{2}{2 + 2x + x^{2}} )^{r} , 2 tan^{−1}(1 + x) − \frac{π}{2} ]

is the coefficient array of the family of orthogonal polynomials P_{n}^{(r)}(x) which satisfy the following three-term recurrence

P_{n}^{(r)}(x) = (x − r − n + 1)P_{n−1}^{(r)}(x) − ((n − 1)r + \binom{n − 1}{2})P_{n−2}^{(r)}(x),

with P_{0}^{(r)}(x) = 1, P_{1}^{(r)}(x) = x − r.

Proof. We have g(x) = ((1 + sin x) sec^{2} x)^{r}, and sec x + tan x − 1. Then using equations (1) and (2) we obtain that the generating function of the production array P_{L} of L is equal to

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

This implies that P_{L} has the form

( \begin{matrix}r & 1 & 0 & 0 & 0 & 0 & ⋯ \\ r & r + 1 & 1 & 0 & 0 & 0 & ⋯ \\ 0 & 2r + 1 & r + 2 & 1 & 0 & 0 & ⋯ \\ 0 & 0 & 3r + 3 & r + 3 & 1 & 0 & ⋯ \\ 0 & 0 & 0 & 4r + 6 & r + 4 & 1 & ⋯ \\ 0 & 0 & 0 & 0 & 5r + 10 & r + 5 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

which proves the assertion.

Corollary 12. The generating function of the moment sequence associated to the family of orthogonal polynomials P_{n}^{(r)} is given by

((1 + sin x) sec^{2} x)^{r} = \frac{1}{1 − rx − \frac{rx^{2}}{1 − (r + 1)x − \frac{(2r + 1)x^{2}}{1 − (r + 2)x − \frac{(3r + 3)x^{2}}{1 − (r + 3)x − \frac{(4r + 6)x^{2}}{1 − ⋯}}}}}.

Corollary 13. The Hankel transform of the moment sequence associated to the family of orthogonal polynomials P_{n}^{(r)} is given by

h_{n} = ∏_{k=0}^{n} k!(r + k/2)^{n−k}.

For r = 1, we get the exponential Riordan array

L = [\frac{1 + sin x}{cos^{2} x}, \frac{sin x + 1 − cos x}{cos x}] ,

which begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & ⋯ \\ 1 & 1 & 0 & 0 & 0 & 0 & ⋯ \\ 2 & 3 & 1 & 0 & 0 & 0 & ⋯ \\ 5 & 11 & 6 & 1 & 0 & 0 & ⋯ \\ 16 & 45 & 35 & 10 & 1 & 0 & ⋯ \\ 61 & 211 & 210 & 85 & 15 & 1 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

The first column is made up of the “shortened” or reduced up-down (or Euler) numbers with e.g.f. (sec x + tan x)′. This matrix is \underline{A147315}. This matrix has production matrix

( \begin{matrix}1 & 1 & 0 & 0 & 0 & 0 & ⋯ \\ 1 & 2 & 1 & 0 & 0 & 0 & ⋯ \\ 0 & 3 & 3 & 1 & 0 & 0 & ⋯ \\ 0 & 0 & 6 & 4 & 1 & 0 & ⋯ \\ 0 & 0 & 0 & 10 & 5 & 1 & ⋯ \\ 0 & 0 & 0 & 0 & 15 & 6 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

and hence the shortened up-down numbers have generating function given by the continued fraction

\frac{1}{1 − x − \frac{x^{2}}{1 − 2x − \frac{3x^{2}}{1 − 3x − \frac{6x^{2}}{1 − 4x − \frac{10x^{2}}{1 − 5x − ⋯}}}}},

and Hankel transform \underline{A154604}

h_{n} = ∏_{k=1}^{n} \binom{k + 1}{2}^{n−k+1} = ∏_{k=0}^{n} \binom{k + 2}{2}^{n−k} = (\frac{1}{2})^{\binom{n+1}{2}} ∏_{k=0}^{n} k!(k + 1)!.

References

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

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2010 Mathematics Subject Classification: Primary 15B36; Secondary 11B83, 11C20, 15B05, 33C45, 42C05. Keywords: Exponential Riordan array, moments, Hankel transform, orthogonal polynomials, Euler numbers, Dowling numbers.

(Concerned with sequences \underline{A000111}, \underline{A000364}, \underline{A001586}, \underline{A007405}, \underline{A039755}, \underline{A055209}, \underline{A091804}, \underline{A147315}, and \underline{A154604}.)

Received September 6 2011; revised versions received August 16 2012; September 4 2012. Published in Journal of Integer Sequences, September 8 2012.

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