Meixner-Type Results for Riordan Arrays and Associated Integer Sequences

Paul Barry School of Science Waterford Institute of Technology Ireland

                     pbarry@wit.ie

Aoife Hennessy Department of Computing, Mathematics and Physics Waterford Institute of Technology Ireland

               aoife.hennessy@gmail.com

Abstract

We determine which (ordinary) Riordan arrays are the coefficient arrays of a family of orthogonal polynomials. In so doing, we are led to introduce a family of polynomials, which includes the Boubaker polynomials, and a scaled version of the Chebyshev polynomials, using the techniques of Riordan arrays. We classify these polynomials in terms of the Chebyshev polynomials of the first and second kinds. We also examine the Hankel transforms of sequences associated with the inverse of the polynomial coefficient arrays, including the associated moment sequences.

1 Introduction

With each Riordan array (A(t), B(t)) we can associate a family of polynomials [19] by

∑_{n=0}^{∞} p_{n}(x)t^{n} = (A(t), B(t)) · \frac{1}{1 − xt} = \frac{A(t)}{1 − xB(t)}.

The question can then be asked as to what conditions must be satisfied by A(t) and B(t) in order to ensure that (p_{n}(x))_{n≥0} be a family of orthogonal polynomials. This can be considered to be a Meixner-type question [22], where the original Meixner result is related to Sheffer sequences (i.e., to exponential generating functions, rather than ordinary generating functions):

∑_{n=0}^{∞} p_{n}(x)t^{n} = A(t) exp(xB(t)).

In providing an answer to this question, we shall introduce a two-parameter family of orthogonal polynomials using Riordan arrays. These polynomials are inspired by the well-known Chebyshev polynomials [25], and the more recently introduced so-called Boubaker polynomials [2, 15, 16]. We shall classify these polynomials in terms of the Chebyshev polynomials of the first and second kinds, and we shall also examine properties of sequences related to the inverses of the coefficient arrays of the polynomials under study. While partly expository in nature, the note assumes a certain familiarity with integer sequences, generating functions, orthogonal polynomials [4, 10, 31], Riordan arrays [26, 30], production matrices [8, 24], and the integer Hankel transform [1, 6, 17]. Many interesting examples of sequences and Riordan arrays can be found in Neil Sloane’s On-Line Encyclopedia of Integer Sequences (OEIS), [28, 29]. 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 Riordan arrays

3. Orthogonal polynomials and Riordan arrays

4. Riordan arrays, production matrices and orthogonal polynomials

5. Chebyshev polynomials and Riordan arrays

6. The Boubaker polynomials

7. The family of Chebyshev-Boubaker polynomials

8. The inverse matrix 𝔅^{−1}

9. A curious relation

10. Acknowledgements

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) = ∑_{k=0}^{∞} a_{k}x^{k} 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}. 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 [11], 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 sometimes write 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 [26, 30], 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 [30]. 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}. Such arrays are also called Appell arrays as they form the elements of the 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)).

In [18, 19] the notation T (f|g) is used to denote the Riordan array

T (f|g) = (\frac{f(x)}{g(x)}, \frac{x}{g(x)}) .

Example 1. The so-called binomial matrix B is the element (\frac{1}{1−x}, \frac{x}{1−x}) of the Riordan group. Thus

B = T (1|1 − x).

This matrix 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” (using the nomenclature of [5] ) 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} ) .

It is often the case that we work with “generalized” Riordan arrays, where we relax some of the defining conditions above. Thus for instance [5] discusses the notion of the “stretched” Riordan array. In this note, we shall encounter “vertically stretched” arrays of the form (g, h) where now f_{0} = f_{1} = 0 with f_{2} ≠ 0. Such arrays are not invertible, but we may explore their left inversion. In this context, standard Riordan arrays as described above are called “proper” Riordan arrays. We note for instance that for any proper Riordan array (g, f), its diagonal sums are just the row sums of the vertically stretched array (g, xf) and hence have g.f. g/(1 − xf).

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}. We have

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

We note 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 [3, 12]. The simplest of these is as follows.

Proposition 4. [12] 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 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) and Z(x) denote the generating functions of these sequences, respectively, we have [20] that

\frac{f(x)}{x} = A(f(x)), g(x) = \frac{d_{0,0}}{1 − xZ(f(x))}.

We therefore deduce that

A(x) = \frac{x}{f̄(x)},

and

Z(x) = \frac{1}{f̄(x)} [1 − \frac{d_{0,0}}{g(f̄(x))}] .

A consequence of this is the following result, which was originally established [19] by Luzón:

Lemma 5. Let D = (g, f) be a Riordan array, whose A-sequence, respectively Z-sequence have generating functions A(x) and Z(x). Then

D^{−1} = (\frac{A − xZ}{d_{0,0}A} , \frac{x}{A}) .

3 Orthogonal polynomials and Riordan arrays

By an orthogonal polynomial sequence (p_{n}(x))_{n≥0} we shall understand [4, 10] 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}} ) ,

where

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

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 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 with 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} 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′ = H ( \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}} .

Of importance to this study are the following results (the first is the well-known “Favard’s Theorem”), which we essentially reproduce from [14].

Theorem 6. [14] (Cf. [32, Théorème 9, p. I-4] or [33, 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}(x), for n ≥ 1,

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

Theorem 7. [14] (Cf. [32, Proposition 1, (7), p. V-5], or [33, 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}(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}  (1)

and

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

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 8. 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. By [13], the array (\frac{1}{1+rx+sx^{2}} , \frac{x}{1+rx+sx^{2}} ) 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.

Thus 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}d_{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}.

We note that in this case the three-term recurrence coefficients α_{n} and β_{n} are constants.

We can strengthen this result as follows. Proposition 9. Every Riordan array of the form

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

is the coefficient array of a family of monic orthogonal polynomials. Proof. We have

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

where (1 − λx − μx^{2}, x) is the array with elements

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

We write

B = (b_{n,k}) = (\frac{1 − λx − μx^{2}}{1 + rx + sx^{2}} , \frac{x}{1 + rx + sx^{2}} ) ,

and

A = (a_{n,k}) = (\frac{1}{1 + rx + sx^{2}} , \frac{x}{1 + rx + sx^{2}} ) ,

where

a_{n+1,k} = a_{n,k−1} − ra_{n,k} − sa_{n−1,k}.  (3)

We now assert that also,

b_{n+1,k} = b_{n,k−1} − rb_{n,k} − sb_{n−1,k}.

This follows since the fact that

B = (1 − λx − μx^{2}, x) · A

tells us that

b_{n+1,k} = a_{n+1,k} − λa_{n,k} − μa_{n−1,k},
b_{n,k−1} = a_{n,k−1} − λa_{n−1,k−1} − μa_{n−2,k−1},
b_{n,k} = a_{n,k} − λa_{n−1,k} − μa_{n−2,k},
b_{n−1,k} = a_{n−1,k} − λa_{n−2,k} − μa_{n−3,k}.

Then using equation (3) and the equivalent equations for a_{n,k} and a_{n−1,k}, we see that

b_{n+1,k} = b_{n,k−1} − rb_{n,k} − sb_{n−1,k}

as required. Noting that

p_{0}(x) = 1, p_{1}(x) = x − r − λ, p_{2}(x) = x^{2} − (2r + λ)x + λr − μ + r^{2} − s, …,

we see that the family of orthogonal polynomials is defined by the α-sequence

α_{0} = r + λ, r, r, r, …

and the β-sequence

β_{1} = s + μ, s, s, s, ….

Proposition 10. The elements in the left-most column of

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

are the moments corresponding to the family of orthogonal polynomials with coefficient array L^{−1}.

Proof. We let

(g, f) = (\frac{1 − λx − μx^{2}}{1 + rx + sx^{2}} , \frac{x}{1 + rx + sx^{2}} ) .

Then

L = (g, f)^{−1} = (\frac{1}{g(f̄)}, f̄) .

Now f̄(x) is the solution to

\frac{u}{1 + rx + sx^{2}} = x,

thus

f̄(x) = \frac{1 − sx − \sqrt{1 − 2sx + (s^{2} − 4r)x^{2}}}{2rx} .

Then

\frac{1}{g(f̄(x))} = \frac{1 + rf̄(x) + s(f̄(x))^{2}}{1 − λf̄(x) − μ(f̄(x))^{2}} .

Simplifying,we find that

\frac{1}{g(f̄(x))} = \frac{2s}{(s + μ)\sqrt{1 − 2rx + (r^{2} − 4s)x^{2}} − (r(s − μ) + 2sλ)x + s − μ}.

We now consider the continued fraction

g̃(x) = \frac{1}{1 − (r + λ)x − \frac{(s + μ)x^{2}}{1 − rx − \frac{sx^{2}}{1 − rx − \frac{sx^{2}}{1 − rx − ·}}}}.

This is equivalent to

g̃(x) = \frac{1}{1 − (r + λ)x − (s + μ)x^{2}h(x)},

where

h(x) = \frac{1}{1 − rx − sx^{2}h(x)}.

Solving for h(x) and subsequently for g̃(x), we find that

g̃(x) = \frac{1}{g(f̄(x))}.

We have in fact the following proposition (see the next section for information on the Chebyshev polynomials).

Proposition 11. 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. We have

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

Thus

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

Now

\frac{1}{1 − 2\frac{x−r}{2\sqrt{s}} \sqrt{s}t + st^{2}} = \frac{1}{1 − (x − r)t + st^{2}}
= (\frac{1}{1 + rt + st^{2}} , \frac{t}{1 + rt + st^{2}} ) · \frac{1}{1 − xt}.

Thus

(\frac{1}{1 + rt + st^{2}} , \frac{t}{1 + rt + st^{2}} ) · \frac{1}{1 − xt} = ∑_{n=0}^{∞}(\sqrt{s})^{n}U_{n} (\frac{x − r}{2\sqrt{s}}) t^{n}

as required.

For another perspective on this result, see [9].

Corollary 12. The Riordan array (\frac{1−λx−μx^{2}}{1+rx+sx^{2}} , \frac{x}{1+rx+sx^{2}} ) is the coefficient array of the generalized Chebyshev polynomials of the second kind given by

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

Proof. We have

U_{n}(x) = [x^{n}]\frac{1}{1 − 2xt + t^{2}}

By the method of coefficients [21] we then have

[x^{n}]\frac{t}{1 − 2xt + t^{2}} = [x^{n−1}]\frac{1}{1 − 2xt + t^{2}} = U_{n−1}(x)

and similarly

[x^{n}]\frac{t^{2}}{1 − 2xt + t^{2}} = [x^{n−2}]\frac{1}{1 − 2xt + t^{2}} = U_{n−2}(x).

A more complete answer to our original question can be found by considering the associated production matrix [7, 8] of a Riordan arrray, which we look at in the next section.

4 Riordan arrays, production matrices and orthogonal polynomials

The concept of a production matrix [7, 8] 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

DA_{P} = A_{P} P

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

In [24, 27] P is called the Stieltjes matrix associated with A_{P}. The sequence formed by the row sums of A_{P} often has combinatorial significance and is called the sequence associated with 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 with a proper Riordan array takes on a special form :

Proposition 13. [8] 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} )

Moreover, columns 0 and 1 of the matrix P are the Z- and A-sequences, respectively, of the Riordan array A_{P}.

Example 14. We calculate the production matrix of the Riordan array

D = (c(x), xc(x)).

We have

f(x) = xc(x) ⇒ f̄(x) = x(1 − x),

and hence

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

Similarly,

Z(x) = \frac{1}{f̄(x)} [1 − \frac{d_{0,0}}{g(f̄(x))}]
= \frac{1}{x(1 − x)} [1 − \frac{1}{c(x(1 − x))}]

= \frac{1}{x(1 − x)} [1 − \frac{1}{\frac{1}{1−x}} ]
= \frac{1}{x(1 − x)} [1 − (1 − x)]
= \frac{1}{1 − x}.

Thus the production matrix of D = (c(x), xc(x)) is the matrix that begins

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

Example 15. We calculate the production matrix of the Riordan array

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

First, we have

f(x) = xc(x^{2}) ⇒ f̄(x) = \frac{x}{1 + x^{2}} ,

and hence

A(x) = \frac{x}{f̄(x)} = 1 + x^{2}.

Next, since

\frac{1}{g(f̄(x))} = \frac{1}{1 + x^{2}} ,

we have

Z(x) = \frac{1}{f̄(x)} [1 − \frac{d_{0,0}}{g(f̄(x))}]
= \frac{1 + x^{2}}{x} [1 − \frac{1}{1 + x^{2}} ]
= \frac{1 + x^{2}}{x} [\frac{1 + x^{2} − 1}{1 + x^{2}}]
= x.

Hence the production matrix of (c(x^{2}), xc(x^{2})) begins

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

We can generalize the last result to give the following important result.

Proposition 16. The Riordan array L where

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

has production matrix (Stieltjes matrix) 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} ) .

Proof. We let

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

By definition of the inverse, we have

f̄(x) = \frac{x}{1 + ax + bx^{2}}

and hence

A(x) = \frac{x}{f̄(x)} = 1 + ax + bx^{2}.

Also by definition of the inverse, we have

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

and hence

Z(x) = \frac{1}{f̄(x)} [1 − \frac{d_{0,0}}{g(f̄(x))}]
= \frac{1 + ax + bx^{2}}{x} [1 − \frac{1 − λx − μx^{2}}{1 + ax + bx^{2}} ]
= \frac{1 + ax + bx^{2}}{x} [1 + ax + bx^{2} − 1 + λx + μx^{2}]
= (a + λ) + (b + μ)x.

Corollary 17. The Riordan array

L = (\frac{1 + (a − a_{1})x + (b − b_{1})x^{2}}{1 + ax + bx^{2}} , \frac{x}{1 + ax + bx^{2}} )^{−1}

has production matrix that begins

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} ) .

Example 18. 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 (see [23]) 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 [23] : Proposition 19. [23] 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.

Of central importance to this note is the following result.

Proposition 20. If L = (g(x), f(x)) is a Riordan array and P = S_{L} is tridiagonal of the form

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} ) ,  (4)

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. (We are indebted to an anonymous reviewer for the form of the proof that follows). The form of the matrix P in (4) is equivalent to saying that A(x) = 1 + ax + bx^{2} and that Z(x) = a_{1} + b_{1}x. Now Lemma 5 tells us that if (d, h) is a Riordan array with A and Z the corresponding A-sequence and Z-sequence, respectively, then

(d, h)^{−1} = (\frac{A − xZ}{d_{0,0}A} , \frac{x}{A}) .

Note that by assumption, d_{0,0} = 1 here. Thus

L^{−1} = (\frac{1 + (a − a_{1})x + (b − b_{1})x^{2}}{1 + ax + bx^{2}} , \frac{x}{1 + ax + bx^{2}} ) = T (1+(a−a_{1})x+(b−b_{1})x^{2}|1+ax+bx^{2}).

Theorem 5 of [19] now yields the required form of the three-term recurrence for the associated polynomials with coefficient array L^{−1}. That these are orthogonal polynomials then follows by Favard’s theorem.

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 21. 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} ) .

We note that if L = (g(x), f(x)) is a Riordan array and P = S_{L} is tridiagonal of the form given in Eq. (4), then the first column of L gives the moment sequence for the weight function associated with the orthogonal polynomials whose coefficient array is L^{−1}.

As pointed out by a referee (to whom we are indebted for this important observation), the main results of the last two sections may be summarized as follows:

Proposition 22. Let L = (d(x), h(x)) be a Riordan array. Then the following are equivalent:

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

2. d(x) = \frac{1−λx−μx^{2}}{1+rx+sx^{2}} and h(x) = \frac{x}{1+rx+sx^{2}} with s ≠ 0.

3. The production matrix of L^{−1} is of the form

P = S_{L^{−1}} = ( \begin{matrix}r_{1} & 1 & 0 & 0 & 0 & 0 & … \\ s_{1} & 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} )

with s ≠ 0.

4. The bivariate generating function of L is of the form

\frac{1 − λx − μx^{2}}{1 + (r − t)x + sx^{2}}

with s ≠ 0.

Under these circumstances, the elements of the left-most column of L^{−1} are the moments associated with the linear functional that defines the family of orthogonal polynomials.

5 Chebyshev polynomials and Riordan arrays

We begin this section by recalling some facts about the Chebyshev polynomials of the first and second kind. Thus the Chebyshev polynomials of the first kind, T_{n}(x), are defined by

T_{n}(x) = \frac{n}{2} ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n − k}{k}\frac{(−1)^{k}}{n − k} (2x)^{n−2k}  (5)

for n > 0, and T_{0}(x) = 1. Similarly, the Chebyshev polynomials of the second kind, U_{n}(x), can be defined by

U_{n}(x) = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n − k}{k}(−1)^{k}(2x)^{n−2k},  (6)

or alternatively as

U_{n}(x) = ∑_{k=0}^{n} \binom{\frac{n+k}{2}}{k}(−1)^{\frac{n−k}{2}} \frac{1 + (−1)^{n−k}}{2} (2x)^{k}.  (7)

In terms of generating functions, we have

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

while

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

The Chebyshev polynomials of the second kind, U_{n}(x), which begin

1, 2x, 4x^{2} − 1, 8x^{3} − 4x, 16x^{4} − 12x^{2} + 1, 32x^{5} − 32x^{3} + 6x, …

have coefficient array

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 0 & 2 & 0 & 0 & 0 & 0 & … \\ −1 & 0 & 4 & 0 & 0 & 0 & … \\ 0 & −4 & 0 & 8 & 0 & 0 & … \\ 1 & 0 & −12 & 0 & 16 & 0 & … \\ 0 & 6 & 0 & −32 & 0 & 32 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) (\underline{A053117})

This is the (generalized) Riordan array

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

We note that the coefficient array of the modified Chebyshev polynomials U_{n}(x/2) which begin

1, x, x^{2} − 1, x^{3} − 2x, x^{4} − 3x^{2} + 1, …,

is given by

( \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 & … \\ 1 & 0 & −3 & 0 & 1 & 0 & … \\ 0 & 3 & 0 & −4 & 0 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) (\underline{A049310})

This is the Riordan array

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

The situation with the Chebyshev polynomials of the first kind is more complicated, since while the coefficient array of the polynomials 2T_{n}(x) − 0^{n}, which begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 0 & 2 & 0 & 0 & 0 & 0 & … \\ −2 & 0 & 4 & 0 & 0 & 0 & … \\ 0 & −6 & 0 & 8 & 0 & 0 & … \\ 2 & 0 & −16 & 0 & 16 & 0 & … \\ 0 & 10 & 0 & −40 & 0 & 32 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

is a (generalized) Riordan array, namely

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

that of T_{n}(x), which begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 0 & 1 & 0 & 0 & 0 & 0 & … \\ −1 & 0 & 2 & 0 & 0 & 0 & … \\ 0 & −3 & 0 & 4 & 0 & 0 & … \\ 1 & 0 & −8 & 0 & 8 & 0 & … \\ 0 & 5 & 0 & −20 & 0 & 16 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) (\underline{A053120})

is not a Riordan array. However the Riordan array

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

which begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 0 & 1 & 0 & 0 & 0 & 0 & … \\ −2 & 0 & 1 & 0 & 0 & 0 & … \\ 0 & −3 & 0 & 1 & 0 & 0 & … \\ 2 & 0 & −4 & 0 & 1 & 0 & … \\ 0 & 5 & 0 & −5 & 0 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) (\underline{A108045})

is the coefficient array for the orthogonal polynomials given by (2 − 0^{n})T_{n}(x/2).

Orthogonal polynomials can also be defined in terms of the three term recurrence that they obey. Thus, for instance,

T_{n}(x) = 2xT_{n−1}(x) − T_{n−2}(x),

with a similar recurrence for U_{n}(x). Of course, we then have

U_{n}(x/2) = xU_{n−1}(x/2) − U_{n−2}(x/2),

for instance. This last recurrence corresponds to the fact that the production matrix of (\frac{1}{1+x^{2}} , \frac{x}{1+x^{2}} )^{−1} = (c(x^{2}), xc(x^{2})) is given by

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

Note that many of the above results can also be found in [18]

6 The Boubaker polynomials

The Boubaker polynomials arose from the discretization of the equations of heat transfer in pyrolysis [2, 15, 16] starting from an assumed solution of the form

\frac{1}{N} e^{\frac{A}{\frac{H}{z} +1}} ∑_{m=0}^{∞} ξ_{m}J_{m}(t)

where J_{m} is the m-th order Bessel function of the first kind. Upon truncation, we get a set of equations

Q_{1}(z)ξ_{0} = ξ_{1}
Q_{1}(z)ξ_{1} = −2ξ_{0} + ξ_{2}
…
Q_{1}(z)ξ_{m} = ξ_{m−1} + ξ_{m+1}
…

with coefficient matrix

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

in which we recognize the production matrix of the Riordan array

(\frac{1 + 3x^{2}}{1 + x^{2}} , \frac{x}{1 + x^{2}} )^{−1} .

The Boubaker polynomials B_{n}(x) are defined to be the family of orthogonal polynomials with coefficient array given by

(\frac{1 + 3x^{2}}{1 + x^{2}} , \frac{x}{1 + x^{2}} ) .

We have B_{0}(x) = 1 and

B_{n}(x) = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n − k}{k}\frac{n − 4k}{n − k} (−1)^{k}x^{n−2k}, n > 0.  (8)

We also have

∑_{n=0}^{∞} B_{n}(x)t^{n} = \frac{1 + 3t^{2}}{1 − xt + t^{2}} .

The connection to Riordan arrays has already been noted in [19]. The matrix (\frac{1+3x^{2}}{1+x^{2}} , \frac{x}{1+x^{2}} ) begins

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

and hence we have

B_{0}(x) = 1
B_{1}(x) = x
B_{2}(x) = x^{2} + 2
B_{3}(x) = x^{3} + 1
B_{4}(x) = x^{4} − 2
B_{5}(x) = x^{5} − x^{3} − 3x, …

We can find an expression for the general term of the Boubaker coefficient matrix (\frac{1+3x^{2}}{1+x^{2}} , \frac{x}{1+x^{2}} ) as follows. We have

(\frac{1 + 3x^{2}}{1 + x^{2}} , \frac{x}{1 + x^{2}} ) = (1 + 3x^{2}, x) · (\frac{1}{1 + x^{2}} , \frac{x}{1 + x^{2}} )
= (3\binom{2}{n − k} − 6\binom{1}{n − k} + 4\binom{0}{n − k}) · (−1)^{\frac{n−k}{2}} \binom{\frac{n+k}{2}}{k}\frac{1 + (−1)^{n−k}}{2} ) ,

where 3\binom{2}{n} − 6\binom{1}{n} + 4\binom{0}{n} represents the general term of the sequence 1, 0, 3, 0, 0, 0, … with g.f. 1 + 3x^{2}. Thus the general term of the Boubaker coefficient array is given by

∑_{j=0}^{n} (3\binom{2}{n − j} − 6\binom{1}{n − j} + 4\binom{0}{n − j}) ((−1)^{\frac{j−k}{2}} \binom{\frac{j+k}{2}}{k}\frac{1 + (−1)^{j−k}}{2} ) .

7 The family of Chebyshev-Boubaker polynomials

Inspired by the foregoing, we now define the Chebyshev-Boubaker polynomials with parameters (r, s) to be the orthogonal polynomials B_{n}(x; r, s) whose coefficient array is the Riordan array

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

That these are orthogonal polynomials is a consequence of the fact that the production array of 𝔅^{−1} is the tridiagonal matrix

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

We immediately note the factorization

𝔅 = (1 + rx + sx^{2}, x) · (\frac{1}{1 + x^{2}} , \frac{x}{1 + x^{2}} ) .  (9)

It is clear that we have

∑_{n=0}^{∞} B_{n}(x; r, s)t^{n} = \frac{1 + rt + st^{2}}{1 − xt + t^{2}} .

We have

B_{n}(x; 0, 0) = U_{n}(x/2),
B_{n}(x; 0, −1) = (2 − 0^{n}) · T_{n}(x/2),
B_{n}(x; 0, 3) = B_{n}(x).

We can characterize B_{n}(x; r, s) in terms of the Chebyshev polynomials as follows.

Proposition 23.

B_{n}(x; r, s) = U_{n}(x/2) + rU_{n−1}(x/2) + sU_{n−2}(x/2).  (10)

Proof. This follows from the definition since (\frac{1}{1+x^{2}} , \frac{x}{1+x^{2}} ) is the coefficient array for U_{n}(x/2).

Proposition 24.

B_{n}(x; r, s) = rU_{n−1}(x/2) + (s + 1)U_{n−2}(x/2) + 2T_{n}(x/2) − 0^{n}.  (11)

Proof. This follows since

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

Proposition 25.

B_{n}(x; r, s) = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n − k}{k}\frac{n − (s + 1)k}{n − k} (−1)^{k}x^{n−2k} + rU_{n−1}(x/2).  (12)

Proof. Indeed, the polynomials defined by

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

are given by

B_{n}(x; 0, s) = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n − k}{k}\frac{n − (s + 1)k}{n − k} (−1)^{k}x^{n−2k}.

This can be shown in a similar fashion to Eq. (8).

We can also use the factorization in Eq. (9) to derive another expression for these polynomials. The general term of the Riordan array (\frac{1}{1+x^{2}} , \frac{x}{1+x^{2}} ) is given by

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

while the general term of the array (1 + rx + sx^{2}, x) is given by f(n − k), where f(n) can be expressed, for instance, as

f(n) = (s − r + 1)\binom{0}{n} + (r − 2s)\binom{1}{n} + s\binom{2}{n}.

Thus the general element of 𝔅 is given by

∑_{j=0} f(n−j)a_{j,k} = ∑_{j=0}((s−r+1)\binom{0}{n − j}+(r−2s)\binom{1}{n − j}+s\binom{2}{n − j})(−1)^{\frac{j−k}{2}} \binom{\frac{j+k}{2}}{k}\frac{1 + (−1)^{j−k}}{2} .

We finish this section by noting that we could have defined a more general family of Chebyshev-Boukaber orthogonal polynomials as follows: Let

B_{r,s,α,β} = (\frac{1 + rx + sx^{2}}{1 + αx + βx^{2}} , \frac{x}{1 + αx + βx^{2}} ) .

Then this array is the coefficient array for the polynomials B(n; r, s; α, β). This is a family of orthogonal polynomials since the production array of B_{r,s,α,β}^{−1} is given by

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

We have

Proposition 26.

B(n; r, s; α, β) = (\sqrt{β})^{n}U_{n} (\frac{x − α}{2\sqrt{β}}) + r(\sqrt{β})^{n−1}U_{n−1} (\frac{x − α}{2\sqrt{β}}) + s(\sqrt{β})^{n−2}U_{n−2} (\frac{x − α}{2\sqrt{β}}) .

8 The inverse matrix 𝔅^{−1}

We recall that the first column of 𝔅^{−1} contains the moment sequence for the weight function defined by the Chebyshev-Boubaker polynomials B(n; r, s). In this section, we shall be interested in studying this sequence, including its Hankel transform, as well as looking at the row sums, and (more briefly) the diagonal sums, of 𝔅^{−1}.

The inverse of the matrix (\frac{1}{1+x^{2}} , \frac{x}{1+x^{2}} ), corresponding to r = s = 0, is the much-studied Catalan related matrix

(c(x^{2}), xc(x^{2})), c(x) = \frac{1 − \sqrt{1 − 4x}}{2x},

where c(x) is the generating function of the Catalan numbers C_{n} = \frac{1}{n+1} \binom{2n}{n} \underline{A000108}. The

inverse of (\frac{1−x^{2}}{1+x^{2}} , \frac{x}{1+x^{2}} ), which corresponds to r = 0, s = −1, is the matrix

(\frac{1}{\sqrt{1 − 4x^{2}}} , xc(x^{2})) ,

which again finds many applications. It is \underline{A108044}.

Using the theory of Riordan arrays, we find that

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

Note also that

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

Thus for instance the generating function of the first column of the inverse matrix is

\frac{c(x^{2})}{1 + rxc(x^{2}) + sx^{2}c(x^{2})^{2}} = \frac{1 − \sqrt{1 − 4x^{2}}}{s + rx + 2(1 − s)x^{2} + (s + rx)\sqrt{1 − 4x^{2}}}  (13)
= \frac{1 + s + 2rx + (s − 1)\sqrt{1 − 4x^{2}}}{2(s + r(s + 1)x + (r^{2} + (s − 1)^{2})x^{2}} .  (14)

We can find expressions for the general term u_{n} of the sequence given by the first column of 𝔅^{−1} and the general term v_{n} of the row sums 𝔅^{−1} as follows. We start with

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

The first matrix has general element

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

while the second matrix is the sequence (Appell) array for the generalized Fibonacci numbers

F_{r,s}(n) = [x^{n}]\frac{1}{1 + rx + sx^{2}} = ∑_{i=0}^{⌊ \frac{n}{2} ⌋} \binom{n − i}{i}(−r)^{n−2i}(−s)^{k}.

Thus the general term of 𝔅^{−1} is given by

T_{n,k} = ∑_{j=0}^{n} \binom{n + 1}{\frac{n−j}{2}} \frac{j + 1}{n + 1} \frac{1 + (−1)^{n−j}}{2} ∑_{i=0}^{⌊ \frac{j−k}{2} ⌋} \binom{j − k − i}{i}(−r)^{j−k−2i}(−s)^{i}.  (15)

Setting k = 0 we obtain

u_{n} = ∑_{j=0}^{n} \binom{n + 1}{\frac{n−j}{2}} \frac{j + 1}{n + 1} \frac{1 + (−1)^{n−j}}{2} ∑_{i=0}^{⌊ \frac{j}{2} ⌋} \binom{j − i}{i}(−r)^{j−2i}(−s)^{i}.  (16)

This last equation translates the fact that

u_{n} = [x^{n}]\frac{c(x^{2})}{1 + rxc(x^{2}) + sx^{2}c(x^{2})^{2}} = [x^{n}](c(x^{2}), xc(x^{2})) · \frac{1}{1 + rx + sx^{2}} .

Note that another expression for u_{n} is given by

u_{n} = ∑_{k=0}^{⌊ \frac{n+1}{2} ⌋} \{\binom{n}{k} − \binom{n}{k − 1}\} ∑_{j=0}^{⌊ \frac{n−2k}{2} ⌋} \binom{n − 2k − j}{j}(−r)^{n−2k−2j}(−b)^{j},

which represents u_{n} as the diagonal sums of the Hadamard product of the reversal of (c(x), xc(x)) and the sequence array of F_{r,s}(n).

The general term v_{n} of the row sums of 𝔅^{−1} is given by

v_{n} = ∑_{k=0}^{n} ∑_{j=0}^{n} \binom{n + 1}{\frac{n−j}{2}} \frac{j + 1}{n + 1} \frac{1 + (−1)^{n−j}}{2} ∑_{i=0}^{⌊ \frac{j−k}{2} ⌋} \binom{j − k − i}{i}(−r)^{j−k−2i}(−s)^{i}.

We can deduce from Eq. (13) that the weight function for the first column {u_{n}} of the inverse is given by

\begin{matrix}−\frac{1}{2π} \frac{(s − 1)\sqrt{4 − x^{2}}}{(1 − s)^{2} + r^{2} + r(s + 1)x + sx^{2}} + α(r, s)δ_{(^{\frac{(1−s)\sqrt{r^{2}−4s}}{2s}} − \frac{r(s+1)}{2s} )} \\ +β(r, s)δ_{(^{\frac{(s−1)\sqrt{r^{2}−4s}}{2s}} − \frac{r(s+1)}{2s} )}, \end{matrix}

for appropriate values of α(r, s) and β(r, s). Thus for r = 1, s = 2, we find that the terms of the first column have integral representation

u_{n} = −\frac{1}{2π} ∫_{−2}^{2} \frac{x^{n}\sqrt{4 − x^{2}}}{2 + 3x + 2x^{2}} dx+(\frac{3}{4} − \frac{1}{4\sqrt{7}}i) (−\frac{3}{4} − \frac{\sqrt{7}}{4}i)^{n}+(\frac{3}{4} + \frac{1}{4\sqrt{7}}i) (−\frac{3}{4} + \frac{\sqrt{7}}{4}i)^{n} ,

while for r = 2, s = 3, we find that the terms of the first column have integral representation

u_{n} = −\frac{1}{π} ∫_{−2}^{2} \frac{x^{n}\sqrt{4 − x^{2}}}{8 + 8x + 3x^{2}} dx+(\frac{2}{3} − \frac{\sqrt{2}}{6}i) (−\frac{4}{3} − \frac{2\sqrt{2}}{3}i)^{n}+(\frac{2}{3} + \frac{\sqrt{2}}{6}i) (−\frac{4}{3} + \frac{2\sqrt{2}}{3}i)^{n} .

Using the techniques of [1, 6] we can prove the following. Proposition 27. The Hankel transform of the first column of

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

is given by h_{n} = (1 − s)^{n}.

We can recouch this result in the following terms: Proposition 28. The moments of the Chebyshev-Boubaker polynomials B_{n}(x; r, s) have Hankel transform equal to (1 − s)^{n}.

Recall that the elements of the first column of 𝔅^{−1} are the moments for the density measure associated with the polynomials B_{n}(x; r, s). We also have Proposition 29. The Hankel transform of the row sums of the Riordan matrix

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

is given by

h_{n} = [x^{n}]\frac{1}{1 + (s − r − 1)x + s^{2}x^{2}} .

Proof. The g.f. of the row sums is given by

\frac{\frac{c(x^{2})}{1+rxc(x^{2})+sx^{2}c(x^{2})^{2}}}{1 − xc(x^{2})} = \frac{1 + s + 2rx + (s − 1)\sqrt{1 − 4x^{2}}}{2(s + r(s + 1)x + (r^{2} + (s − 1)^{2})x^{2}} \frac{1 − 2x + \sqrt{1 − 4x^{2}}}{2(1 − 2x)} .

The result again follows from the techniques of [1, 6].

We note that the g.f. of the row sums may be written as

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

Thus the row sums of the inverse matrix are a convolution of

[x^{n}]\frac{1}{1 + rxc(x^{2}) + sx^{2}c(x^{2})^{2}}

and the central binomial coefficients \binom{n}{⌊ \frac{n}{2} ⌋} \underline{A001405}, or alternatively a convolution of

[x^{n}]\frac{c(x^{2})}{1 + rxc(x^{2}) + sx^{2}c(x^{2})^{2}}

and

\binom{n − 1}{⌊\frac{n−1}{2}⌋} + 0^{n}.

Example 30. The Hankel transforms of the row sums of the inverse matrices

(\frac{1 + x − x^{2}}{1 + x^{2}} , \frac{x}{1 + x^{2}} )^{−1} and (\frac{1 + 3x + x^{2}}{1 + x^{2}} , \frac{x}{1 + x^{2}} )^{−1}

are both given by F_{2n+2}.

In the case of the matrix

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

the row sums are expressible as

([x^{n}]\frac{\sqrt{1 − 4x^{2}} − x}{1 − 5x^{2}})∗(\binom{n − 1}{⌊\frac{n−1}{2}⌋} + 0^{n}) = ((−1)^{n} ∑_{k=0}^{⌊ \frac{n+1}{2} ⌋} (\binom{n}{k} − \binom{n}{k − 1}) F_{n−2k+1})∗(\binom{n − 1}{⌊\frac{n−1}{2}⌋} + 0^{n}) ,

where the first element of the convolution is (−1)^{n} \underline{A098615}(n). Note that these constituent sequences have Hankel transforms of 2^{n} and 1, 0, −1, 0, 1, 0, −1, 0, …, respectively. Alternatively the row sums are given by

([x^{n}]\frac{1 − x − 4x^{2} + (1 − x)\sqrt{1 − 4x^{2}}}{2(1 − 5x^{2})} ) ∗ \binom{n}{⌊\frac{n}{2}⌋}.

In this case, the Hankel transforms of the constituent elements of the convolution are given by 1, 1, 1, 1, 0, 0, −1, −1, −1 − 1, 0, …, and the all 1’s sequence. For the case of the matrix

(\frac{1 + 3x + x^{2}}{1 + x^{2}} , \frac{x}{1 + x^{2}} )^{−1} ,

we note that

\frac{c(x^{2})}{1 + 3xc(x^{2}) + x^{2}c(x^{2})^{2}} = \frac{1}{1 + 3x}

and so the row sums of the inverse in this case are simply given by

∑_{k=0}^{n}(−3)^{n−k} (\binom{k − 1}{⌊\frac{k−1}{2}⌋} + 0^{k}) .

We finish this section by noting that the diagonal sums of 𝔅^{−1} are also of interest. They have generating function

\frac{1 + rx − 2(1 − s)x^{2} − (1 + rx)\sqrt{1 − 4x^{2}}}{2x^{2}(s + rx(1 + s) + (r^{2} + (s − 1)^{2})x^{2})} .

For instance, the diagonal sums of

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

which begin

1, 1, 4, 6, 18, 32, 85, 165, 411, 839, 2013, …,

have as Hankel transform the 12-period sequence with g.f. \frac{1+3x+x^{2}−x^{3}}{1−x^{2}+x^{4}} which begins

1, 3, 2, 2, 1, −1, −1, −3, −2, −2, −1, 1, 1, 3, 2, 2, 1, −1, −1, −3, −2, …

Similarly, the diagonal sums of the matrix

(\frac{1 − 2x − 3x^{2}}{1 + x^{2}} , \frac{x}{1 + x^{2}} )^{−1} ,

which begin

1, 2, 9, 26, 94, 300, 1025, 3370, 11322, …,

have as Hankel transform the sequence with g.f. \frac{1+5x+3x^{2}−x^{3}}{(1+x^{2})^{2}} and general term

(1 − n) cos (\frac{πn}{2}) + (3n + 2) sin (\frac{πn}{2}) .

We can conjecture that the Hankel transform of the diagonal sums of 𝔅^{−1} in the general case is given by

[x^{n}]\frac{1 + (2 − s)x − sx^{2} − x^{3}}{1 + (r^{2} − 2)x^{2} + x^{4}} .

9 A curious relation

The third column of the Boubaker coefficient matrix (\frac{1+3x^{2}}{1+x^{2}} , \frac{x}{1+x^{2}} ) has general term

t_{n} = ∑_{j=0}^{n} (3\binom{2}{n − j} − 6\binom{1}{n − j} + 4\binom{0}{n − j}) (−1)^{\frac{j−2}{2}} \binom{\frac{j+2}{2}}{2}\frac{1 + (−1)^{j−2}}{2} .

This sequence begins

0, 0, 1, 0, 0, 0, −3, 0, 8, 0, −15, 0, 24, 0, −35, 0, 48, 0, −63, 0, 80, …

Now the sequence t_{2n+2} is therefore given by

1, 0, −3, 8, −15, 24, −35, 48, …

This is \underline{A131386}, with general term (1 − n^{2})(−1)^{n}. The interested reader may wish to verify that

t_{2n+2} = \frac{1}{2π} ℜ ∫_{−2}^{2} (\frac{1 + x}{1 − x})^{n} \sqrt{4 − x^{2}} dx

(here, ℜ returns the real part of a complex number).

10 Acknowledgements

The authors would like to thank the anonymous referees for their careful reading and cogent suggestions which have hopefully led to a clearer paper. We have also been happy to change the working title of the paper at the suggestion of a referee, as the new title more properly captures the thrust of the work.

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2010 Mathematics Subject Classification: Primary 42C05; Secondary 11B83, 33C45, 11B39, 11C20, 15B05, 15B36. Keywords: Chebyshev polynomials, Boubaker polynomials, integer sequence, orthogonal polynomials, Riordan array, production matrix, Hankel determinant, Hankel transform.

(Concerned with sequences \underline{A000007}, \underline{A000045}, \underline{A000108}, \underline{A001405}, \underline{A007318}, \underline{A009766}, \underline{A033184}, \underline{A049310}, \underline{A053117}, \underline{A053120}, \underline{A053121}, \underline{A098615}, \underline{A108044}, \underline{A108045}, and \underline{A131386}.)

Received May 14 2010; revised version received September 8 2010; October 4 2010. Published in Journal of Integer Sequences, December 6 2010.

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