The γ-Vectors of Pascal-like Triangles Defined by Riordan Arrays

Paul Barry School of Science Waterford Institute of Technology Ireland

                     pbarry@wit.ie

Abstract

We define and characterize the γ-matrix associated with Pascal-like matrices that are defined by ordinary and exponential Riordan arrays. We also define and characterize the γ-matrix of the reversions of these triangles, in the case of ordinary Riordan arrays. We are led to the γ-matrices of a one-parameter family of generalized Narayana triangles. Thus these matrices generalize the matrix of γ-vectors of the associahedron. The principal tools used are the bivariate generating functions of the triangles and Jacobi continued fractions.

1 Introduction

A polynomial P_{n}(x) = ∑_{k=0}^{n} a_{n,k}x^{k} of degree n is said to be reciprocal if

P_{n}(x) = x^{n}P_{n}(1/x).

Thus we have

[x^{k}]P_{n}(x) = a_{n,k} = [x^{k}]x^{n}P_{n}(1/x).

Now

[x^{k}]x^{n}P_{n}(1/x) = [x^{k−n}] ∑_{i=0} a_{n,i} \frac{1}{x^{i}}
= [x^{k−n}] ∑_{i=0} a_{n,i}x^{−i}
= a_{n,n−k}.

Thus P_{n}(x) = ∑_{k=0}^{n} a_{n,k}x^{k} defines a family of reciprocal polynomials if and only if a_{n,k} = a_{n,n−k}. We shall call a lower-triangular matrix (a_{n,k}) Pascal-like if

1. a_{n,k} = a_{n,n−k}

2. a_{n,0} = a_{n,n} = 1.

Such a matrix will then be the coefficient array of a family of monic reciprocal polynomials. We have the following well-known result [7]

Proposition 1. Let P_{n}(x) be a reciprocal polynomial of degree n. Then there exists a unique polynomial γ_{n} of degree ⌊\frac{n}{2}⌋ with the property

P_{n}(x) = (1 + x)^{n}γ_{n} (\frac{x}{(1 + x)^{2}} ) .

If P_{n}(x) has integer coefficients then so does γ_{n}(x).

By this means, we can associate with every Pascal-like matrix (a_{n,k}) a matrix (γ_{n,k}) so that for all n, we have

P_{n}(x) = ∑_{k=0}^{n} a_{n,k}x^{k} = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} γ_{n,k}x^{k}(1 + x)^{n−2k}.

We shall call this matrix the γ-matrix associated with the coefficient array (a_{n,k}) of the family of polynomials P_{n}(x).

We can characterize the matrix (a_{n,k}) in terms of the γ-matrix (γ_{n,k}) as follows. Before we do this, we shall change our notation somewhat. In algebraic topology, it is customary to use the notation h(x) for palindromic (reciprocal) polynomials [9, 15]. Thus we shall set h_{n}(x) = ∑_{k=0}^{n} h_{n,k}x^{k}, where (h_{n,k}) now denotes a Pascal-like matrix. We shall denote by h(x, y) the bivariate generating function of this matrix.

Proposition 2. For a Pascal-like matrix (h_{n,k}) we have

h_{n,k} = ∑_{i=0}^{⌊ \frac{n}{2} ⌋} \binom{n − 2i}{k − i}γ_{n,i}.

Proof. We have

h_{n,k} = [x^{k}] ∑_{i=0}^{n} h_{n,i}x^{i}

= [x^{k}] ∑_{i=0}^{⌊ \frac{n}{2} ⌋} γ_{n,i}x^{i}(1 + x)^{n−2i}

= ∑_{i=0}^{⌊ \frac{n}{2} ⌋} γ_{n,i}[x^{k}]x^{i}(1 + x)^{n−2i}

= ∑_{i=0}^{⌊ \frac{n}{2} ⌋} γ_{n,i}[x^{k−i}] ∑_{j=0}^{n−2i} \binom{n − 2i}{j}x^{j}

= ∑_{i=0}^{⌊ \frac{n}{2} ⌋} γ_{n,i}\binom{n − 2i}{k − i}.

Example 3. The identity

\binom{n}{k} = ∑_{i=0}^{⌊ \frac{n}{2} ⌋} \binom{n − 2i}{k − i}δ_{i,0}

shows that the matrix that begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} )

is the γ-matrix for the binomial matrix B = (\binom{n}{k}) \underline{A007318}. Here, we have used the Annnnnn number of the On-Line Encyclopedia of Integer Sequences [13, 14] for the binomial matrix (Pascal’s triangle).

When (γ_{n,k}) is the γ-matrix for (h_{n,k}), we shall say the (γ_{n,k}) generates, or is the generator of, the matrix (h_{n,k}).

Example 4. The matrix that begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 & 0 \end{matrix} )

with γ_{n,0} = 1, γ_{n,⌊ \frac{n}{2} ⌋} = 1, and 0 otherwise, generates the matrix (h_{n,k}) that begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 3 & 1 & 0 & 0 & 0 \\ 1 & 3 & 3 & 1 & 0 & 0 \\ 1 & 4 & 7 & 4 & 1 & 0 \\ 1 & 5 & 10 & 10 & 5 & 1 \end{matrix} ) .

2 Pascal-like matrices defined by Riordan arrays

We now wish to characterize the γ-matrices that are generators for the family of Pascal-like matrices that are determined by the one-parameter family of Riordan arrays

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

We shall also determine the (generalized) γ-matrices associated with the reversion of these triangles. We recall that an ordinary Riordan array (g(x), f(x)) is defined [1, 10, 11] by two power series

g(x) = 1 + g_{1}x + g_{2}x^{2} + ⋯,
f(x) = x + f_{2}x^{2} + f_{3}x^{3} + ⋯, ,

where the (n, k)-th element of the resulting lower-triangular matrix is given by

a_{n,k} = [x^{n}]g(x)f(x)^{k}.

Such matrices are invertible. When they have integer entries, the inverse again is an integer matrix (note that we have a_{n,n} = 1 in our case because g_{0} = 1 and f_{1} = 1). The bivariate generating function of the Riordan array (g, f) is given by

\frac{g(x)}{1 − yf(x)}.

Matrices defined in a similar manner but with f(x) replaced by ϕ(x) = x^{2} + ϕ_{3}x^{3} + … are called “stretched” Riordan arrays [5]. They are not invertible but they do possess left inverses.

Example 5. The stretched Riordan array (\frac{1}{1−x}, x^{2}) begins

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

It is the γ-matrix for the Pascal-like triangle that begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 3 & 1 & 0 & 0 & 0 & 0 \\ 1 & 4 & 4 & 1 & 0 & 0 & 0 \\ 1 & 5 & 9 & 5 & 1 & 0 & 0 \\ 1 & 6 & 14 & 14 & 6 & 1 & 0 \\ 1 & 7 & 20 & 29 & 20 & 7 & 1 \end{matrix} ) .

Example 6. The matrix \binom{n−k}{k} is the stretched Riordan array (\frac{1}{1−x}, \frac{x^{2}}{1−x}) that begins

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

It generates the Pascal-like matrix that begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 3 & 1 & 0 & 0 & 0 & 0 \\ 1 & 5 & 5 & 1 & 0 & 0 & 0 \\ 1 & 7 & 13 & 7 & 1 & 0 & 0 \\ 1 & 9 & 25 & 25 & 9 & 1 & 0 \\ 1 & 11 & 41 & 63 & 41 & 11 & 1 \end{matrix} ) .

We shall see that this is the Riordan array (\frac{1}{1−x}, \frac{x(1+x)}{1−x}), which is \underline{A008288}, the triangle of Delannoy numbers.

The bivariate generating function of the stretched Riordan array (g(x), ϕ(x)) is given by

\frac{g(x)}{1 − yϕ(x)}.

We have the following proposition [4]. Proposition 7. The Riordan array (\frac{1}{1−x}, \frac{x(1+rx)}{1−x}) is Pascal-like (for any r ∈ ℤ).

This is clear since in this case we have

h_{n,k} = ∑_{j=0}^{k} \binom{k}{j}\binom{n − j}{n − k − j}r^{j} = ∑_{j=0}^{k} \binom{k}{j}\binom{n − k}{n − k − j}(r + 1)^{j}.

We can now characterize the γ-matrices that generate these Pascal-like matrices. Proposition 8. The γ-matrices that generate the Pascal-like matrices (\frac{1}{1−x}, \frac{x(1+rx)}{1−x}) defined by ordinary Riordan arrays are given by the stretched Riordan arrays

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

with (n, k)-th term

γ_{n,k} = \binom{n − k}{k}r^{k}.

Proof. The generating function of the Pascal-like matrix (\frac{1}{1−x}, \frac{x(1+rx)}{1−x}) is given by

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

Similarly, the generating function of the matrix (\binom{n−k}{k}r^{k}) is given by

γ(x, y) = \frac{1}{1 − x} \frac{1}{1 − y \frac{rx^{2}}{1−x}} = \frac{1}{1 − x − rx^{2}y} .

We now have

h(x, y) = γ ((1 + y)x, \frac{y}{(1 + y)^{2}} ) .

We recall that for a generating function f(x), its INVERT(α) transform is the generating function

\frac{f(x)}{1 + αxf(x)}.

Note that

\frac{\frac{v}{1+αxv}}{1 − αx\frac{v}{1+αxv}} = v,

and thus the inverse of the INVERT(α) transform is the INVERT(−α) transform.

Corollary 9. The generating function h(x, y) of the Pascal-like matrix (\frac{1}{1−x}, \frac{x(1+rx)}{1−x}) is the INVERT(y) transform of the generating function γ(x, y) of the corresponding γ-matrix.

Proof. A direct calculation shows that for γ(x, y) = \frac{1}{1−x−rx^{2}y} we have

\frac{γ(x, y)}{1 − yxγ(x, y)} = \frac{1}{1 − (y + 1)x − rx^{2}y} = h(x, y).

Equivalently, we can say that the generating function of the γ-matrix is the INVERT(−y) transform of the generating function of the corresponding Pascal-like matrix.

We make the following observation, which will be relevant when we discuss a family of generalized Narayana triangles. The γ-matrix corresponding to the signed Pascal-like matrix

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

has generating function

\frac{1}{1 + x + rx^{2}y} .

This is the matrix with general term (−1)^{n−k}r^{k}\binom{n−k}{k}. By a signed Pascal-like matrix in this case we mean that a_{n,k} = a_{n,n−k} but we now have a_{n,0} = a_{n,n} = (−1)^{n}.

We close this section by recalling the formula

γ_{n} = (1 + x)^{n}γ_{n} (\frac{x}{(1 + x)^{2}} ) .

We now note that the inverse of the Riordan array

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

is given by

(1, xc(x)^{2}) ,

where

c(x) = \frac{1 − \sqrt{1 − 4x}}{2x}

is the generating function of the Catalan numbers C_{n} = \frac{1}{n+1} \binom{2n}{n} \underline{A000108}. In fact, we have the following result [9].

Proposition 10. (Zeilberger’s Lemma).

γ_{n,k} = [x^{k}]\frac{h_{n}(xc(x)^{2})}{c(x)^{n}}.

We can use this result to find an explicit formula for γ_{n,k} in terms of h_{n,k}. We let α_{n,k} be the general (n, k)-th element of the Riordan array (1, xc(x)^{2}) [8]. We have

α_{n,k} = \begin{cases}1, & if n = 0 and k = 0; \\ \binom{2n−1}{n−k} \frac{2k}{n+k}, & otherwise; \end{cases}

or, equivalently,

α_{n,k} = \binom{2n − 1}{n − k}\frac{2k + 0^{n+k}}{n + k + 0^{n+k}} = \binom{2n − 2}{n − k} − \binom{2n − 2}{n − k − 2}.

We let β_{n,k} be the general (n, k)-th term of the Riordan array (1, \frac{x}{c(x)}). We have β_{n,n} = 1, and

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

otherwise. This is essentially \underline{A271875}. Then we have the following result.

Corollary 11. We have

γ_{n,k} = ∑_{i=0}^{k} (∑_{j=0}^{n} h_{n,j}α_{i,j}) β_{n+k−i,n}.

Proof. We have

[x^{k}][x^{k}]\frac{h_{n}(xc(x)^{2})}{c(x)^{n}} = ∑_{i=0}^{n}[x^{i}] ∑_{j=0}^{n} h_{n,j}(xc(x^{2}))^{j}[x^{k−i}]\frac{1}{c(x)^{n}}

= ∑_{i=0}^{k} (∑_{j=0}^{n} h_{n,j}[x^{i}](xc(x)^{2})^{j}) [x^{k−1+n}]\frac{x^{n}}{c(x)^{n}}

= ∑_{i=0}^{k} (∑_{j=0}^{n} h_{n,j}α_{i,j}) β_{n+k−i,n}.

This gives us the following formula:

γ_{n,k} = ∑_{i=0}^{k} ∑_{j=0}^{n} h_{n,j}\binom{2i − 1}{i − j}\frac{2j + 0^{i+j}}{i + j + 0^{i+j}} · \begin{cases}1, & if i = k; \\ ∑_{m=0}^{k−i} \frac{m(−1)^{m}}{k−i} \binom{n−1+m}{m}\binom{2(k−i)}{k−i−m}, & otherwise; \end{cases}

which we can also write as

γ_{n,k} = ∑_{i=0}^{k} ∑_{j=0}^{n} h_{n,j}\binom{2i − 1}{i − j}\frac{2j + 0^{i+j}}{i + j + 0^{i+j}} If [i = k, 1, ∑_{m=0}^{k−i} \frac{m(−1)^{m}}{k − i} \binom{n − 1 + m}{m}\binom{2(k − i)}{k − i − m}] .

Example 12. If we take (h_{n,k}) to be the triangle of Eulerian numbers \underline{A008292} that begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 4 & 1 & 0 & 0 & 0 & 0 \\ 1 & 11 & 11 & 1 & 0 & 0 & 0 \\ 1 & 26 & 66 & 26 & 1 & 0 & 0 \\ 1 & 57 & 302 & 302 & 57 & 1 & 0 \\ 1 & 120 & 1191 & 2416 & 1191 & 120 & 1 \end{matrix} )

we find that the γ-matrix (γ_{n,k}) is the triangle \underline{A101280} that begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 & 0 & 0 & 0 \\ 1 & 8 & 0 & 0 & 0 & 0 & 0 \\ 1 & 22 & 16 & 0 & 0 & 0 & 0 \\ 1 & 52 & 136 & 0 & 0 & 0 & 0 \\ 1 & 114 & 720 & 272 & 0 & 0 & 0 \end{matrix} ) .

This is the triangle of γ-vectors for the permutahedra (of type A). It also gives the number of permutations of n objects with k descents such that every descent is a peak [12]. Example 13. We consider the Pascal-like matrix (h_{n,k}) = (\frac{1}{1−x}, x) that begins

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

We note that the row elements are constant. We have that

γ_{n,k} = ∑_{i=0}^{k} ∑_{j=0}^{n} \binom{2i − 1}{i − j}\frac{2j + 0^{i+j}}{i + j + 0^{i+j}} If [i = k, 1, ∑_{m=0}^{k−i} \frac{m(−1)^{m}}{k − i} \binom{n − 1 + m}{m}\binom{2(k − i)}{k − i − m}] .

We find that the γ-matrix in this case begins

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

This is the matrix (\binom{n−k}{k}(−1)^{k}). Thus

∑_{i=0}^{⌊ \frac{n}{2} ⌋} \binom{n − 2i}{k − i}\binom{n − i}{i}(−1)^{i} = If[k ≤ n, 1, 0].

3 Stretched Riordan arrays as γ-matrices

Every stretched Riordan array of the form

(\frac{1}{1 − x}, x^{2}g(x)) ,

where

g(x) = 1 + g_{1}x + g_{2}x^{2} + ⋯

can be used to generate a Pascal-like matrix. Thus with each power series g(x) above we can associate a Pascal-like matrix whose γ-matrix is given by this stretched Riordan array.

In this section, we shall concentrate on the case when g(x) = \frac{1+rx}{1−x}.

Example 14. For r = 1, we obtain the γ-matrix that begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 3 & 0 & 0 & 0 & 0 & 0 \\ 1 & 5 & 1 & 0 & 0 & 0 & 0 \\ 1 & 7 & 5 & 0 & 0 & 0 & 0 \\ 1 & 9 & 13 & 1 & 0 & 0 & 0 \end{matrix} ) .

The corresponding Pascal-like matrix then begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 3 & 1 & 0 & 0 & 0 & 0 \\ 1 & 6 & 6 & 1 & 0 & 0 & 0 \\ 1 & 9 & 17 & 9 & 1 & 0 & 0 \\ 1 & 12 & 36 & 36 & 12 & 1 & 0 \\ 1 & 15 & 64 & 101 & 64 & 15 & 1 \end{matrix} ) .

The row sums of this matrix, which begin

1, 2, 5, 14, 37, 98, 261, …

give \underline{A077938}, with generating function

\frac{1}{1 − 2x − x^{2} − 2x^{3}} .

The diagonal sums, which begin

1, 1, 2, 4, 8, 16, 31, …

are the Pentanacci numbers \underline{A001591} with generating function

\frac{1}{1 − x − x^{2} − x^{3} − x^{4} − x^{5}} .

We have the following proposition. Proposition 15. The Pascal-like triangle that begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 3 & 1 & 0 & 0 & 0 & 0 \\ 1 & r + 5 & r + 5 & 1 & 0 & 0 & 0 \\ 1 & 2r + 7 & 4r + 13 & 2r + 7 & 1 & 0 & 0 \\ 1 & 3r + 9 & 11r + 25 & 11r + 25 & 3r + 9 & 1 & 0 \\ 1 & 4r + 11 & r^{2} + 22r + 41 & 2r^{2} + 36r + 63 & r^{2} + 22r + 41 & 4r + 11 & 1 \end{matrix} )

with γ-matrix given by the stretched Riordan array (\frac{1}{1−x}, \frac{x^{2}(1+rx)}{1−x}), has row sums with generating function

\frac{1}{1 − 2x − x^{2} − 2rx^{3}} ,

and diagonal sums given by the generalized Pentanacci numbers with generating function

\frac{1}{1 − x − x^{2} − x^{3} − rx^{4} − rx^{5}} .

4 Reverting triangles

Let h(x, y) be the generating function of the lower-triangular matrix h_{n,k}, with h_{0,0} = 1. By the reversion of this triangle, we shall mean the triangle whose generating function h^{∗}(x, y) is given by

h^{∗}(x, y) = \frac{1}{x}Rev_{x}(xh(x, y)).

Procedurally, this means that we solve the equation

uh(u, y) = x

and then we divide the solution u(x, y) that satisfies u(0, y) = 0 by x.

Proposition 16. The generating function of the reversion of the Pascal-like matrix defined by the Riordan array (\frac{1}{1−x}, \frac{x(1+rx)}{1−x}) is given by

h^{∗}(x, y) = \frac{1}{1 + x(y + 1)}c (\frac{−rx^{2}y}{(1 + x(y + 1))^{2}} ) ,

where

c(x) = \frac{1 − \sqrt{1 − 4x}}{2x}

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

Proof. Solving the equation

\frac{u}{1 − u(y + 1) − ru^{2}y} = x

gives us

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

Thus

h^{∗}(x, y) = \frac{1}{1 + x(y + 1)}c (\frac{−rx^{2}y}{(1 + x(y + 1))^{2}} ) .

We note that we can now calculate an expression for the terms of the reverted triangle, since, using the language of Riordan arrays, we have

h^{∗}(x, y) = (\frac{1}{1 + y(x + 1)}, \frac{−rx^{2}y}{(1 + x(y + 1))^{2}} ) · c(x).

Proposition 17. We have

[x^{n}][y^{i}]h^{∗}(x, y) = h_{n,i}^{∗} = ∑_{k=0}^{⌊ \frac{n}{2} ⌋}(−1)^{n}(−r)^{k}\binom{n}{2k}C_{k}\binom{n − 2k}{i − k}.

The γ-matrix of the reverted triangle (h_{n,k}^{∗}) is given by

γ_{n,k}^{∗} = (−1)^{n}(−r)^{k}\binom{n}{2k}C_{k}.

The γ-matrix (γ_{n,k}^{∗}) of the reverted triangle (h_{n,k}^{∗}) is the reversion of the triangle γ_{n,k}. Proof. The expression for h_{n,k}^{∗} results from a direct calculation. Reverting the expression γ(x, y) = \frac{1}{1−x−rx^{2}y} in the sense above gives us

γ^{∗}(x, y) = \frac{1}{1 + x}c (\frac{−rx^{2}y}{(1 + x)^{2}} ) ,

from which we deduce the other statements. Example 18. For r = −1, 0, 1, the triangles (h_{n,k}) begin, respectively,

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 \end{matrix} ) , ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 2 & 1 & 0 & 0 & 0 & 0 \\ 1 & 3 & 3 & 1 & 0 & 0 & 0 \\ 1 & 4 & 6 & 4 & 1 & 0 & 0 \\ 1 & 5 & 10 & 10 & 5 & 1 & 0 \\ 1 & 6 & 15 & 20 & 15 & 6 & 1 \end{matrix} ) ,

and

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 3 & 1 & 0 & 0 & 0 & 0 \\ 1 & 5 & 5 & 1 & 0 & 0 & 0 \\ 1 & 7 & 13 & 7 & 1 & 0 & 0 \\ 1 & 9 & 25 & 25 & 9 & 1 & 0 \\ 1 & 11 & 41 & 63 & 41 & 11 & 1 \end{matrix} ) .

The corresponding reverted triangles (h_{n,k}^{∗}), are, respectively,

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ −1 & −1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 3 & 1 & 0 & 0 & 0 & 0 \\ −1 & −6 & −6 & −1 & 0 & 0 & 0 \\ 1 & 10 & 20 & 10 & 1 & 0 & 0 \\ −1 & −15 & −50 & −50 & −15 & −1 & 0 \\ 1 & 21 & 105 & 175 & 105 & 21 & 1 \end{matrix} ) , ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ −1 & −1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 2 & 1 & 0 & 0 & 0 & 0 \\ −1 & −3 & −3 & −1 & 0 & 0 & 0 \\ 1 & 4 & 6 & 4 & 1 & 0 & 0 \\ −1 & −5 & −10 & −10 & −5 & −1 & 0 \\ 1 & 6 & 15 & 20 & 15 & 6 & 1 \end{matrix} ) ,

and

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ −1 & −1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ −1 & 0 & 0 & −1 & 0 & 0 & 0 \\ 1 & −2 & −4 & −2 & 1 & 0 & 0 \\ −1 & 5 & 10 & 10 & 5 & −1 & 0 \\ 1 & −9 & −15 & −15 & −15 & −9 & 1 \end{matrix} ) .

Note that for r = −1, the reverted triangle is (−1)^{n} times the Narayana triangle \underline{A001263}. The corresponding γ-matrices (γ_{n,k}) are given by, respectively,

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & −1 & 0 & 0 & 0 & 0 & 0 \\ 1 & −2 & 0 & 0 & 0 & 0 & 0 \\ 1 & −3 & 1 & 0 & 0 & 0 & 0 \\ 1 & −4 & 3 & 0 & 0 & 0 & 0 \\ 1 & −5 & 6 & −1 & 0 & 0 & 0 \end{matrix} ) , ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ) ,

and

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

The corresponding reverted γ-matrices (γ_{n,k}^{∗}) are then, respectively,

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ −1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ −1 & −3 & 0 & 0 & 0 & 0 & 0 \\ 1 & 6 & 2 & 0 & 0 & 0 & 0 \\ −1 & −10 & −10 & 0 & 0 & 0 & 0 \\ 1 & 15 & 30 & 5 & 0 & 0 & 0 \end{matrix} ) , ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ −1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ −1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ −1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ) ,

and

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ −1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & −1 & 0 & 0 & 0 & 0 & 0 \\ −1 & 3 & 0 & 0 & 0 & 0 & 0 \\ 1 & −6 & 2 & 0 & 0 & 0 & 0 \\ −1 & 10 & −10 & 0 & 0 & 0 & 0 \\ 1 & −15 & 30 & −5 & 0 & 0 & 0 \end{matrix} ) .

It is interesting to represent the generating functions of the (γ_{n,k}^{∗}) and the (h_{n,k}^{∗}) triangles as Jacobi continued fractions. We have

Proposition 19. The generating function h^{∗}(x, y) can be expressed as the Jacobi continued fraction

J (−(y + 1), −(y + 1), −(y + 1), …; −ry, −ry, −ry, …).

The generating function γ^{∗}(x, y) can be expressed as the Jacobi continued fraction

J (−1, −1, −1, …; −ry, −ry, −ry, …).

Proof. We solve the continued fraction equation

u = \frac{1}{1 + (y + 1)x + rx^{2}u}

to retrieve the generating function h^{∗}(x, y). Similarly, we solve the continued fraction equation

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

to retrieve the generating function γ^{∗}(x, y).

Note that we have used the notation J (a, b, c, …; r, s, t, …) to denote the Jacobi continued fraction [2, 16]

\frac{1}{1 − ax − \frac{rx^{2}}{1 − bx − \frac{sx^{2}}{1 − cx − \frac{tx^{2}}{1 − ⋯}}}}.

We can now express the relationship between the generating functions h^{∗}(x, y) and γ^{∗}(x, y) in terms of repeated binomial transforms.

Corollary 20. The generating function h^{∗}(x, y) is the (−y)-th binomial transform of the γ generating function γ^{∗}(x, y):

h^{∗}(x, y) = \frac{1}{1 + xy} γ^{∗} (\frac{x}{1 + xy} , y) .

Equivalently, the γ generating function γ^{∗}(x, y) is the y-th binomial transform of the generating function h^{∗}(x, y):

γ^{∗}(x, y) = \frac{1}{1 − xy} h^{∗} (\frac{x}{1 − xy} , y) .

This reflects the general assertion that the reversion of an INVERT transform is a binomial transform.

5 The γ-vectors of generalized Narayana numbers

The Riordan array (\frac{1}{1+x}, \frac{−x(1+rx)}{1+x}), with bivariate generating function

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

has a γ-matrix with generating function

\frac{1}{1 + x + rx^{2}y} .

We shall call elements of the reversions of the Riordan array (\frac{1}{1+x}, \frac{−x(1+rx)}{1+x}) r-Narayana

numbers. The Narayana numbers N_{n,k} = \frac{1}{k+1} \binom{n+1}{k}\binom{n}{k} are then the 1-Narayana numbers. The bivariate generating function for the r-Narayana numbers is given by

\frac{1}{1 − x(y + 1)}c (\frac{rx^{2}y}{(1 − x(y + 1))^{2}} ) .

The bivariate generating function for the γ-matrix of the r-Narayana numbers is then obtained by reverting the generating function \frac{1}{1+x+rx^{2}y}. We thus obtain the following result. Proposition 21. The γ-matrix for the r-Narayana numbers has generating function

\frac{1}{1 − x}c (\frac{rx^{2}y}{(1 − x)^{2}} ) .

This is the matrix that begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & r & 0 & 0 & 0 & 0 & 0 \\ 1 & 3r & 0 & 0 & 0 & 0 & 0 \\ 1 & 6r & 2r^{2} & 0 & 0 & 0 & 0 \\ 1 & 10r & 10r^{2} & 0 & 0 & 0 & 0 \\ 1 & 15r & 30r^{2} & 5r^{3} & 0 & 0 & 0 \end{matrix} ) ,

with general term

\binom{n}{2k}r^{k}C_{k}.

For r = −1, 0, 1, the matrices (\frac{1}{1+x}, \frac{−x(1+rx)}{1+x}) begin, respectively,

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ −1 & −1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 3 & 1 & 0 & 0 & 0 & 0 \\ −1 & −5 & −5 & −1 & 0 & 0 & 0 \\ 1 & 7 & 13 & 7 & 1 & 0 & 0 \\ −1 & −9 & −25 & −25 & −9 & −1 & 0 \\ 1 & 11 & 41 & 63 & 41 & 11 & 1 \end{matrix} ) , ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ −1 & −1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 2 & 1 & 0 & 0 & 0 & 0 \\ −1 & −3 & −3 & −1 & 0 & 0 & 0 \\ 1 & 4 & 6 & 4 & 1 & 0 & 0 \\ −1 & −5 & −10 & −10 & −5 & −1 & 0 \\ 1 & 6 & 15 & 20 & 15 & 6 & 1 \end{matrix} ) ,

and

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

The corresponding matrices of r-Narayana numbers are, respectively,

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & −2 & −4 & −2 & 1 & 0 & 0 \\ 1 & −5 & −10 & −10 & −5 & 1 & 0 \\ 1 & −9 & −15 & −15 & −15 & −9 & 1 \end{matrix} ) , ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 2 & 1 & 0 & 0 & 0 & 0 \\ 1 & 3 & 3 & 1 & 0 & 0 & 0 \\ 1 & 4 & 6 & 4 & 1 & 0 & 0 \\ 1 & 5 & 10 & 10 & 5 & 1 & 0 \\ 1 & 6 & 15 & 20 & 15 & 6 & 1 \end{matrix} ) ,

and

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 3 & 1 & 0 & 0 & 0 & 0 \\ 1 & 6 & 6 & 1 & 0 & 0 & 0 \\ 1 & 10 & 20 & 10 & 1 & 0 & 0 \\ 1 & 15 & 50 & 50 & 15 & 1 & 0 \\ 1 & 21 & 105 & 175 & 105 & 21 & 1 \end{matrix} ) .

This last matrix, as expected, is the Narayana triangle \underline{A001263}. The corresponding γmatrices for these r-Narayana triangles are, respectively,

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & −1 & 0 & 0 & 0 & 0 & 0 \\ 1 & −3 & 0 & 0 & 0 & 0 & 0 \\ 1 & −6 & 2 & 0 & 0 & 0 & 0 \\ 1 & −10 & 10 & 0 & 0 & 0 & 0 \\ 1 & −15 & 30 & −5 & 0 & 0 & 0 \end{matrix} ) , ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ) ,

and

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 3 & 0 & 0 & 0 & 0 & 0 \\ 1 & 6 & 2 & 0 & 0 & 0 & 0 \\ 1 & 10 & 10 & 0 & 0 & 0 & 0 \\ 1 & 15 & 30 & 5 & 0 & 0 & 0 \end{matrix} ) .

This last matrix is \underline{A055151}. The rows of this triangle are the γ-vectors of the n-dimensional (type A) associahedra [9]. We have seen that its elements are given by

γ_{n,k} = ∑_{i=0}^{k} ∑_{j=0}^{n} N_{n,j}\binom{2i − 1}{i − j}\frac{2j + 0^{i+j}}{i + j + 0^{i+j}} If (k = i, 1, ∑_{m=0}^{k−i} \frac{m(−1)^{m}}{k − i} \binom{n − 1 + m}{m}\binom{2(k − i)}{k − i − m}) ,

where N_{n,k} denotes the (n, k)-th Narayana number \underline{A001263}.

The relationship between the γ-matrix and the r-Narayana numbers can be further clarified as follows.

Proposition 22. The generating function of the r-Narayana numbers can be expressed as the Jacobi continued fraction

J ((y + 1), (y + 1), (y + 1), …; ry, ry, ry, …).

The generating function of the corresponding γ-matrix can be expressed as the Jacobi continued fraction

J (1, 1, 1, …; ry, ry, ry, …).

Corollary 23. The generating function of the r-Narayana numbers is the y-th binomial transform of the generating function of the corresponding γ-matrix.

h^{∗}(x, y) = \frac{1}{1 − xy} γ^{∗} (\frac{x}{1 − xy} , y) .

Equivalently, the γ generating function γ^{∗}(x, y) is the (−y)-th binomial transform of the generating function h^{∗}(x, y):

γ^{∗}(x, y) = \frac{1}{1 + xy} h^{∗} (\frac{x}{1 + xy} , y) .

6 Pascal-like triangles defined by exponential Riordan arrays

We recall that an exponential Riordan array [g(x), f(x)] [1, 6] is defined by two exponential generating functions

g(x) = 1 + g_{1} \frac{x}{1!} + g_{2} \frac{x}{2!} + ⋯,

and

f(x) = \frac{x}{1!} + f_{2} \frac{x^{2}}{2!} + ⋯,

with its (n, k)-th term a_{n,k} given by

a_{n,k} = \frac{n!}{k!}[x^{n}]g(x)f(x)^{k}.

In the context of Pascal-like matrices, we have that the exponential Riordan array

[e^{x}, x(1 + rx/2)] ,

with general term

h_{n,k} = \frac{n!}{k!} ∑_{j=0}^{k} \frac{r^{j}}{(n − k − j)!2^{j}} ,

is a Pascal-like matrix [3]. This matrix begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & r + 2 & 1 & 0 & 0 & 0 & 0 \\ 1 & 3r + 3 & 3r + 3 & 1 & 0 & 0 & 0 \\ 1 & 6r + 4 & 3r^{2} + 12r + 6 & 6r + 4 & 1 & 0 & 0 \\ 1 & 10r + 5 & 15r^{2} + 30r + 10 & 15r^{2} + 30r + 10 & 10r + 5 & 1 & 0 \\ 1 & 15r + 6 & 45r^{2} + 60r + 15 & 15r^{3} + 90r^{2} + 90r + 20 & 45r^{2} + 60r + 15 & 15r + 6 & 1 \end{matrix} ) .

We have the following result.

Proposition 24. The γ-matrix of the Pascal-like exponential Riordan array [e^{x}, x(1 + rx/2)] is the matrix with general term

\binom{n}{2k}r^{k}(2k − 1)!!

In fact, the generating function of the exponential Riordan array [e^{x}, x(1 + rx/2)] is given by

J (y + 1, y + 1, y + 1, …; ry, 2ry, 3ry, …)

while that of its γ-matrix is given by

J (1, 1, 1, …; ry, 2ry, 3ry, …).

Proposition 25. The generating function of the γ-matrix of the Pascal-like exponential Riordan array [e^{x}, x(1 + rx/2)] has generating function

e^{x(1+rxy/2)}.

Proof. By the theory of exponential Riordan arrays, the generating function of the Riordan array [e^{x}, x(1 + rx/2)] is given by

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

Taking the (−y)-th binomial transform of this, we obtain

e^{x(1+rxy/2)}.

Example 26. For r = 1, we get the γ-matrix that begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 3 & 0 & 0 & 0 & 0 & 0 \\ 1 & 6 & 3 & 0 & 0 & 0 & 0 \\ 1 & 10 & 15 & 0 & 0 & 0 & 0 \\ 1 & 15 & 45 & 15 & 0 & 0 & 0 \end{matrix} ) .

This is \underline{A100861}, the triangle of Bessel numbers that count the number of k-matchings of the complete graph K(n). The corresponding Pascal-like matrix begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 3 & 1 & 0 & 0 & 0 & 0 \\ 1 & 6 & 6 & 1 & 0 & 0 & 0 \\ 1 & 10 & 21 & 10 & 1 & 0 & 0 \\ 1 & 15 & 55 & 55 & 15 & 1 & 0 \\ 1 & 21 & 120 & 215 & 120 & 21 & 1 \end{matrix} ) .

This is \underline{A100862}, which counts the number of k-matchings of the corona K′(n) of the complete graph K(n) and the complete graph K(1).

Example 27. For r = 2, we obtain the γ-matrix that begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 & 0 & 0 & 0 \\ 1 & 6 & 0 & 0 & 0 & 0 & 0 \\ 1 & 12 & 12 & 0 & 0 & 0 & 0 \\ 1 & 20 & 60 & 0 & 0 & 0 & 0 \\ 1 & 30 & 180 & 120 & 0 & 0 & 0 \end{matrix} ) .

This is \underline{A059344}, where row n consists of the nonzero coefficients of the expansion of 2^{n}x^{n} in terms of Hermite polynomials with decreasing subscripts. The corresponding Pascal-like matrix begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 4 & 1 & 0 & 0 & 0 & 0 \\ 1 & 9 & 9 & 1 & 0 & 0 & 0 \\ 1 & 16 & 42 & 16 & 1 & 0 & 0 \\ 1 & 25 & 130 & 130 & 25 & 1 & 0 \\ 1 & 36 & 315 & 680 & 315 & 36 & 1 \end{matrix} ) .

The row sums of this matrix are given by \underline{A000898}, the number of symmetric involutions of [2n] (Deutsch).

7 Conclusion

It is the case that the set of Pascal-like matrices defined by Riordan arrays is a restricted one. Nevertheless, we hope that this note indicates that they have interesting properties, including in particular their generating γ-matrices. In the case of Pascal-like matrices defined by ordinary Riordan arrays, we have seen that be reverting them, we find additional (signed) Pascal-like triangles, including triangles of Narayana type. The γ-matrices of these new triangles are again the reversions of the original triangles’ γ-matrices.

We have also shown that stretched Riordan arrays play a useful role, and in particular can lead to further (non-Riordan) Pascal-like matrices. We have also found it useful to use Riordan array techniques to find an explicit closed form formula for the elements γ_{n,k} of the γ-matrix in terms of h_{n,k}.

References

[1] P. Barry, Riordan Arrays: A Primer, Logic Press, 2017.

[2] P. Barry, Continued fractions and transformations of integer sequences, J. Integer Sequences, 12 (2009), Article 09.7.6.

[3] P. Barry, On a family of generalized Pascal triangles defined by exponential Riordan array, J. Integer Sequences, 10 (2007), Article 07.3.5.

[4] P. Barry, On integer-sequence-based constructions of generalized Pascal triangles, J. Integer Sequences, 9 (2006), Article 06.2.4.

[5] C. Corsani, D. Merlini, and R. Sprugnoli, Left-inversion of combinatorial sums, Discrete Math., 180 (1998), 107–122.

[6] E. Deutsch and L. Shapiro, Exponential Riordan arrays, Lecture Notes, Nankai Univer-

    sity, 2004, available electronically at http://www.combinatorics.net/ppt2004/Louis
    %20W.%20Shapiro/shapiro.htm.

[7] S. R. Gal, Real root conjecture fails for five and higher dimensional spheres, Discrete Comput. Geom., 34 (2005), 269–284.

[8] Tian-Xiao He and L. W. Shapiro, Fuss-Catalan matrices, their weighted sums, and stabilizer subgroups of the Riordan group, Linear Algebra Appl., 552 (2017), 25–42.

[9] K. Petersen, Eulerian Numbers, Birkhäuser, 2015.

[10] L. Shapiro, A survey of the Riordan group, available electronically at http://www.

    combinatorics.cn/activities/Riordan%20Group.pdf. Center for Combinatorics,

Nankai University, 2018.

[11] L. W. Shapiro, S. Getu, W.-J. Woan, and L. C. Woodson, The Riordan group, Discr. Appl. Math. 34 (1991), 229–239.

[12] L. W. Shapiro, W.-J. Woan, and S. Getu, Runs, slides and moments, SIAM J. Alg. Discrete Methods, 4 (1983), 459–466.

[13] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences. Published electroni-

    cally at https://oeis.org, 2018.

[14] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, Notices Amer. Math. Soc., 50 (2003), 912–915.

[15] R. P. Stanley, f-vectors and h-vectors of simplicial posets, J. Pure Appl. Algebra, 71 (1991), 319–331.

[16] H. S. Wall, Analytic Theory of Continued Fractions, AMS Chelsea Publishing, 2001.

2010 Mathematics Subject Classification: Primary 11B83; Secondary 33C45, 42C-5, 15B36, 15B05, 14N10, 11C20. Keywords: Gamma vector, Pascal-like triangle, Riordan array, Narayana number, Eulerian number, associahedron, permutahedron.

(Concerned with sequences \underline{A000108}, \underline{A000898}, \underline{A001263}, \underline{A001591}, \underline{A007318}, \underline{A008288}, \underline{A008292}, \underline{A055151}, \underline{A059344}, \underline{A077938}, \underline{A100861}, \underline{A100862}, \underline{A101280}, and \underline{A271875}.)

Received April 13 2018; revised version received December 19 2018. Published in Journal of Integer Sequences, December 19 2018.

Return to Journal of Integer Sequences home page.