The Vertical Recursive Relation of Riordan Arrays and Its Matrix Representation

Tian-Xiao He Department of Mathematics Illinois Wesleyan University Bloomington, IL 61702-2900 USA

                      the@iwu.edu

Abstract

It is known that the entries of a Riordan array satisfy horizontal recursive relations represented by the A- and Z-sequences. In this paper, we study a vertical recursive relation approach to Riordan arrays. This vertical recursive approach gives a way to represent the entries of a Riordan array (g, f) in terms of a recursive linear combination of the coefficients of g. We also give a matrix representation of the vertical recursive relation. The set of all those matrices forms a group, called the quasi-Riordan group. We present extensions of the horizontal recursive relation and the vertical recursive relation in terms of c- and C-Riordan arrays, with illustrations by using the rook triangle and the Laguerre triangle. These extensions represent a way to study nonlinear recursive relations of the entries of some triangular matrices from linear recursive relations of the entries of Riordan arrays. In addition, the matrix representation of the vertical recursive relation of Riordan arrays provides transforms between lower order and higher order finite Riordan arrays, where the mth order Riordan array is defined by (g, f)_{m} = (d_{n,k})_{m≥n,k≥0}. Furthermore, the vertical relation approach to Riordan arrays provides a unified approach to construct identities.

1 Introduction

Riordan matrices are infinite, lower triangular matrices defined by the generating function of their columns. With matrix multiplication, they form a group, called the Riordan group (see Shapiro, Getu, Woan and Woodson [29]).

More formally, let us consider the set of formal power series ring F = K[[t]], where 𝕂 is the field of ℝ or ℂ. The order of f(t) ∈ F, f(t) = ∑_{k=0}^{∞} f_{k}t^{k} (f_{k} ∈ 𝕂), is the minimum number r ∈ ℕ_{0} such that f_{r} ≠ 0, where ℕ_{0} = ℕ ∪ {0} and ℕ is the positive integer set. We let F_{r} denote the set of formal power series of order r. Let g(t) ∈ F_{0} and f(t) ∈ F_{1}; the pair (g, f) defines the (proper) Riordan matrix D = (d_{n,k})_{n,k∈ℕ_{0}} = (g, f) where

d_{n,k} = [t^{n}]g(t)f(t)^{k}  (1)

or, in other words, having gf^{k} as the generating function of the kth column of (g, f). The first fundamental theorem of Riordan matrices concerns the action of the proper Riordan matrices on the formal power series presented by

(g(t), f(t))h(t) = g(t)(h ∘ f)(t),

which can be abbreviated as (g, f)h = gh(f). Thus we immediately see that the usual row-by-column product of two Riordan matrices is also a Riordan matrix:

(g_{1}, f_{1})(g_{2}, f_{2}) = (g_{1}g_{2}(f_{1}), f_{2}(f_{1})).  (2)

The Riordan matrix I = (1, t) is the identity matrix because its entries are d_{n,k} = [t^{n}]t^{k} = δ_{n,k}.

Let (g (t) , f(t)) be a Riordan matrix. Then its inverse is

(g (t) , f(t))^{−1} = (\frac{1}{g(\overline{f}(t))}, \overline{f}(t), ) ,  (3)

where \overline{f}(t) is the compositional inverse of f(t), i.e., (f ∘ \overline{f})(t) = (\overline{f} ∘ f)(t) = t. In this way, the set R of all proper Riordan matrices forms a group (see [29]) called the Riordan group.

Here is a list of six important subgroups of the Riordan group (see [29, 31, 1, 30]):

• the Appell subgroup {(g(z), z)}.

• the Lagrange (associated) subgroup {(1, f(z))}.

• the k-Bell subgroup {(g(z), z(g(z))^{k})}, where k is a fixed positive integer.

• the hitting-time subgroup {(zf′(z)/f(z), f(z))}.

• the derivative subgroup {(f′(z), f(z))}.

• the checkerboard subgroup {(g(z), f(z))}, where g is an even function and f is an odd function.

The 1-Bell subgroup is referred to as the Bell subgroup for short, and the Appell subgroup can be considered as the 0-Bell subgroup if we allow k = 0 to be included in the definition of the k-Bell subgroup.

Let G be a group, and let H and N be two subgroups of G with N normal. Then the following statements are equivalent:

• NH = G and N ∩ H = e.

• Every g ∈ G can be written uniquely as g = nh, where n ∈ N and h ∈ H.

• Define ψ : H → G/N by ψ(h) = Nh, h ∈ H. Then ψ is an isomorphism.

If these conditions hold, we write G = N ⋊ H and say that G is expressed as a semidirect product of N and H. Since for every (g, f) ∈ R we have

(g, f) = (g, t)(1, f),

where (g, t) ∈ A and (1, f) ∈ L and A and L are the Appell subgroup, the normal subgroup of R, and the Lagrange subgroup, respectively.

An infinite lower triangular matrix [d_{n,k}]_{n,k∈ℕ_{0}} is a Riordan matrix if and only if a unique sequence A = (a_{0} ≠ 0, a_{1}, a_{2}, …) exists such that for every n, k ∈ ℕ_{0} we have

d_{n+1,k+1} = a_{0}d_{n,k} + a_{1}d_{n,k+1} + ⋯ + a_{n−k}d_{n,n}.  (4)

This is equivalent to

f(t) = tA(f(t)) or t = f̄(t)A(t).  (5)

Here A(t) is the generating function of the A-sequence. The first formula of (5) is also called the second fundamental theorem of Riordan matrices. Moreover, there exists a unique sequence Z = (z_{0}, z_{1}, z_{2}, …) such that every element in column 0 can be expressed as the linear combination

d_{n+1,0} = z_{0}d_{n,0} + z_{1}d_{n,1} + ⋯ + z_{n}d_{n,n},  (6)

or equivalently,

g(t) = \frac{1}{1 − tZ(f(t))},  (7)

in which and throughout we always assume g(0) = g_{0} = 1, a usual hypothesis for proper Riordan matrices with normalization. From (7), we may obtain the equation.

Z(t) = \frac{g(f̄(t)) − 1}{f̄(t)g(f̄(t))}.

A- and Z-sequence characterizations of Riordan matrices were introduced, developed, and/or studied in Merlini, Rogers, Sprugnoli, and Verri [24], Roger [28], Sprugnoli and the author [17], Cheon and Jin [5], Cheon, Luzón, Morón, Prieto Martinez, and Song [7], [13], Jean-Louis and Nkwanta [19], Luzón, Morón, and Prieto-Martinez [21, 22], etc. In [17] the expressions of the A- and Z-sequences of the product depend on the analogous sequences of the two given factors.

The Catalan numbers, C_{n} = \binom{2n}{n}/(n + 1), are a sequence of integers that occur in many counting situations. The books Enumerative Combinatorics, Volume 2 [33] and the more recent Catalan Numbers by Richard Stanley [34] give a wealth of information about them.

The generating function of the Catalan numbers is denoted by C(z), which can be written as

C(t) = ∑_{n=0}^{∞} C_{n}t^{n} = \frac{1 − \sqrt{1 − 4t}}{2t},

which is equivalent to the functional equation C(t) = 1 + tC(t)^{2}. It can be shown that [11]

C(t)^{k} = ∑_{n=0}^{∞} \frac{k}{2n + k} \binom{2n + k}{n}t^{n}.

A Dyck path of length 2n is a path in the plane lattice ℤ × ℤ from the origin (0, 0) to (2n, 0), made up with steps (1, 1) and (1, −1). The other requirement is that a path can never go below the x -axis. We refer to n as the semilength of the path. It is well known that the number of Dyck paths of semilength n is the n-th Catalan number C_{n}. A partial Dyck path is a Dyck path without requiring that the end point be on the x-axis. Hence we have that

C(n, k) := [t^{n}]C(t)^{k} = \frac{k}{2n + k} \binom{2n + k}{n}  (8)

is the number of the partial Dyck paths from (0, 0) to (2n + k, k).

For a positive integer m, an m-Dyck path is a path from the origin to (mn, 0) using the steps (1, 1) and (1, 1 − m) and again not going below the x-axis. We refer to mn as the length of the path. A partial m-Dyck path is defined as an m-partial Dyck path. It is well known that the number of m-Dyck paths of length mn is (see, for example, [11])

F_{m}(n, 1) = \frac{1}{mn + 1}\binom{mn + 1}{n},

the Fuss-Catalan numbers. For m = 2, the Fuss-Catalan numbers are the Catalan numbers F_{2}(n, 1). More generally, the Fuss-Catalan numbers are

F_{m}(n, r) := \frac{r}{mn + r}\binom{mn + r}{n},  (9)

which are named after N. I. Fuss and E. C. Catalan (see [9, 11, 20, 25, 26, 12, 15, 18]). The Fuss-Catalan numbers have many combinatorial applications (see, for example, Shapiro and the author [18]).

The generating function F_{m}(t) for the Fuss-Catalan numbers, {F_{m}(n, 1)}_{n≥0} is called the generalized binomial series in [11], and it satisfies the function equation F_{m}(t) = 1+tF_{m}(t)^{m}. Hence from Lambert’s formula for the Taylor expansion of the powers of F_{m}(t) (see [11]), we have that

F_{m}^{r} ≡ F_{m}(t)^{r} = ∑_{n≥0} \frac{r}{mn + r}\binom{mn + r}{n}t^{n}  (10)

for all r ∈ ℝ. The key case(10) leads to the following formula for F_{m}(t):

F_{m}(t) = 1 + tF_{m}^{m}(t).  (11)

Actually,

1 + tF_{m}^{m}(t) = 1 + ∑_{n≥0} \frac{m}{mn + m}\binom{mn + m}{n}t^{n+1}

= 1 + ∑_{n≥1} \frac{m}{mn}\binom{mn}{n − 1}t^{n}

= ∑_{n≥0} \frac{1}{mn + 1}\binom{mn + 1}{n}t^{n} = F_{m}(t).

For the cases m = 1 and 2, we have F_{1} = 1/(1 − t) and F_{2} = C(z), respectively. When m = 3, the Fuss-Catalan numbers (F_{3})_{n} form the sequence \underline{A001764} (see [32]),

1, 1, 3, 12, 55, 273, 1428, …,

the so-called ternary numbers. The ternary numbers count the number of 3-Dyck paths or ternary paths. The generating function of the ternary numbers is denoted by T (t) = ∑_{n=0}^{∞} T_{n}t^{n} with T_{n} = \frac{1}{3n+1} \binom{3n+1}{n}, and is given equivalently by the equation

T (t) = 1 + tT (t)^{3}.

2 A vertical recursive relation view of Riordan arrays

In a Riordan array (g, f) = (d_{n,k})_{n,k≥0}, the first column (0th column), with its generating function g(t), usually possesses an interesting combinatorial interpretation or represents an important sequence, while the other columns might be considered as the compositions of the first column in the view of the following recurrence relation:

d_{n,k} = [t^{n}]gf^{k} = ∑_{j=1}^{n} f_{j}[t^{n−j}]gf^{k−1} = ∑_{j=1}^{n−k+1} f_{j}d_{n−j,k−1}  (12)

for n, k ≥ 1, where f(t) = ∑_{j≥1} f_{j}t^{j}. For instance,

d_{1,1} = f_{1}d_{0,0},
d_{2.1} = f_{2}d_{0,0} + f_{1}d_{1,0}, d_{2,2} = f_{1}d_{1,1},
d_{3,1} = f_{3}d_{0,0} + f_{2}d_{1,0} + f_{1}d_{2.0}, d_{3,2} = f_{2}d_{2,1} + f_{1}d_{1,1}, d_{3,3} = f_{1}d_{2,2}, …

Example 1. For the Pascal triangle (1/(1−t), t/(1−t)), from (12) we obtain the well-known recursive relation

\binom{n}{k} = ∑_{j=1}^{n−k+1} \binom{n − j}{k − 1}.  (13)

Recurrence relation (12) provides a resource for constructing identities and an algorithm for computing powers and multiplications of formal power series. The latter approach may simplify the corresponding computation process by using Faà di Bruno’s formula.

Theorem 2. Let (g, f) = (d_{n,k})_{n,k≥0} be a Riordan array, and let C(n, k) and F_{m}(n, r) be defined by (8) and (9), respectively. Then we obtain the recurrence relation (12). In particular, for g = F_{m} we have

\frac{k + 1}{m(n − k) + k + 1}\binom{m(n − k) + k + 1}{n − k}

= ∑_{j=0}^{n−k} \frac{k}{(mj + 1)(m(n − j − k) + k)}\binom{mj + 1}{j}\binom{m(n − j − k) + k}{n − k − j}.  (14)

If m = 2, then g = F_{2} = C and (14) becomes

\frac{k + 1}{2n − k + 1}\binom{2n − k + 1}{n − k}

= ∑_{j=0}^{n−k} \frac{k}{(2j + 1)(2(n − j − k) + k)}\binom{2j + 1}{j}\binom{2(n − j − k) + k}{n − k − j},  (15)

or equivalently,

C(n − k, k + 1) = ∑_{j=0}^{n−k} C(j, 1)C(n − j − k, k),  (16)

where C(m, ℓ) are defined by (8), and C(j, 1) = [t^{j}]C(t), the jth Catalan number.

Theorem 3. Let (g, f) = (d_{n,k})_{n,k≥0} be a Riordan array. Then its entries d_{n,k}, n, k ≥ 0, can be represented recursively by

d_{n,k} = ∑_{i_{k}=1}^{n−(k−1)} f_{i_{k}} ∑_{i_{k−1}=1}^{n−(k−2)−i_{k}} f_{i_{k−1}} ∑_{i_{k−2}=1}^{n−(k−3)−i_{k}−i_{k−1}} f_{i_{k−2}} ⋯

∑_{i_{1}=1}^{n−i_{k}−i_{k−1}−i_{2}} f_{i_{1}}g_{n−i_{1}−i_{2}−···−i_{k}}.  (17)

Let A and B be m × m and n × n matrices, respectively. Then we define the direct sum of A and B by

A ⊕ B = [\begin{matrix}A & 0 \\ 0 & B \end{matrix}]_{(m+n)×(m+n)} .  (18)

Definition 4. Let g ∈ F_{0} with g(0) = 1 and f ∈ F_{1}. We let [g, f] denote the following array, called a quasi-Riordan array.

[g, f] := (g, f, tf, t^{2}f, …),  (19)

where g, f, tf, t^{2}f ⋯ are the generating functions of the 0th, 1st, 2nd, 3rd, ⋯, columns of the matrix [g, f], respectively. It is clear that [g, f] can be written as

[g, f] = (\begin{matrix}g(0) & 0 \\ (g − g(0))/t & (f, t) \end{matrix}) ,  (20)

where (f, t) = (f, tf, t^{2}f, t^{3}f, …). Particularly, if f = tg, then the quasi-Riordan array [g, tg] = (g, t), an Appell-type Riordan array.

Note that [g, f] defined by (19) is not the Riordan-Krylov matrix defined in [6], which relationship is worth being investigated. Paul Barry pointed out a connection between quasi-Riordan arrays and almost-Riordan arrays (cf. Barry [2] and Barry and Pantelidis [3]) in a personal communication, which is being studied by Barry, Pantelidis, and the author.

Theorem 5. Let (g, f) = (d_{n,k})_{n,k≥0} be a Riordan array, and let ([1] ⊕ (g, f)) and [g, f] be defined by (17) and (19), respectively. Then (g, f) has the expression

(g, f) = [g, f]([1] ⊕ (g, f)).  (21)

Proof. We write the formal power series g and f in the Riordan array (g, f) as g = ∑_{n≥0} g_{n}t^{n} and f = ∑_{n≥1} f_{n}t^{n}. Then

(\begin{matrix}g(0) & 0 \\ (g − g(0))/t & (f, t) \end{matrix}) ([1] ⊕ (g, f))

= (\begin{matrix}d_{0,0} \\ d_{1,0} & f_{1} \\ d_{2,0} & f_{2} & f_{1} \\ d_{3,0} & f_{3} & f_{2} & f_{1} \\ d_{4,0} & f_{4} & f_{3} & f_{2} & f_{1} \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix}) (\begin{matrix}1 \\ 0 & d_{0,0} \\ 0 & d_{1,0} & d_{1,1} \\ 0 & d_{2,0} & d_{2,1} & d_{2,2} \\ 0 & d_{3,0} & d_{3,1} & d_{3,2} & d_{3,3} \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix})

= ( \begin{matrix}d_{0,0} \\ d_{1,0} & f_{1}d_{0,0} \\ d_{2,0} & f_{2}d_{0,0} + f_{1}d_{1,0} & f_{1}d_{1,1} \\ d_{3,0} & f_{3}d_{0,0} + f_{2}d_{1,0} + f_{1}d_{2,0} & f_{2}d_{1,1} + f_{1}d_{2,1} & f_{1}d_{2,2} \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

= (\begin{matrix}d_{0,0} \\ d_{1,0} & d_{1,1} \\ d_{2,0} & d_{2,1} & d_{2,2} \\ d_{3,0} & d_{3,1} & d_{3,2} & d_{3,3} \\ d_{4,0} & d_{4,1} & d_{4,2} & d_{4,3} & d_{4,4} \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix}) ,  (22)

where the last step follows (12), which completes the proof of (19).

For an integer r ≥ 0 we let h|_{r} := ∑_{n=0}^{r} h_{n}t^{n} denote the r-th truncations of a power series h = ∑_{n≥0} h_{n}t^{n}.

Corollary 6. Let (g, f) = (d_{n,k})_{n,k≥0} be a Riordan array with g = ∑_{n≥0} g_{n}t^{n} and f = ∑_{n≥1} f_{n}t^{n}, and let (g, f)_{n} be the nth order leading principle submatrix of (g, f). Then we have the recursive relation of (g, f)_{n} in the following form:

(g, f)_{n} = [g, f]_{n}([1] ⊕ (g, f)_{n−1}),  (23)

where [g, f]_{n} is the nth order leading principle submatrix of the quasi-Riordan array [g, f] defined by (17), namely

[g, f]_{n} = (\begin{matrix}g(0) & 0 \\ ((g − g(0))/t)|_{n−1} & (f, t)_{n−1} \end{matrix}) ,  (24)

where the n − 1st truncation of (g − g(0)/t)|_{n−1} is (g − g(0)/t)|_{n−1} = ∑_{k=1}^{n−1} g_{k}t^{k−1}, and (f, t)_{n−1} = (f, tf, t^{2}f, …, t^{n−1}f). We call [g, f]_{n} the recursive matrix of the Riordan array (g, f).

Mao, Mu, and Wang [23] use (24) to give another interesting criterion for the total positivity of Riordan arrays.

In the next section, we will show that the set of all quasi-Riordan arrays forms a group, called the quasi-Riordan group.

3 The quasi-Riordan group

Let R_{r} denote the set of all quasi-Riordan arrays defined by (19). In this section, we show that R_{r} is a group with respect to regular matrix multiplication.

Proposition 7. Let [g, f], [d, h] ∈ R_{r}, and let u = ∑_{n≥0} u_{n}t^{n} ∈ F_{0}. Then the first fundamental theorem for quasi-Riordan arrays (FFTQRA) is

[g, f]u = gu_{0} + \frac{f}{t} (u − u_{0}),  (25)

which implies that

[g, f][d, h] = [g + \frac{f}{t} (d − 1), \frac{fh}{t}] .  (26)

Hence, [1, t] is the identity of R_{r}. Proof. The FFTQRA (25) can be proved as follows.

[g, f]u = (g, f, tf, t^{2}f, ⋯) (u_{0} ; u_{1} ; u_{2} ; ⋮)

= gu_{0} + f ∑_{n≥1} u_{n}t^{n−1} = gu_{0} + \frac{f}{t} (u − u_{0}).

By using FFTQRA and noting d(0) = 1 and h(0) = 0, we have [g, f][d, h] = (g, f, tf, t^{2}f, ⋯)(d, h, th, t^{2}h, ⋯)

= (g + \frac{f}{t} (d − 1), \frac{f}{t} h, \frac{f}{t} th, \frac{f}{t} t^{2}h, ⋯) ,

which implies (26).

Substituting [g, f] = [1, t] and [d, h] = [1, t] into (26), we obtain, respectively, [1, t][g, f] = [1 + (g − 1), f] = [g, f] and

[g, f][1, t] = [g + \frac{f}{t} (1 − 1), \frac{f}{t} t] = [g, f],

which implies [1, t], the identity matrix, is the identity of R_{r}. Theorem 8. The set of all quasi-Riordan arrays R_{r} is a group, called the quasi-Riordan group, with respect to the multiplication represented in (26). Proof. From (26) of Proposition 7, R_{r} is closed with respect to the multiplication. (26) also shows [1, t] is the identity of R_{r}. For any [g, f] ∈ R_{r}, its inverse is

[g, f]^{−1} = [1 + \frac{t}{f} (1 − g), \frac{t^{2}}{f} ] ,  (27)

since

[1 + \frac{t}{f} (1 − g), \frac{t^{2}}{f} ] [g, f] = [1 + \frac{t}{f} (1 − g) + \frac{t^{2}/f}{t} (g − 1), \frac{t^{2}}{f} \frac{f}{t} ] = [1, t].  (28)

Finally, the associative law is satisfied for any [g, f], [d, h], and [u, v] ∈ R_{r} because

([g, f][d, h])[u, v] = [g + \frac{f}{t} (d − 1), \frac{fh}{t}] [u, v]

= [g + \frac{f}{t} (d − 1) + \frac{fh}{t^{2}}(u − 1), \frac{fhv}{t^{2}} ]

and

[g, f]([d, h])[u, v] = [g, f] [d + \frac{h}{t}(u − 1), \frac{hv}{t} ]

= [g + \frac{f}{t} (d + \frac{h}{t}(u − 1) − 1) , \frac{fhv}{t^{2}} ]

= [g + \frac{f}{t} (d − 1) + \frac{fh}{t^{2}}(u − 1), \frac{fhv}{t^{2}} ]

shows ([g, f][d, h])[u, v] = [g, f]([d, h][u, v]).

Example 9. Consider the quasi-Riordan array [1/(1 − t), t/(1 − t)], which is an Appell-type Riordan array (1/(1 − t), t) with its inverse (1 − t, t). From (27) its inverse in R_{r} is

[\frac{1}{1 − t}, \frac{t}{1 − t}]^{−1} = [1 + \frac{t}{f} (1 − g), \frac{t^{2}}{f} ]

= [1 + \frac{t}{t/(1 − t)} (1 − \frac{1}{1 − t}) , \frac{t^{2}}{t/(1 − t)}] = [1 − t, t(1 − t)]

= (1 − t, t) = (\begin{matrix}1 \\ −1 & 1 \\ 0 & −1 & 1 \\ 0 & 0 & −1 & 1 \\ ⋮ & ⋮ & ⋮ & ... \end{matrix}) .

Theorem 10. A quasi-Riordan array [g, f] is a Riordan array if and only if f = tg, i.e., when [g, f] is an Appell-type Riordan array. Hence, A = {[g, tg] : g ∈ F_{0}, g(0) = 1} is a subgroup of R_{r}, which is called the Appell quasi-Riordan subgroup.

Proof. Let [g, tg], [d, td] ∈ R_{r}. Then

[g, tg][d, td] = [g + \frac{tg}{t} (d − 1), \frac{t^{2}gd}{t}] = [gd, tgd].

A is closed under multiplication. In addition, the inverse of [g, f]

[g, tg]^{−1} = [1 + \frac{t}{tg} (1 − g), \frac{t^{2}}{tg} ] = [\frac{1}{g}, \frac{t}{g} ]

is also in A.

Theorem 11. Let R_{r} be the quasi-Riordan group. Then every conjugate of an element [g, f] ∈ R_{r} is in the set R(f)_{r} := {[d, f] : d ∈ F_{0}, d(0) = 1}. Hence, A = {[g, tg] : g ∈ F_{0}, g(0) = 1} is a normal subgroup of R_{r}.

Proof. Let [d, h] ∈ R_{r}. Then for an arbitrarily fixed [g, f] ∈ R_{r}, we have

[d, h][g, f][d, h]^{−1} = [d, h][g, f] [1 + \frac{t}{h}(1 − d), \frac{t^{2}}{h} ]

= [d + \frac{h}{t}(g − 1) − \frac{f}{t} (d − 1), f] ∈ R(f)_{r}.

In particular, if f = tg in [g, f], then

[d, h][g, tg][d, h]^{−1} = [d + \frac{h}{t}(g − 1) − g(d − 1), tg] ,

which implies A = {[g, tg] : g ∈ F_{0}, g(0) = 1} is a normal subgroup of R_{r}.

4 A vertical recursive relation view of (c)- and (C)- Riordan arrays

In this section, we extend the vertical recursive relation to (c) Riordan arrays. We start from the following definition of the (c)-class of Riordan arrays.

Definition 12. Let (g, f) = (d_{n,k})_{n,k≥0} be a Riordan array, and let c = (c_{0}, c_{1}, c_{2}, …) be a sequence satisfying c_{0} = 1 and c_{k} ≠ 0 for all k = 1, 2, . . .. We define

(g, f)_{c} = (\frac{c_{n}}{c_{k}} d_{n,k})_{n,k≥0} .  (29)

Then the set

{R}_{c} := {(g, f)_{c} : (g, f) = (d_{n,k})_{n,k≥0} ∈ R}  (30)

is called the (c)-class of R or the set of the (c)-Riordan arrays. Since we may change c_{n} and c_{k} respectively to 1/c_{n} and 1/c_{k},

(\frac{c_{k}}{c_{n}} d_{n,k})_{n,k≥0} = (\frac{1/c_{n}}{1/c_{k}} d_{n,k})_{n,k≥0}

is in the (1/c)-class of R, where (1/c) = (1/c_{0}, 1/c_{1}, 1/c_{2}, …).

Let (g, f) = (d_{n,k})_{n,k≥0} be a Riordan array, and let C = (c_{n,k})_{n,k≥0} be a lower triangular matrix satisfying c_{n,0} = 1 and c_{n,k} ≠ 0 for all 0 ≤ k ≤ n and c_{n,k} = 0 for all k > n. We define

(g, f)_{C} = (\frac{c_{n,n}}{c_{n,k}} d_{n,k})_{n,k≥0} .  (31)

The set

{R}_{C} := {(g, f)_{C} : (g, f) = (d_{n,k})_{n,k≥0} ∈ R}  (32)

is called the (C)-class of R, or the set of the (C)-Riordan arrays. Similarly, (\frac{c_{n,k}}{c_{n,n}} d_{n,k})_{n,k≥0}

is in the (1/C)-class of R, where (1/C) = (1/c_{n,k})_{n,k≥0}.

Wang and Wang [35] defined the (c)-Riordan arrays by using the generalized formal power series. Gould and the author [10] claimed that for a sequence (c) the (c)-class of R, or the set of the (c)-Riordan arrays {R}_{c}, forms a group. More precisely, we have the following theorem.

Theorem 13. [10] Let c = (c_{0}, c_{1}, c_{2}, …) be a sequence satisfying c_{0} = 1 and c_{k} ≠ 0 for all k = 1, 2, …, and let {R}_{c} be the (c)-class defined by (30). Then {R}_{c} forms a group with respect to the regular matrix multiplication. We let R_{c} denote this group and called the (c)-Riordan group with respect to the sequence (c).

Proof. If (g_{1}, f_{1})_{c}, (g_{2}, f_{2})_{c} ∈ {(g, f)}_{c}, and define (g_{1}, f_{1}) = (d_{n,k})_{n,k≥0}, (g_{2}, f_{2}) = (e_{n,k})_{n,k≥0}, and (g_{1}, f_{1})(g_{2}, f_{2}) = (g_{1}g_{2}(f_{1}), f_{2}(f_{1})) = (h_{n,k})_{n,k≥0}, then (d_{n,k})_{n,k≥0}(e_{n,k})_{n,k≥0} = (h_{n,k})_{n,k≥0}, and

(g_{1}, f_{1})_{c}(g_{2}, f_{2})_{c} = ( ∑_{j=0}^{min n,k} (\frac{c_{n}}{c_{j}} d_{n,j}) (\frac{c_{j}}{c_{k}} e_{j,k}))_{n,k≥0}

= (\frac{c_{n}}{c_{k}} ∑_{j=0}^{min n,k}(d_{n,j}e_{j,k}))_{n,k}

= (\frac{c_{n}}{c_{k}} h_{n,k})_{n,k≥0} = (g_{1}g_{2}(f_{1}), f_{2}(f_{1}))_{c}.

Hence, we may find (1.t)_{c} is the identity of R_{c} and the inverse of an element (g, f)_{c} ∈ R_{c} is (1/g(\overline{f}), \overline{f}), where \overline{f} is the compositional inverse of f. Finally, we have

((g, f)_{c}(d, h)_{c}) (u, v)_{c} = (gd(f)u(h(v)), v(h(f)))
= (g, f)_{c} ((d, h)_{c}(u, v)_{c}) ,

completing the proof.

More material on (c)-Riordan arrays can be found in [14, 30].

Example 14. Let (g, f) = (1/(1−t), t/(1−t)). Then (1/(1−ℓt), t/(1−ℓt)) = (1/(1−t), t/(1− t))_{c} with c = (c_{n})_{n≥0} = (ℓ^{n})_{n≥0}, where ℓ ∈ 𝕂. The (c)-Riordan array (1/(1 − ℓt), t/(1 − ℓt))_{c} with c = (c_{n})_{n≥0} = (ℓ^{n})_{n≥0} begins

[ \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ ℓ & 1 & 0 & 0 & 0 & 0 & 0 \\ ℓ^{2} & ℓ^{2} & 1 & 0 & 0 & 0 & 0 \\ ℓ^{3} & ℓ^{2} + ℓ^{3} & ℓ^{2} + ℓ & 1 & 0 & 0 & 0 \\ ℓ^{4} & 2ℓ^{3} + ℓ^{4} & 2ℓ^{2} + 2ℓ^{3} & ℓ^{2} + 2ℓ & 1 & 0 & 0 \\ ℓ^{5} & 3ℓ^{4} + ℓ^{5} & 4ℓ^{3} + 3ℓ^{4} & 4ℓ^{2} + 3ℓ^{3} & ℓ^{2} + 3ℓ & 1 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ]

Example 15. Let (g, f) = (1/(1 − t), t/(1 − t)) = (\binom{n}{k})_{n,k≥0}, and let c = (n!)_{n≥0}. Then

(r_{n,k})_{n,k≥0} = (\frac{n!}{k!}\binom{n}{k})_{n,k≥0} = ((n − k)! (\binom{n}{k})^{2}) ∈ \{(1/(1 − t), t/(1 − t))\}_{c}, (33)

where c = (c_{n})_{n≥0} = (n!)_{n≥0}. The (c)-Riordan array (r_{n,k})_{n,k≥0} begins

(r_{n,k})_{n,k≥0} = [ \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 2 & 4 & 1 & 0 & 0 & 0 & 0 \\ 6 & 18 & 9 & 1 & 0 & 0 & 0 \\ 24 & 96 & 72 & 16 & 1 & 0 & 0 \\ 120 & 600 & 600 & 200 & 5 & 1 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ] .

The matrix (r_{n,k})_{n,k≥0} is called a rook matrix. The polynomial r_{n}(x) = ∑_{k=0}^{n} r(n, k)x^{n−k} is called the rook polynomial of nth order (cf. Fielder [8] and Riordan [27, §7.2, 7.3, and 7.4]).

In general, consider a board that represents a full or a part chess board. Let m be the number of squares present in the board. Two pawns or rooks placed on a board are said to be in non capturing positions if they are not in same row or same column. For 2 ≤ k ≤ m, let r_{k} denote the number of ways in which k rooks can be placed on a board such that no two rooks capture each other. Then the polynomial 1 + r_{1}x + r_{2}x^{2} + ⋯ + r_{m}x^{m} is called the (general) rook polynomial for the board considered. If the board is denoted by C, then the corresponding polynomial is denoted by r(C, x). The rook polynomials are defined for m ≥ 2. If m = 1, then the board contains only one square so that r_{k} = 0 for k ≥ 2, and r(C, x) = 1 + x. Hence, the coefficient r_{n,k} of the roof polynomial r_{n}(x) = ∑_{k=0}^{n} r(n, k)x^{n−k} is referred to as the number of ways in which k rooks can be placed on a n × n full chess board such that no two rooks capture each other. The rook polynomials for 1 × 1, 2 × 2, 3 × 3, 4 × 4, and 5 × 5 full boards are, respectively,

r_{1}(x) = x + 1,

r_{2}(x) = 2x^{2} + 4x^{1} + 1,
r_{3}(x) = 6x^{3} + 18x^{2} + 9x + 1,
r_{4}(x) = 24x^{4} + 96x^{3} + 72x^{2} + 16x + 1,
r_{5}(x) = 120x^{5} + 600x^{4} + 600x^{3} + 200x^{2} + 5x + 1.

In a given board C, suppose we choose a particular square and let (∗) denote the square. Let D denote the board obtained from C by deleting the row and column containing the square (∗), and let E be the board obtained from C by deleting only the square (∗). Then the rook polynomial for the board C is given by r(C, x) = xr(D, x) + r(E, x). This is known as expansion formula for r(C, x).

We let C_{n+1} denote the (n + 1) × (n + 1) full board. We suppose n × n full board C_{n} is located at the left upper corner of C_{n+1}, and let E_{n} denote the board obtained from the board C_{n+1} by deleting only the square at the right lower corner of C_{n+1}. Then, we have

r(E_{n}, x) = r(C_{n+1}, x) − xr(C_{n}, x) = r_{n+1}(x) − xr_{n}(x)

= ∑_{k=0}^{n+1} \frac{(n + 1)!}{k!}\binom{n + 1}{k}x^{n+1−k} − x ∑_{k=0}^{n} \frac{n!}{k!}\binom{n}{k}x^{n−k}

= ∑_{k=0}^{n+1} \frac{n!}{k!} ((n + 1)\binom{n + 1}{k} − \binom{n}{k}) x^{n+1−k}

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

= ∑_{k=0}^{n} \frac{n!}{k!} \frac{n^{2} + n + k}{n − k + 1} \binom{n}{k}x^{n+1−k} + 1

= ∑_{k=0}^{n} \frac{n^{2} + n + k}{n − k + 1} (n − k)! (\binom{n}{k})^{2} x^{n+1−k} + 1.  (34)

We define r(E_{n}, x) = ∑_{k=0}^{n+1} E_{n,k}x^{n+1−k} and call it the remainder polynomial, where

E_{n,k} = \frac{n^{2} + n + k}{n − k + 1} (n − k)! (\binom{n}{k})^{2}  (35)

for 0 ≤ k ≤ n and E_{n,n+1} = 1. Then, the lower triangle matrix (E_{n,k})_{n,k≥0}, called the remainder triangle, begins

(E_{n,k})_{n,k≥0} = [ \begin{matrix}0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 3 & 1 & 0 & 0 & 0 & 0 \\ 4 & 14 & 8 & 1 & 0 & 0 & 0 \\ 18 & 78 & 63 & 15 & 1 & 0 & 0 \\ 96 & 504 & 528 & 184 & 4 & 1 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ] .

Here is a recursive relation related to the rook triangle (r_{n,k})_{n,k≥0} and the triangle (E_{n,k})_{n,k≥0}.

Proposition 16. Let (r_{n,k})_{n,k≥0} and (E_{n,k})_{n,k≥0} be the rook triangles and the remainder triangle defined in Example 15, and let r_{n}(x) = ∑_{k=0}^{n} r(n, k)x^{n−k} and r(E_{n}, x) = ∑_{k=0}^{n} E_{n,k}x^{n−k} be the rook polynomial and the remainder polynomial, respectively. Then, there exists the recursive relation

r_{n+1}(x) = xr_{n}(x) + r(E_{n}, x)  (36)

for n ≥ 0, which can be expressed as a matrix form

\overline{(r_{n,k})} = (r_{n,k}) + (E_{n,k}),  (37)

where n ≥ 0 and \overline{(r_{n,k})} is obtained from (r_{n,k}) by deleting its first row. Furthermore, we have the expansion formula for r_{n+1}(x) as

r_{n+1}(x) = ∑_{k=0}^{n} x^{n−k}r(E_{k}, x) + x^{n+1}.  (38)

Proof. Equation (36) comes from the expansion formula. Substituting r_{n}(x) = ∑_{k=0}^{n} r(n, k)x^{n−k} and r(E_{n}, x) = ∑_{k=0}^{n} E_{n,k}x^{n−k} into (36) and comparing the coefficients of the same powers of x on the both sides yields

r_{n+1,k} = r_{n,k} + E_{n,k},

which can be combined as the matrix form (37) for n, k ≥ 0. From (36) we have

r_{n+1}(x) = xr_{n}(x) + r(E_{n}, x)
xr_{n}(x) = x^{2}r_{n−1}(x) + xr(E_{n−1}, x)
x^{2}r_{n−1}(x) = x^{3}r_{n−2}(x) + x^{2}r(E_{n−2}, x)
⋮
x^{n}r_{1}(x) = x^{n+1}r_{0}(x) + x^{n}r(E_{0}, x).

Adding up the above system and cancelling the same terms from the both sides yields (38).

Example 17. Let (g, f) = (1/(1 − t), t/(1 − t)) = (\binom{n}{k})_{n,k≥0}. Then

(L_{n,k})_{n,k≥0} = (\frac{(−1)^{n}/(n)_{n}}{(−1)^{k}/(n)_{k}} \binom{n}{k})_{n,k≥0}
= (\frac{(−1)^{n−k}}{(n − k)!}\binom{n}{k}) ∈ \{(1/(1 − t), t/(1 − t))\}_{C},  (39)

where C = (c_{n,k})_{n,k≥0} = ((−1)^{n}/(n)_{k})_{n,k≥0}. The (C)-Riordan array (L_{n,k})_{n,k≥0} begins

[ \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ −1 & 1 & 0 & 0 & 0 & 0 & 0 \\ \frac{1}{2} & −2 & 1 & 0 & 0 & 0 & 0 \\ −\frac{1}{6} & \frac{3}{2} & −3 & 1 & 0 & 0 & 0 \\ \frac{1}{24} & −\frac{2}{3} & 3 & −4 & 1 & 0 & 0 \\ −\frac{1}{120} & \frac{5}{24} & −\frac{5}{3} & 5 & −5 & 1 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ] .

The matrix (L_{n,k})_{n,k≥0} is called the Laguerre matrix. The polynomial L_{n}(n, k) = ∑_{k=0}^{n} L_{n,k}x^{n−k} is called the Laguerre polynomial of order n (cf., for example, Hsu, Shiue, and the author [16] and Riordan [27, pp. 164–170]).

Comparing (33) and (39) and noting that

r_{n,k} = \frac{n!}{k!}\binom{n}{k},

we immediately obtain

r_{n,n−k} = \frac{n!}{(n − k)!}\binom{n}{k} = \frac{n!}{(n − k)!}(−1)^{n−k}(n − k)!L_{n,k} = (−1)^{n−k}n!L_{n,k},  (40)

which implies

∑_{k=0}^{n} r_{n,n−k}x^{n−k} = n! ∑_{k=0}^{n} L_{n,k}(−x)^{n−k} = n!L_{n}(−x)  (41)

for 0 ≤ k ≤ n. The left-hand side of (41) can be written as

∑_{k=0}^{n} r_{n,n−k}x^{n−k} = ∑_{k=0}^{n} r_{n,k}x^{k} = x^{n} ∑_{k=0}^{n} r_{n,k} (\frac{1}{x})^{n−k} = x^{n}r_{n} (\frac{1}{x}) .

Hence, we have the well-known formula

r_{n}(x) = n!x^{n}L_{n} (−\frac{1}{x})

for n ≥ 0.

Definition 18. We may extend the rook triangle and the rook polynomials to a general case by defining the following generalized rook triangle and generalized rook polynomials:

(r̂_{n,k})_{n,k≥0} = (\frac{n!}{k!}d_{n,k})_{n,k≥0} = ((n − k)!\binom{n}{k}d_{n,k}) ∈ \{(g(t), f(t)\}_{c},  (42)

where c = (c_{n})_{n≥0} = (n!)_{n≥0} and (g, f) = (d_{n,k})_{n,k≥0}. In addition, the polynomial

r̂_{n}(x) = ∑_{k=0}^{n} r̂_{n,k}x^{n−k}  (43)

is referred to as the generalized rook polynomial of degree n.

We may also extend the Laguerre triangle and the Laguerre polynomials to a general case by defining the following generalized Riordan type Laguerre triangle and generalized Riordan type Laguerre polynomials:

(L̂_{n,k})_{n,k≥0} = (\frac{(−1)^{n}/(n)_{n}}{(−1)^{k}/(n)_{k}} d_{n,k})_{n,k≥0}
= (\frac{(−1)^{n−k}}{(n − k)!}\binom{n}{k}d_{n,k}) ∈ \{(g(t), f(t)\}Ĉ,  (44)

where Ĉ = (c_{n,k})_{n,k≥0} = (n!)_{n≥0} and (g, f) = (d_{n,k})_{n,k≥0}. In addition, the polynomial

L̂_{n}(x) = ∑_{k=0}^{n} L̂_{n,k}x^{n−k}  (45)

is referred to as the generalized Laguerre polynomial of degree n.

Similar to the relationship between the rook triangle and the Laguerre triangle and the relationship between the rook polynomials and the Laguerre polynomials, we may extend the relationships to the general cases as follows.

Theorem 19. Let the generalized rook triangle and the rook polynomials and the generalized Riordan type Laguerre triangle and the generalized Riordan type Laguerre polynomials be defined in Definition 18. Then we have

r̂_{n,n−k} = (−1)^{n−k}n!L̂_{n,k}  (46)

for 0 ≤ k ≤ n and

r̂_{n}(x) = n!x^{n}L̂_{n} (−\frac{1}{x})

for n ≥ 0.

From Examples 15 and 47, we have a relationship between

Theorem 20. Let (g, f) = (d_{n,k})_{n,k≥0} be a Riordan array, and let (g, f)_{c} and (g, f)Ĉ be defined in Definition 12. Then (g, f)_{c} = (d_{n,k}^{(c)})_{n,k≥0} and (g, f)Ĉ = d_{n,k≥0}^{(Ĉ)} satisfy the horizontal recursive relations, which are extensions of A- and Z-sequence representation (4) and (6) of (g, f) to (g, f)_{c} and (g, f)Ĉ, respectively:

d_{n+1,k+1}^{(c)} = \frac{c_{n+1}}{c_{n}c_{k+1}} ∑_{j=0}^{n−k} a_{j}c_{k+j}d_{n,k+j}^{(c)}  (47)

for k ≥ 0 and

d_{n+1,0}^{(c)} = \frac{c_{n+1}}{c_{n}} ∑_{j=0}^{n} z_{j}c_{j}d_{n,j}^{(c)}  (48)

as well as

d_{n+1,k+1}^{(Ĉ)} = \frac{c_{n+1,n+1}}{c_{n,n}c_{n+1,k+1}} ∑_{j=0}^{n−k} a_{j}c_{n,k+j}d_{n,k+j}^{(Ĉ)}  (49)

for k ≥ 0 and

d_{n+1,0}^{(Ĉ)} = \frac{c_{n+1,n+1}}{c_{n,n}} ∑_{j=0}^{n} z_{j}c_{n,j}d_{n,j}^{(Ĉ)}.  (50)

Theorem 21. Let (g, f) = (d_{n,k})_{n,k≥0} be a Riordan array, and let (g, f)_{c} and (g, f)Ĉ be defined in Definition 12. Then (g, f)_{c} = (d_{n,k}^{(c)})_{n,k≥0} and (g, f)Ĉ = d_{n,k≥0}^{(Ĉ)} satisfy the vertical recursive relations, which are extensions of vertical recursive relation (12) of the entries of (g, f) to (g, f)_{c} and (g, f)Ĉ, namely,

d_{n,k}^{(c)} = \frac{c_{n}}{c_{k}} ∑_{j=1}^{n−k+1} f_{j} \frac{c_{k−1}}{c_{n−j}} d_{n−j,k−1}^{(c)}  (51)

and

d_{n,k}^{(Ĉ)} = \frac{c_{n,n}}{c_{n,k}} ∑_{j=1}^{n−k+1} f_{j} \frac{c_{n−j,k−1}}{c_{n−j,n−j}} d_{n−j,k−1}^{(Ĉ)},  (52)

respectively.

Example 22. It is known that the A- and Z- sequences of (1/(1−t), t/(1−t)) are (1, 1, 0, …) and (1, 0, 0, …), respectively. Then for the rook triangle represented in Example 15 with d_{n,k} = \binom{n}{k}, from (47) and (48), we obtain the horizontal recursive relations for the entries of the rook triangle,

r_{n,k} = nr_{n−1,k} + \frac{n}{k}r_{n−1,k−1}  (53)

for k ≥ 1 and

r_{n,0} = nr_{n−1,0}.  (54)

From (51) and noting (13) we obtain the vertical recursive relation for the entries of the rook triangle

r_{n,k} = ∑_{j=1}^{n−k+1} \frac{(n)_{j}}{k} r_{n−j,k−1}  (55)

for k ≥ 1.

From (33), one may obtain an identity equivalent to (53); namely,

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

Similarly, from (55), we get

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

Example 23. Similarly, for the Laguerre triangle represented in Example 17 with d_{n,k} = \binom{n}{k}, from (47) and (48), we obtain the horizontal recursive relations for the entries of the Laguerre triangle; namely,

L_{n,k} = L_{n−1,k−1} − \frac{1}{n − k} L_{n−1,k}  (56)

for k ≥ 1 and

L_{n,0} = −\frac{1}{n}L_{n−1,0}.  (57)

From (51) and noting (13) we obtain the vertical recursive relation for the entries of the Laguerre triangle; namely,

L_{n,k} = \frac{1}{(n − k)!} ∑_{j=1}^{n−k+1}(−1)^{j−1}(n − k − j + 1)!L_{n−j,k−1}  (58)

for k ≥ 1. It can be seen that (58) is equivalent to (13).

5 Acknowledgments

The authors wish to express their sincere thanks to the referee for helpful comments and remarks that lead to an improved and revised version of the original manuscript.

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2020 Mathematics Subject Classification: Primary 05A15. Secondary 05A05, 15B36, 15A06, 05A19, 11B83.

Keywords: Riordan array, Riordan group, quasi-Riordan array, vertical recursive relation, Asequence, Z-sequence, rook triangle, Laguerre triangle, Fuss-Catalan number, Fuss-Catalan matrix, Catalan number, Catalan matrix, (c)-Riordan array, (C)-Riordan array.

(Concerned with sequence \underline{A001764}.)

Received September 30 2022; revised versions received October 1 2022; October 3 2022; November 14 2022; November 15 2022; November 16 2022; November 17 2022. Published in Journal of Integer Sequences, November 18 2022.

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