A Note on a Family of Generalized Pascal Matrices Defined by Riordan Arrays

Paul Barry School of Science Waterford Institute of Technology Ireland

                     pbarry@wit.ie

Abstract

We study the properties of a parameterized family of generalized Pascal matrices, defined by Riordan arrays. In particular, we characterize the central elements of these lower triangular matrices, which are analogues of the central binomial coefficients. We then specialize to the value 2 of the parameter, and study the inverse of the matrix in question, and in particular we study the sequences given by the first column and row sums of the inverse matrix. Links to moments and orthogonal polynomials are examined, and Hankel transforms are calculated. We study the effect of the powers of the binomial matrix on the family. Finally we posit a conjecture concerning determinants related to the Christoffel-Darboux bivariate quotients defined by the polynomials whose coefficient arrays are given by the generalized Pascal matrices.

1 Introduction

Pascal’s triangle [19] holds a special place in mathematics. It has been generalized in many directions [7, 8, 11, 13, 36, 37]. One approach to generalizing Pascal’s matrix is to use Riordan arrays [2]. This is very apt, as Pascal’s triangle, as a mathematical object, is a Riordan array. In this note, we investigate some properties of the two-parameter generalized Pascal triangle M^{(m)}(a, b) given by the Riordan array

M^{(m)}(a, b) = (\frac{1}{1 − ax}, \frac{x(1 + bx)}{(1 − ax)^{m}} ) .

The binomial matrix B = (\frac{1}{1−x}, \frac{x}{1−x}) whose elements coincide with Pascal’s triangle, is the case a = 1, b = 0 and m = 1 or M^{(1)}(1, 0). It is also given by M^{(2)}(1, −1). The matrix M^{(2)}(1, 1) \underline{A114188} begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & ⋯ \\ 1 & 1 & 0 & 0 & 0 & 0 & ⋯ \\ 1 & 4 & 1 & 0 & 0 & 0 & ⋯ \\ 1 & 9 & 7 & 1 & 0 & 0 & ⋯ \\ 1 & 16 & 26 & 10 & 1 & 0 & ⋯ \\ 1 & 25 & 70 & 52 & 13 & 1 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ,

and factorizes as

B · (\binom{k}{n − k}2^{n−k})_{0≤n,k≤∞} .

M^{(2)}(−1, −2) is \underline{A110522}. Our investigations lie in four principal directions. Firstly, we give characterizations of the central elements of the array M^{(m)}(a, b). These elements generalize the central binomial coefficients \binom{2n}{n} \underline{A000984} of Pascal’s triangle. Secondly, we look at the specialized matrix with m = 2, and in particular we establish properties of the inverse matrix. These investigations cover such things as the Hankel transform of associated sequences, and links with moments and orthogonal polynomials. We then look at the images of M^{(m)}(a, b) under the powers B^{k} of the binomial matrix. Finally we study matrices and determinants related to Christoffel-Darboux-type bivariate generating functions related to the matrices M^{(m)}(a, b).

2 Preliminaries on integer sequences, Riordan arrays and Hankel transforms

In this section we define terms used later to discuss integer sequences, Riordan arrays, production matrices, orthogonal polynomials and Hankel transforms. Readers familiar with these areas and the links between them may skip this section.

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 relationship 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 [21], 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.

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}.

The Riordan group [28, 31], 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 [31]. We often require in addition that f_{1} = 1. 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.

If M is the matrix (g, f), and a = (a_{0}, a_{1}, …)^{T} is an integer sequence (expressed as an infinite column vector) 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}).

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

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

The production matrix of a matrix L is the matrix P = L^{−1}L̄, where L̄ is the matrix L with its top line removed. A Riordan array L whose inverse has a suitable tri-diagonal production matrix is the coefficient array of a family of orthogonal polynomials [4, 5]. Orthogonal polynomials [15, 20] are often associated with the three-term recurrences that can be used to define them.

An important feature of Riordan arrays is that they have a number of sequence characterizations [14, 22]. The simplest of these is as follows.

Proposition 2. ; [22, 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 = [α_{0}, α_{1}, α_{2}, …] and Z = [z_{0}, z_{1}, z_{2}, …] with α_{0} ≠ 0, z_{0} ≠ 0 such that

• d_{n+1,k+1} = ∑_{j=0}^{∞} α_{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 α_{0}, α_{1}, α_{2}, … and z_{0}, z_{1}, z_{2}, … are called the A-sequence and the Zsequence of the Riordan array M = (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{d_{0,0}}{g(f̄(x))}) .  (1)

Example 3. We can deduce from the last result that the A-sequence and the Z-sequence of the inverse Riordan array

(g(x), f(x))^{−1}

have generating functions \frac{x}{f(x)} and \frac{1}{f(x)}(1 − g(x)), respectively.

Thus in the case of the inverse matrix

M^{(m)}(a, b)^{−1} = (\frac{1}{1 − ax}, \frac{x(1 + bx)}{(1 − ax)^{m}} )^{−1} ,

we see that the corresponding generating functions are given by

A(x) = \frac{(1 − ax)^{m}}{1 + bx} , Z(x) = \frac{−a(1 − ax)^{m−1}}{1 + bx} .

Many interesting examples of sequences and Riordan arrays can be found in Neil Sloane’s On-Line Encyclopedia of Integer Sequences (OEIS), [29, 30]. Sequences are frequently referred to by their OEIS number. For instance, the binomial matrix B (“Pascal’s triangle”) is \underline{A007318}.

Given a sequence a_{n}, we denote by h_{n} the general term of the sequence with h_{n} = |a_{i+j}|_{0≤i,j≤n}. The sequence h_{n} is called the Hankel transform of a_{n} [24, 25]. A well known example of Hankel transform is that of the Catalan numbers, C_{n} = \frac{1}{n+1} \binom{2n}{n}, where we find that h_{n} = 1 for all n. Hankel determinants occur naturally in many branches of mathematics, from combinatorics [10] to number theory [26] and to mathematical physics [33].

There are a number of known ways of calculating Hankel transforms of sequences [12, 16, 23, 24, 27, 33]. One involves the theory of orthogonal polynomials, whereby we seek to represent the sequence under study as moments of a density function. Standard techniques of orthogonal polynomials then allow us to compute the desired Hankel transforms [6, 17]. These techniques are based on the following results (the first is the well-known “Favard’s Theorem”), which we essentially reproduce from [24].

Theorem 4. [24] (cf. [34, Théorème 9, p. I-4]; [35, 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 5. [24] (cf. [34, Proposition 1, (7), p. V-5]; [35, 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 of μ_{n}, which is the sequence with general term h_{n} = |μ_{i+j}|_{0≤i,j≤n}, is then given by

h_{n} = μ_{0}^{n+1}β_{1}^{n}β_{2}^{n−1} ⋯ β_{n−1}^{2}β_{n}.

Other methods of proving Hankel transform evaluations include lattice path methods (the Lindström-Gessel-Viennot theorem) [1, 32] and Dodgson condensation (Desnanot-Jacobi adjoint matrix theorem) [1, 9, 18].

3 Central elements

We have studied the matrices M^{(1)}(a, b) previously [2]. These are Pascal-like matrices (i.e., they are lower-triangular and have T_{n,k} = T_{n,n−k} where T_{n,k} is the n, k-th term, along with T_{n,0} = T_{n,n} = 1 for all n) and their central coefficients (the (2n, n)-elements) are of particular interest. In this section we propose to study the central elements of the general M^{(m)}(a, b) matrices. A first task is to find an expression for the general (n, k)-th element of M^{(m)}(a, b).

The general term T_{n,k}^{(m)} of the Riordan array M^{(m)}(a, b) = (\frac{1}{1−ax}, \frac{x(1+bx)}{(1−ax)^{m}} ) can be calculated as follows.

T_{n,k}^{(m)}(a, b) = [x^{n}](1 + bx)^{k}x^{k}(1 − ax)^{−(mk+1)}
= [x^{n−k}] ∑_{j=0}^{k} \binom{k}{j}b^{j}x^{j} ∑_{i=0}^{∞} \binom{−(mk + 1)}{i}(−1)^{i}a^{i}x^{i}

= [x^{n−k}] ∑_{j} \binom{k}{j}b^{j}x^{j} ∑_{i} \binom{mk + i}{i}a^{i}x^{i}

= ∑_{j=0}^{k} \binom{k}{j}\binom{mk + n − k − j}{n − k − j}b^{j}a^{n−k−j}
= ∑_{j=0}^{k} \binom{k}{j}\binom{n + (m − 1)k − j}{n − k − j}b^{j}a^{n−k−j}.

Note that by interchanging k − j and j in the last summation we get

T_{n,k}^{(m)}(a, b) = ∑_{j=0}^{k} \binom{k}{j}\binom{n + (m − 2)k + j}{n − 2k + j}b^{k−j}a^{n−2k+j}.

Alternatively, we have

T_{n,k}^{(m)}(a, b) = ∑_{j=0}^{k} \binom{k}{j}\binom{n + (m − 2)k + j}{mk}b^{k−j}a^{n−2k+j}.

We restate these results as a proposition. Proposition 6. The general term T_{n,k}^{(m)}(a, b) of the generalized Pascal matrix

M^{(m)}(a, b) = (\frac{1}{1 − ax}, \frac{x(1 + bx)}{(1 − ax)^{m}} )

is given by

T_{n,k}^{(m)}(a, b) = ∑_{j=0}^{k} \binom{k}{j}\binom{n + (m − 1)k − j}{n − k − j}b^{j}a^{n−k−j} = ∑_{j=0}^{k} \binom{k}{j}\binom{n + (m − 2)k + j}{n − 2k + j}b^{k−j}a^{n−2k+j}
= ∑_{j=0}^{k} \binom{k}{j}\binom{n + (m − 2)k + j}{mk}b^{k−j}a^{n−2k+j}.

Example 7. We have M^{(2)}(1, −1) = B. This implies that

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

from which we can deduce that

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

We are now in a position to calculate an expression for the central terms T_{2n,n}^{(m)}(a, b). We have

T_{2n,n}^{(m)}(a, b) = ∑_{j=0}^{n} \binom{n}{j}\binom{2n + (m − 1)n − j}{2n − n − j}b^{j}a^{2n−n−j}

= ∑_{j=0}^{n} \binom{n}{j}\binom{(m + 1)n − j}{n − j}b^{j}a^{n−j}.

Exchanging n − j for j in the above summation we also get

T_{2n,n}^{(m)}(a, b) = ∑_{j=0}^{n} \binom{n}{j}\binom{nm + j}{j}a^{j}b^{n−j}.

Note that when b = 0, only the term with n = j survives in the above sum. Thus in the case b = 0, we find that the central terms of (\frac{1}{1−ax}, \frac{x}{1−ax}) are given by

T_{2n,n}^{(m)}(a, 0) = \binom{(m + 1)n}{n}a^{n}.

We gather these results in the following proposition.

Proposition 8. The central terms of the Riordan array (\frac{1}{1−ax}, \frac{x(1+bx)}{(1−ax)^{m}} ) are given by

T_{2n,n}^{(m)}(a, b) = ∑_{j=0}^{n} \binom{n}{j}\binom{(m + 1)n − j}{n − j}b^{j}a^{n−j} = ∑_{j=0}^{n} \binom{n}{j}\binom{nm + j}{j}a^{j}b^{n−j}.

For the case b = 0, the central terms of (\frac{1}{1−ax}, \frac{x}{1−ax}) are given by

T_{2n,n}^{(m)}(a, 0) = \binom{(m + 1)n}{n}a^{n}.

Example 9. We take the case of b = −a. Then we have

M^{(m)} = (\frac{1}{1 − ax}, \frac{x(1 − ax)}{(1 − ax)^{m}} ) = (\frac{1}{1 − ax}, \frac{x}{(1 − ax)^{m−1}} ) .

The proposition then implies that

\binom{mn}{n}a^{n} = ∑_{j=0}^{n} \binom{n}{j}\binom{mn + j}{j}a^{j}(−a)^{n−j}.

We conclude that

\binom{mn}{n} = ∑_{j=0}^{n} \binom{n}{j}\binom{mn + j}{j}(−1)^{n−j}.  (2)

We have the following alternative characterization of T_{2n,n}^{(m)}(a, b). Proposition 10.

T_{2n,n}^{(m)}(a, b) = [x^{n}]((1 + ax)^{m+1} + bx(1 + ax)^{m})^{n} = [x^{n}](1 + ax)^{mn}(1 + (a + b)x)^{n}. Proof. Let

S_{n} = [x^{n}]((1 + ax)^{m+1} + bx(1 + ax)^{m})^{n}.

Then

S_{n} = [x^{n}] ∑_{j=0}^{n} \binom{n}{j}(1 + ax)^{(m+1)j}(bx)^{n−j}(1 + ax)^{m(n−j)}
= [x^{n}] ∑_{j=0}^{n} \binom{n}{j}(1 + ax)^{mn+j}(bx)^{n−j}
= [x^{n}] ∑_{j=0}^{n} \binom{n}{j} ∑_{i} \binom{mn + j}{i}a^{i}x^{i}b^{n−j}x^{j}
= ∑_{j=0}^{n} \binom{n}{j}\binom{mn + j}{j}a^{j}b^{n−j}
= T_{2n,n}^{(m)}(a, b).

We next define a polynomial P_{n}(x; a, b) of degree n as follows.

P_{n}(x; a, b) = (1 + ax)^{n} + bx(1 + ax)^{n−1}.

Proposition 11. Let

P_{n}(x; a, b) = ∑_{k=0} p_{n,k}x^{n−k}.

Then the coefficient array (p_{n,k})_{0≤n,k≤∞} coincides with the Riordan array

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

Proof. We let p̃_{n,k} denote the (n, k)-th element of (\frac{1+bx}{1−ax}, \frac{x}{1−ax}). Then

p̃_{n,k} = [x^{n}](1 + bx)x^{k}(1 − ax)^{−(k+1)}
= [x^{n−k}](1 + bx) ∑_{j=0}^{∞} \binom{−(k + 1)}{j}(−1)^{j}a^{j}x^{j}
= [x^{n−k}](1 + bx) ∑_{j=0} \binom{k + j}{j}a^{j}x^{j}
= [x^{n−k}] ∑_{j} \binom{k + j}{j}a^{j}x^{j} + b[x^{n−k−1}] ∑_{j} \binom{k + j}{j}a^{j}x^{j}
= \binom{k + n − k}{n − k}a^{n−k} + b\binom{k + n − k − 1}{n − k − 1}a^{n−k−1}
= \binom{n}{k}a^{n−k} + b\binom{n − 1}{n − k − 1}a^{n−k−1}.

We thus have

∑_{k=0}^{n} p̃_{n,k}x^{n−k} = (1 + ax)^{n} + bx(1 + ax)^{n−1}.

We note that the decomposition

(\frac{1 + bx}{1 − ax}, \frac{x}{1 − ax}) = (\frac{1}{1 − ax}, \frac{x}{1 − ax}) + b (\frac{x}{1 − ax}, \frac{x}{1 − ax})

shows again that

p̃_{n,k} = \binom{n}{k}a^{n−k} + b\binom{n − 1}{n − k − 1}a^{n−k−1}.

Corollary 12. Let T_{2n,n}^{(m)}(a, b) be the central elements of the Riordan array (\frac{1}{1−ax}, \frac{x(1+bx)}{(1−ax)^{m}} ) and let P_{n}(x; a, b) be the family of polynomials with coefficient array (\frac{1+bx}{1−ax}, \frac{x}{1−ax}). Then

T_{2n,n}^{(m)}(a, b) = [x^{n}](P_{m+1}(x; a, b))^{n}.

Thus the central terms T_{2n,n}^{(m)}(a, b) are defined by the elements of the (m + 1)-th row of the Riordan array (\frac{1+bx}{1−ax}, \frac{x}{1−ax}).

Example 13. The (2, 1)-Pascal triangle (\frac{1+x}{1−x}, \frac{x}{1−x}) \underline{A029653} begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & ⋯ \\ 2 & 1 & 0 & 0 & 0 & 0 & ⋯ \\ 2 & 3 & 1 & 0 & 0 & 0 & ⋯ \\ 2 & 5 & 4 & 1 & 0 & 0 & ⋯ \\ 2 & 7 & 9 & 5 & 1 & 0 & ⋯ \\ 2 & 9 & 16 & 14 & 6 & 1 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

We then have, since a = b = 1 in this case,

T_{2n,n}^{(2)}(1, 1) = ∑_{k=0}^{n} \binom{n}{k}\binom{2n + k}{k} = [x^{n}](1 + 4x + 5x^{2} + 2x^{3})^{n},

T_{2n,n}^{(3)}(1, 1) = ∑_{k=0}^{n} \binom{n}{k}\binom{3n + k}{k} = [x^{n}](1 + 5x + 9x^{2} + 7x^{3} + 2x^{4})^{n},

T_{2n,n}^{(4)}(1, 1) = ∑_{k=0}^{n} \binom{n}{k}\binom{4n + k}{k} = [x^{n}](1 + 6x + 14x^{2} + 16x^{3} + 9x^{4} + 2x^{5})^{n},

for instance (\underline{A114496}, \underline{A156886}, and \underline{A156887}, respectively).

Example 14. We have

T_{2n,n}^{(2)}(−1, 1) = [x^{n}](1 − x)^{2n} = (−1)^{n}\binom{2n}{n}.

In general, we have

T_{2n,n}^{(m)}(a, −a) = [x^{n}](1 + ax)^{mn} = \binom{mn}{n}a^{n}.

4 The matrices M^{(2)}(a, b), moments and orthogonal polynomials

In this section, we look at some properties of M^{(2)}(a, b) and of the inverse Riordan array

M^{(2)}(a, b)^{−1} = (\frac{1}{1 − ax}, \frac{x(1 + bx)}{(1 − ax)^{2}})^{−1} .

The presence of the term (1−ax)^{2} in the denominator is an indicator that links to orthogonal polynomials may be possible [4], and indeed, we show that this is the case. We will in particular study the first column elements and the row sum elements of the inverse matrix, and show that they constitute moments for appropriate families of orthogonal polynomials. We indicate the form of the defining densities on the real number line that define these families. We take advantage of these links to orthogonal polynomials (and the corresponding continued faction expressions for the generating functions) to calculate the Hankel transforms of these moment sequences.

To begin this study, we have

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

If we write this inverse matrix as M^{(2)}(a, b)^{−1} = (g(x), f(x)), then

g(x) = ∑_{k=0}^{∞} g_{n}x^{n} = \frac{a\sqrt{1 + 4x(a + b)} − a − 2b}{2(a^{2}x − b)}

is the g.f. of the first column sequence g_{n}.

Proposition 15. The Hankel transform of the first column elements of the inverse array is given by

h_{n} = a^{n}(a + b)^{n^{2}}.

Proof. We consider the continued fraction [35]

g̃(x) = \frac{1}{1 + ax − \frac{a(a + b)x^{2}}{1 + 2(a + b)x − \frac{(a + b)^{2}x^{2}}{1 + 2(a + b)x − \frac{(a + b)^{2}x^{2}}{1 − ⋯}}}}.  (3)

We then have

g̃(x) = \frac{1}{1 + ax − a(a + b)x^{2}u}

where u satisfies the equation

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

Solving for u and substituting into g̃(x), we find that

g̃(x) = \frac{a\sqrt{1 + 4x(a + b)} − a − 2b}{2(a^{2}x − b)} = g(x),

the g.f. of the first column elements. It follows that

h_{n} = \frac{a^{n}}{(a + b)^{n}} (a + b)^{2\binom{n+1}{2}} = a^{n}(a + b)^{n^{2}}.

Proposition 16. We have the following expression for g(x), the g.f. of the first column elements of M^{(2)}(a, b)^{−1}.

g(x) = \frac{1}{1 + axc(−(a + b)x)} = \frac{1}{1 + \frac{ax}{1 + \frac{(a + b)x}{1 + \frac{(a + b)x}{1 + ⋯}}}},

where

c(x) = \frac{1 − \sqrt{1 − 4x}}{2x}

is the g.f for the Catalan numbers C_{n} = \frac{1}{n+1} \binom{2n}{n}.

Proof. The continued fraction is an equivalent form of (3). Substituting the expression for c(x) into \frac{1}{1+axc(−(a+b)x)} and simplifying shows that this is indeed g(x). We can recover the continued fraction expression again by noting that

c(x) = \frac{1}{1 − \frac{x}{1 − \frac{x}{1 − ⋯}}} .

Corollary 17. We have

M^{(2)}(a, b)^{−1} = (\frac{1}{1 + axc(−(a + b)x)}, \frac{xc(−(a + b)x)}{1 + axc(−(a + b)x)}) .

We can express the first column elements g_{n} as moments in the following way:

g_{n} = \frac{1}{2π} ∫_{−4(a+b)}^{0} x^{n} \frac{a\sqrt{−x(x + 4(a + b))}}{x(bx − a^{2})} dx,

where we have taken the case a + b > 0.

Example 18. We let a = 2, b = 3. Then the sequence g_{n} has g.f. \frac{\sqrt{1+20x}−4}{4x−3} and begins

1, −2, 14, −148, 1886, −26652, ….

In this case we have

g_{n} = \frac{1}{π} ∫_{−20}^{0} x^{n} \frac{\sqrt{−x(x + 20)}}{x(3x − 4)} dx.

The first column elements g_{n} of the inverse array can be expressed as follows.

g_{n} = ∑_{k=0}^{n}(−1)^{n−k}\binom{2n}{n − k}\frac{2k + 1}{n + k + 1}(a + b)^{n−k}b^{k}.

An alternative expression is

g_{n} = ∑_{k=0}^{n} ∑_{j=0}^{n} \binom{2n}{j}\binom{j}{k}(−1)^{j} \frac{n − 2j + 1}{2n − j + 1}a^{k}b^{n−k}.

The first column elements begin

1, −a, a(2a + b), −a(5a^{2} + 6ab + 2b^{2}), a(14a^{3} + 28a^{2}b + 20ab^{2} + 5b^{3}), ….

We can represent this as

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & ⋯ \\ 0 & −1 & 0 & 0 & 0 & 0 & ⋯ \\ 0 & b & 2 & 0 & 0 & 0 & ⋯ \\ 0 & −2b^{2} & −6b & −5 & 0 & 0 & ⋯ \\ 0 & 5b^{3} & 20b^{2} & 28b & 14 & 0 & ⋯ \\ 0 & −14b^{4} & −70b^{3} & −135b^{2} & −120b & −42 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ( 1 ; a ; a^{2} ; a^{3} ; a^{4} ; a^{5} ; ⋮ ) .

Here, the infinite matrix has bivariate generating function

\frac{1}{1 + \frac{xy}{1 + \frac{bx + xy}{1 + \frac{bx + xy}{1 + ⋯}}}},

or in other words, the matrix is given by the Deleham construct [3]

[0, −b, −b, −b, …] Δ [−1, −1, −1, …].

See \underline{A094385} and \underline{A157491}. Proposition 19. The Hankel transform h_{n}^{∗} of the shifted sequence g_{n+1} is given by

h_{n}^{∗} = (−a)^{n+1}(a + b)^{n(n+1)}.

Proof. The generating function of g_{n+1} is given by

\frac{g(x) − 1}{x} = \frac{−ac(−(a + b)x)}{1 + axc(−(a + b)x)}.

By the theory of Hankel transforms, g_{n+1} then has the same Hankel transform as

[x^{n}](−ac(−(a + b)x))

(this is an example of invariance under the INVERT transform [25]). The result follows from this.

We now turn our attention to the row sum elements s_{n} of the inverse array.

Proposition 20. The Hankel transform of the row sums sequence of the inverse Riordan

array (\frac{1}{1−ax}, \frac{x(1+bx)}{(1−ax)^{2}})^{−1} is given by

h_{n} = (a − 1)^{n}(a + b)^{n^{2}}.

Proof. The generating function of the row sum sequence of the inverse array is given by \frac{g(x)}{1−f(x)}, where the inverse array is expressed as (g(x), f(x)). We find that this g.f. is given by

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

But this can be expressed as the following continued fraction.

s(x) = \frac{1}{1 + (a − 1)x − \frac{(a − 1)(a + b)x^{2}}{1 + 2(a + b)x − \frac{(a + b)^{2}x^{2}}{1 + 2(a + b)x − \frac{(a + b)^{2}x^{2}}{1 − ⋯}}}}.  (4)

Thus we obtain

h_{n} = (a − 1)^{n}(a + b)^{n^{2}}.

We note that the generating function s(x) of the row sum sequence can be written as

s(x) = \frac{1}{1 + (a − 1)xc(−(a + b)x)} = (1, xc(−(a + b)x)) · \frac{1}{1 − (1 − a)x}.

We now recall that the general element of the Riordan array (1, xc(λx)) is given by

\binom{2n − k − 1}{n − k}\frac{k + 0^{n+k}}{n + 0^{nk}} λ^{n−k}.

Thus we obtain the following expression for the row sum elements.

s_{n} = ∑_{k=0}^{n} \binom{2n − k − 1}{n − k}\frac{k + 0^{n+k}}{n + 0^{nk}} (−a − b)^{n−k}(1 − a)^{k}.

We can express the row sum elements s_{n} as moments as follows

s_{n} = \frac{1}{2π} ∫_{−4(a+b)}^{0} x^{n} \frac{(a − 1)\sqrt{−x(x + 4(a + b))}}{x(x(1 + b) − (a − 1)^{2})} dx,

where we take the case a + b > 0. Example 21. We let a = b = −1. Then s_{n} = 2^{n}C_{n} and we obtain

s_{n} = \frac{1}{π} ∫_{0}^{8} x^{n} \frac{\sqrt{x(8 − x)}}{4x} dx.

As with the first column elements, we have the following alternative continued fraction form of the generating function s(x). Proposition 22. We have the following expression for s(x), the g.f. of the row sum elements of M^{(2)}(a, b)^{−1}.

s(x) = \frac{1}{1 + (a − 1)xc(−(a + b)x)} = \frac{1}{1 + \frac{(a − 1)x}{1 + \frac{(a + b)x}{1 + \frac{(a + b)x}{1 + ⋯}}}},

where

c(x) = \frac{1 − \sqrt{1 − 4x}}{2x}

is the g.f for the Catalan numbers C_{n} = \frac{1}{n+1} \binom{2n}{n}. Proposition 23. The Hankel transform h_{n}^{∗} of the shifted sequence s_{n+1} is given by

h_{n}^{∗} = (1 − a)^{n+1}(a + b)^{n(n+1)}.

Proof. The generating function of s_{n+1} is given by

\frac{s(x) − 1}{x} = \frac{−(a − 1)c(−(a + b)x)}{1 + (a − 1)xc(−(a + b)x)}.

By the theory of Hankel transforms, s_{n+1} then has the same Hankel transform as

[x^{n}](−(a − 1)c(−(a + b)x)) = [x^{n}](1 − a)c(−(a + b)x)

(this is an example of invariance under the INVERT transform [25]). The result follows from this.

When b = 0, the production array of the inverse matrix is given by

( \begin{matrix}−a & 1 & 0 & 0 & 0 & 0 & ⋯ \\ a^{2} & −2a & 1 & 0 & 0 & 0 & ⋯ \\ 0 & a^{2} & −2a & 1 & 0 & 0 & ⋯ \\ 0 & 0 & a^{2} & −2a & 1 & 0 & ⋯ \\ 0 & 0 & 0 & a^{2} & −2a & 1 & ⋯ \\ ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

This corresponds to the original matrix, (\frac{1}{1−ax}, \frac{x}{(1−ax)^{2}} ), being the coefficient array of the family of orthogonal polynomials Q_{n}(x) defined by

Q_{n}(x) = (x + 2a)Q_{n−1}(x) − a^{2}Q_{n−2}(x), Q_{0}(x) = 1, Q_{1}(x) = x + a.

The equation (3) tells us that the first column elements of the inverse matrix M^{(2)}(a, b)^{−1} are the moments for a family of orthogonal polynomials. The production matrix of the inverse of the coefficient array of those polynomials is then given by

( \begin{matrix}−a & 1 & 0 & 0 & 0 & 0 & ⋯ \\ a(a + b) & −2(a + b) & 1 & 0 & 0 & 0 & ⋯ \\ 0 & (a + b)^{2} & −2(a + b) & 1 & 0 & 0 & ⋯ \\ 0 & 0 & (a + b)^{2} & −2(a + b) & 1 & 0 & ⋯ \\ 0 & 0 & 0 & (a + b)^{2} & −2(a + b) & 1 & ⋯ \\ ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

The form of this production matrix implies that the matrix of coefficients of the sought-for orthogonal polynomials is given by

(\frac{1 − bx}{1 − (a + b)x}, \frac{x}{(1 − (a + b)x)^{2}} ) .

This gives us

Proposition 24. The first column elements of the inverse Riordan array

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

are the moments of the family of orthogonal polynomials whose coefficient array is given by the Riordan array

(\frac{1 − bx}{1 − (a + b)x}, \frac{x}{(1 − (a + b)x)^{2}} ) .

Corollary 25. The family of orthogonal polynomials R_{n}(x) for which the first column elements g_{n} of (\frac{1}{1−ax}, \frac{x(1+bx)}{(1−ax)^{2}})^{−1} are the moments is defined by

R_{n}(x) = (x + 2(a + b))R_{n−1}(x) − (a + b)^{2}R_{n−2}(x), R_{0}(x) = 1, R_{1}(x) = x + a, with coefficient array (\frac{1−bx}{1−(a+b)x}, \frac{x}{(1−(a+b)x)^{2}} ).

Using equation (4) in a similar way, we have the following result. Proposition 26. The row sum elements of the inverse Riordan array

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

are the moments of the family of orthogonal polynomials whose coefficient array is given by the Riordan array

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

Corollary 27. The family of orthogonal polynomials S_{n}(x) for which the row sum elements s_{n} of (\frac{1}{1−ax}, \frac{x(1+bx)}{(1−ax)^{2}})^{−1} are the moments is defined by

S_{n}(x) = (x + 2(a + b))S_{n−1}(x) − (a + b)^{2}S_{n−2}(x), S_{0}(x) = 1, S_{1}(x) = x + a − 1, with coefficient array (\frac{1−(b+1)x}{1−(a+b)x}, \frac{x}{(1−(a+b)x)^{2}} ).

We close this section with the following observation. The Hankel transform of the first column elements of the matrix product

M^{(2)}(a, b)^{−1} · M^{(1)}(ã, b̃)

is given by

h_{n} = (a − ã)^{n}(a + b)^{n^{2}}.

The generating function of the first column elements of the matrix product is then given by

\frac{1}{1 + (a − ã)x − \frac{(a − ã)(a + b)x^{2}}{1 + 2(a + b)x − \frac{(a + b)^{2}x^{2}}{1 + 2(a + b)x − \frac{(a + b)^{2}x^{2}}{1 − ⋯}}}}.

An equivalent and simpler form is

\frac{1}{1 + \frac{(a − ã)x}{1 + \frac{(a + b)x}{1 + \frac{(a + b)x}{1 + ⋯}}}}.

5 Images of M^{(2)}(a, b)

In this section we look at the image of M^{(2)}(a, b) under the c-fold binomial transformation. Thus we let

B_{c} = (\frac{1}{1 − cx}, \frac{x}{1 − cx}) .

Then

B_{c} = B^{c}

where B = (\frac{1}{1−x}, \frac{x}{1−x}) is the standard binomial array. In particular B_{1} = B.

Proposition 28. The image of M^{(2)}(a, b) under the c-fold binomial transformation B_{c} is M^{(2)}(a + c, b − c):

B_{c} · M^{(2)}(a, b) = M^{(2)}(a + c, b − c).  (5)

Proof. We have

(\frac{1}{1 − cx}, \frac{x}{1 − cx}) · (\frac{1}{1 − ax}, \frac{x(1 + bx)}{(1 − ax)^{2}}) = (\frac{1}{1 − cx} \frac{1}{1 − \frac{ax}{1−cx}} , \frac{\frac{x}{1−cx}(1 + b\frac{x}{1−cx})}{(1 − a\frac{x}{1−cx})^{2}})

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

Example 29. The image of the Riordan array (\frac{1}{1−x}, \frac{x(1+(b−1)x)}{1−x}) under the inverse binomial transformation is the Riordan array (1, x(1+bx)). This latter matrix has inverse (1, xc(−bx)). Thus we have

(1, xc(−bx)) · B^{−1} = (\frac{1}{1 − x}, \frac{x(1 + (b − 1)x)}{1 − x})^{−1} ,

that is,

(1, xc(−bx)) · B^{−1} = M^{(2)}(−1, b − 1)^{−1}.

Example 30. We have

B^{−1} · M^{(2)}(a, −1) = M^{(2)}(a − 1, 0) = (\frac{1}{1 − (a − 1)x}, \frac{x}{(1 − (a − 1)x)^{2}} ) .

Thus the image of M^{(2)}(a, −1) under B^{−1} is the coefficient array of the family of orthogonal polynomials P_{n}(x; a) defined by

P_{n}(x; a) = (x − 2(1 − a))P_{n−1}(x; a) − (a − 1)^{2}P_{n−2}(x; a), P_{0}(x; a) = 1, P_{1}(x; a) = x + a − 1.

More generally, we see that

B_{b} · M^{(2)}(a, b) = M^{(2)}(a + b, 0),

where M^{(2)}(a + b, 0) is the coefficient array of the family of orthogonal polynomials P_{n}(x) defined by

P_{n}(x; a) = (x + 2(a + b))P_{n−1}(x; a) − (a + b)^{2}P_{n−2}(x; a), P_{0}(x; a) = 1, P_{1}(x; a) = x + a + b.

Equivalently

M^{(2)}(a, b) = B_{b}^{−1}M^{(2)}(a + b, 0) = B^{−b}M^{(2)}(a + b, 0).

We finish this section by remarking that the property (5) can be generalized. For instance, we define

M̃^{(3)}(a, b, c) = (\frac{1}{1 − ax}, \frac{x(1 + bx + cx^{2})}{(1 − ax)^{3}}) .

We can then show that

B_{d}M̃^{(3)}(a, b, c) = M̃^{(3)}(a + d, b − 2d, c − bd + d^{2}).

Similarly, we get

B_{e}M̃^{(4)}(a, b, c, d) = M̃^{(4)}(a + e, b − 3e, −c + 2be − 3e^{2}, d − ce + be^{2} − e^{3}).

6 A Christoffel-Darboux conjecture

We can regard the matrix M^{(m)}(a, b) as the coefficient array of a family of polynomials P_{n}(x; m, a, b). We have

P_{n}(x; m, a, b) = ∑_{k=0}^{n} T_{n,k}^{(m)}(a, b)x^{k}.

For each n, the Christoffel-Darboux quotient

\frac{P_{n+1}(x; m, a, b)P_{n}(y; m, a, b) − P_{n}(x; m, a, b)P_{n+1}(y; m, a, b)}{x − y}

is the bi-variate generator of a (n + 1) × (n + 1) matrix D_{n}(m, a, b). We then have the following conjecture.

Conjecture 31. We have

|D_{n}(2, a, b)| = a^{\binom{n+1}{2}}(a + b)^{n}(a + 2b)^{\binom{n}{2}}.

It is interesting to compare the matrices defined by the quotient above with those defined by

∑_{j=0}^{n} P_{j}(x; m, a, b)P_{j}(y; m, a, b).

In the case of monic orthogonal polynomials, such expressions lead to the same matrices. However, M^{(m)}(a, b) is not in general the coefficient array of a family of monic orthogonal polynomials, and hence the two family of matrices will in general differ. In the case m = 2, a = b = 1, we can conjecture the following.

Conjecture 32. The matrix D̃_{n}(2, 1, 1) with generating function

∑_{k=0}^{n} P_{j}(x; 2, 1, 1)P_{j}(y; 2, 1, 1)

is given by

M_{n}^{(2)}(a, b)^{t}M_{n}^{(2)}(a, b).

We then have

|D̃_{n}(2, 1, 1)| = 1 for n ≥ 0.

In the above the notation M_{n} denotes the matrix formed from the first (n + 1) rows and columns of M.

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2010 Mathematics Subject Classification: Primary 11B83; Secondary 05A10, 05A15, 11C08, 11C20, 15B05. Keywords: Riordan array, Pascal’s triangle, central binomial coefficient, orthogonal polynomial, moments, Hankel determinant, Hankel transform, sequence.

(Concerned with sequences \underline{A000007}, \underline{A000045}, \underline{A000108}, \underline{A000984}, \underline{A007318}, \underline{A029653}, \underline{A094385}, \underline{A110522}, \underline{A114188}, \underline{A114496}, \underline{A156886}, \underline{A156887}, \underline{A157491}.)

Received October 12 2012; revised version received May 3 2013. Published in Journal of Integer Sequences, May 16 2013.

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