Exponential Almost-Riordan Arrays

Abstract

The group of the almost-Riordan arrays with exponential generating functions is defined. The subgroups of the exponential almost-Riordan group are presented. Also, some isomorphisms between the exponential almost-Riordan group and the exponential Riordan group are considered. Then, the production matrix for the exponential almost-Riordan array is obtained.

1 Introduction

Generating functions are a very useful tool in combinatorics, analysis, and other areas of mathematics. By using the properties of generating functions, some combinatorial identities can be obtained more easily.

For the integer sequence $\{a_n\}_{n\ge 0}$, the formal power series

$$\begin{aligned} f(x)=a_{0}+a_{1}x+a_{2}{x^2}+a_{3}{x^3}+\dots . \end{aligned}$$

represents an ordinary generating function. The coefficient of $x^n$ in the formal power series f(x) is

$$\begin{aligned} a_n=[x^n]f(x) \end{aligned}$$

(1.1)

in [12]. Additionally, the following identities are valid:

$$\begin{aligned}{}[x^n](x^kf(x))=[x^{n-k}]f(x) \end{aligned}$$

(1.2)

and

$$\begin{aligned}{}[x^n]f'(x)=(n+1)[x^{n+1}]f(x) \end{aligned}$$

(1.3)

in [18]. The detailed information regarding the generating functions and coefficient extraction can be found in [6, 12, 17, 18, 23, 27]. Also, the following power series

$$\begin{aligned} g(x)=a_{0}+a_{1}x+a_{2}\frac{x^2}{2!}+a_{3}\frac{x^3}{3!}+\dots \end{aligned}$$

represents the ordinary generating function of the sequence $\{\frac{a_n}{n!}\}_{n\ge 0}$. This generating function is known as the exponential generating function of the sequence $\{a_n\}_{n\ge 0}$. Then, the coefficient $a_n$ of $x^n$ in g(x) is denoted by

$$\begin{aligned} a_n=n![x^n]g(x) \end{aligned}$$

(1.4)

in [6]. Sometimes, the properties of the ordinary generating function of the sequence $\{a_n\}_{n\ge 0}$ can be complex. In such cases, we can consider the generating function of the sequence $\{\frac{a_n}{n!}\}_{n\ge 0}$. For instance, the ordinary generating function of factorial numbers sequence is as follows:

$$\begin{aligned} \sum _{n=0}^{\infty }n!x^n=\int _{n=0}^{\infty }\frac{e^{-t}}{1-tx}dt. \end{aligned}$$

The exponential generating function of factorial numbers sequence is given as

$$\begin{aligned} \sum _{n=0}^{\infty }n!\frac{x^n}{n!}=\frac{1}{1-x}. \end{aligned}$$

Obviously, the exponential generating function of factorial numbers sequence is simpler than the ordinary generating function for factorial numbers sequence. Similarly, the ordinary generating function for the first kind of Stirling numbers sequence is complicated and divergent. However, the exponential generating function of the same sequence is more useful than the ordinary generating function.

Riordan arrays play a crucial role in deriving combinatorial identities and solving combinatorial sums. Renzo Sprugnoli has contributed significantly to the utilization of Riordan arrays in computing combinatorial sums [21, 22]. Also, Sprugnoli has investigated sums involving binomial coefficients and Stirling numbers by using the fundamental theorem of Riordan arrays in [20]. Riordan arrays are infinite lower triangular matrices defined by ordinary generating functions. There are many studies about Riordan arrays in literature, and some of these can be found in [6, 8, 11, 13,14,15,16, 19, 23, 26].

We consider the following formal power series:

$$\begin{aligned} g(x)=g_{0}+g_{1}x+g_{2}\frac{x^2}{2!}+g_{3}\frac{x^3}{3!}+\dots \end{aligned}$$

and

$$\begin{aligned} f(x)=f_{0}+f_{1}x+f_{2}\frac{x^2}{2!}+f_{3}\frac{x^3}{3!}+\dots \end{aligned}$$

with $g_{0}\ne 0$, $f_{0}=0$ and $f_{1}\ne 0$. The functions g(x) and f(x) represent the exponential generating functions of the sequences $\{g_n\}_{n\ge 0}$ and $\{f_n\}_{n\ge 0}$, respectively. The exponential generating function of the kth column of the exponential Riordan arrays is defined as follows:

$$\begin{aligned} g(x)\frac{f(x)^k}{k!},\hspace{5.0pt}\hspace{5.0pt}k=0,1,2,\dots . \end{aligned}$$

Additionally, the exponential Riordan arrays are denoted as pairs of exponential generating functions, represented as [g(x), f(x)]. The multiplication operation between two exponential Riordan arrays is defined as follows:

$$\begin{aligned}{}[g(x),f(x)][h(x),l(x)]=[g(x)h(f(x)),l(f(x))]. \end{aligned}$$

(1.5)

The set of exponential Riordan arrays is a group with the multiplication operation defined in (1.5). This group is known as the exponential Riordan group. The identity element of the exponential Riordan group is defined as

$$\begin{aligned} I=[1,x] \end{aligned}$$

(1.6)

and the inverse of [g(x), f(x)] is given by

$$\begin{aligned}{}[g(x),f(x)]^{-1}=\left[ \frac{1}{g(\overline{f}(x))},\overline{f}(x)\right] \end{aligned}$$

(1.7)

where $\overline{f}(x)$ is the compositional inverse of f(x) in [10].

Let R be the exponential Riordan matrix. Then, we have

$$\begin{aligned} P=R^{-1}\overline{R} \end{aligned}$$

(1.8)

which P, $R^{-1}$ and $\overline{R}$ represent the production matrix of the matrix R, the inverse of the matrix R and the version of the matrix R with the 0th row removed, respectively in [10]. The characterization and production matrices of exponential Riordan arrays are provided in [6, 10, 11].

Proposition 1.1

[11]. Let $D=(d_{n,k})_{n,k\ge 0}=[g(x),f(x)]$ be an exponential Riordan matrix with $g_0\ne 0,f_0\ne 0$. Let

$$\begin{aligned} c(y)=c_0+c_1y+c_2y^2+\dots ,\hspace{5.0pt}r(y)=r_0+r_1y+r_2y^2+\dots \end{aligned}$$

(1.9)

be two formal power series such that

$$\begin{aligned} \begin{aligned} r(xf(x))&=(xf(x))',\\ c(xf(x))&=\frac{g'(x)}{g(x)}. \end{aligned} \end{aligned}$$

(1.10)

Then

$$\begin{aligned} d_{n+1,0}&=\sum _{i}i!c_{i}d_{n,i} , \end{aligned}$$

(1.11)

$$\begin{aligned} d_{n+1,k}&=r_0d_{n,k-1}+\frac{1}{k!}\sum _{i\ge k}i!(c_{i-k}+kr_{i-k+1})d_{n,i} \end{aligned}$$

(1.12)

or, defining $c_{-1}=0$,

$$\begin{aligned} d_{n+1,k}=\frac{1}{k!}\sum _{i\ge k-1}i!(c_{i-k}+kr_{i-k+1})d_{n,i}. \end{aligned}$$

(1.13)

Conversely, starting from the sequences defined by (1.9), the infinite array $(d_{n,k})_{n,k\ge 0}$ defined by (1.13) is an exponential Riordan matrix.

In [11], the production matrix P of the exponential Riordan matrix D is given as follows:

$$\begin{aligned} P= \begin{pmatrix} c_0&{} \quad r_0&{} \quad 0&{} \quad 0&{} \quad 0&{} \quad \ldots \\ 1!c_1&{} \quad \frac{1!}{1!}(c_0+r_1)&{} \quad r_0&{} \quad 0&{} \quad 0&{} \quad \ldots \\ 2!c_2&{} \quad \frac{2!}{1!}(c_1+r_2)&{} \quad \frac{2!}{2!}(c_0+2r_1)&{} \quad r_0&{} \quad 0&{} \quad \ldots \\ 3!c_3&{} \quad \frac{3!}{1!}(c_2+r_3)&{} \quad \frac{3!}{2!}(c_1+2r_2)&{} \quad \frac{3!}{3!}(c_0+3r_1)&{} \quad r_0&{} \quad \ldots \\ 4!c_4&{} \quad \frac{4!}{1!}(c_3+r_4)&{} \quad \frac{4!}{2!}(c_2+2r_3)&{} \quad \frac{4!}{3!}(c_1+3r_2)&{} \quad \frac{4!}{4!}(c_0+4r_1)&{} \quad \ldots \\ \vdots &{} \quad \vdots &{} \quad \vdots &{} \quad \vdots &{} \quad \vdots &{} \quad \ddots \end{pmatrix}. \nonumber \\ \end{aligned}$$

(1.14)

The detailed information about the exponential Riordan arrays can be found in [2,3,4, 6, 9].

Subgroups in group theory are significant topic, and there has been extensive research on the subgroups of the Riordan group. The subgroups of the Riordan group are investigated and isomorphisms among these subgroups are given by Jean-Louis and Nkwanta in [15]. Some generalization of the Riordan arrays are examined. The generalized Riordan arrays are defined using the generalized generating functions, and their properties are investigated in [26]. The generalization forms of the Riordan subgroups are defined in [8, 16].

Another generalization of the Riordan arrays is the almost-Riordan arrays. Let’s now present the definition and some properties of the almost-Riordan arrays, as introduced by Barry in [5].

Let’s consider the following formal power series:

$$\begin{aligned}{} & {} a(x)=a_{0}+a_{1}x+a_{2}x^2+\dots , \\{} & {} g(x)=g_{0}+g_{1}x+g_{2}x^2+\dots \end{aligned}$$

and

$$\begin{aligned} f(x)=f_{0}+f_{1}x+f_{2}x^2+\dots \end{aligned}$$

with $a_{0}\ne 0$, $g_{0}\ne 0$, $f_{0}=0$ and $f_{1}\ne 0$. The notation for first order almost-Riordan arrays is (a(x)|g(x), f(x)). The generating function of the kth column of a first order almost-Riordan array is

$$\begin{aligned}{} & {} a(x),\hspace{5.0pt}\hspace{5.0pt}for\hspace{5.0pt}k=0, \end{aligned}$$

(1.15)

$$\begin{aligned}{} & {} xg(x)(f(x))^{k-1},\hspace{5.0pt}\hspace{5.0pt}for\hspace{5.0pt}k=1,2,3,\dots . \end{aligned}$$

(1.16)

Additionally, the multiplication of two almost-Riordan arrays is defined as follows:

$$\begin{aligned} (a(x)|g(x),f(x))(b(x)|h(x),l(x))=((a(x)|g(x),f(x))b(x)\big \vert g(x)h(f(x)),l(f(x))) \end{aligned}$$

(1.17)

where the operation (a(x)|g(x), f(x))b(x) is given as

$$\begin{aligned} (a(x)|g(x),f(x))b(x)=b_0a(x)+xg(x)\frac{b(f(x))-b_0}{f(x)}. \end{aligned}$$

(1.18)

The set of the first order almost-Riordan arrays is a group with the multiplication defined in (1.17). The identity element of this group is

$$\begin{aligned} I=(1|1,x) \end{aligned}$$

(1.19)

and the inverse of the almost-Riordan arrays is given as follows:

$$\begin{aligned} (a(x)|g(x),f(x))^{-1}=\left( \frac{1}{a_0}\left( 1-\frac{x}{g(\overline{f}(x))}\frac{a(\overline{f}(x))-a_0}{\overline{f}(x)}\right) \bigg \vert \frac{1}{g(\overline{f}(x))},\overline{f}(x)\right) . \end{aligned}$$

(1.20)

The sequence characterizations of the almost-Riordan arrays are provided in [1]. The pseduo-involutions and involutions in the almost-Riordan arrays are studied in [7, 25].

Based on the preceding studies, we introduce the exponential almost-Riordan arrays. Also, we examine the subgroups of the exponential almost-Riordan group and provide some isomorphisms among them. Furthermore, the production matrix of the exponential almost-Riordan arrays is presented in this paper.

2 Exponential almost-Riordan arrays

In this section, we define the exponential almost-Riordan arrays and give some row sums of them. Additionally, we introduce the exponential almost-Riordan group.

Definition 2.1

Let’s consider the following exponential generating functions:

$$\begin{aligned}{} & {} a(x)=a_{0}+a_{1}x+a_{2}\frac{x^2}{2!}+a_{3}\frac{x^3}{3!}+\dots , \\{} & {} g(x)=g_{0}+g_{1}x+g_{2}\frac{x^2}{2!}+g_{3}\frac{x^3}{3!}+\dots \end{aligned}$$

and

$$\begin{aligned} f(x)=f_{0}+f_{1}x+f_{2}\frac{x^2}{2!}+f_{3}\frac{x^3}{3!}+\dots \end{aligned}$$

with $a_{0}\ne 0$, $g_{0}\ne 0$, $f_{0}=0$ and $f_{1}\ne 0$. The notation for the exponential almost-Riordan arrays is [a(x)|g(x), f(x)]. The exponential generating function of the kth column of the exponential almost-Riordan arrays is defined as follows:

$$\begin{aligned} a(x),&\hspace{5.0pt}&for\hspace{5.0pt}k=0, \end{aligned}$$

(2.1)

$$\begin{aligned} \frac{1}{(k-1)!}\int _{0}^{x}g(t)f^{k-1}(t)dt,&\hspace{5.0pt}\hspace{5.0pt}&for\hspace{5.0pt}k=1,2,3,\dots . \end{aligned}$$

(2.2)

Example 2.2

The exponential almost-Riordan array $\left[ \frac{1}{1-x}|\frac{1}{1-x},\ln (\frac{1}{1-x})\right]$ is given as

$$\begin{aligned} \begin{pmatrix} 1&{} \quad 0&{} \quad 0&{} \quad 0&{} \quad 0&{} \quad 0&{} \quad 0&{} \quad 0&{} \quad \ldots \\ 1&{} \quad 1&{} \quad 0&{} \quad 0&{} \quad 0&{} \quad 0&{} \quad 0&{} \quad 0&{} \quad \ldots \\ 2&{} \quad 1&{} \quad 1&{} \quad 0&{} \quad 0&{} \quad 0&{} \quad 0&{} \quad 0&{} \quad \ldots \\ 6&{} \quad 2&{} \quad 3&{} \quad 1&{} \quad 0&{} \quad 0&{} \quad 0&{} \quad 0&{} \quad \ldots \\ 24&{} \quad 6&{} \quad 11&{} \quad 6&{} \quad 1&{} \quad 0&{} \quad 0&{} \quad 0&{} \quad \ldots \\ 120&{} \quad 24&{} \quad 50&{} \quad 35&{} \quad 10&{} \quad 1&{} \quad 0&{} \quad 0&{} \quad \ldots \\ 720&{} \quad 120&{} \quad 274&{} \quad 225&{} \quad 85&{} \quad 15&{} \quad 1&{} \quad 0&{} \quad \ldots \\ 5040&{} \quad 720&{} \quad 1764&{} \quad 1624&{} \quad 735&{} \quad 175&{} \quad 21&{} \quad 1&{} \quad \ldots \\ \vdots &{} \quad \vdots &{} \quad \vdots &{} \quad \vdots &{} \quad \vdots &{} \quad \vdots &{} \quad \vdots &{} \quad \vdots &{} \quad \ddots \end{pmatrix} \end{aligned}$$

where the column 0th is composed of the factorial numbers which is the sequence A000142 in OEIS [24].

Proposition 2.3

Let $D=(d_{n,k})_{n,k\ge 0}=[a(x)|g(x),f(x)]$ be an exponential almost-Riordan array. Then, the elements of D are as follows:

$$\begin{aligned} d_{n,k}=&n![x^n]a(x),\hspace{5.0pt}&for\hspace{5.0pt}k=0, \end{aligned}$$

(2.3)

$$\begin{aligned} d_{n,k}=&\frac{(n-1)!}{(k-1)!}[x^{n-1}]g(x)f^{k-1}(x),\hspace{5.0pt}&for\hspace{5.0pt}k\ge 1. \end{aligned}$$

(2.4)

Proof

From (1.4) and (2.1), the equation (2.3) is clear. Considering (1.4) and (2.2), we have

$$\begin{aligned} d_{n,k}=\frac{n!}{(k-1)!}[x^n]F(x) \end{aligned}$$

where $F(x)=\int _{0}^{x}g(t)f^{k-1}(t)dt.$ From (1.3), we get

$$\begin{aligned}{}[x^n]F(x)=\frac{1}{n}[x^{n-1}]F'(x)=\frac{1}{n}[x^{n-1}]g(x)f^{k-1}(x). \end{aligned}$$

Then, we obtain

$$\begin{aligned} d_{n,k}=\frac{(n-1)!}{(k-1)!}[x^{n-1}]g(x)f^{k-1}(x). \end{aligned}$$

$\square$

Example 2.4

Let’s take the following exponential almost-Riordan array:

$$\begin{aligned} D=\left[ \frac{\alpha e^{\alpha x}-\beta e^{\beta x}}{\alpha -\beta }\bigg \vert e^x(1+x),x\right] \end{aligned}$$

where $\alpha =\frac{1+\sqrt{5}}{2}$ and $\beta =\frac{1-\sqrt{5}}{2}$. From (2.3), we have

$$\begin{aligned} d_{n,0}=n![x^n]\left( \frac{\alpha e^{\alpha x}-\beta e^{\beta x}}{\alpha -\beta }\right) =n![x^n]\sum _{n=0}^{\infty }F_{n+1}\frac{x^n}{n!}. \end{aligned}$$

Using (1.4), we obtain $d_{n,0}=F_{n+1}$ that $F_{n}$ is the nth Fibonacci number. By using (2.4), we get

$$\begin{aligned} d_{n,k}=\frac{(n-1)!}{(k-1)!}[x^{n-1}]e^x(1+x)x^{k-1}. \end{aligned}$$

From (1.2), we have

$$\begin{aligned} d_{n,k}=\frac{(n-1)!}{(k-1)!}[x^{n-k}](e^x+xe^x). \end{aligned}$$

Considering the formal power series of $e^x$, we obtain

$$\begin{aligned} d_{n,k}=(n-k+1)\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) , \end{aligned}$$

which is the sequence A093375 in OEIS [24]. The matrix D is given as follows:

$$\begin{aligned} D= \begin{pmatrix} 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 1&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 2&{}\quad 2&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 3&{}\quad 3&{}\quad 4&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 5&{}\quad 4&{}\quad 9&{}\quad 6&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 8&{}\quad 5&{}\quad 16&{}\quad 18&{}\quad 8&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 13&{}\quad 6&{}\quad 25&{}\quad 40&{}\quad 30&{}\quad 10&{}\quad 1&{}\quad 0&{}\quad \ldots \\ 21&{}\quad 7&{}\quad 36&{}\quad 75&{}\quad 80&{}\quad 45&{}\quad 12&{}\quad 1&{}\quad \ldots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{pmatrix}. \end{aligned}$$

Proposition 2.5

Let [a(x)|g(x), f(x)] be an exponential almost-Riordan array, and let h(x) be an exponential generating function of the sequence $\{h_n\}_{n\ge 0}$. Then,

$$\begin{aligned}{}[a(x)|g(x),f(x)]h(x)=h_0a(x)+\int _0^xg(t)h'(f(t))dt \end{aligned}$$

(2.5)

where $h'(x)$ is the first order derivative of h(x).

Proof

If we consider the product of [a(x)|g(x), f(x)] and h(x), we obtain

$$\begin{aligned}{} & {} {}[a(x)|g(x),f(x)]h(x)\\{} & {} \qquad =h_0a(x)+h_1\int _0^xg(t)dt+h_2\int _0^xg(t)f(t)dt+h_3\int _0^xg(t)\frac{f^2(t)}{2!}dt+\dots . \end{aligned}$$

Then, we have

$$\begin{aligned} \begin{aligned}{}[a(x)|g(x),f(x)]h(x)&=h_0a(x)+\int _0^xg(t)\left( h_1+h_2f(t)+h_3\frac{f^2(t)}{2!}+\dots \right) dt\\&=h_0a(x)+\int _0^xg(t)h'(f(t))dt. \end{aligned} \end{aligned}$$

$\square$

For example, let’s consider $[e^{2x}|e^x,x]$ and $h(x)=e^x$, we have

$$\begin{aligned}{}[e^{2x}|e^x,x]e^x=e^{2x}+\int _0^xe^{2t}dt=\frac{1}{2}(3e^{2x}-1). \end{aligned}$$

The sequence of this exponential generating function is A003945 in OEIS [24].

Proposition 2.6

Let $D=(d_{n,k})_{n,k\ge 0}=[a(x)|g(x),f(x)]$ be an exponential almost-Riordan array, and let h(x) be the exponential generating function of the sequence $\{h_n\}_{n\ge 0}$. Then,

$$\begin{aligned} \sum _{k=0}^{n}d_{n,k}h_k=h_0a_n+n![x^n]\int _0^xg(t)h'(f(t)) dt. \end{aligned}$$

(2.6)

Specially, taking $h(x)=e^x$ in (2.6), we obtain the row sums for the exponential almost-Riordan array. Similarly, taking $h(x)=e^{-x}$ in (2.6), the alternating row sums for the exponential almost-Riordan array are obtained, as stated in the following corollary.

Corollary 2.7

The row sums and the alternating row sums for an exponential almost-Riordan array $D=(d_{n,k})_{n,k\ge 0}=[a(x)|g(x),f(x)]$ are given by the following expressions:

$$\begin{aligned} \sum _{k=0}^{n}d_{n,k}=a_n+n![x^n]\int _0^xg(t)e^{f(t)}dt \end{aligned}$$

(2.7)

and

$$\begin{aligned} \sum _{k=0}^{n}(-1)^kd_{n,k}=a_n-n![x^n]\int _0^xg(t)e^{-f(t)}dt. \end{aligned}$$

(2.8)

Example 2.8

Let’s consider the exponential almost-Riordan array $D=[1|2-e^x,x]$. The matrix D is given as

$$\begin{aligned}D=\left( \begin{matrix} 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 0&{}\quad -1&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 0&{}\quad -1&{}\quad -2&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 0&{}\quad -1&{}\quad -3&{}\quad -3&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 0&{}\quad -1&{}\quad -4&{}\quad -6&{}\quad -4&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 0&{}\quad -1&{}\quad -5&{}\quad -10&{}\quad -10&{}\quad -5&{}\quad 1&{}\quad 0&{}\quad \ldots \\ 0&{}\quad -1&{}\quad -6&{}\quad -15&{}\quad -20&{}\quad -15&{}\quad -6&{}\quad 1&{}\quad \ldots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{matrix} \right) . \end{aligned}$$

Firstly, let’s find the row sums of the matrix $D=(d_{n,k})_{n,k\ge 0}$. Using (2.7), we have

$$\begin{aligned} \sum _{k=0}^{n}d_{n,k}=a_n+n![x^n]\int _0^x(2-e^t)e^tdt. \end{aligned}$$

Then, we obtain

$$\begin{aligned} \sum _{k=0}^{n}d_{n,k}=a_n+n![x^n]\left( 2\sum _{n=1}^{\infty }\frac{x^n}{n!}-\frac{1}{2}\sum _{n=1}^{\infty }\frac{2^nx^n}{n!}\right) . \end{aligned}$$

Hence, we have

$$\begin{aligned} \sum _{k=0}^{n}d_{n,k}= \left\{ \begin{array}{ll} 1, &{} n=0 \\ 2-2^{n-1}, &{} n\ge 1 \end{array} \right. \end{aligned}$$

where is the sequence A122958 in OEIS [24].

Now, let’s find the alternating row sums of the matrix $D=(d_{n,k})_{n,k\ge 0}$. By utilizing the equation (2.8), we obtain

$$\begin{aligned} \sum _{k=0}^{n}(-1)^kd_{n,k}=a_n-n![x^n]\int _0^x(2-e^t)e^{-t}dt. \end{aligned}$$

Considering the formal power series of $e^{-x}$, we find

$$\begin{aligned} \sum _{k=0}^{n}(-1)^kd_{n,k}= \left\{ \begin{array}{ll} 1, &{} n=0 \\ -1, &{} n=1 \\ 2(-1)^n,&{}n\ge 2 \end{array} \right. . \end{aligned}$$

Specially, if we take $h(x)=e^x(x+1)$ in (2.6), we obtain the weighted row sums for the exponential almost-Riordan array. Similarly, taking $h(x)=e^{-x}(1-x)$ in (2.6), the alternating weighted row sums for the exponential almost-Riordan array are obtained as stated in the following corollary:

Corollary 2.9

The weighted row sums and the alternating weighted row sums for an exponential almost-Riordan array $D=(d_{n,k})_{n,k\ge 0}=[a(x)|g(x),f(x)]$ are given as follows:

$$\begin{aligned} \sum _{k=0}^{n}(k+1)d_{n,k}=a_n+n![x^n]\int _0^x(f(t)+2)g(t)e^{f(t)}dt \end{aligned}$$

(2.9)

and

$$\begin{aligned} \sum _{k=0}^{n}(-1)^k(k+1)d_{n,k}=a_n+n![x^n]\int _0^x(f(t)-2)g(t)e^{-f(t)}dt. \end{aligned}$$

(2.10)

Example 2.10

Let’s consider the exponential almost-Riordan array given by $D=\left[ \frac{1}{1-x}\bigg \vert e^{x^2},2x\right]$. The matrix D is obtained as

$$\begin{aligned} D= \begin{pmatrix} 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 1&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 2&{}\quad 0&{}\quad 2&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 6&{}\quad 2&{}\quad 0&{}\quad 4&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 24&{}\quad 0&{}\quad 12&{}\quad 0&{}\quad 8&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 120&{}\quad 12&{}\quad 0&{}\quad 48&{}\quad 0&{}\quad 16&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 720&{}\quad 0&{}\quad 120&{}\quad 0&{}\quad 160&{}\quad 0&{}\quad 32&{}\quad 0&{}\quad \ldots \\ 5040&{}\quad 120&{}\quad 0&{}\quad 720&{}\quad 0&{}\quad 480&{}\quad 0&{}\quad 64&{}\quad \ldots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{pmatrix}. \end{aligned}$$

Let’s calculate the weighted row sums of the matrix D. Using (2.9), we have

$$\begin{aligned} \sum _{k=0}^{n}(k+1)d_{n,k}=a_n+n![x^n]\int _0^x(2t+2)e^{t^2+2t}dt. \end{aligned}$$

Considering the formal power series of $e^{x^2+2x}$, we find the weighted row sums of the matrix D as follows:

$$\begin{aligned} \sum _{k=0}^{n}(k+1)d_{n,k}= \left\{ \begin{array}{ll} 1, &{} n=0 \\ n!+d_n, &{} n\ge 1 \\ \end{array} \right. \end{aligned}$$

where $\{d_n\}$ is the sequence A000898 in OEIS [24].

Now, we find the alternating weighted row sums of the matrix D. If we use (2.10), we get

$$\begin{aligned} \sum _{k=0}^{n}(-1)^k(k+1)d_{n,k}=a_n+n![x^n]\int _0^x(2t-2)e^{t^2-2t}dt. \end{aligned}$$

From the formal power series of $e^{x^2-2x}$, we obtain the alternating weighted row sums of the matrix D as follows:

$$\begin{aligned} \sum _{k=0}^{n}(-1)^k(k+1)d_{n,k}= \left\{ \begin{array}{ll} 1, &{} n=0 \\ n!+(-1)^nd_n, &{} n\ge 1 \\ \end{array} \right. \end{aligned}$$

where $\{d_n\}$ is the sequence A000898 in OEIS [24].

The multiplication operation of two exponential almost-Riordan arrays is defined as follows:

$$\begin{aligned} \left[ h_0a(x)+\int _0^xg(t)h'(f(t))dt\bigg \vert g(x)h(f(x)),l(f(x))\right] . \end{aligned}$$

(2.11)

Example 2.11

Let $D_1$ and $D_2$ be the exponential almost-Riordan arrays as follows:

$$\begin{aligned} D_1=\left[ \frac{1}{1-x}\bigg \vert 2-e^{x},x\right] \hspace{5.0pt}\hspace{5.0pt}{and} \hspace{5.0pt}\hspace{5.0pt}D_2=\left[ 1\bigg \vert 1-x,x\left( 1-\frac{x}{2}\right) \right] . \end{aligned}$$

Then, we have

$$\begin{aligned} D_1D_2=\left[ \frac{1}{1-x}\bigg \vert (2-e^{x})(1-x),x\left( 1-\frac{x}{2}\right) \right] . \end{aligned}$$

Namely, the matrix

$$\begin{aligned} D_1D_2=\left( \begin{matrix} 1&{}{}\quad 0&{}{}\quad 0&{}{}\quad 0&{}{}\quad 0&{}{}\quad 0&{}{}\quad 0&{}{}\quad 0&{}{}\quad \ldots \\ 1&{}{}\quad 1&{}{}\quad 0&{}{}\quad 0&{}{}\quad 0&{}{}\quad 0&{}{}\quad 0&{}{}\quad 0&{}{}\quad \ldots \\ 2&{}{}\quad -2&{}{}\quad 1&{}{}\quad 0&{}{}\quad 0&{}{}\quad 0&{}{}\quad 0&{}{}\quad 0&{}{}\quad \ldots \\ 6&{}{}\quad 1&{}{}\quad -5&{}{}\quad 1&{}{}\quad 0&{}{}\quad 0&{}{}\quad 0&{}{}\quad 0&{}{}\quad \ldots \\ 24&{}{}\quad 2&{}{}\quad 9&{}{}\quad -9&{}{}\quad 1&{}{}\quad 0&{}{}\quad 0&{}{}\quad 0&{}{}\quad \ldots \\ 120&{}{}\quad 3&{}{}\quad 2&{}{}\quad 33&{}{}\quad -14&{}{}\quad 1&{}{}\quad 0&{}{}\quad 0&{}{}\quad \ldots \\ 720&{}{}\quad 4&{}{}\quad -5&{}{}\quad -40&{}{}\quad 85&{}{}\quad -20&{}{}\quad 1&{}{}\quad 0&{}{}\quad \ldots \\ 5040&{}{}\quad 5&{}{}\quad -21&{}{}\quad -30&{}{}\quad -245&{}{}\quad 180&{}{}\quad -27&{}{}\quad 1&{}{}\quad \ldots \\ \vdots &{}{}\quad \vdots &{}{}\quad \vdots &{}{}\quad \vdots &{}{}\quad \vdots &{}{}\quad \vdots &{}{}\quad \vdots &{}{}\quad \vdots &{}{}\quad \ddots \end{matrix} \right) \end{aligned}$$

is equal to

$$\begin{aligned}&\left( \begin{array}{rrrrrrrrr} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 2 &{}\quad -1 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 6 &{}\quad -1 &{}\quad -2 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 24 &{}\quad -1 &{}\quad -3 &{}\quad -3 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 120 &{}\quad -1 &{}\quad -4 &{}\quad -6 &{}\quad -4 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 720 &{}\quad -1 &{}\quad -5 &{}\quad -10 &{}\quad -10 &{}\quad -5 &{}\quad 1 &{}\quad 0 &{}\quad \ldots \\ 5040 &{}\quad -1 &{}\quad -6 &{}\quad -15 &{}\quad -20 &{}\quad -15 &{}\quad -6 &{}\quad 1 &{}\quad \ldots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{array}\right) \\&\left( \begin{array}{rrrrrrrrr} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 0 &{}\quad -1 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 0 &{}\quad 0 &{}\quad -3 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 0 &{}\quad 0 &{}\quad 3 &{}\quad -6 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 15 &{}\quad -10 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad -15 &{}\quad 45 &{}\quad -15 &{}\quad 1 &{}\quad 0 &{}\quad \ldots \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad -105 &{}\quad 105 &{}\quad -21 &{}\quad 1 &{}\quad \ldots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{array} \right) . \end{aligned}$$

Theorem 2.12

The set of the exponential almost-Riordan arrays is a group with the multiplication defined in (2.11), and denoted by $\mathcal {R}_e^a$.

Proof

The set $\mathcal {R}_e^a$ is closed and associative for multiplication in (2.11). The identity element of this set is [1|1, x]. Additionally, the inverse of the exponential almost-Riordan arrays is defined as follows:

$$\begin{aligned}{}[a(x)|g(x),f(x)]^{-1}=\left[ \frac{1}{a_0}\left( 1-\int _0^x\frac{a'(\overline{f}(t))}{g(\overline{f}(t))}dt\right) \bigg \vert \frac{1}{g(\overline{f}(x))},\overline{f}(x)\right] \end{aligned}$$

(2.12)

where $\overline{f}(x)$ is the compositional inverse of f(x). $\square$

Example 2.13

Let’s consider an exponential almost-Riordan array

$D=\left[ 1\bigg \vert 1-x,x\left( 1-\frac{x}{2}\right) \right]$. The matrix D is given as

$$\begin{aligned} D=\left( \begin{matrix} 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 0&{}\quad -1&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 0&{}\quad 0&{}\quad -3&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 0&{}\quad 0&{}\quad 3&{}\quad -6&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 0&{}\quad 0&{}\quad 0&{}\quad 15&{}\quad -10&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 0&{}\quad 0&{}\quad 0&{}\quad -15&{}\quad 45&{}\quad -15&{}\quad 1&{}\quad 0&{}\quad \ldots \\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad -105&{}\quad 105&{}\quad -21&{}\quad 1&{}\quad \ldots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{matrix} \right) . \end{aligned}$$

Using (2.12), we have $D^{-1}=\left[ 1\bigg \vert \frac{1}{\sqrt{1-2x}},1-\sqrt{1-2x}\right] .$ The matrix $D^{-1}$ is as follows:

$$\begin{aligned} D^{-1}= \begin{pmatrix} 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 0&{}\quad 1&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 0&{}\quad 3&{}\quad 3&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 0&{}\quad 15&{}\quad 15&{}\quad 6&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 0&{}\quad 105&{}\quad 105&{}\quad 45&{}\quad 10&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad \ldots \\ 0&{}\quad 945&{}\quad 945&{}\quad 420&{}\quad 105&{}\quad 15&{}\quad 1&{}\quad 0&{}\quad \ldots \\ 0&{}\quad 10395&{}\quad 10395&{}\quad 4725&{}\quad 1260&{}\quad 210&{}\quad 21&{}\quad 1&{}\quad \ldots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{pmatrix}. \end{aligned}$$

3 Subgroups and isomorphisms

In this section, we consider the subgroups of the exponential almost-Riordan group $\mathcal {R}_e^a$ and provide the isomorphisms between these subgroups.

Proposition 3.1

The set of the elements in the form [a(x)|g(x), x] is a subgroup of $\mathcal {R}^{a}_{e}$.

Proof

Let [a(x)|g(x), x] and [b(x)|h(x), x] be the elements in the set. Using (2.11) and (2.12), we have

$$\begin{aligned}{} & {} [a(x)|g(x),x][b(x)|h(x),x] =\left[ b_0a(x)+\int _0^xg(t)b'(t)dt\bigg \vert g(x)h(x),x\right] \end{aligned}$$

and

$$\begin{aligned}{}[a(x)|g(x),x]^{-1}=\left[ \frac{1}{a_0}\left( 1-\int _0^x\frac{a'(t)}{g(t)}dt\right) \bigg \vert \frac{1}{g(x)},x\right] . \end{aligned}$$

Therefore, the set of the exponential almost-Riordan arrays in the form $[a(x)|g(x),x]$ is a subgroup of $\mathcal {R}_e^a$. $\square$

For example, an exponential almost-Riordan array belonging to the subgroup, denoted as $\left[ e^\frac{x}{1-x}\bigg \vert \frac{1}{1-x},x\right] ,$ is given as follows:

$$\begin{aligned} \begin{pmatrix} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 3 &{}\quad 1 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 13 &{}\quad 2 &{}\quad 2 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 73 &{}\quad 6 &{}\quad 6 &{}\quad 3 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 501 &{}\quad 24 &{}\quad 24 &{}\quad 12 &{}\quad 4 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 4051 &{}\quad 120 &{}\quad 120 &{}\quad 60 &{}\quad 20 &{}\quad 5 &{}\quad 1 &{}\quad 0 &{}\quad \ldots \\ 37633 &{}\quad 720 &{}\quad 720 &{}\quad 360 &{}\quad 120 &{}\quad 30 &{}\quad 6 &{}\quad 1 &{}\quad \ldots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{pmatrix} \end{aligned}$$

where the column 0th consists of the elements of the sequence A000262 in OEIS [24].

Proposition 3.2

The set of the elements in the form [a(x)|1, f(x)] is a subgroup of $\mathcal {R}_e^a$.

Proof

Let [a(x)|1, f(x)] and [b(x)|1, l(x)] be the elements in the set. Using (2.11) and (2.12), we obtain

$$\begin{aligned}{}[a(x)|1,f(x)][b(x)|1,l(x)]=\left[ b_0a(x)+\int _0^xb'(f(t))dt\bigg \vert 1,l(f(x))\right] . \end{aligned}$$

and

$$\begin{aligned}{}[a(x)|1,f(x)]^{-1}=\left[ \frac{1}{a_0}\left( 1-\int _0^xa'(\overline{f}(t))dt\right) \bigg \vert 1,\overline{f}(x)\right] . \end{aligned}$$

Namely, the set constitutes a subgroup of the exponential almost-Riordan group. $\square$

For example, an exponential almost-Riordan array belonging to the subgroup, denoted as $\left[ e^x\big \vert 1,ln(\frac{1}{1-x})\right]$, is provided as follows:

$$\begin{aligned} \begin{pmatrix} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 0 &{}\quad 1 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 0 &{}\quad 2 &{}\quad 3 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 0 &{}\quad 6 &{}\quad 11 &{}\quad 6 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 0 &{}\quad 24 &{}\quad 50 &{}\quad 35 &{}\quad 10 &{}\quad 1 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 0 &{}\quad 120 &{}\quad 274 &{}\quad 225 &{}\quad 85 &{}\quad 15 &{}\quad 1 &{}\quad \ldots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{pmatrix}. \end{aligned}$$

Proposition 3.3

The set of the elements in the form $[a(x)|f'(x),f(x)]$ is a subgroup of $\mathcal {R}_e^a$.

Proof

Let $[a(x)|f'(x),f(x)]$ and $[b(x)|l'(x),l(x)]$ be the elements of the set. Using (2.11) and (2.12), we get

$$\begin{aligned}{}[a(x)|f'(x),f(x)][b(x)|l'(x),l(x)]=\left[ b(f(x))+b_0a(x)-b_0\bigg \vert f'(x)l'(f(x)),l(f(x))\right] \end{aligned}$$

and

$$\begin{aligned}{} & {} {}[a(x)|f'(x),f(x)]^{-1}\\{} & {} \qquad =\left[ \frac{1}{a_0}\left( 1-\int _0^x\frac{a'(\overline{f}(t))}{f'(\overline{f}(t))}dt\right) \bigg \vert \frac{1}{f'(\overline{f}(x))},\overline{f}(x)\right] . \end{aligned}$$

Therefore, the set is a subgroup of the exponential almost-Riordan group $\mathcal {R}_e^a$. $\square$

For example, the exponential almost-Riordan array belonging to the subgroup, denoted as $\left[ e^{e^{x}-1}\big \vert {e^x},e^x-1\right]$, is given as follows:

$$\begin{aligned} \begin{pmatrix} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 2 &{}\quad 1 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 5 &{}\quad 1 &{}\quad 3 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 15 &{}\quad 1 &{}\quad 7 &{}\quad 6 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 52 &{}\quad 1 &{}\quad 15 &{}\quad 25 &{}\quad 10 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 203 &{}\quad 1 &{}\quad 31 &{}\quad 90 &{}\quad 65 &{}\quad 15 &{}\quad 1 &{}\quad 0 &{}\quad \ldots \\ 877 &{}\quad 1 &{}\quad 63 &{}\quad 301 &{}\quad 350 &{}\quad 140 &{}\quad 21 &{}\quad 1 &{}\quad \ldots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{pmatrix}. \end{aligned}$$

where the column 0th is composed of the sequence A000110 in OEIS [24].

Theorem 3.4

The set of the elements in the form $\left[ 1|f'(x),f(x)\right]$ is a subgroup of $\mathcal {R}_e^a$, and it’s isomorphic to the associated subgroup of the exponential Riordan group.

Proof

It follows from Proposition 3.3 that the set of the exponential almost-Riordan arrays in the form $[1|f'(x),f(x)]$ constitutes a subgroup of $\mathcal {R}_e^a$. We consider the map $\varphi$ such that

$$\begin{aligned} \varphi \left( [1|f'(x),f(x)]\right) =[1,f(x)]. \end{aligned}$$

Let’s show that $\varphi$ is a homomorphism. We get

$$\begin{aligned} \begin{aligned} \varphi \left( [1|f'(x),f(x)][1|l'(x),l(x)]\right) =&\varphi ([1|f'(x)l'(f(x)),l(f(x))])\\ =&[1,l(f(x))]\\ =&[1,f(x)][1,l(x)]\\ =&\varphi ([1|f'(x),f(x)])\varphi ([1|l'(x),l(x)]). \end{aligned} \end{aligned}$$

Because $\varphi$ is one to one and onto, $\varphi$ is an isomorphism. $\square$

Theorem 3.5

The set of the elements in the form [1|1, f(x)] is a subgroup of $\mathcal {R}_e^a$ and it’s isomorphic to the subgroup in the form $[1|f'(x),f(x)]$ of $\mathcal {R}_e^a$.

Proof

It follows from the Proposition 3.2 that the set of the exponential almost-Riordan arrays in the form [1|1, f(x)] constitutes a subgroup. Let’s consider the map $\varphi$ such that

$$\begin{aligned} \varphi ([1|f'(x),f(x)])=[1|1,f(x)]. \end{aligned}$$

Let’s show that $\varphi$ is a homomorphism. We have

$$\begin{aligned} \begin{aligned} \varphi \left( [1|f'(x),f(x)][1|l'(x),l(x)]\right) =&\varphi ([1|f'(x)l'(f(x)),l(f(x))])\\ =&[1|1,l(f(x))]\\ =&[1|1,f(x)][1|1,l(x)]\\ =&\varphi ([1|f'(x),f(x)])\varphi ([1|l'(x),l(x)]). \end{aligned} \end{aligned}$$

For $\varphi$ is one to one and onto, $\varphi$ is an isomorphism. $\square$

Proposition 3.6

$D=[1|f'(x),f(x)]$ is an involution if and only if $f(x)=\overline{f}(x)$.

Proof

Let D be an involution. Then

$$\begin{aligned} D^2=[1|f'(x),f(x)][1|f'(x),f(x)]=[1|1,x]. \end{aligned}$$

Using (2.11), we obtain

$$\begin{aligned}{}[1|f'(x)f'(f(x)),f(f(x))]=[1|1,x]. \end{aligned}$$

Namely, $f(x)=\overline{f}(x)$. Conversely, let’s take $f(x)=\overline{f}(x)$. Then, we find

$$\begin{aligned} D^{2}=[1|f'(x)f'(f(x)),f(f(x))]=[1|1,x]. \end{aligned}$$

$\square$

Proposition 3.7

The set of the elements in the form $\left[ a(x)\bigg \vert \frac{xf'(x)}{f(x)},f(x)\right]$ is a subgroup of $\mathcal {R}_e^a$.

Proof

Let’s take elements $\left[ a(x)\bigg \vert \frac{xf'(x)}{f(x)},f(x)\right]$ and $\left[ b(x)\bigg \vert \frac{xl'(x)}{l(x)},l(x)\right]$ from the set. Using (2.11) and (2.12), we have

$$\begin{aligned}{} & {} \left[ a(x)\bigg \vert \frac{xf'(x)}{f(x)},f(x)\right] \left[ b(x)\bigg \vert \frac{xl'(x)}{l(x)},l(x)\right] \\{} & {} \quad =\left[ b_0a(x)+\int _0^x\frac{tf'(t)b'(f(t))}{f(t)}dt\bigg \vert \frac{xf'(x)l'(f(x))}{l(f(x))},l(f(x))\right] \end{aligned}$$

and

$$\begin{aligned} \left[ a(x)\bigg \vert \frac{xf'(x)}{f(x)},f(x)\right] ^{-1}=\left[ \frac{1}{a_0}\left( 1-\int _0^x\frac{ta'(\overline{f}(t))}{\overline{f}(t)f'(\overline{f}(t))}dt\right) \bigg \vert \frac{x(\overline{f}(x))'}{\overline{f}(x)},\overline{f}(x)\right] . \end{aligned}$$

Therefore, the set is a subgroup of $\mathcal {R}_e^a$. $\square$

For example, an exponential almost-Riordan array belonging to the subgroup, denoted as $\left[ e^{-x}\big \vert 1+\frac{x}{2+x},x(1+\frac{x}{2})\right] ,$ is given as follows:

$$\begin{aligned} \left( \begin{matrix} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ -1 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad \frac{1}{2} &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ -1 &{}\quad -\frac{1}{2} &{}\quad 2 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad \frac{3}{4} &{}\quad 0 &{}\quad \frac{9}{2} &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ -1 &{}\quad -\frac{3}{2} &{}\quad 0 &{}\quad 6 &{}\quad 8 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad \frac{15}{4} &{}\quad 0 &{}\quad 0 &{}\quad 25 &{}\quad \frac{25}{2} &{}\quad 1 &{}\quad 0 &{}\quad \ldots \\ -1 &{}\quad -\frac{45}{4} &{}\quad 0 &{}\quad 0 &{}\quad 30 &{}\quad \frac{135}{2} &{}\quad 18 &{}\quad 1 &{}\quad \ldots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{matrix} \right) . \end{aligned}$$

Theorem 3.8

The set of the elements in the form $\left[ 1\bigg \vert \frac{xf'(x)}{f(x)},f(x)\right]$ is a subgroup of $\mathcal {R}_e^a$ and it’s isomorphic to the subgroup in the form $[1|f'(x),f(x)]$ of $\mathcal {R}_e^a$.

Proof

From Proposition 3.7, the set of $\left[ 1\bigg \vert \frac{xf'(x)}{f(x)},f(x)\right]$ is a subgroup. Let’s consider the map $\varphi$ such that

$$\begin{aligned} \varphi \left( \left[ 1\bigg |\frac{xf'(x)}{f(x)},f(x)\right] \right) =[1|f'(x),f(x)]. \end{aligned}$$

$\varphi$ is one to one and onto. Then, we obtain

$$\begin{aligned} \begin{aligned}&\varphi \left( \left[ 1\bigg |\frac{xf'(x)}{f(x)},f(x)\right] \left[ 1\bigg |\frac{xl'(x)}{l(x)},l(x)\right] \right) \\&\qquad = \varphi \left( \left[ 1\bigg |\frac{xf'(x)l'(f(x))}{l(f(x))},l(f(x))\right] \right) \\&\qquad =[1|f'(x)l'(f(x)),l(f(x))]\\&\qquad =[1|f'(x),f(x)][1|l'(x),l(x)]\\&\qquad = \varphi \left( \left[ 1\bigg |\frac{xf'(x)}{f(x)},f(x)\right] \right) \varphi \left( \left[ 1\bigg |\frac{xl'(x)}{l(x)},l(x)\right] \right) . \end{aligned} \end{aligned}$$

Consequently, $\varphi$ is an isomorphism. $\square$

Proposition 3.9

The set of the elements in the forms [a(x)|g(x), xg(x)] or $[a(x)|\frac{f(x)}{x},f(x)]$ is a subgroup of $\mathcal {R}_e^a$.

For example, an exponential almost-Riordan array belonging to the subgoup, denoted as $\left[ e^x\big \vert \frac{1}{1-x},\frac{x}{1-x}\right]$, is given as follows:

$$\begin{aligned} \begin{pmatrix} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 1 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 2 &{}\quad 4 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 6 &{}\quad 18 &{}\quad 9 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 24 &{}\quad 96 &{}\quad 72 &{}\quad 16 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 120 &{}\quad 600 &{}\quad 600 &{}\quad 200 &{}\quad 25 &{}\quad 1 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 720 &{}\quad 4320 &{}\quad 5400 &{}\quad 2400 &{}\quad 450 &{}\quad 36 &{}\quad 1 &{}\quad \ldots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{pmatrix}. \end{aligned}$$

Theorem 3.10

The set of the elements in the form $\left[ 1\bigg \vert \frac{f(x)}{x},f(x)\right]$ is a subgroup of $\mathcal {R}_e^a$ and it’s isomorphic to the subgroup of the form $\left[ 1|1,f(x)\right]$ of $\mathcal {R}_e^a$.

Proof

Considering Proposition 3.9, the set of exponential almost-Riordan arrays in the form $\left[ 1\bigg \vert \frac{f(x)}{x},f(x)\right]$ is a subgroup of $\mathcal {R}_e^a$. Let’s take the map $\varphi$ such that

$$\begin{aligned} \varphi ([1|1,f(x)])=\left[ 1\bigg |\frac{f(x)}{x},f(x)\right] . \end{aligned}$$

It’s clear that $\varphi$ is one to one and onto. Also, we have

$$\begin{aligned} \begin{aligned} \varphi \left( [1|1,f(x)][1|1,l(x)]\right) =&\varphi ([1|1,l(f(x))])\\ =&\left[ 1\bigg |\frac{l(f(x))}{x},l(f(x))\right] \\ =&\left[ 1\bigg |\frac{f(x)}{x},f(x)\right] \left[ 1\bigg |\frac{l(x)}{x},l(x)\right] \\ =&\varphi ([1|1,f(x)])\varphi ([1|1,l(x)]). \end{aligned} \end{aligned}$$

$\square$

Now, we give the definitions of the stochastic and stabilizer subgroups of $\mathcal {R}_e^a$.

Proposition 3.11

The following subset of the group $\mathcal {R}_e^a$ is a subgroup, known as the stochastic subgroup.

$$\begin{aligned} \mathfrak {D}=\left\{ \left[ a(x)\bigg \vert \frac{e^x-a'(x)}{e^{f(x)}},f(x)\right] :a_0=1,a_1\ne 1,f_0=0,f_1\ne 0\right\} . \end{aligned}$$

Proof

Firstly, we show that the row sums equal to 1. Using (2.5), we get

$$\begin{aligned} \left[ a(x)\bigg \vert \frac{e^x-a'(x)}{e^{f(x)}},f(x)\right] e^x=a(x)+\int _0^x(e^t-a'(t))dt=e^x. \end{aligned}$$

Let $\left[ a(x)\big \vert \frac{e^x-a'(x)}{e^{f(x)}},f(x)\right]$ and $\left[ b(x)\big \vert \frac{e^x-b'(x)}{e^{l(x)}},l(x)\right]$ be two elements of the set $\mathfrak {D}$. Using the multiplication defined in (2.11), we have

$$\begin{aligned} \left[ b_0a(x)+\int _0^x\frac{e^t-a'(t)}{e^{f(t)}}b'(f(t))dt\bigg \vert \frac{e^x-a'(x)}{e^{f(x)}}\frac{e^{f(x)}-b'(f(x))}{e^{l(f(x))}},l(f(x))\right] . \end{aligned}$$

If we consider (2.12), we find

$$\begin{aligned}{} & {} \left[ a(x)\bigg \vert \frac{e^x-a'(x)}{e^{f(x)}},f(x)\right] ^{-1}\\{} & {} \qquad =\left[ 1-\int _0^x\frac{a'(\overline{f}(t))}{e^{\overline{f}(t)}-a'(\overline{f}(t))}e^tdt\bigg \vert \frac{e^x}{e^{\overline{f}(x)}-a'(\overline{f}(x))},\overline{f}(x)\right] . \end{aligned}$$

Thus, the set $\mathfrak {D}$ is a subgroup of $\mathcal {R}_e^a$. $\square$

For example, an exponential almost-Riordan array belonging to the stochastic subgroup, denoted as $[e^{e^x-1-x}|1-e^{e^x-1-x}(1-e^{-x}),x]$, is given as follows:

$$\begin{aligned} \left( \begin{matrix} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad -1 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 1 &{}\quad -2 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 4 &{}\quad -4 &{}\quad 3 &{}\quad -3 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 11 &{}\quad 3 &{}\quad -16 &{}\quad 6 &{}\quad -4 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 41 &{}\quad -21 &{}\quad 15 &{}\quad -40 &{}\quad 10 &{}\quad -5 &{}\quad 1 &{}\quad 0 &{}\quad \ldots \\ 162 &{}\quad -10 &{}\quad -126 &{}\quad 45 &{}\quad -80 &{}\quad 15 &{}\quad -6 &{}\quad 1 &{}\quad \ldots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{matrix} \right) \end{aligned}$$

where the elements of the column 0th is the elements of the sequence A000296 in OEIS [24]. Also, we can see that the row sums of this matrix equal to 1.

Theorem 3.12

The set of the elements in the form $[1\big \vert e^{x-f(x)},f(x)]$ is a subgroup of $\mathcal {R}_e^a$. The map $\varphi$, defined such that

$$\begin{aligned} \varphi \left( [1|e^{x-f(x)},f(x)]\right) =[e^{x-f(x)},f(x)], \end{aligned}$$

is an isomorphism from the subgroup of the group $\mathcal {R}_e^a$ to exponential Riordan group in the form $[e^{x-f(x)},f(x)]$.

Proof

It follows from Proposition 3.11 that, the set of the exponential almost-Riordan arrays in the form $[1|e^{x-f(x)},f(x)]$ constitutes a subgroup. Let’s show that $\varphi$ is a homomorphism. We have

$$\begin{aligned} \begin{aligned} \varphi \left( [1|e^{x-f(x)},f(x)][1|e^{x-l(x)},l(x)]\right) =&\varphi ([1|e^{x-l(f(x))},l(f(x))])\\ =&[e^{x-l(f(x))},l(f(x))]\\ =&[e^{x-f(x)},f(x)][e^{x-l(x)},l(x)]\\ =&\varphi ([1|e^{x-f(x)},f(x)])\varphi ([1|e^{x-l(x)},l(x)]). \end{aligned} \end{aligned}$$

Also, $\varphi$ is one to one and onto. $\square$

Theorem 3.13

The sets of the elements in the forms $[1|f'(x),f(x)]$ and $[1|e^{x-f(x)},f(x)]$ are isomorphic subgroups.

Proof

Let’s consider the map $\varphi$ such that

$$\begin{aligned} \varphi ([1|f'(x),f(x)])=[1|e^{x-f(x)},f(x)]. \end{aligned}$$

Clearly, $\varphi$ is a homomorphism, one to one and onto. $\square$

Proposition 3.14

$D=[1|e^{x-f(x)},f(x)]$ is an involution if and only if $f(x)=\overline{f}(x)$.

Proof

Let D be an involution.Then

$$\begin{aligned} D^2=[1|e^{x-f(x)},f(x)][1|e^{x-f(x)},f(x)]=[1|1,x]. \end{aligned}$$

Using (2.11), we get

$$\begin{aligned}{}[1|e^{x-f(f(x))},f(f(x))]=[1|1,x]. \end{aligned}$$

Thus, we have $f(x)=\overline{f}(x)$. Conversely, let’s take $f(x)=\overline{f}(x)$, we find

$$\begin{aligned} D^{2}=[1|e^{x-f(f(x))},f(f(x))]=[1|1,x]. \end{aligned}$$

$\square$

Let h(x) be the exponential generating function of the sequence $\{h_n\}_{n\ge 0}$. A column vector whose elements are determined by the generating function h(x) must satisfy the following condition in order to an exponential almost-Riordan array to stabilize it.

$$\begin{aligned}{}[a(x)|g(x),f(x)]h(x)=h(x). \end{aligned}$$

(3.1)

Now, we define the stabilizer subgroup of the exponential almost-Riordan group $\mathcal {R}_e^a$.

Proposition 3.15

The following subset of the group $\mathcal {R}_e^a$ is a subgroup, known as the stabilizer subgroup.

$$\begin{aligned} \mathfrak {B}=\left\{ \left[ a(x)\bigg \vert \frac{h'(x)-h_0a'(x)}{h'(f(x))},f(x)\right] :a_0=1,a_1h_0\ne h_1,h_1\ne 0,f_0=0, f_1\ne 0\right\} . \end{aligned}$$

Proof

Using (2.5), we get

$$\begin{aligned} \left[ a(x)\bigg \vert \frac{h'(x)-h_0a'(x)}{h'(f(x))},f(x)\right] h(x)=h_0a(x)+\int _0^x(h'(t)-h_0a'(t))dt=h(x). \end{aligned}$$

Let $\left[ a(x)\big \vert \frac{h'(x)-h_0a'(x)}{h'(f(x))},f(x)\right]$ and $\left[ b(x)\big \vert \frac{h'(x)-h_0b'(x)}{h'(l(x))},l(x)\right]$ be two elements of the set $\mathfrak {B}$. Using (2.11), we get

$$\begin{aligned} \left[ a(x)+\int _0^x\frac{h'(t)-h_0a'(t)}{h'(f(t))}b'(f(t))dt\bigg \vert \frac{h'(x)-h_0a'(x)}{h'(f(x))}\frac{h'(f(x))-h_0b'(f(x))}{h'(l(f(x)))},l(f(x))\right] . \end{aligned}$$

From (2.12), we obtain the inverse as

$$\begin{aligned}{} & {} \left[ a(x)\bigg \vert \frac{h'(x)-h_0a'(x)}{h'(f(x))},f(x)\right] ^{-1}\\{} & {} \quad =\left[ 1-\int _0^x\frac{a'(\overline{f}(t))h'(t)}{h'({\overline{f}(t)})-h_0a'(\overline{f}(t))}dt\bigg \vert \frac{h'(x)}{h'({\overline{f}(x)})-h_0a'(\overline{f}(x))},\overline{f}(x)\right] . \end{aligned}$$

Thus, the set $\mathfrak {B}$ is a subgroup of $\mathcal {R}_e^a$. $\square$

For example, let’s take $h(x)=2e^x+1$ and $\left[ e^x\big |-\frac{1}{2}e^{x-e^x+1},e^x-1\right]$. Hence, we obtain

$$\begin{aligned} \left( \begin{matrix} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad -\frac{1}{2} &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 0 &{}\quad -\frac{1}{2} &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad \frac{1}{2} &{}\quad -\frac{1}{2} &{}\quad -\frac{1}{2} &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad \frac{1}{2} &{}\quad 1 &{}\quad -\frac{3}{2} &{}\quad -\frac{1}{2} &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad -1 &{}\quad \frac{9}{2} &{}\quad -\frac{1}{2} &{}\quad -3 &{}\quad -\frac{1}{2} &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad -\frac{9}{2} &{}\quad \frac{9}{2} &{}\quad \frac{25}{2} &{}\quad -\frac{15}{2} &{}\quad -5 &{}\quad -\frac{1}{2} &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad -\frac{9}{2} &{}\quad -25 &{}\quad 52 &{}\quad 10 &{}\quad -25 &{}\quad -\frac{15}{2} &{}\quad -\frac{1}{2} &{}\quad \ldots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{matrix} \right) \end{aligned}$$

It’s noted that the multiplication of this matrix and $(3,2,2,2,\dots )^T$ is $(3,2,2,2,\dots )^T$.

4 Production matrix

In this section, we present the production matrix of the exponential almost-Riordan arrays.

Proposition 4.1

Let $D=(d_{n,k})_{n,k\ge 0}= [a(x)|g(x),f(x)]$ be an exponential almost-Riordan array and $P=(p_{n,k})_{n,k\ge 0}$ be the production matrix of the matrix D. Then, we have

$$\begin{aligned}&\begin{aligned} d_{n+1,0}=&p_{0,0}d_{n,0}+p_{1,0}d_{n,1}+p_{2,0}d_{n,2}+\dots \end{aligned} \end{aligned}$$

(4.1)

$$\begin{aligned}&\begin{aligned} d_{n+1,1}=&p_{0,1}d_{n,0}+p_{1,1}d_{n,1}+p_{2,1}d_{n,2}+\dots \end{aligned} \end{aligned}$$

(4.2)

$$\begin{aligned}&\begin{aligned} d_{n+1,k+1}=&p_{k,k+1}d_{n,k}+p_{k+1,k+1}d_{n,k+1}+p_{k+2,k+1}d_{n,k+2}+\dots . \end{aligned} \end{aligned}$$

(4.3)

Proposition 4.2

Let the matrix P be the production matrix of $D=[a(x)|g(x),f(x)]$. The exponential generating function of the kth column of the matrix P is given as follows:

$$\begin{aligned}&\begin{aligned} P_0(x)=&\frac{a_1}{a_0}+\int _0^x\frac{1}{g(\overline{f}(t))}\left( a''(\overline{f}(t))-\frac{a_1}{a_0}a'(\overline{f}(t))\right) dt, \end{aligned} \end{aligned}$$

(4.4)

$$\begin{aligned}&\begin{aligned} P_1(x)=&\frac{g_0}{a_0}+\int _0^x\frac{1}{g(\overline{f}(t))}\left( g'(\overline{f}(t))-\frac{g_0}{a_0}a'(\overline{f}(t))\right) dt, \end{aligned} \end{aligned}$$

(4.5)

and

$$\begin{aligned} \begin{aligned} P_{k+1}(x)=&\int _0^x\frac{t^{k-1}}{(k-1)!}\left( \frac{t}{k}\frac{g'(\overline{f}(t))}{g(\overline{f}(t))}+f'(\overline{f}(t))\right) dt \end{aligned} \end{aligned}$$

(4.6)

for $k\ge 1$.

Proof

Considering (4.1), (2.1) and (2.2), we have

$$\begin{aligned} a'(x)=p_{0,0}a(x)+p_{1,0}\int _0^xg(t)dt+p_{2,0}\int _0^xg(t)f(t)dt+\dots . \end{aligned}$$

Then, we get

$$\begin{aligned} a''(x)=p_{0,0}a'(x)+p_{1,0}g(x)+p_{2,0}g(x)f(x)+\dots . \end{aligned}$$

Hence, we have

$$\begin{aligned} \frac{a''(x)-p_{0,0}a'(x)}{g(x)}=P_0'(f(x)). \end{aligned}$$

Considering (1.8), we find $p_{0,0}=\frac{a_1}{a_0}$. From the previous equation, the equation (4.4) is obtained.

Using (4.2), (2.1) and (2.2), we obtain

$$\begin{aligned} g(x)=p_{0,1}a(x)+p_{1,1}\int _0^xg(t)dt+p_{2,1}\int _0^xg(t)f(t)dt+\dots . \end{aligned}$$

Hence,

$$\begin{aligned} g'(x)=p_{0,1}a'(x)+p_{1,1}g(x)+p_{2,1}g(x)f(x)+\dots . \end{aligned}$$

Then, we have

$$\begin{aligned} \frac{g'(x)-p_{0,1}a'(x)}{g(x)}=P_1'(f(x)). \end{aligned}$$

Considering (1.8), we find $p_{0,1}=\frac{g_0}{a_0}$. Therefore, the equation (4.5) is found.

Considering (4.3) and (2.2), we have

$$\begin{aligned}{} & {} g(x)\frac{f^{k}(x)}{k!}=p_{k,k+1}\left( \int _0^xg(t) \frac{f^{k-1}(t)}{(k-1)!}dt\right) +p_{k+1,k+1} \left( \int _0^xg(t)\frac{f^{k}(t)}{k!}dt\right) \\{} & {} \quad +p_{k+2,k+1}\left( \int _0^xg(t)\frac{f^{k+1}(t)}{(k+1)!}dt\right) +\dots . \end{aligned}$$

Then, we get

$$\begin{aligned}{} & {} \frac{g'(x)f^k(x)}{k!}+\frac{g(x)f'(x)f^{k-1}(x)}{(k-1)!}\\{} & {} \quad =g(x)\left( p_{k,k+1}\frac{f^{k-1}(x)}{(k-1)!}+p_{k+1,k+1} \frac{f^{k}(x)}{k!}+p_{k+2,k+1}\frac{f^{k+1}(x)}{(k+1)!}+\dots \right) . \end{aligned}$$

Thus, we have

$$\begin{aligned} \frac{f^{k-1}(x)}{(k-1)!}\left( \frac{g'(x)}{g(x)}\frac{f(x)}{k}+f'(x)\right) =P_{k+1}'(f(x)). \end{aligned}$$

Hence, the equation (4.6) is clear. $\square$

Corollary 4.3

Let $P=(p_{n,k})_{n,k\ge 0}$ be the production matrix of the exponential almost-Riordan array [a(x)|g(x), f(x)]. For $n\ge 1$,

$$\begin{aligned}&\begin{aligned} p_{n,0}=&c_{n-1}-\frac{a_1}{a_0}r_{n-1}, \end{aligned} \end{aligned}$$

(4.7)

$$\begin{aligned}&\begin{aligned} p_{n,1}=&z_{n-1}-\frac{g_0}{a_0}r_{n-1}, \end{aligned} \end{aligned}$$

(4.8)

and for $k\ge 2$,

$$\begin{aligned} p_{n,k}=\left( {\begin{array}{c}n-1\\ k-1\end{array}}\right) z_{n-k}+\left( {\begin{array}{c}n-1\\ k-2\end{array}}\right) s_{n-k+1} \end{aligned}$$

(4.9)

where $r_n,c_n,z_n$ and $s_n$ are the nth elements of the following exponential generating functions, respectively.

$$\begin{aligned}&\begin{aligned} r(x)=&\sum _{i=0}^\infty r_i\frac{x^i}{i!}=\frac{a'(\overline{f}(x))}{g(\overline{f}(x))}, \end{aligned} \end{aligned}$$

(4.10)

$$\begin{aligned}&\begin{aligned} c(x)=&\sum _{i=0}^\infty c_i\frac{x^i}{i!}=\frac{a''(\overline{f}(x))}{g(\overline{f}(x))}, \end{aligned} \end{aligned}$$

(4.11)

$$\begin{aligned}&\begin{aligned} Z(x)=&\sum _{i=0}^\infty z_i\frac{x^i}{i!}=\frac{g'(\overline{f}(x))}{g(\overline{f}(x))}, \end{aligned} \end{aligned}$$

(4.12)

$$\begin{aligned}&\begin{aligned} S(x)=&\sum _{i=0}^\infty s_i\frac{x^i}{i!}=f'(\overline{f}(x)). \end{aligned} \end{aligned}$$

(4.13)

Proof

Considering (1.4) and (4.4), we get

$$\begin{aligned} p_{n,0}=n![x^n]\int _0^x\left( c(t)-\frac{a_1}{a_0}r(t)\right) dt. \end{aligned}$$

Using (1.3), we have

$$\begin{aligned} p_{n,0}=(n-1)![x^{n-1}]\left( c(x)-\frac{a_1}{a_0}r(x)\right) . \end{aligned}$$

From (1.4), the equation (4.7) is found. By the similar way, the equations (4.8) and (4.9) are obtained. $\square$

According to Corollary 4.3, the production matrix of the exponential almost Riordan array [a(x)|g(x), f(x)] is given as follows:

$$\begin{aligned} P=\left( \begin{array}{lllllll} \frac{a_{1}}{a_{0}} &{}\quad \frac{g_{0}}{a_{0}} &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ c_{0}-\frac{a_{1}}{a_{0}}r_{0} &{}\quad z_{0}-\frac{g_{0}}{a_{0}}r_{0} &{}\quad s_{0} &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ c_{1}-\frac{a_{1}}{a_{0}}r_{1} &{}\quad z_{1}-\frac{g_{0}}{a_{0}}r_{1} &{}\quad z_{0}+s_{1} &{}\quad s_{0} &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ c_{2}-\frac{a_{1}}{a_{0}}r_{2} &{}\quad z_{2}-\frac{g_{0}}{a_{0}}r_{2} &{}\quad 2z_{1}+s_{2} &{}\quad z_{0}+2s_{1} &{}\quad s_{0} &{}\quad 0 &{}\quad \ldots \\ c_{3}-\frac{a_{1}}{a_{0}}r_{3} &{}\quad z_{3}-\frac{g_{0}}{a_{0}}r_{3} &{}\quad 3z_{2}+s_{3} &{}\quad 3z_{1}+3s_{2} &{}\quad z_{0}+3s_{1} &{}\quad s_{0} &{}\quad \ldots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{array} \right) . \end{aligned}$$

(4.14)

Corollary 4.4

Let $D=(d_{n,k})_{n,k\ge 0}=[a(x)|g(x),f(x)]$ be an exponential almost-Riordan array. Then,

$$\begin{aligned}&\begin{aligned} d_{n+1,0}=&\frac{a_1}{a_0}a_n+\sum _{i=1}^n\left( c_{i-1}-\frac{a_1}{a_0}r_{i-1}\right) d_{n,i}, \end{aligned} \end{aligned}$$

(4.15)

$$\begin{aligned}&\begin{aligned} d_{n+1,1}=&\frac{g_0}{a_0}a_n+\sum _{i=1}^n\left( z_{i-1}-\frac{g_0}{a_0}r_{i-1}\right) d_{n,i}, \end{aligned} \end{aligned}$$

(4.16)

and

$$\begin{aligned} d_{n+1,k}=\sum _{i=k-1}^n\left( \left( {\begin{array}{c}i-1\\ k-1\end{array}}\right) z_{i-k}+\left( {\begin{array}{c}i-1\\ k-2\end{array}}\right) s_{i-k+1}\right) d_{n,i} \end{aligned}$$

(4.17)

for $k\ge 2$.

Proof

Considering the Proposition 4.1 and Corollary 4.3, the result is clear. $\square$

Example 4.5

Let’s consider an exponential almost-Riordan array

$D=\left[ \frac{1}{1-x}\bigg |e^{-x},x(1+\frac{x}{2})\right]$, the matrix D is given as

$$\begin{aligned} D=\left( \begin{matrix} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 1 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 2 &{}\quad -1 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 6 &{}\quad 1 &{}\quad -1 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 24 &{}\quad -1 &{}\quad 0 &{}\quad 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 120 &{}\quad 1 &{}\quad 2 &{}\quad -3 &{}\quad 2 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad \ldots \\ 720 &{}\quad -1 &{}\quad -5 &{}\quad 5 &{}\quad -5 &{}\quad 5 &{}\quad 1 &{}\quad 0 &{}\quad \ldots \\ 5040 &{}\quad 1 &{}\quad 9 &{}\quad 0 &{}\quad -5 &{}\quad 0 &{}\quad 9 &{}\quad 1 &{}\quad \ldots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots \end{matrix} \right) . \end{aligned}$$

Now, we find the production matrix of the matrix D. By using (4.10), we have

$$\begin{aligned} r(x)=\frac{e^{\sqrt{2x+1}-1}}{(2-\sqrt{2x+1})^2} \end{aligned}$$

which is exponential generating function of the sequence

$$\begin{aligned} 1,3,8,25,87,386,1663,11313,39560\dots . \end{aligned}$$

If we use (4.11), we get

$$\begin{aligned} c(x)=\frac{2e^{\sqrt{2x+1}-1}}{(2-\sqrt{2x+1})^3} \end{aligned}$$

which is exponential generating function of the sequence

$$\begin{aligned} 2,8,30,122,548,2802,15638,100760\dots . \end{aligned}$$

Similarly, we obtain $Z(x)=-1.$ From (4.13), we find $S(x)=\sqrt{2x+1}$ which is the exponential generating function of the following sequence

$$\begin{aligned} 1,1,-1,3,-15,105,-945,10395,\dots . \end{aligned}$$

Then, the production matrix is obtained as follows:

$$\begin{aligned} P_D=\left( \begin{matrix} 1&{}1&{}0&{}0&{}0&{}0&{}0&{}0&{}\ldots \\ 1&{}-2&{}1&{}0&{}0&{}0&{}0&{}0&{}\ldots \\ 5&{}-3&{}0&{}1&{}0&{}0&{}0&{}0&{}\ldots \\ 22&{}-8&{}-1&{}1&{}1&{}0&{}0&{}0&{}\ldots \\ 97&{}-25&{}3&{}-3&{}2&{}1&{}0&{}0&{}\ldots \\ 461&{}-87&{}-15&{}12&{}-6&{}3&{}1&{}0&{}\ldots \\ 2416&{}-386&{}105&{}-75&{}30&{}-10&{}4&{}1&{}\ldots \\ \vdots &{}\vdots &{}\vdots &{}\vdots &{}\vdots &{}\vdots &{}\vdots &{}\vdots &{}\ddots \end{matrix} \right) . \end{aligned}$$