Extensions of Riordan Arrays and Their Applications

Abstract

The Riordan group of Riordan arrays was first described in 1991, and since then, it has provided useful tools for the study of areas such as combinatorial identities, polynomial sequences (including families of orthogonal polynomials), lattice path enumeration, and linear recurrences. Useful extensions of the idea of a Riordan array have included almost Riordan arrays, double Riordan arrays, and their generalizations. After giving a brief overview of the Riordan group, we define two further extensions of the notion of Riordan arrays, and we give a number of applications for these extensions. The relevance of these applications indicates that these new extensions are worthy of study. The first extension is that of the reverse symmetrization of a Riordan array, for which we give two applications. The first application of this symmetrization is to the study of a family of Riordan arrays whose symmetrizations lead to the famous Robbins numbers as well as to numbers associated with the 20 vertex model of mathematical physics. We provide closed-form expressions for the elements of these arrays, and we also give a canonical Catalan factorization for them. We also describe an alternative family of Riordan arrays whose symmetrizations lead to the same integer sequences. The second application of this symmetrization process is to the area of the enumeration of lattice paths. We remain with the applications to lattice paths for the second extension of Riordan arrays that we introduce, which is the interleaved Riordan array. The methods used include generating functions, linear algebra, weighted compositions, and linear recurrences. In the case of the symmetrization process applied to Riordan arrays, we focus on the principal minor sequences of the resulting square matrices in the context of integrable lattice models.

1. Introduction with Preliminaries on Riordan Arrays

The Riordan group was first described in [1], and its field of applications has grown since. Initially applied in the area of combinatorial identities, it has found further applications in the combinatorics of trees and lattice paths, orthogonal polynomials, and more recently in the description of sequences related to some integrable lattice models. Here, we focus on two areas of application, which are those of sequences that arise in the study of integrable lattice models and sequences that arise from the study of paths on the integer lattice. In order to carry out these explorations, we define two suitable extensions of the notion of Riordan arrays. These extensions, namely the symmetrization of Riordan arrays and the interleaving of Riordan arrays, are defined in terms of the generating functions involved. The symmetrization process is new to this context, and its applications are novel, while the interleaving concept has antecedents in work on the double Riordan group. The resulting arrays have matrix representations, which allow us to apply linear algebra techniques as well as purely algebraic techniques. This contributes to the versatility of the methods deployed in each context. In order to define the extensions to Riordan arrays that we have in mind, and to set our notation, we first give a brief overview of Riordan arrays in the next section.

Since its description in 1991, the Riordan group has found applications in many areas where combinatorics play a part. The elements of the Riordan group are easy to define, and they allow for an interplay between algebra and linear algebra, which provides a variety of tools with which to approach problems. Even for classical problems, new perspectives can be had through the lens of the Riordan group, which can lead to further generalizations. When applying Riordan arrays in practice, certain fruitful directions of generalization have emerged, including the group of almost Riordan arrays (of which the Riordan group is a subgroup), and the double Riordan group and its extensions. We explore two other extensions, and we justify their investigation by giving a number of areas of application for these extensions. Prior to this, we will briefly review the elements of the theory of Riordan arrays [2,3]. To define Riordan arrays, we first set the following:

$${\mathcal{F}}_r=\{f\in \mathcal{R}\left[\left[x\right]\right]|f\left(x\right)=\sum _{k=r}^{\infty }{f}_k{x}^k\}.$$

Here, x is a “dummy variable” or indeterminate, and $\mathcal{R}$ is any ring in which the operations we will define make sense. Often, it can be any of the fields $\mathbb{Q},\mathbb{R}$, or $\mathbb{C}$. When looking at combinatorial applications, it can be the ring of integers $\mathbb{Z}$ (in this case, we demand that the diagonals of the matrices consist of all 1 s). We now let $g\left(x\right)\in {\mathcal{F}}_0$ and $f\left(x\right)\in {\mathcal{F}}_1$. This means that if $g\left(x\right)={\sum }_{n=0}^{\infty }{g}_n{x}^n$, then $g_0\ne 0$. Thus, $g\left(x\right)$ has a multiplicative inverse ${\displaystyle \frac{1}{g\left(x\right)}}\in {\mathcal{F}}_0$. If $f\left(x\right)={\sum }_{n=0}^{\infty }{f}_n{x}^n$, then we have $f_0=0$ and $f_1\ne 0$. This ensures that $f\left(x\right)$ will have a compositional inverse $\overline{f}\in {\mathcal{F}}_1$, also denoted by $f^{⟨-1⟩}$, which is the solution $v\left(x\right)$ of the equation $f\left(v\right)=x$ with $v\left(0\right)=0$. Based on definition, we have $\overline{f}\left(f\left(x\right)\right)=x$ and $f\left(\overline{f}\left(x\right)\right)=x$. A Riordan array is then defined to be an element $(g,f)\in {\mathcal{F}}_0\times {\mathcal{F}}_1$. The term “array” signifies the fact that every element $(g,f)\in {\mathcal{F}}_0\times {\mathcal{F}}_1$ has a matrix representation ${\left(t_{n,k}\right)}_{0\le n,k\le \infty }$ given by the following:

$$t_{n,k}=\left[x^n\right]g\left(x\right)f{\left(x\right)}^k.$$

Here, $\left[x^n\right]$ is the functional on $\mathcal{R}\left[\right[x\left]\right]$ that returns the coefficient of $x^n$ of an element in $\mathcal{R}\left[\right[x\left]\right]$ [4]. With this notation, we can turn the set ${\mathcal{F}}_0\times {\mathcal{F}}_1$ into a group using the product operation:

$$\left(g\right(x),f(x\left)\right)·\left(u\right(x),v(x\left)\right)=\left(g\right(x\left)u\right(f\left(x\right)),v(f\left(x\right)).$$

The inverse for this operation is given by the following:

$${(g\left(x\right),f\left(x\right))}^{-1}=\left({\displaystyle \frac{1}{g\left(\overline{f}\left(x\right)\right)}},\overline{f}\left(x\right)\right).$$

The identity element is given by $(1,x)$, which corresponds to the identity matrix.

When passing to the matrix representation, these operations correspond to matrix multiplication and taking the matrix inverse, respectively. Note that in the matrix representation, the generating functions of the columns are given by the geometric progression $g\left(x\right)f{\left(x\right)}^k$ in $\mathcal{R}\left[\right[x\left]\right]$. The operation of a Riordan array on a power series $h\left(x\right)={\sum }_{n=0}^{\infty }{h}_n{x}^n$ is given by the following:

$$\left(g\right(x),f(x\left)\right)·h\left(x\right)=g\left(x\right)h\left(f\right(x\left)\right).$$

This weighted composition rule is called the fundamental theorem of Riordan arrays. In matrix terms, this is equivalent to multiplying the column vector ${(h_0,h_1,h_2,\dots )}^T$ by the matrix $\left(t_{n,k}\right)$. Because $f\in {\mathcal{F}}_1$, Riordan arrays have lower-triangular matrix representations. As an abuse of language, we use the term “Riordan array” interchangeably to denote either the pair $(g,f)$ or the matrix $\left(t_{n,k}\right)$, letting the context indicate which is being referred to at the time. The bivariate generating function of the array $(g,f)$ is given by the following:

$${\displaystyle \frac{g\left(x\right)}{1-yf\left(x\right)}}.$$

Thus, we have

$$\left[x^n\right]g\left(x\right)f{\left(x\right)}^k=\left[x^n{y}^k\right]{\displaystyle \frac{g\left(x\right)}{1-yf\left(x\right)}}.$$

To see this, we have

$$\begin{array}{cc}\hfill \left[x^n{y}^k\right]{\displaystyle \frac{g\left(x\right)}{1-yf\left(x\right)}}& =\left[x^n{y}^k\right]g\left(x\right)\sum _{i=0}^{\infty }{y}^{i}f{\left(x\right)}^i\hfill \\ \hfill & =\left[x^n\right]g\left(x\right)\left[y^k\right]\sum _{i=0}^{\infty }{y}^{i}f{\left(x\right)}^i\hfill \\ \hfill & =\left[x^n\right]g\left(x\right)f{\left(x\right)}^k.\hfill \end{array}$$

Setting $y=0$ and $y=1$, respectively, in the generating function, yields the generating functions of the row sums and the diagonal sums of the matrix $\left(t_{n,k}\right)$. That is, the row sums of the Riordan array $\left(g\right(x),f(x\left)\right)$ have generating function ${\displaystyle \frac{g\left(x\right)}{1-f\left(x\right)}}$, while the diagonal sums have generating function ${\displaystyle \frac{g\left(x\right)}{1-xf\left(x\right)}}$.

The production matrix of a Riordan array takes a special simple form, where for a Riordan array with matrix representative M, its production matrix is given by

$$M^{-1}\overline{M},$$

where $\overline{M}$ is the matrix M with its top row removed. For a lower-triangular matrix M to represent a Riordan array, its production matrix must be of the following form:

$$\left(\begin{array}{ccccccc}\hfill z_0& \hfill a_0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill z_1& \hfill a_1& \hfill a_0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill z_2& \hfill a_2& \hfill a_1& \hfill a_0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill z_3& \hfill a_3& \hfill a_2& \hfill a_1& \hfill a_0& \hfill 0& \hfill \cdots \\ \hfill z_4& \hfill a_4& \hfill a_3& \hfill a_2& \hfill a_1& \hfill a_0& \hfill \cdots \\ \hfill z_5& \hfill a_5& \hfill a_4& \hfill a_3& \hfill a_2& \hfill a_1& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

An almost Riordan array (of first order) [5] is a lower-triangular matrix $A={\left(a_{i,j}\right)}_{0\le i,j\le \infty }$ such that the first column $a_{n,0}$ has a generating function $a\left(x\right)$ and the embedded matrix ${\left(a_{i,j}\right)}_{1\le i,j\le \infty }$ is a Riordan matrix $\left(g\right(x),f(x\left)\right)$. We write $\left(a\right(x),g(x),f(x\left)\right)$ or $\left(a\right(x);g(x),f(x\left)\right)$ for such an almost Riordan array. The almost Riordan array $\left(a\right(x),g(x),f(x\left)\right)$ has its bivariate generating function given by

$$B(x,y)=a\left(x\right)+xy{\displaystyle \frac{g\left(x\right)}{1-yf\left(x\right)}}.$$

The fundamental theorem of Riordan arrays extends to the group of almost Riordan arrays, in the sense that we can describe the action of an almost Riordan array on a power series in a weighted composition form. Note that almost Riordan arrays have been generalized to double almost Riordan arrays [6].

Many examples of Riordan arrays can be found in the On-Line Encyclopedia of Integer Sequences (OEIS) [7,8]. Sequences that we shall encounter in this note will be referred to by their OEIS number where relevant.

2. Main Results

The main results of this paper occur in two areas. The first area is that of significant number sequences arising from integrable lattice models. The second area is that of number sequences arising from the enumeration of lattice paths. The main techniques involve the extension of the notion of a Riordan array. Thus, we define the symmetrization of a Riordan array, and we show that the principal minors of two particular such symmetrizations yield the Robbins numbers for the six-vertex model and the corresponding numbers for the twenty-vertex model. We then show that symmetrizations of appropriate Riordan arrays can also be used to enumerate certain lattice paths. We announce a conjecture concerning a general family of symmetric matrices defined by rational generating functions and associated Riordan arrays and lattice paths. Finally, we define the notion of interleaved Riordan arrays and give examples of their relationship with lattice paths where the step set alternates between levels.

3. The Square Symmetrization of Riordan Arrays

In this section, we introduce the first of the extensions that we wish to explore. This is the symmetrization of a Riordan array. Two areas of application of this notion will be given. The first is where we use the principal minor sequences of the resulting square matrices to enumerate items of interest both in combinatorics and mathematical physics. The second area of application is that of lattice paths.

Given a Riordan array $\left(g\right(x),f(x\left)\right)$, we have seen that its generating function is the bivariate power series

$$B(x,y)={\displaystyle \frac{g\left(x\right)}{1-yf\left(x\right)}}.$$

By the square symmetrization of the Riordan array $\left(g\right(x),f(x\left)\right)$, we mean the matrix whose generating function is given by

$$\mathfrak{S}(x,y)=B\left(xy,{\displaystyle \frac{1}{y}}\right)+B\left(xy,{\displaystyle \frac{1}{x}}\right)-g\left(xy\right).$$

It is clear that

$$\mathfrak{S}(x,y)=\mathfrak{S}(y,x);$$

hence, the square symmetrization of a Riordan array is a symmetric matrix.

We sometimes call this the “reverse” symmetrization of the Riordan matrix $\left(t_{n,k}\right)$ in question, because $B\left(xy,{\displaystyle \frac{1}{y}}\right)$ is the generating function of the reversed array $\left(t_{n,n-k}\right)$.

In terms of the matrix representation of Riordan arrays, the symmetrization process is easily described. We start with a lower-triangular matrix $\left(t_{n,k}\right)$ that represents $\left(g\right(x),f(x\left)\right)$, and we reverse it to obtain the lower-triangular matrix $\left(t_{n,n-k}\right)$. To this we add its own transpose, while we subtract the diagonal matrix whose diagonal entries are the diagonal entries of the reversed array (this ensures that the diagonal only appears once in the symmetrized matrix).

If $B(x,y)$ is the generating function of an almost Riordan array, then again $\mathfrak{S}(x,y)=B\left(xy,{\displaystyle \frac{1}{y}}\right)+B\left(xy,{\displaystyle \frac{1}{x}}\right)-g\left(xy\right)$ will be the generating function of its symmetrization.

Example 1.

We take the example of the Riordan array $(g\left(x\right),f\left(x\right))=\left({\displaystyle \frac{1}{{(1-x)}^2}},{\displaystyle \frac{x}{1-x}}\right)$ whose matrix realization begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 2& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 3& \hfill 3& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 4& \hfill 6& \hfill 6& \hfill 1& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 5& \hfill 10& \hfill 10& \hfill 5& \hfill 1& \hfill 0& \hfill \cdots \\ \hfill 6& \hfill 15& \hfill 20& \hfill 15& \hfill 6& \hfill 1& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

This has the bivariate generating function

$$B(x,y)={\displaystyle \frac{1}{(1-x)(1-x(1+y\left)\right)}}.$$

Then, $B\left(xy,{\displaystyle \frac{1}{y}}\right)={\displaystyle \frac{1}{(1-xy)(1-x(1+y\left)\right)}}$ is the bivariate generating function of the reversed array, which begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 1& \hfill 2& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 1& \hfill 3& \hfill 3& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 1& \hfill 4& \hfill 6& \hfill 4& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 1& \hfill 5& \hfill 10& \hfill 10& \hfill 5& \hfill 0& \hfill \cdots \\ \hfill 1& \hfill 6& \hfill 15& \hfill 20& \hfill 15& \hfill 6& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

The symmetrization now consists of adding to this matrix its transpose while subtracting the diagonal matrix that begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 0& \hfill 2& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 0& \hfill 0& \hfill 3& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 0& \hfill 0& \hfill 0& \hfill 4& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 5& \hfill 0& \hfill \cdots \\ \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 6& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

We thus obtain the matrix with generating function

$$B\left(xy,{\displaystyle \frac{1}{y}}\right)+B\left(xy,{\displaystyle \frac{1}{x}}\right)-{\displaystyle \frac{1}{{(1-xy)}^2}}={\displaystyle \frac{1-3xy+x^2{y}^2}{{(1-xy)}^2(1-y(1+x))(1-x(1+y))}},$$

which begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill \cdots \\ \hfill 1& \hfill 2& \hfill 3& \hfill 4& \hfill 5& \hfill 6& \hfill \cdots \\ \hfill 1& \hfill 3& \hfill 3& \hfill 6& \hfill 10& \hfill 15& \hfill \cdots \\ \hfill 1& \hfill 4& \hfill 6& \hfill 4& \hfill 10& \hfill 20& \hfill \cdots \\ \hfill 1& \hfill 5& \hfill 10& \hfill 10& \hfill 5& \hfill 15& \hfill \cdots \\ \hfill 1& \hfill 6& \hfill 15& \hfill 20& \hfill 15& \hfill 6& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

The principal minor sequence of this array is the following sequence of determinants:

$$\left|1\right|,\left|\begin{array}{cc}1& 1\\ 1& 2\end{array}\right|,\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 3& 3\end{array}\right|,\dots .$$

4. Interleaved Riordan Arrays

A double Riordan array [3,9] is a lower triangular matrix defined by three power series $(g,f_1,f_2)$, where $g\left(x\right)$ is an even power series (only even powers of the indeterminate x appear in its development), while $f_1\left(x\right)$ and $f_2\left(x\right)$ are odd power series (only odd powers of the indeterminate x appear in their developments). The columns of the resulting matrix then have their generating functions given by the following:

$$g,g{f}_1,g{f}_1{f}_2,g{f}_1^2{f}_2,g{f}_1^2{f}_2^2,\cdots ,g{f}_1^{\lfloor {\displaystyle \frac{n+1}{2}}\rfloor }{f}_2^{\lfloor {\displaystyle \frac{n}{2}}\rfloor },\dots .$$

Such matrices form a group called the double Riordan group.

For arbitrary $g\in {\mathcal{F}}_0$ and $f_1,f_2\in {\mathcal{F}}_1$, the lower-triangular matrix whose n-th column has generating $g{f}_1^{\lfloor {\displaystyle \frac{n+1}{2}}\rfloor }{f}_2^{\lfloor {\displaystyle \frac{n}{2}}\rfloor }$ is in general not a Riordan matrix. We shall call such a lower-triangular matrix an interleaved Riordan array, or just an interleaved array. While such matrices do not form a group, there is still a fundamental theorem which determines the action of such a matrix on a power series, which we will exploit in the context of the enumeration of special lattice paths. In the special case where $(g,f_1,f_2)=(g,f,x)$, we have the following fundamental theorem.

Theorem 1.

Given a power series $h\left(x\right)$, we let $h_e\left(x\right)={\displaystyle \frac{h\left(x\right)+h(-x)}{2}}$ and $h_o\left(x\right)={\displaystyle \frac{h\left(x\right)-h(-x)}{2}}$ be the even and odd parts of $h\left(x\right)$, respectively. Moreover, let ${\tilde{h}}_o\left(x\right)=h_o\left(x\right)/x$. Then, we have the following:

$$(g,f,x)·h\left(x\right)=g\left(x\right)h_e\left(\sqrt{xf\left(x\right)}\right)+g\left(x\right)f\left(x\right){\tilde{h}}_0\left(\sqrt{xf\left(x\right)}\right).$$

Proof. 

This follows from the corresponding result for double Riordan arrays. Specifically, we can decompose the operation of $(g,f,x)$ on $h\left(x\right)$ as the sum of

$$(g,0,gfx,0,g{f}^2{x}^2,0,\dots )=g(1,0,xf,0,x^2{f}^2,0\dots )$$

on the vector ${(h_0,0,h_2,0,h_4,0,\dots )}^T$ and the action of

$$(0,gf,0,g{f}^{2}x,0,g{f}^3{x}^2,0,\dots )=gf(0,1,0,xf,0,x^2{f}^2,\dots )$$

on $(0,h_1,0,h_3,0,h_5,0,\dots )$. Gathering terms, where $h_e\left(x\right)={\displaystyle \frac{1}{1-x^2{y}^2}}$ and $h_o={\displaystyle \frac{xy}{1-x^2{y}^2}}$, yields the result. □

Applying this result to the power series ${\displaystyle \frac{1}{1-xy}}$, we see that the bivariate generating function of the interleaved matrix $(g,f,x)$ is given as follows:

$$B(x,y)=g\left(x\right){\displaystyle \frac{1+yf\left(x\right)}{1-x{y}^{2}f\left(x\right)}}.$$

We note that the powers of $g,f,$ and x appear in the development of $(g,f,x)$ as in the following array.

$$\left(\begin{array}{ccccccccc}g& g& g& g& g& g& g& g& \dots \\ 1& f& f& f^2& f^2& f^3& f^3& f^4& \dots \\ 1& 1& x& x& x^2& x^2& x^3& x^3& \dots \end{array}\right).$$

It is evident that every double Riordan array is an interleaved Riordan array.

Another variant on the idea of an interleaved Riordan array is to use the following development array.

$$\left(\begin{array}{ccccccccc}g& 1& g& 1& g& 1& g& 1& \dots \\ 1& f& f& f^2& f^2& f^3& f^3& f^4& \dots \\ 1& 1& x& x& x^2& x^2& x^3& x^3& \dots \end{array}\right).$$

Thus, the n-th column of such an interleaved array would have generating function $g^{e\left(n\right)}{f}^{\lfloor {\displaystyle \frac{n+1}{2}}\rfloor }{x}^{\lfloor {\displaystyle \frac{n}{2}}\rfloor }$, where

$$e\left(n\right)={\displaystyle \frac{1+{(-1)}^n}{2}}$$

gives the sequence $1,0,1,0,1,0,\dots$, and hence $g^{e\left(n\right)}$ evaluates to $g,1,g,1,g,1,\dots$. We shall use the notation $\left(\right(g,f,x\left)\right)$ for this type of interleaved Riordan array. We have the following generalization of the fundamental theorem of Riordan arrays for these interleaved arrays.

Theorem 2.

Given a power series $h\left(x\right)$, we let $h_e\left(x\right)={\displaystyle \frac{h\left(x\right)+h(-x)}{2}}$ and $h_o\left(x\right)={\displaystyle \frac{h\left(x\right)-h(-x)}{2}}$ be the even and odd parts of $h\left(x\right)$, respectively. Moreover, let ${\tilde{h}}_o\left(x\right)=h_o\left(x\right)/x$. Then, we have

$$\left((g,f,x)\right)·h\left(x\right)=g\left(x\right)h_e\left(\sqrt{xf\left(x\right)}\right)+f\left(x\right){\tilde{h}}_0\left(\sqrt{xf\left(x\right)}\right).$$

Proof. 

We can decompose the operation of $\left(\right(g,f,x\left)\right)$ on $h\left(x\right)$ as the sum of

$$(g,0,gfx,0,g{f}^2{x}^2,0,\dots )=g(1,0,xf,0,x^2{f}^2,0\dots )$$

on the vector ${(h_0,0,h_2,0,h_4,0,\dots )}^T$ and the action of

$$(0,f,0,f^{2}x,0,f^3{x}^2,0,\dots )=f(0,1,0,xf,0,x^2{f}^2,\dots )$$

on $(0,h_1,0,h_3,0,h_5,0,\dots )$. Gathering terms, where $h_e\left(x\right)={\displaystyle \frac{1}{1-x^2{y}^2}}$ and $h_o={\displaystyle \frac{xy}{1-x^2{y}^2}}$, yields the result. □

Applying this result to the power series ${\displaystyle \frac{1}{1-xy}}$, we see that the bivariate generating function of the interleaved matrix $\left(\right(g,f,x\left)\right)$ is given by

$$B(x,y)={\displaystyle \frac{g\left(x\right)+yf\left(x\right)}{1-x{y}^{2}f\left(x\right)}}.$$

5. Riordan Arrays, Symmetrization, and the Robbins Numbers

One of the outstanding mathematical stories of the last century (brilliantly recounted in [10]) was that of the Robbins numbers (OEIS sequence A005130) which linked the six-vertex integrable lattice model [11] to alternating sign matrices and plane partitions, thus building a bridge between integrable models and combinatorics. The Robbins numbers $A_n$ are the sequence of integers that begins as follows:

$$1,1,2,7,42,429,7436,218348,10850216,911835460,129534272700,31095744852375,\dots$$

with

$$A_n=\prod _{k=0}^{n-1}{\displaystyle \frac{(3k+1)!}{(n+k)!}}.$$

Among other combinatorial objects, the Robbins numbers count $n\times n$ alternating sign matrices. An alternating sign matrix (ASM) is a matrix with entries drawn from the set $\{-1,0,1\}$, such that 1s and $-1$s alternate in each column and each row (when a $-1$ occurs), and such that the first and last non-zero entry in each row and column is 1. They extend the partially ordered set of permutation matrices into a lattice. For instance, the 7 $3\times 3$ alternating sign matrices are as follows.

$$\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right),\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right),\left(\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right),\left(\begin{array}{ccc}0& 1& 0\\ 1& -1& 1\\ 0& 1& 0\end{array}\right),$$

$$\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right),\left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right),\left(\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right).$$

Every permutation matrix is an alternating sign matrix; thus, there are $A_n-n!$ (OEIS sequence A321511) alternating sign matrices which have at least one $-1$ among their elements.

A plane partition is a two-dimensional array of nonnegative integers ${\pi }_{i,j}$ (with positive integer indices i and j) that is non-increasing in both indices.

The corresponding numbers $B_n$ (OEIS A358069) for the 20-vertex integrable lattice model (with domain wall boundary conditions) [12,13] begin as follows:

$$1,3,23,433,19705,2151843,561696335,349667866305,518369549769169,\dots .$$

Following [14], we have the following example of a Riordan array whose symmetrization leads to the Robbins numbers $A_{n+1}$.

Example 2.

We consider the Riordan array

$$\left({\displaystyle \frac{1}{1+x+x^2}},{\displaystyle \frac{x}{1+x}}\right).$$

The generating function of this array is given by

$$B(x,y)={\displaystyle \frac{\frac{1}{1+x+x^2}}{1-y\frac{x}{1+x}}}.$$

Thus, we have

$$B(x,y)={\displaystyle \frac{1+x}{(1+x+x^2)(1+x-xy)}}.$$

Then, we have

$$\mathfrak{S}(x,y)={\displaystyle \frac{1}{(1-y+xy)(1-x+xy)}}.$$

The resulting symmetric matrix begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill \cdots \\ \hfill 1& \hfill -1& \hfill -2& \hfill -3& \hfill -4& \hfill -5& \hfill \cdots \\ \hfill 1& \hfill -2& \hfill 0& \hfill 2& \hfill 5& \hfill 9& \hfill \cdots \\ \hfill 1& \hfill -3& \hfill 2& \hfill 1& \hfill -1& \hfill -6& \hfill \cdots \\ \hfill 1& \hfill -4& \hfill 5& \hfill -1& \hfill -1& \hfill 0& \hfill \cdots \\ \hfill 1& \hfill -5& \hfill 9& \hfill -6& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

Then, the principal minor sequence of this matrix

$$\left|1\right|=1,\left|\begin{array}{cc}1& 1\\ 1& -1\end{array}\right|=-2,\left|\begin{array}{ccc}1& 1& 1\\ 1& -1& -2\\ 1& -2& 0\end{array}\right|=-7,\dots ,$$

is given by ${(-1)}^{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2}}\right)}{A}_{n+1}$.

Our main result concerning the Robbins numbers $A_n$ and the numbers $B_n$ is the following.

Proposition 1.

We consider the square symmetrization of the Riordan array family

$${\mathfrak{R}}_r=\left({\displaystyle \frac{1}{(1-rx)\sqrt{1-4x}}},xc\left(x\right)\right),$$

where $c\left(x\right)={\displaystyle \frac{1-\sqrt{1-4x}}{2x}}$ is the generating function of the Catalan numbers. Then, the principal minor sequences of ${\mathfrak{R}}_1$ and of ${\mathfrak{R}}_2$ are given by $A_{n+1}$ and $B_n$, respectively.

Proof. 

For the Riordan array $\left({\displaystyle \frac{1}{(1-x)\sqrt{1-4x}}},xc\left(x\right)\right)$ of the proposition, we have the following:

$$B(x,y)={\displaystyle \frac{(y-2)\sqrt{1-4x}-4xy+y}{2(1-x)(4x-1)(1-y+x{y}^2)}}.$$

Then we find that

$$\mathfrak{S}(x,y)={\displaystyle \frac{1}{(1-xy)(1-x-y)}}.$$

It is a classical result that the Robbins numbers $A_{n+1}$ are given by the principal minor sequence of the matrix

$$\left({\displaystyle \genfrac{}{}{0pt}{}{n+k}{k}}\right)-{\delta }_{n,k+1},$$

which has the generating function

$${\displaystyle \frac{1}{1-x-y}}-{\displaystyle \frac{y}{1-xy}}.$$

Multiplying the matrix with this generating function by the unipotent Riordan array

$$\left({\displaystyle \frac{1}{(1-y)(\frac{1}{1-y}-y)}},y\right)$$

produces a matrix with the generating function $\mathfrak{S}(x,y)$. Since the Riordan array $\left({\displaystyle \frac{1}{(1-y)(\frac{1}{1-y}-y)}},y\right)$ is unipotent, all of its principal minors are 1, which establishes the result. Here, we have used the fact that the determinant of a product is the product of the determinants.

We now consider the Riordan array

$$\left({\displaystyle \frac{1}{(1-2x)\sqrt{1-4x}}},xc\left(x\right)\right).$$

We have

$$B(x,y)={\displaystyle \frac{(y-2)\sqrt{1-4x}-4xy+y}{2(1-x)(4x-1)(1-y+x{y}^2)}},$$

and we find that this simplifies to give

$$\mathfrak{S}(x,y)={\displaystyle \frac{1}{(1-2xy)(1-x-y)}}.$$

We must show that the principal minor sequence of the matrix with this generating function is $B_n$. From [12], we know that $B_n$ is given by the principal minor sequence of the matrix with generating function

$${\displaystyle \frac{2y}{(1-y)(1-x-y-xy)}}+{\displaystyle \frac{1}{1-xy}}={\displaystyle \frac{(1-x)(1+y^2)}{(1-y)(1-xy)(1-x-xy)}}.$$

Using the fundamental theorem of Riordan arrays (multiplying the corresponding matrix by appropriate unipotent Riordan arrays), we obtain the symmetric matrix with generating function

$$g(x,y)={\displaystyle \frac{(1-x)(1-y)}{(1-xy)(1-x-y-xy)}},$$

which begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 0& \hfill 3& \hfill 2& \hfill 2& \hfill 2& \hfill 2& \hfill \cdots \\ \hfill 0& \hfill 2& \hfill 9& \hfill 12& \hfill 16& \hfill 20& \hfill \cdots \\ \hfill 0& \hfill 2& \hfill 12& \hfill 35& \hfill 62& \hfill 98& \hfill \cdots \\ \hfill 0& \hfill 2& \hfill 16& \hfill 62& \hfill 161& \hfill 320& \hfill \cdots \\ \hfill 0& \hfill 2& \hfill 20& \hfill 98& \hfill 320& \hfill 803& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

We now multiply by the Riordan array $\left({\displaystyle \frac{1}{1+x}},{\displaystyle \frac{x}{1+x}}\right)$ on the left and by its transpose on the right to obtain the matrix with generating function

$${\displaystyle \frac{1}{(1+x)(1+y)}}g\left({\displaystyle \frac{x}{1+x}},{\displaystyle \frac{y}{1+y}}\right)={\displaystyle \frac{1}{(1-2xy)(1+x+y)}}.$$

Multiplying this new matrix on the left and on the right by the Riordan array matrix $(1,-x)$ now gives us the matrix with generating function

$${\displaystyle \frac{1}{(1-2xy)(1-x-y)}}.$$

As all matrix transformations (except for the last) involve unipotent matrices, the principal minor sequences remain the same. For the last transformation, the matrix $(1,-x)$ has principal minor sequence ${(-1)}^{\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2}}\right)}$, which when multiplied by itself yields the unit sequence as required. We conclude that the matrices with generating functions ${\displaystyle \frac{2y}{(1-y)(1-x-y-xy)}}+{\displaystyle \frac{1}{1-xy}}$ and ${\displaystyle \frac{1}{(1-2xy)(1-x-y)}}$ have the same principal minor sequence, namely $B_n$. □

5.1. Further Observations on Principal Minor Sequences

For general r, the principal minor sequence arising from the Riordan array $({\displaystyle \frac{1}{(1-rx)\sqrt{1-4x}}},$$xc\left(x\right))$ begins as follows:

$$1,r+1,r^3+2r^2+3·r+1,r^6+3r^5+7r^4+13r^3+11r^2+6r+1,\dots .$$

For $r=0,\dots ,5$ we exhibit the first few elements of the corresponding sequences in matrix form.

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill \cdots \\ \hfill 1& \hfill 2& \hfill 7& \hfill 42& \hfill 429& \hfill 7436& \hfill \cdots \\ \hfill 1& \hfill 3& \hfill 23& \hfill 433& \hfill 19705& \hfill 2151843& \hfill \cdots \\ \hfill 1& \hfill 4& \hfill 55& \hfill 2494& \hfill 365953& \hfill 171944344& \hfill \cdots \\ \hfill 1& \hfill 5& \hfill 109& \hfill 9993& \hfill 3791001& \hfill 5898286349& \hfill \cdots \\ \hfill 1& \hfill 6& \hfill 191& \hfill 31306& \hfill 26094301& \hfill 109913708076& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

We have

$${\left({\displaystyle \frac{1}{(1-rx)\sqrt{1-4x}}},xc\left(x\right)\right)}^{-1}=\left((1-2x)(1-rx+r{x}^2),x(1-x)\right).$$

For $r=0$ and $r=1$, the principal minor sequences of the symmetrization of this matrix begin as follows, respectively,

$$1,-3,-13,81,144,-2017,-1757,79513,22704,\dots$$

and

$$1,-4,-33,427,5046,-56241,-316626,7178034,26671624,\dots .$$

We are not aware of any combinatorial interpretation of these numbers.

5.2. A Second Family of Riordan Arrays for $A_n$ and $B_n$

We consider the family of Riordan arrays ${\tilde{\mathfrak{R}}}_r$ defined by

$$\begin{array}{cc}\hfill {\tilde{\mathfrak{R}}}_r& =\left({\displaystyle \frac{1}{(1-x)\sqrt{1-2(r+2)x+r^2{x}^2}}},{\displaystyle \frac{1-rx-\sqrt{1-2(r+2)x+r^2{x}^2}}{2}}\right)\hfill \\ & ={\left((1-x+rx+x^2){\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{x(1-x)}{1+x}}\right),{\displaystyle \frac{x(1-x)}{1+rx}}\right)}^{-1}.\hfill \end{array}$$

Proposition 2.

The Robbins numbers $A_{n+1}$ and the numbers $B_n$ are given by the principal minor sequences of the symmetrizations of ${\tilde{\mathfrak{R}}}_0$ and ${\tilde{\mathfrak{R}}}_1$, respectively.

Proof. 

We find that the generating function of the symmetrization of ${\tilde{\mathfrak{R}}}_r$ is given by

$${\displaystyle \frac{1}{(1-xy)(1-x-y-rxy)}}.$$

The result follows from this. □

The principal minor sequence of ${\tilde{\mathfrak{R}}}_2$ is the sequence that begins with

$$1,4,55,2494,365953,171944344,\dots ,$$

for example. In general, the principal minor sequence of the symmetrization of ${\mathfrak{R}}_r$ is that of the symmetrization of ${\tilde{\mathfrak{R}}}_{(r-1)}$.

5.3. The Riordan Arrays ${\mathfrak{R}}_1$ and ${\mathfrak{R}}_2$

The Riordan array ${\mathfrak{R}}_1$ is given by the following:

$${\mathfrak{R}}_1=\left({\displaystyle \frac{1}{(1-x)\sqrt{1-4x}}},xc\left(x\right)\right)={\left({(1-x)}^3-x^3,x(1-x)\right)}^{-1}.$$

It begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 3& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 9& \hfill 4& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 29& \hfill 14& \hfill 5& \hfill 1& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 99& \hfill 49& \hfill 20& \hfill 6& \hfill 1& \hfill 0& \hfill \cdots \\ \hfill 351& \hfill 175& \hfill 76& \hfill 27& \hfill 7& \hfill 1& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

The general term of this matrix can be expressed as

$$t_{n,k}=\sum _{j=0}^{n-k}\left({\displaystyle \genfrac{}{}{0pt}{}{k+2j}{j}}\right)$$

for $k\le n$ and 0 otherwise. It embeds into the Riordan array ${\left({\displaystyle \frac{{(1-x)}^3-x^3}{(1-x)}},x(1-x)\right)}^{-1}$, which is the OEIS sequence A361654. The $(n,k)$-th element of this matrix gives the number of nonempty subsets of $\{1,\dots ,2n-1\}$ with median n and minimum k.

The symmetrization of ${\mathfrak{R}}_1$ has its general $(n,k)$-th term given by the following:

$$s_{n,k}=[k\le n]\sum _{j=0}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n-k+2j}{j}}\right)+[k>n]\sum _{j=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{k-n+2j}{j}}\right).$$

Here, we use the Iverson bracket $\left[P\right]$, which evaluates to 1 if P is true and 0 otherwise [15]. This symmetric matrix begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill \cdots \\ \hfill 1& \hfill 3& \hfill 4& \hfill 5& \hfill 6& \hfill 7& \hfill \cdots \\ \hfill 1& \hfill 4& \hfill 9& \hfill 14& \hfill 20& \hfill 27& \hfill \cdots \\ \hfill 1& \hfill 5& \hfill 14& \hfill 29& \hfill 49& \hfill 76& \hfill \cdots \\ \hfill 1& \hfill 6& \hfill 20& \hfill 49& \hfill 99& \hfill 175& \hfill \cdots \\ \hfill 1& \hfill 7& \hfill 27& \hfill 76& \hfill 175& \hfill 351& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

We then have

$$\left|1\right|=1,\left|\begin{array}{cc}1& 1\\ 1& 3\end{array}\right|=2,\left|\begin{array}{ccc}1& 1& 1\\ 1& 3& 4\\ 1& 4& 9\end{array}\right|=7,\dots .$$

The matrix ${\mathfrak{R}}_2$ is given by the following:

$${\mathfrak{R}}_2=\left({\displaystyle \frac{1}{(1-2x)\sqrt{1-4x}}},xc\left(x\right)\right)={\left({(1-x)}^4-x^4,x(1-x)\right)}^{-1}.$$

It begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 4& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 14& \hfill 5& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 48& \hfill 20& \hfill 6& \hfill 1& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 166& \hfill 75& \hfill 27& \hfill 7& \hfill 1& \hfill 0& \hfill \cdots \\ \hfill 584& \hfill 276& \hfill 110& \hfill 35& \hfill 8& \hfill 1& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

Its general term $t_{n,k}$ is given by

$$t_{n,k}=\sum _{j=0}^{n-k}{2}^{n-j-k}\left({\displaystyle \genfrac{}{}{0pt}{}{k+2j}{j}}\right)$$

for $k\le n$ and 0 otherwise. Its symmetrization begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill \cdots \\ \hfill 1& \hfill 4& \hfill 5& \hfill 6& \hfill 7& \hfill 8& \hfill \cdots \\ \hfill 1& \hfill 5& \hfill 14& \hfill 20& \hfill 27& \hfill 35& \hfill \cdots \\ \hfill 1& \hfill 6& \hfill 20& \hfill 48& \hfill 75& \hfill 110& \hfill \cdots \\ \hfill 1& \hfill 7& \hfill 27& \hfill 75& \hfill 166& \hfill 276& \hfill \cdots \\ \hfill 1& \hfill 8& \hfill 35& \hfill 110& \hfill 276& \hfill 584& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

We then have the following:

$$\left|1\right|=1,\left|\begin{array}{cc}1& 1\\ 1& 4\end{array}\right|=3,\left|\begin{array}{ccc}1& 1& 1\\ 1& 4& 5\\ 1& 5& 14\end{array}\right|=23,\dots .$$

In general, the Riordan array ${\mathfrak{R}}_r$ will have general term ${\sum }_{j=0}^{n-k}{r}^{n-j-k}\left({\displaystyle \genfrac{}{}{0pt}{}{k+2j}{j}}\right)$ for $k\le n$ and 0 otherwise.

We have a canonical factorization of the arrays ${\mathfrak{R}}_r$ in terms of the Catalan matrix $\left(c\right(x),xc(x\left)\right)$ (OEIS A033184) as follows.

$${\mathfrak{R}}_r=(c\left(x\right),xc\left(x\right))·\left({\displaystyle \frac{1-x}{(1-2x)(1-rx+r{x}^2)}},x\right).$$

The significance of the above results is that there exist families of simply defined Riordan arrays whose symmetrizations appear to be closely linked to important outputs of certain integrable lattice models. On the one hand, this points to those lattice models being related to many objects of combinatorial interest—in this case, to elements of the Riordan group. On the other hand, it shows that Riordan arrays can play important roles in unexpected and important areas seemingly far removed from their origin. Of course, the Riordan array $\left({\displaystyle \frac{1}{1-x}},{\displaystyle \frac{x}{1-x}}\right)$, that is, Pascal’s triangle $\left(\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\right)$ (OEIS A007318) has always played an important role in the analysis of lattice path models.

6. Symmetrization of Riordan Arrays and Lattice Paths in the First Quadrant

We now turn to a second application of the notion of the symmetrization of a Riordan array, namely lattice path enumeration. In order to motivate the following discussion, we begin with two examples that illustrate the direction that we will take.

Example 3.

We regard the bivariate generating function

$$\mathfrak{S}(x,y)={\displaystyle \frac{1}{1-x-y-x^2{y}^2-x^{2}y-x{y}^2}}$$

as the generating function of an algebraically defined square matrix (with integer entries). Clearly, we have

$$\mathfrak{S}(x,y)=\mathfrak{S}(y,x);$$

hence, this is the generating function of a symmetric integer matrix. The matrix in question begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill \cdots \\ \hfill 1& \hfill 2& \hfill 4& \hfill 6& \hfill 8& \hfill 10& \hfill \cdots \\ \hfill 1& \hfill 4& \hfill 11& \hfill 21& \hfill 35& \hfill 53& \hfill \cdots \\ \hfill 1& \hfill 6& \hfill 21& \hfill 52& \hfill 108& \hfill 196& \hfill \cdots \\ \hfill 1& \hfill 8& \hfill 35& \hfill 108& \hfill 269& \hfill 573& \hfill \cdots \\ \hfill 1& \hfill 10& \hfill 53& \hfill 196& \hfill 573& \hfill 1414& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

This is the “square symmetrization” of the lower-triangular matrix that begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 2& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 11& \hfill 4& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 52& \hfill 21& \hfill 6& \hfill 1& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 269& \hfill 108& \hfill 35& \hfill 8& \hfill 1& \hfill 0& \hfill \cdots \\ \hfill 1414& \hfill 573& \hfill 196& \hfill 53& \hfill 10& \hfill 1& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

Now this lower-triangular invertible matrix is the Riordan array

$$(g\left(x\right),f\left(x\right))=\left({\displaystyle \frac{1}{(1+x)\sqrt{1-6x+x^2}}},{\displaystyle \frac{1-x-\sqrt{1-6x+x^2}}{2}}\right).$$

We may write

$$f\left(x\right)={\displaystyle \frac{x}{1+x}}c\left({\displaystyle \frac{x}{{(1-x)}^2}}\right)={\displaystyle \frac{x(1+x)}{(1-x^2)}}c\left(x{\displaystyle \frac{{(1+x)}^2}{{(1-x^2)}^2}}\right).$$

This is an application of the fundamental theorem of Riordan arrays. Here, we have

$$c\left(x\right)={\displaystyle \frac{1-\sqrt{1-4x}}{2x}},$$

the generating function of the Catalan numbers $C_n={\displaystyle \frac{1}{n+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)$ (OEIS A000108). In addition, the generating function $g\left(x\right)$ may be derived from the form of $f\left(x\right)$.

The symmetric matrix enumerates lattice paths in the first quadrant with step set

$$\left\{\right(1,0),(0,1),(1,2),(2,1),(2,2\left)\right\}.$$

In particular, the sequence OEIS A026933, which begins with

$$1,2,11,52,269,\dots ,$$

enumerates lattice paths from $(0,0)$ to $(n,n)$ for the given step set. The elements of the matrix $t_{n,k}$ satisfy the recurrence

$$t_{n,k}=t_{n-1,k}+t_{n,k-1}+t_{n-1,k-2}+t_{n-2,k-1}+t_{n-2,k-2},$$

where we have the initial conditions of $t_{n,k}=0$ if $n<0$ or $k<0$, and $t_{0,0}=1$. Using the notion of the cross section of a recurrence [16], we can go from the above two-dimensional recurrence to the one-dimensional recurrence

$$a_n=a_{n-1}+y{a}_n+y^2{a}_{n-1}+y{a}_{n-2}+y^2{a}_{n-2},$$

or

$$a_n={\displaystyle \frac{1}{1-y}}\left(a_{n-1}+y^2{a}_{n-1}+y{a}_{n-2}+y^2{a}_{n-2}\right),$$

with initial conditions $a_n=0$ for $n<0$, and $a_0=1$. The sequence $a_n=a_n\left(y\right)$ of rational functions begins as follows:

$$1,{\displaystyle \frac{1}{1-y}},{\displaystyle \frac{1+y^2}{{(1-y)}^2}},{\displaystyle \frac{1+y+2y^2-y^3+y^4}{{(1-y)}^3}},{\displaystyle \frac{1+2y+3y^2+3y^4-2y^5+y^6}{{(1-y)}^4}},\dots .$$

These are then the generating functions of the columns of the symmetric matrix. The second section gives us the recurrence

$$b_n=x^2{b}_{n-1}+x{b}_{n-2}+x^2{b}_{n-2}+x{b}_n+b_{n-1},$$

or

$$b_n={\displaystyle \frac{1}{1-x}}\left(x^2{b}_{n-1}+x{b}_{n-2}+x^2{b}_{n-2}+b_{n-1}\right),$$

with initial conditions $b_n=0$ if $n<0$, and $b_0=1$. We find the same sequence of rational functions (in x this time), reflecting the symmetry of the situation.

Example 4.

We consider the bi-variate generating function

$$\mathfrak{S}(x,y)={\displaystyle \frac{1-x^{2}y-x{y}^2}{1-x-y}}.$$

We again have $\mathfrak{S}(x,y)=\mathfrak{S}(y,x)$; therefore, this is the generating function of a symmetric matrix. This matrix begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill \cdots \\ \hfill 1& \hfill 2& \hfill 2& \hfill 3& \hfill 4& \hfill 5& \hfill \cdots \\ \hfill 1& \hfill 2& \hfill 4& \hfill 7& \hfill 11& \hfill 16& \hfill \cdots \\ \hfill 1& \hfill 3& \hfill 7& \hfill 14& \hfill 25& \hfill 41& \hfill \cdots \\ \hfill 1& \hfill 4& \hfill 11& \hfill 25& \hfill 50& \hfill 91& \hfill \cdots \\ \hfill 1& \hfill 5& \hfill 16& \hfill 41& \hfill 91& \hfill 182& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

This is the “square symmetrization” of the lower-triangular matrix that begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 2& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 4& \hfill 2& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 14& \hfill 7& \hfill 3& \hfill 1& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 50& \hfill 25& \hfill 11& \hfill 4& \hfill 1& \hfill 0& \hfill \cdots \\ \hfill 182& \hfill 91& \hfill 41& \hfill 16& \hfill 5& \hfill 1& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

This lower-triangular invertible matrix is now the almost Riordan array of first order, given as follows:

$$\left({\displaystyle \frac{1-x}{\sqrt{1-4x}}}+x,{\displaystyle \frac{(1-x)(\sqrt{1-4x}+4x-1)}{2x(1-4x)}},xc\left(x\right)\right).$$

The sequence $1,2,4,14,50,\dots$ is essentially 2 times the sequence OEIS A097613. Its general terms are given by

$$2\left(\left({\displaystyle \genfrac{}{}{0pt}{}{2n-3}{n-1}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{2n-2}{n}}\right)\right)-0^n.$$

The sequence counts the number of Dyck $(2n-1)$-paths with maximum pyramid size n. Denoting the above matrix by S, we define its principal minor sequence to be

$$m_n={\left|S\right|}_{0\le i,j\le n}.$$

Then, the principal minor sequence of S is given as follows:

$$1,1,2,2,-1,-5,-5,1,8,8,-1,\cdots .$$

This can be compared with OEIS A187307, which begins as follows:

$$1,2,2,-1,-5,-5,1,8,8,-1,-11,-11,\dots .$$

This latter sequence is the Hankel transform [17] of the alternating sum of the Motzkin numbers.

We note that for the symmetric matrix of the first example, its principal minor sequence is the all 1s sequence. This follows from its factorization as $D{D}^t$, where D is the Delannoy triangle [18] given by the Riordan array $\left({\displaystyle \frac{1}{1-x}},{\displaystyle \frac{x(1+x)}{1-x}}\right)$ which begins as follows:

$$D=\left(\begin{array}{ccccccc}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 1& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 1& \hfill 3& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 1& \hfill 5& \hfill 5& \hfill 1& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 1& \hfill 7& \hfill 13& \hfill 7& \hfill 1& \hfill 0& \hfill \cdots \\ \hfill 1& \hfill 9& \hfill 25& \hfill 25& \hfill 9& \hfill 1& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

We remark that in the examples above, the presence of $1-x-y$ in the denominator is responsible for having 1s in the first row and first column of the resulting symmetric matrices.

Example 5.

We use an example from [19] to illustrate the case of a symmetric matrix arising from a lattice path enumeration which is not the symmetrization of a Riordan array. Thus, we consider the case of the knight’s walk, which is a walk that can start anywhere on the lines $x=0$ or 1, $y=0$ or 1, and take only two kinds of steps: $(-1,2)$ and $(2,-1)$, remaining in the region $x\ge 2$, $y\ge 2$ once they have left their starting point. Let $t_{m,n}$ denote the number of such walks ending at $(n,k)$. We have

$$t_{n,k}=t_{n+1,k-2}+t_{n-2,k+1}ifn,k\ge 2,$$

and $t_{n,k}=1$ otherwise. We find that the symmetric matrix $\left(t_{n,k}\right)$ begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill \cdots \\ \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill \cdots \\ \hfill 1& \hfill 1& \hfill 2& \hfill 2& \hfill 3& \hfill 3& \hfill \cdots \\ \hfill 1& \hfill 1& \hfill 2& \hfill 2& \hfill 4& \hfill 5& \hfill \cdots \\ \hfill 1& \hfill 1& \hfill 3& \hfill 4& \hfill 6& \hfill 10& \hfill \cdots \\ \hfill 1& \hfill 1& \hfill 3& \hfill 5& \hfill 10& \hfill 14& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

This is the symmetrization of the matrix that begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 1& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 2& \hfill 1& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 2& \hfill 2& \hfill 1& \hfill 1& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 6& \hfill 4& \hfill 3& \hfill 1& \hfill 1& \hfill 0& \hfill \cdots \\ \hfill 14& \hfill 10& \hfill 5& \hfill 3& \hfill 1& \hfill 1& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

While this is a lower-triangular invertible matrix, it is not a Riordan array, as can be seen from the form of its production matrix.

6.1. A Conjecture for Special Symmetric Matrices

We have the following conjecture for a special class of symmetric matrices.

Conjecture 1.

Let S be the symmetric matrix whose bi-variate generating function is given as follows:

$$\mathfrak{S}(x,y)={\displaystyle \frac{1}{1-{\sum }_{i=0}^r{\alpha }_i(x^i{y}^{i+1}+x^{i+1}{y}^i)-{\sum }_{j=1}^s{\beta }_j{x}^i{y}^i}}.$$

Then, S is the square symmetrization of the Riordan array $\left(g\right(x),f(x\left)\right)$, where

$$f\left(x\right)=x{\displaystyle \frac{1+{\sum }_{i=0}^r{\alpha }_i{x}^i}{1-{\sum }_{j=1}^s\beta x^i}}c\left(x{\left({\displaystyle \frac{1+{\sum }_{i=0}^r{\alpha }_i{x}^i}{1-{\sum }_{j=1}^s{\beta }_i{x}^i}}\right)}^2\right),$$

and

$$g\left(x\right)=1-\sum _{j=1}^s{\beta }_j{x}^j-2x\left(1+\sum _{i=0}^s{\alpha }_i{x}^i\right)f\left(x\right)/x.$$

The matrix S enumerates paths from $(0,0)$ to $(n,k)$ in the positive quadrant with steps $(i,i)$ for $i=1\dots s$ and steps $(j,j+1)$ and $(j+1,j)$ for $j=1\dots r$, where the step $(i,i)$ can have ${\beta }_i$ colors, and the steps $(j,j+1)$ and $(j+1,j)$ can each have ${\alpha }_i$ colors. Elements $s_{n,k}$ of the matrix S obey the recurrence equation

$$s_{n,k}=s_{n-1,k}+s_{n,k-1}+\sum _{i=1}^s{\beta }_i{s}_{n-i,k-i}+\sum _{j=1}^s{\alpha }_i(s_{n-i,k-i-1}+s_{n-i-1,k-i}).$$

Example 6.

We consider the symmetric matrix with the following bi-variate generating function:

$$\mathfrak{S}(x,y)={\displaystyle \frac{1}{1-x-y-2xy-5x^2{y}^2-3(x{y}^2+x^{2}y)-4(x^2{y}^3+x^3{y}^2)}}.$$

We obtain the matrix S that begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill \cdots \\ \hfill 1& \hfill 4& \hfill 10& \hfill 16& \hfill 22& \hfill 28& \hfill \cdots \\ \hfill 1& \hfill 10& \hfill 39& \hfill 99& \hfill 195& \hfill 327& \hfill \cdots \\ \hfill 1& \hfill 16& \hfill 99& \hfill 364& \hfill 992& \hfill 2196& \hfill \cdots \\ \hfill 1& \hfill 22& \hfill 195& \hfill 992& \hfill 3581& \hfill 10153& \hfill \cdots \\ \hfill 1& \hfill 28& \hfill 327& \hfill 2196& \hfill 10153& \hfill 36032& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

The elements $s_{n,k}$ of this matrix then satisfy the recurrence

$$\begin{array}{cc}\hfill s_{n,k}=s_{n-1,k}+s_{n,k-1}& +2s_{n-1,k-1}+5s_{n-2,k-2}\hfill \\ \hfill +3(s_{n-1,k-2}& +s_{n-2,k-1})+4(s_{n-2,k-3}+s_{n-3,k-2}),\hfill \end{array}$$

with the boundary conditions $s_{n,k}=0$ if $n<0$ or $k<0$, and $s_{n,0}=s_{0,k}=1$. This matrix is then the symmetrization of the Riordan array $\left(g\right(x),f(x\left)\right)$, where

$$\begin{array}{cc}\hfill f\left(x\right)& =x{\displaystyle \frac{1+3x+4x^2}{1-2x-5x^2}}c\left(x{\left({\displaystyle \frac{1+3x+4x^2}{1-2x-5x^2}}\right)}^2\right)\hfill \\ & ={\displaystyle \frac{1-2x-5x^2-\sqrt{1-8x-30x^2-71x^4-64x^5}}{2(1+3x+4x^2)}},\hfill \end{array}$$

and

$$g\left(x\right)={\displaystyle \frac{1}{\sqrt{1-8x-30x^2-71x^4-64x^5}}}.$$

Although we cannot prove the conjecture in full generality, each individual case can be proven by using $b(x,y)={\displaystyle \frac{g\left(x\right)}{1-yf\left(x\right)}}$ and simplification to show that

$$\mathfrak{S}(x,y)=b(xy,1/x)+b(xy,1/y)-b(xy,0)$$

has the required form.

6.2. The Symmetrization of a Pascal-like Riordan Array

By a Pascal-like array, we mean a lower triangular matrix $\left(t_{n,k}\right)$ where $t_{n,k}=t_{n,n-k}$ and $t_{n,0}=t_{n,n}=1$. We consider the Pascal-like Riordan array $D=\left({\displaystyle \frac{1}{1-x}},{\displaystyle \frac{x(1+x)}{1-x}}\right)$, which is the centrally symmetric Delannoy triangle seen above. We have $b(x,y)={\displaystyle \frac{1}{1-x-xy-x^{2}y}}$ and its symmetrization has generating function

$$\mathfrak{S}(x,y)={\displaystyle \frac{1-3xy-x^2{y}^2-x^3{y}^3}{(1-xy)(1-x-xy-x^{2}y)(1-y-xy-x{y}^2)}}.$$

The denominator expands to give

$$1-x-y-2xy+4x^2{y}^2-2x^3{y}^3-x^4{y}^4+x{y}^2+x^{2}y+x^2{y}^3+x^3{y}^2-(x^3{y}^4+x^4{y}^3).$$

The symmetrization of D begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill \cdots \\ \hfill 1& \hfill 1& \hfill 3& \hfill 5& \hfill 7& \hfill 9& \hfill \cdots \\ \hfill 1& \hfill 3& \hfill 1& \hfill 5& \hfill 13& \hfill 25& \hfill \cdots \\ \hfill 1& \hfill 5& \hfill 5& \hfill 1& \hfill 7& \hfill 25& \hfill \cdots \\ \hfill 1& \hfill 7& \hfill 13& \hfill 7& \hfill 1& \hfill 9& \hfill \cdots \\ \hfill 1& \hfill 9& \hfill 25& \hfill 25& \hfill 9& \hfill 1& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

Noting that all Pascal-like (ordinary) Riordan arrays can be described by $\left({\displaystyle \frac{1}{1-x}},{\displaystyle \frac{x(1+rx)}{1-x}}\right)$ for appropriate r values, we can see that they will follow a similar pattern. Setting the numerator in $\mathfrak{S}(x,y)$ above to 1, we obtain the generating function

$$\tilde{\mathfrak{S}}(x,y)={\displaystyle \frac{1}{(1-xy)(1-x-xy-x^{2}y)(1-y-xy-x{y}^2)}}.$$

This turns out to be the generating function of the symmetrization of the Riordan array

$$\left({\displaystyle \frac{1}{(1-x)(1-3x-x^2-x^3)}},{\displaystyle \frac{x(1+x)}{1-x}}\right)=\left({\displaystyle \frac{1}{1-4x+2x^2+x^4}},{\displaystyle \frac{x(1+x)}{1-x}}\right).$$

6.3. Further Remarks on the Numerator

We have seen that in the case of the numerator of the bivariate generating function not being equal to 1, we do not in general obtain a Riordan array.

Example 7.

We consider the symmetric matrix S with generating function

$${\displaystyle \frac{1-(x^{2}y+x{y}^2)}{1-x-y-xy-x^2{y}^2-(x^{2}y+x{y}^2)}}.$$

It turns out that S is the symmetrization of an almost Riordan array of first order $\left(a\right(x),g(x),f(x\left)\right)$, where the Riordan array $\left(g\right(x),f(x\left)\right)$ is of the form

$$\left(g\right(x),f(x\left)\right)=\left(G\right(x),f(x\left)\right)\left(h\right(x),x),$$

where $\left(G\right(x),f(x\left)\right)$ is the Riordan array whose symmetrization has

$${\displaystyle \frac{1}{1-x-y-xy-x^2{y}^2-(x^{2}y+x{y}^2)}}$$

as generating function. For this case, we have

$$\begin{array}{cc}\hfill f\left(x\right)& =x{\displaystyle \frac{1+x}{1-x-x^2}}c\left({\left(x{\displaystyle \frac{1+x}{1-x-x^2}}\right)}^2\right)={\displaystyle \frac{1-x-x^2-\sqrt{1-6x-9x^2-2x^3+x^4}}{2(1+x)}},\hfill \\ \hfill G\left(x\right)& ={\displaystyle \frac{1}{\sqrt{1-6x-9x^2-2x^3+x^4}}},\hfill \\ \hfill h\left(x\right)& ={\displaystyle \frac{1+x+x^4+x\sqrt{1+6x+3x^2-2x^3+x^4}}{1-x^2}}.\hfill \end{array}$$

In contrast to this, the symmetric matrix with generating function

$${\displaystyle \frac{1-xy}{1-x-y-xy-x^2{y}^2-(x^{2}y+x{y}^2)}}$$

is the symmetrization of the Riordan array

$$\left(\right(1-x\left)G\right(x),f(x\left)\right).$$

7. Interleaved Matrices and Lattice Paths

We now investigate applications of the second Riordan array extension, namely interleaved Riordan arrays, to lattice paths. Common lattice paths that are associated with Riordan arrays (some ordinary, some exponential) are the Dyck, Motzkin, and Schroeder paths. A Dyck path is a path from $(0,0)$ to $(2n,0)$, with a step set including a diagonal step up and a diagonal step down; that is, $(1,1)$ and $(1,-1)$. A Motzkin path is a path from $(0,0)$ to $(n,0)$ with step set $\left\{\right(1,0),(1,1),(1,-1\left)\right\}$. A Schroeder path is a path from $(0,0)$ to $(2n,0)$ with step set $\left\{\right(2,0),(1,1),(1,-1\left)\right\}$. We may also consider Motzkin–Schroeder paths (from $(0,0)$ to $(n,0)$) with step set

$$\left\{\right(1,0),(2,0),(1,1),(1,-1\left)\right\}.$$

Lower-triangular matrices enter the picture when we want to enumerate such paths that go from $(0,0)$ to $(n,k)$ for $k\le n$. Such paths are traditionally called left-factors of lattice paths.

Example 8.

We consider Schroeder paths with no level steps at even level. Thus, at odd levels, the step set is $\left\{\right(2,0),(1,1),(1,-1\left)\right\}$, whilst at even levels, the step set is $\left\{\right(1,1),(1,-1\left)\right\}$. It is this alternating of step sets that leads to the interleaving in the left-factor matrix. We find that the left-factor matrix is the interleaved Riordan array

$$\left(\left({\displaystyle \frac{1-x^2-\sqrt{1-6x^2+5x^4}}{2x^2}},{\displaystyle \frac{1-3x^2-\sqrt{1-6x^2+5x^4}}{2x^3}},x\right)\right),$$

which begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 1& \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 0& \hfill 3& \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 3& \hfill 0& \hfill 4& \hfill 0& \hfill 1& \hfill 0& \hfill \cdots \\ \hfill 0& \hfill 10& \hfill 0& \hfill 6& \hfill 0& \hfill 1& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

The row sums of this matrix give the number of left factors of such paths. We obtain the sequence OEIS A026392 that begins as follows:

$$1,1,2,4,8,17,34,75,150,339,678,\dots .$$

With $g\left(x\right)$ and $f\left(x\right)$ above, we find that this sequence has generating function

$${\displaystyle \frac{g\left(x\right)+f\left(x\right)}{1-xf\left(x\right)}}={\displaystyle \frac{(1+2x)\sqrt{1-6x^2+5x^4}+5x^2-1}{2x(1-5x^2)}}.$$

The recurrence for this matrix is

$$\begin{array}{ccc}\hfill t_{n,k}& =t_{n-1,k-1}+t_{n-1,k+1}\hfill & \hfill \mathit{for}\mathit{even}k\mathit{values},\\ \hfill t_{n,k}& =t_{n-1,k-1}+t_{n-2,k}+t_{n-1,k+1}\hfill & \hfill \mathit{for}\mathit{odd}k\mathit{values}.\end{array}$$

Example 9.

We now consider a more general case. Thus, we assume that for even levels, we have a colored Dyck path step set

$$\{2\ast (1,1),3\ast (1,-1\left)\right\}$$

where the up steps can have 2 colors, and the down steps can have 3 colors. For odd levels, we assume that we have the Motzkin–Schroeder step set of

$$\left\{\right(1,0),(2,0),(1,1),(1,-1\left)\right\}.$$

The left-factor matrix $\left(t_{n,k}\right)$ can then be calculated as follows. For $k<0$, $n<0$, and $n>k$, we have $t_{n,k}=0$. For $n=0$, we have $t_{n,k}=0^k$. Otherwise, for even k values, we have

$$t_{n,k}=2t_{n-1,n-1}+3t_{n-1,k+1},$$

while for odd k values, we have

$$t_{n,k}=t_{n-1,k-1}+t_{n-1,k}+t_{n-2,k}+t_{n-1,k+1}.$$

We find that the matrix $\left(t_{n,k}\right)$ begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 3& \hfill 1& \hfill 2& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 3& \hfill 7& \hfill 2& \hfill 2& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 21& \hfill 13& \hfill 20& \hfill 4& \hfill 4& \hfill 0& \hfill \cdots \\ \hfill 39& \hfill 61& \hfill 38& \hfill 30& \hfill 8& \hfill 4& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

This is the interleaved Riordan array $\left(\right(g\left(x\right),f\left(x\right),2x\left)\right)$, where

$$\begin{array}{cc}\hfill g\left(x\right)& ={\displaystyle \frac{1-x-x^2}{1-x-2x^2}}c\left({\displaystyle \frac{2x^2(1-x-x^2)}{{(1-x-2x^2)}^2}}\right),\hfill \\ \hfill f\left(x\right)& ={\displaystyle \frac{x}{1-x-6x^2}}c\left({\displaystyle \frac{6x^4}{{(1-x-6x^2)}^2}}\right)\hfill \end{array}$$

The generating function of this matrix is given by ${\displaystyle \frac{g\left(x\right)+yf\left(x\right)}{1-2x{y}^{2}f\left(x\right)}}$. Setting $y=1$, we obtain the generating function of the row sums, which enumerate the left factors. We obtain the generating function

$${\displaystyle \frac{1-11x^2-6x^3-(1+5x)\sqrt{1-2x-11x^2+12x^3+12x^4}}{4x(1-x-11x^2)}},$$

or equivalently,

$${\displaystyle \frac{1+3x-3x^2-3x^3}{1-11x^2-6x^3}}c\left({\displaystyle \frac{-2x(1-x-11x^2)(1+3x-3x^2-3x^3)}{{(1-11x^2-6x^3)}^2}}\right).$$

The corresponding sequence begins as follows:

$$1,1,6,14,62,180,718,2320,8850,\dots .$$

Example 10.

More generally, for lattice paths with the step sets $\{a\ast (1,1),b\ast (1,-1\left)\right\}$ (even levels) and $\left\{\right(c\ast (1,0),d\ast (2,0),(1,1),(1,-1)\}$ (odd levels), we can conjecture the following generating function for $g\left(x\right)$, which enumerates such paths from $(0,0)$ to $(n,0)$.

$$g\left(x\right)={\displaystyle \frac{1-cx-d{x}^2}{1-cx-(b-a+d)x^2}}c\left({\displaystyle \frac{a{x}^2(1-cx-d{x}^2)}{{(1-cx-(b-a+d)x^2)}^2}}\right).$$

The left factors for these paths obey the recurrence, for even k values,

$$t_{n,k}=a{t}_{n-1,k-1}+b{t}_{n-1,k+1},$$

while for odd k values, we have

$$t_{n,k}=t_{n-1,k-1}+c{t}_{n-1,k}+d{t}_{n-2,k}+t_{n-1,k+1}.$$

The left factor matrix in the case $(a=1,b=2,c=1,d=2)$ begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 2& \hfill 1& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 2& \hfill 6& \hfill 1& \hfill 1& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 12& \hfill 11& \hfill 8& \hfill 2& \hfill 1& \hfill 0& \hfill \cdots \\ \hfill 22& \hfill 43& \hfill 15& \hfill 13& \hfill 2& \hfill 1& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

We have the following:

$$\begin{array}{cc}\hfill g\left(x\right)& ={\displaystyle \frac{1-x-2x^2}{1-x-3x^2}}c\left({\displaystyle \frac{x^2(1-x-2x^2)}{{(1-x-3x^2)}^2}}\right),\hfill \\ \hfill f\left(x\right)& ={\displaystyle \frac{x}{1-x-5x^2}}c\left({\displaystyle \frac{2x^4}{{(1-x-5x^2)}^2}}\right).\hfill \end{array}$$

The generating function for the sequence that enumerates the left factors is then given as follows:

$${\displaystyle \frac{(1+x)(1+x-4x^2)}{1-8x^2-5x^3}}c\left({\displaystyle \frac{-x(1+x)(1+x-4x^2)(1-x-8x^2)}{{(1-8x^2-5x^3)}^2}}\right).$$

This expands to give the sequence that begins as follows:

$$1,1,4,10,34,96,310,926,2944,\dots .$$

The expansion of $g\left(x\right)$, which begins with

$$1,0,2,2,12,22,86,204,686,1842,5848,16746,\dots ,$$

has the interesting feature that this sequence and its partial sum sequence have the same Hankel transform.

An Interesting Simple Interleaved Riordan Array

The lattice path interleaved Riordan arrays above have had a relatively complicated nature. To finish this note, we look at a more elementary example, where the inverse of the matrix is of interest, in that it leads us back to sequences which count alternating sign matrices. Thus, we consider the simple interleaved Riordan array $\left(\right(1-x,x(1-x),x\left)\right)$, which begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill -1& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 0& \hfill -1& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 0& \hfill 0& \hfill -2& \hfill 1& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 0& \hfill 0& \hfill 1& \hfill -2& \hfill 1& \hfill 0& \hfill \cdots \\ \hfill 0& \hfill 0& \hfill 0& \hfill 1& \hfill -3& \hfill 1& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

The inverse of this matrix (which is not an interleaved Riordan array) begins as follows:

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 1& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 1& \hfill 1& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 2& \hfill 2& \hfill 2& \hfill 1& \hfill 0& \hfill 0& \hfill \cdots \\ \hfill 3& \hfill 3& \hfill 3& \hfill 2& \hfill 1& \hfill 0& \hfill \cdots \\ \hfill 7& \hfill 7& \hfill 7& \hfill 5& \hfill 3& \hfill 1& \hfill \cdots \\ \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill ⋮& \hfill \ddots \end{array}\right).$$

The initial column of this matrix, which begins with

$$1,1,1,2,3,7,12,30,55,143,273,\cdots ,$$

is OEIS A047749, which counts the number of $(n+1,n+2)$ core partitions into even parts [20,21]. Multiplying the column vector ${(1,0,1,0,1,0,\dots )}^T$ by this inverse matrix yields the sequence that begins as follows:

$$1,1,2,4,7,17,31,80,152,404,790,\dots .$$

The sequence OEIS A299293, which enumerates $(n+1,n+2)$ core partitions into odd parts, begins as follows:

$$1,2,4,7,17,31,80,152,404,790,\dots .$$

We note that applying the matrix to the vector $\left({(-1)}^n\right)$ yields the following sequence:

$$1,0,1,1,2,4,7,17,31,80,152,\dots .$$

We remark that the Hankel transform of the sequence OEIS A047749 begins as follows:

$$1,0,-1,0,9,0,-676,0,417316,0,-2105433225,0,86576511622500,0,\dots .$$

Taking the square root of the absolute value of every second term, we obtain the sequence OEIS A005156, which begins as follows:

$$1,1,3,26,646,45885,9304650,5382618660,\dots .$$

This sequence counts the number of alternating sign $(2n+1)\times (2n+1)$ matrices symmetric about the vertical axis, thus leading us back to alternating sign matrices. Similarly, the Hankel transform of OEIS A005156$(n+1)$, beginning with $1,1,2,3,7,12,30,55,\dots$, is given by OEIS A005161$(n+1)$, where the sequence OEIS A005161 counts the number of alternating sign $(2n+1)\times (2n+1)$ matrices symmetric with respect to both horizontal and vertical axes. The Hankel transform of OEIS A047749$(n+2)$ begins as follows:

$$1,-1,-4,9,121,-676,-28900,417316,55190041,-2105433225,-847246611600,\dots .$$

Taking the square root of the absolute values of the terms, we obtain the following sequence:

$$1,1,2,3,11,26,170,646,7429,45885,920460,\dots .$$

This is the interleaving of the sequences

$$1,2,11,170,7429,\dots$$

and

$$1,3,26,646,45885,\dots .$$

The sequence $1,2,11,170,7429,\dots$ is OEIS A051255$(n+1)$, where OEIS A051255 counts the number of cyclically symmetric transpose complement plane partitions in a $2n\times 2n\times 2n$ box.

8. Conclusions

Riordan arrays have been shown to be suited to the exploration of a wide number of combinatorial problems. Their versatility is in part due to the ease of their description in algebraic terms and the simplicity of their application in their matrix representation. The fundamental theorem of Riordan arrays (specifying the operation of applying an array to a power series, mirrored by applying a matrix to a vector) gives us a powerful descriptive and analytical tool. This tool has been extended to almost Riordan arrays and to double Riordan arrays. We have also shown that this tool plays a role for interleaved Riordan arrays, and we have been able to apply it to special “interleaved” lattice paths to obtain generating functions for the total number of left factors. Thus, while we may no longer have a group structure, the existence of this action on power series allows us to derive significant results. We have also shown that the idea of the reverse symmetrization of Riordan arrays can be applied to a number of different areas, including that of lattice paths. For integrable lattice models, the symmetrization process leads to square matrices whose principal minor sequences become the main focus. In tandem with the Riordan array approach, we have also shown that two-dimensional recurrences can be used to describe the terms of the matrices that represent Riordan and generalized Riordan arrays that we have met, particularly in the case of lattice paths. Of interest in this context are the cross sections of these recurrences, whose solutions are related to row and column generating functions.