On Higher-Order Generalized Fibonacci Hybrid Numbers with q-Integer Components: New Properties, Recurrence Relations, and Matrix Representations

Abstract

Many properties of special numbers, such as sum formulas, symmetric properties, and their relationships with each other, have been studied in the literature with the help of the Binet formula and generating function. In this paper, higher-order generalized Fibonacci hybrid numbers with q-integer components are defined through the utilization of q-integers and higher-order generalized Fibonacci numbers. Several special cases of these newly established hybrid numbers are presented. The article explores the integration of q-calculus and hybrid numbers, resulting in the derivation of a Binet-like formula, novel identities, a generating function, a recurrence relation, an exponential generating function, and sum properties of hybrid numbers with quantum integer coefficients. Furthermore, new identities for these types of hybrids are obtained using two novel special matrices. To substantiate the findings, numerical examples are provided, generated with the assistance of Maple.

1. Introduction

Integer sequences possess a significance that transcends the realm of pure and applied mathematics, infiltrating diverse scientific domains such as physics and engineering. Among these sequences, the Fibonacci sequence stands out as particularly noteworthy, deriving its name from the esteemed Italian mathematician Leonardo Pisano, widely recognized as Fibonacci. This sequence demonstrates remarkable versatility, finding extensive applications across multiple disciplines, including mathematics, physics, and engineering. The symmetry properties of Fibonacci numbers are deeply intertwined with their mathematical structure and have far-reaching implications in number theory. One notable symmetry is that the ratio of consecutive Fibonacci numbers converges to the golden ratio, which is a fundamental constant in mathematics and nature. The subject matter has elicited considerable scholarly interest, prompting extensive investigation by diverse researchers [1,2]. The Fibonacci and Lucas sequences are defined by the following recurrence relations for $n\ge 2$:

$$F_n=F_{n-1}+F_{n-2}$$

and

$$L_n=L_{n-1}+L_{n-2}$$

with initial conditions $F_0=0$, $F_1=1,$$L_0=2$ and $L_1=1$, respectively. The Binet formulas for the Fibonacci numbers and Lucas numbers are given as follows:

$$F_n={\displaystyle \frac{{\alpha }^n-{\beta }^n}{\alpha -\beta }}$$

and

$$L_n={\alpha }^n+{\beta }^n,$$

where $\alpha$ and $\beta$ are the zeros of the characteristic equation $t^2-t-1=0.$ The concept of Fibonacci divisors, also referred to in the scholarly literature as higher-order Fibonacci numbers, was initially explored by Pashaev and Nalci (for further details, see [3] and also [4]). These higher-order Fibonacci numbers, or Fibonacci divisors, which are conjugate to $F_s$, are defined for integer values of $s\ge 1$ as follows:

$$F_n^{\left(s\right)}={\displaystyle \frac{F_{ns}}{F_s}}={\displaystyle \frac{{\left({\alpha }^s\right)}^n-{\left({\beta }^s\right)}^n}{{\alpha }^s-{\beta }^s}}.$$

(1)

Since $F_{ns}$ is divisible by $F_s$, the ratio of ${\displaystyle \frac{F_{ns}}{Fs}}$ is always an integer. Thus, all higher-order Fibonacci numbers $F_n^{\left(s\right)}$ are integers and $F_n^{\left(1\right)}=F_n$. In [5], Pashaev elucidated the essential properties of higher-order Fibonacci numbers and concurrently illustrated their application through various physical examples.

The ring $\mathbb{K}$ of hybrid numbers is defined by

$$\mathbb{K}=\left\{h_1+h_2\mathbf{i}+h_3\mathsf{ϵ}+h_4\mathbf{h}:h_1,h_2,h_3,h_4\in \mathbb{R}\right\},$$

where the hybrid units $\mathbf{i}$, $\mathsf{ϵ}$, $\mathbf{h}$ satisfy the relations ${\mathbf{i}}^2=-1, {\mathsf{ϵ}}^2=0, {\mathbf{h}}^2=1, \mathbf{ih}=\mathbf{hi}=\mathsf{ϵ}+\mathbf{i}$. This is introduced by Özdemir [6]. This number system is a general form of complex, hyperbolic, and dual number systems. Here, $\mathbf{i}$ is complex unit, $\mathsf{ϵ}$ is dual unit, and $\mathbf{h}$ is hyperbolic unit. These units are called hybrid units. In recent years, researchers from a variety of disciplines have studied this number system and used it in various fields of applied and computational science. For some applications of hybrid numbers, see [7,8] and references therein.

The conjugate of a hybrid number $h=h_1+h_2\mathbf{i}+h_3\mathsf{ϵ}+h_4\mathbf{h}$ is expressed by

$$\overline{h}=h_1-h_2\mathbf{i}-h_3\mathsf{ϵ}-h_4\mathbf{h}.$$

From the definition of hybrid numbers, the multiplication table of the hybrid units is given in the following table (Table 1):

Table 1. Multiplication table for $\mathbb{K}$.

.	1	i	$\mathsf{ϵ}$	h
$\mathbf{1}$	$\mathbf{1}$	$\mathbf{i}$	$\mathsf{ϵ}$	$\mathbf{h}$
$\mathbf{i}$	$\mathbf{i}$	$-\mathbf{1}$	$\mathbf{1}-\mathbf{h}$	$\mathsf{ϵ}+\mathbf{i}$
$\mathsf{ϵ}$	$\mathsf{ϵ}$	$\mathbf{h}+\mathbf{1}$	0	$-\mathsf{ϵ}$
$\mathbf{h}$	$\mathbf{h}$	$-\mathsf{ϵ}-\mathbf{i}$	$\mathsf{ϵ}$	$\mathbf{1}$

As can be seen from the table above, the multiplication of hybrid numbers is not commutative. Let $h=h_1+h_2\mathbf{i}+h_3\mathsf{ϵ}+h_4\mathbf{h}$ and $k=k_1+k_2\mathbf{i}+k_3\mathsf{ϵ}+k_4\mathbf{h}$ be two hybrid numbers. Some arithmetic operations related to hybrid numbers are as follows:

$\left(i\right)$

$h=k,$ if and only if, $h_1=k_1,$$h_2=k_2,$$h_3=k_3,$$h_4=k_4$(equality).

$\left(ii\right)$

$h+k=\left(h_1+k_1\right)+\left(h_2+k_2\right)\mathbf{i}+\left(h_3+k_3\right)\mathsf{ϵ}+\left(h_4+k_4\right)\mathbf{h}$(addition).

$\left(iii\right)$

$h-k=\left(h_1-k_1\right)+\left(h_2-k_2\right)\mathbf{i}+\left(h_3-k_3\right)\mathsf{ϵ}+\left(h_4-k_4\right)\mathbf{h}$(subtraction).

$\left(iv\right)$

$\alpha h=\alpha h_1+\alpha h_2\mathbf{i}+\alpha h_3\mathsf{ϵ}+\alpha h_4\mathbf{h}$(multiplication by scalar $\alpha \in \mathbb{R}$).

$\left(v\right)$

The multiplication of hybrid numbers h and k is given by

$$\begin{array}{ccc}h.g\hfill & =& \left(h_1{k}_1-h_2{k}_2+h_2{k}_3+h_3{k}_2+h_4{k}_4\right)+\left(h_1{k}_2+h_2{k}_1+h_2{k}_4-h_4{k}_2\right)\mathbf{i}\hfill \\ & & +\left(h_1{k}_3+h_2{k}_4+h_3{k}_1-h_3{k}_4-h_4{k}_2+h_4{k}_3\right)\mathsf{ϵ}\hfill \\ & & +\left(h_1{k}_4-h_2{k}_3+h_3{k}_2+h_4{k}_1\right)\mathbf{h}.\hfill \end{array}$$

(2)

Hybrid sequences whose components are terms of various integer sequences have been studied by many researchers. For instance, in [9], Szynal-Liana and Wloch examined the Fibonacci hybrid numbers and obtained some combinatorial properties of these numbers. In [10], Cerda-Morales introduced generalized hybrid Fibonacci numbers and obtained some of their properties. In [11], Kızılateş examined the q-Fibonacci and the q-Lucas hybrid numbers and obtained some combinatorial properties of these numbers. Kızılateş [12] also introduced and studied the Horadam hybrid polynomials called Horadam hybrinomials. In [13], with the help of the Fibonacci divisor numbers, Kızılateş and Kone introduced and examined the Fibonacci divisor hybrid numbers. By using higher-order generalized Fibonacci polynomials, Kızılateş et al. [14] studied the higher-order generalized Fibonacci hybrid polynomials called higher-order generalized Fibonacci hybrinomials and obtained some special cases and algebraic properties of the higher-order generalized Fibonacci hybrinomials. The q-integer [15] is expressed as

$${\left[m\right]}_q={\displaystyle \frac{1-q^m}{1-q}}=1+q+q^2+\cdots +q^{m-1},$$

for $m\in \mathbb{N}$ and $q\in \mathbb{R}\setminus \left\{1\right\}$. Some properties of q-numbers are as follows:

• ${[-m]}_q=-q^{-m}{\left[m\right]}_q.$
• ${[m+p]}_q={\left[p\right]}_q+q^p{\left[m\right]}_q.$
• ${[m-p]}_q={\left[m\right]}_q-q^{m-p}{\left[p\right]}_q.$
• ${\left[mp\right]}_q={\left[m\right]}_{q^p}{\left[p\right]}_q.$
• ${[m+2]}_{q^p}=\left(1+q^p\right){[m+1]}_{q^p}-q^p{\left[m\right]}_{q^p},$
  where ${\left[m\right]}_{q^p}={\displaystyle \frac{1-q^{pm}}{1-q^p}}.$

For more detailed information, interested readers should consult the remarkable monograph cited in [16]. Pashaev elucidated critical aspects of higher-order Fibonacci numbers while demonstrating their applicability in various physical contexts. For instance, in [5], through the utilization of quantum calculus, the infinite hierarchy of Golden quantum oscillators with the integer spectrum determined by Fibonacci divisors, the hierarchy of Golden coherent states, and associated Fock–Bargman representations in the space of complex analytic functions were derived.

The main aim of this paper is to extend the results obtained by Kızılateş [11] for hybrid numbers to higher-order generalized Fibonacci hybrid numbers with quantum integer components by considering a parameter $s.$ According to this definition, several families of hybrid numbers can be derived, which may or may not be documented in the existing literature. The structure of this paper is as follows: The second section introduces q-integer higher-order generalized hybrid numbers, presenting a Binet-like formula for these numbers and exploring numerous algebraic properties derived from this formula. Additionally, numerical examples are provided to validate the results, utilizing Maple for demonstration. The third section examines significant properties of these numbers through the application of two previously defined special matrices. The final section and Remark include the Maple code for the product of two hybrid numbers, as obtained from the examples, and offers suggestions for future research.

2. The Higher-Order Generalized Fibonacci Hybrid Numbers with Quantum Integer Components

In this section, we will first define the higher-order generalized Fibonacci hybrid numbers with quantum integer components and provide a large number of results involving them.

Definition 1.

Let $a$, $s\in \mathbb{Z}$, $\delta \in \mathbb{R}$, and $q\in \mathbb{R}\setminus \left\{1\right\}$, and let $\mathit{i}$, ε, and $\mathit{h}$ be units of ${\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)$. Then, we define higher-order generalized Fibonacci hybrid numbers ${\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)$ with quantum integer components as follows:

$${\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)={\delta }^{s\left(a-1\right)}{\left[a\right]}_{q^s}+{\delta }^{sa}{\left[a+1\right]}_{q^s}\mathit{i}+{\delta }^{s\left(a+1\right)}{\left[a+2\right]}_{q^s}\mathsf{\epsilon }+{\delta }^{s\left(a+2\right)}{\left[a+3\right]}_{q^s}\mathit{h}.$$

(3)

The aforementioned definition is notably comprehensive, encompassing the variables s, $\delta$, and q, and generalizing to various hybrid numbers documented in the literature. Several examples of these are enumerated below.

• For $s=1,$ Equation (3) becomes the q-Fibonacci-type hybrid number [11].
• For $\delta ={\displaystyle \frac{1+\sqrt{5}}{2}}$, $q=-{\displaystyle \frac{1}{{\delta }^2}}$, Equation (3) becomes the higher-order Fibonacci hybrid number [13].
• For $\delta ={\displaystyle \frac{1+\sqrt{5}}{2}}$, $q=-{\displaystyle \frac{1}{{\delta }^2}}$, and $s=1$, Equation (3) becomes the Fibonacci hybrid number [9].
• For $\delta =2$ and $q=-{\displaystyle \frac{1}{2}}$, Equation (3) becomes the higher-order Jacobsthal hybrid number.
• For $\delta =2$, $q=-{\displaystyle \frac{1}{2}}$, and $s=1$, Equation (3) becomes the Jacobsthal hybrid number [17].
• For $\delta =1+\sqrt{2}$ and $q=-{\displaystyle \frac{1}{{\left(1+\sqrt{2}\right)}^2}}$, Equation (3) becomes the higher-order Pell hybrid number.
• For $\delta =1+\sqrt{2}$, $q=-{\displaystyle \frac{1}{{\left(1+\sqrt{2}\right)}^2}}$, and $s=1,$ Equation (3) becomes the Pell hybrid number [18].

We note that the following identity holds:

$${\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)+\overline{{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)}=2{\delta }^{s\left(a-1\right)}{\left[a\right]}_{q^s}.$$

Now, we present the Binet-like formula for these types of newly established hybrid number sequences. Note that the Binet-like formula will play a key role in some of the results obtained in this section.

Theorem 1.

For a, $s\in \mathbb{Z}$, $\delta \in \mathbb{R}$, and $q\in \mathbb{R}\setminus \left\{1\right\},$ the Binet-like formula of ${\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)$ is expressed as follows:

$${\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)={\displaystyle \frac{{\delta }^{s\left(a-1\right)}\left(\mathfrak{h}-q^{sa}\mathfrak{g}\right)}{1-q^s}},$$

(4)

where $\mathfrak{h}=1+{\delta }^s\mathit{i}+{\delta }^{2s}\mathsf{\epsilon }+{\delta }^{3s}\mathit{h}$ and $\mathfrak{g}=1+{\left(\delta q\right)}^s\mathit{i}+{\left(\delta q\right)}^{2s}\mathsf{\epsilon }+{\left(\delta q\right)}^{3s}\mathit{h}$.

Proof. 

From the definition of ${\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)$, we obtain

$$\begin{array}{ccc}{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)\hfill & =& {\delta }^{s\left(a-1\right)}{\left[a\right]}_{q^s}+{\delta }^{sa}{\left[a+1\right]}_{q^s}\mathit{i}+{\delta }^{s\left(a+1\right)}{\left[a+2\right]}_{q^s}\mathsf{\epsilon }+{\delta }^{s\left(a+2\right)}{\left[a+3\right]}_{q^s}\mathit{h}\hfill \\ & =& {\displaystyle \frac{{\delta }^{s\left(a-1\right)}}{1-q^s}}\left(1-q^{sa}+{\delta }^s\left(1-q^{s\left(a+1\right)}\right)\mathit{i}+{\delta }^{2s}\left(1-q^{s\left(a+2\right)}\right)\mathsf{\epsilon }+{\delta }^{3s}\left(1-q^{s\left(a+3\right)}\right)\mathit{h}\right)\hfill \\ & =& {\displaystyle \frac{{\delta }^{s\left(a-1\right)}}{1-q^s}}\left(1+{\delta }^s\mathit{i}+{\delta }^{2s}\mathsf{\epsilon }+{\delta }^{3s}\mathit{h}-q^{sa}\left(1+{\left(\delta q\right)}^s\mathit{i}+{\left(\delta q\right)}^{2s}\mathsf{\epsilon }+{\left(\delta q\right)}^{3s}\mathit{h}\right)\right)\hfill \\ & =& {\displaystyle \frac{{\delta }^{s\left(a-1\right)}\left(\mathfrak{h}-q^{sa}\mathfrak{g}\right)}{1-q^s}}.\hfill \end{array}$$

Thus, our claim is proved. □

Example 1.

Substituting $\delta ={\displaystyle \frac{1+\sqrt{5}}{2}}$, $q=-{\displaystyle \frac{1}{{\delta }^2}}={\displaystyle \frac{-3+\sqrt{5}}{2}}$, $s=2$, and $a=2$ in (3), we have a second higher-order Fibonacci hybrid number as follows:

$${\mathbb{H}}_2^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right)=3+8\mathit{i}+21\mathsf{\epsilon }+55\mathit{h}.$$

On the other hand, by virtue of the Binet-like formula, we have

$$\begin{array}{ccc}\mathfrak{h}\hfill & =& 1+{\displaystyle \frac{3+\sqrt{5}}{2}}\mathit{i}+{\displaystyle \frac{7+3\sqrt{5}}{2}}\mathsf{\epsilon }+\left(9+4\sqrt{5}\right)\mathit{h},\hfill \\ \mathfrak{g}\hfill & =& 1+{\displaystyle \frac{3-\sqrt{5}}{2}}\mathit{i}+{\displaystyle \frac{7-3\sqrt{5}}{2}}\mathsf{\epsilon }+\left(9-4\sqrt{5}\right)\mathit{h}.\hfill \end{array}$$

(5)

From the right-side of Equation (4), we find that

$$\begin{array}{ccc}& & {\mathbb{H}}_2^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right)\hfill \\ & =& {\displaystyle \frac{{\delta }^2\left(\mathfrak{h}-q^4\mathfrak{g}\right)}{1-q^2}}\hfill \\ & =& {\displaystyle \frac{{\left(\frac{1+\sqrt{5}}{2}\right)}^2\left(\left(1+\frac{3+\sqrt{5}}{2}\mathit{i}+\frac{7+3\sqrt{5}}{2}\mathsf{\epsilon }+\left(9+4\sqrt{5}\right)\mathit{h}\right)-{\left(\frac{-3+\sqrt{5}}{2}\right)}^4\left(1+\frac{3-\sqrt{5}}{2}\mathit{i}+\frac{7-3\sqrt{5}}{2}\mathsf{\epsilon }+\left(9-4\sqrt{5}\right)\mathit{h}\right)\right)}{1-{\left(\frac{-3+\sqrt{5}}{2}\right)}^2}}\hfill \\ & & \\ & =& 3+8\mathit{i}+21\mathsf{\epsilon }+55\mathit{h}.\hfill \end{array}$$

Hence, Equality (4) is satisfied.

Theorem 2.

For a, $s\in \mathbb{Z}$, $\delta \in \mathbb{R}$, and $q\in \mathbb{R}\setminus \left\{1\right\}$, the following equation holds:

$${\mathbb{H}}_{-a}^{\left(-s\right)}\left(\delta ,q\right)=-{\left({\delta }^{2}q\right)}^s{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right).$$

Proof. 

By the aid of (4), we have

$$\begin{array}{ccc}{\mathbb{H}}_{-a}^{\left(-s\right)}\left(\delta ,q\right)\hfill & =& {\displaystyle \frac{{\delta }^{-s\left(-a-1\right)}\left(\mathfrak{h}-q^{sa}\mathfrak{g}\right)}{1-q^{-s}}}\hfill \\ & =& {\displaystyle \frac{{\delta }^{s\left(a+1\right)}\left(\mathfrak{h}-q^{sa}\mathfrak{g}\right)}{1-q^{-s}}}\hfill \\ & =& -q^s{\displaystyle \frac{{\delta }^{2s}{\delta }^{s\left(a-1\right)}\left(\mathfrak{h}-q^{sa}\mathfrak{g}\right)}{-q^s\left(1-q^{-s}\right)}}\hfill \\ & =& -{\left({\delta }^{2}q\right)}^s{\displaystyle \frac{{\delta }^{s\left(a-1\right)}\left(\mathfrak{h}-q^{sa}\mathfrak{g}\right)}{1-q^s}}\hfill \\ & =& -{\left({\delta }^{2}q\right)}^s{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right).\hfill \end{array}$$

The proof is completed. □

Theorem 3.

For a, $s\in \mathbb{Z}$, $\delta \in \mathbb{R}$, and $q\in \mathbb{R}\setminus \left\{1\right\}$, the following equation holds:

$${\mathbb{H}}_a^{\left(-s\right)}\left(\delta ,q\right)=-{\left({\delta }^{-2\left(a-1\right)}q\right)}^s{\displaystyle \frac{\mathfrak{h}-q^{-sa}\mathfrak{g}}{\mathfrak{h}-q^{sa}\mathfrak{g}}}{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right).$$

Proof. 

Using (4), we obtain

$$\begin{array}{ccc}{\mathbb{H}}_a^{\left(-s\right)}\left(\delta ,q\right)\hfill & =& {\displaystyle \frac{{\delta }^{-s\left(a-1\right)}\left(\mathfrak{h}-q^{-sa}\mathfrak{g}\right)}{1-q^{-s}}}\hfill \\ & =& {\displaystyle \frac{\left(1-q^s\right){\delta }^{-s\left(a-1\right)}\left(\mathfrak{h}-q^{-sa}\mathfrak{g}\right)}{\left(1-q^{-s}\right)\left(1-q^s\right)}}\hfill \\ & =& -{\displaystyle \frac{q^s{\delta }^{-s\left(a-1\right)}\left(\mathfrak{h}-q^{-sa}\mathfrak{g}\right){\delta }^{s\left(a-1\right)}\left(\mathfrak{h}-q^{sa}\mathfrak{g}\right)}{{\delta }^{s\left(a-1\right)}\left(\mathfrak{h}-q^{sa}\mathfrak{g}\right)\left(1-q^s\right)}}\hfill \\ & =& -{\left({\delta }^{-2\left(a-1\right)}q\right)}^s{\displaystyle \frac{\mathfrak{h}-q^{-sa}\mathfrak{g}}{\mathfrak{h}-q^{sa}\mathfrak{g}}}{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right).\hfill \end{array}$$

The proof is completed. □

Theorem 4.

For a, $s\in \mathbb{Z}$, $\delta \in \mathbb{R}$, and $q\in \mathbb{R}\setminus \left\{1\right\}$, the following equation holds:

$${\mathbb{H}}_{-a}^{\left(s\right)}\left(\delta ,q\right)={\delta }^{-2sa}{\displaystyle \frac{\left(\mathfrak{h}-q^{-sa}\mathfrak{g}\right)}{\mathfrak{h}-q^{sa}\mathfrak{g}}}{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right).$$

Proof. 

By virtue of the Equation (4), we obtain

$$\begin{array}{ccc}{\mathbb{H}}_{-a}^{\left(s\right)}\left(\delta ,q\right)\hfill & =& {\displaystyle \frac{{\delta }^{s\left(-a-1\right)}\left(\mathfrak{h}-q^{-sa}\mathfrak{g}\right)}{1-q^s}}\hfill \\ & =& {\delta }^{-2sa}{\displaystyle \frac{\left(\mathfrak{h}-q^{-sa}\mathfrak{g}\right)}{\mathfrak{h}-q^{sa}\mathfrak{g}}}{\displaystyle \frac{{\delta }^{s\left(a-1\right)}\left(\mathfrak{h}-q^{sa}\mathfrak{g}\right)}{1-q^s}}\hfill \\ & =& {\delta }^{-2sa}{\displaystyle \frac{\left(\mathfrak{h}-q^{-sa}\mathfrak{g}\right)}{\mathfrak{h}-q^{sa}\mathfrak{g}}}{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right).\hfill \end{array}$$

The proof is completed. □

This section will present the generating function of the hybrid sequence under consideration. It is widely recognized that the concept of generating functions constitutes a powerful analytical tool in combinatorial mathematics.

Theorem 5.

For a, $s\in \mathbb{Z}$, $b\in \mathbb{N}\cup \left\{0\right\}$, $\delta \in \mathbb{R}$, and $q\in \mathbb{R}\setminus \left\{1\right\}$, the generating function of ${\mathbb{H}}_{a+b}^{\left(s\right)}\left(\delta ,q\right)$ is given as

$$\sum _{a=0}^{\infty }{\mathbb{H}}_{a+b}^{\left(s\right)}\left(\delta ,q\right)t^a={\displaystyle \frac{{\delta }^{s\left(b-1\right)}}{1-q^s}}\left({\displaystyle \frac{\mathfrak{h}-q^{sb}\mathfrak{g}-{\left(\delta q\right)}^s\left(\mathfrak{h}-q^{s\left(b-1\right)}\mathfrak{g}\right)t}{1-{\delta }^s\left(1+q^s\right)t+{\left({\delta }^{2}q\right)}^s{t}^2}}\right).$$

(6)

Proof. 

Applying Equation (4) and using the series expansion yields

$$\begin{array}{ccc}\sum _{a=0}^{\infty }{\mathbb{H}}_{a+b}^{\left(s\right)}\left(\delta ,q\right)t^a\hfill & =& \sum _{a=0}^{\infty }{\displaystyle \frac{{\delta }^{s\left(a+b-1\right)}\left(\mathfrak{h}-q^{s\left(a+b\right)}\mathfrak{g}\right)}{1-q^s}}{t}^a\hfill \\ & =& {\displaystyle \frac{{\delta }^{s\left(b-1\right)}}{1-q^s}}\left(\mathfrak{h}\sum _{a=0}^{\infty }{\left({\delta }^{s}t\right)}^a-q^{sb}\mathfrak{g}\sum _{a=0}^{\infty }{\left({\left(\delta q\right)}^{s}t\right)}^a\right)\hfill \\ & =& {\displaystyle \frac{{\delta }^{s\left(b-1\right)}}{1-q^s}}\left(\mathfrak{h}{\displaystyle \frac{1}{1-{\delta }^{s}t}}-q^{sb}\mathfrak{g}{\displaystyle \frac{1}{1-{\left(\delta q\right)}^{s}t}}\right)\hfill \\ & =& {\displaystyle \frac{{\delta }^{s\left(b-1\right)}}{1-q^s}}\left({\displaystyle \frac{\mathfrak{h}\left(1-{\left(\delta q\right)}^{s}t\right)-q^{sb}\mathfrak{g}\left(1-{\delta }^{s}t\right)}{\left(1-{\delta }^{s}t\right)\left(1-{\left(\delta q\right)}^{s}t\right)}}\right)\hfill \\ & =& {\displaystyle \frac{{\delta }^{s\left(b-1\right)}}{1-q^s}}\left({\displaystyle \frac{\mathfrak{h}-{\left(\delta q\right)}^s\mathfrak{h}t-q^{sb}\mathfrak{g}+{\left(\delta q^b\right)}^s\mathfrak{g}t}{1-{\left(\delta q\right)}^{s}t-{\delta }^{s}t+{\left({\delta }^{2}q\right)}^s{t}^2}}\right)\hfill \\ & =& {\displaystyle \frac{{\delta }^{s\left(b-1\right)}}{1-q^s}}\left({\displaystyle \frac{\mathfrak{h}-q^{sb}\mathfrak{g}-{\left(\delta q\right)}^s\left(\mathfrak{h}-q^{s\left(b-1\right)}\mathfrak{g}\right)t}{1-{\delta }^s\left(1+q^s\right)t+{\left({\delta }^{2}q\right)}^s{t}^2}}\right),\hfill \end{array}$$

which gives (6). □

We can easily obtain the following conclusion if we take $b=0$ in (6).

Corollary 1.

For $a\in \mathbb{N}\cup \left\{0\right\}$, $s\in \mathbb{Z}$, $\delta \in \mathbb{R}$, and $q\in \mathbb{R}\setminus \left\{1\right\}$, the generating function of ${\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)$ is expressed by

$$\sum _{a=0}^{\infty }{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)t^a={\displaystyle \frac{1}{{\delta }^s\left(1-q^s\right)}}\left({\displaystyle \frac{\mathfrak{h}-\mathfrak{g}-{\delta }^s\left(q^s\mathfrak{h}-\mathfrak{g}\right)t}{1-{\delta }^s\left(1+q^s\right)t+{\left({\delta }^{2}q\right)}^s{t}^2}}\right).$$

Theorem 6.

For $a\in \mathbb{N}\cup \left\{0\right\}$, $s\in \mathbb{Z}$, $\delta \in \mathbb{R}$, and $q\in \mathbb{R}\setminus \left\{1\right\}$, the exponential generating function of ${\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)$ is given as follows:

$$\sum _{a=0}^{\infty }{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right){\displaystyle \frac{t^a}{a!}}={\displaystyle \frac{\mathfrak{h}{e}^{{\delta }^{s}t}-\mathfrak{g}{e}^{{\left(\delta q\right)}^{s}t}}{{\delta }^s\left(1-q^s\right)}}.$$

Proof. 

By virtue of the Equation (4), we have

$$\begin{array}{ccc}\sum _{a=0}^{\infty }{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right){\displaystyle \frac{t^a}{a!}}\hfill & =& \sum _{a=0}^{\infty }\left({\displaystyle \frac{{\delta }^{s\left(a-1\right)}\left(\mathfrak{h}-q^{sa}\mathfrak{g}\right)}{1-q^s}}\right){\displaystyle \frac{t^a}{a!}}\hfill \\ & =& {\displaystyle \frac{1}{{\delta }^s\left(1-q^s\right)}}\left(\mathfrak{h}\sum _{a=0}^{\infty }{\displaystyle \frac{{\left({\delta }^{s}t\right)}^a}{a!}}-\mathfrak{g}\sum _{a=0}^{\infty }{\displaystyle \frac{{\left({\left(\delta q\right)}^{s}t\right)}^a}{a!}}\right)\hfill \\ & =& {\displaystyle \frac{\mathfrak{h}{e}^{{\delta }^{s}t}-\mathfrak{g}{e}^{{\left(\delta q\right)}^{s}t}}{{\delta }^s\left(1-q^s\right)}}.\hfill \end{array}$$

So, the conclusion is proved. □

Theorem 7.

For $a,s\in \mathbb{N}$, $\delta \in \mathbb{R}$, and $q\in \mathbb{R}\setminus \left\{1\right\}$, ${\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)$ satisfies the following recurrence relation:

$${\mathbb{H}}_{a+1}^{\left(s\right)}\left(\delta ,q\right)={\delta }^s\left(1+q^s\right){\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)-{\left({\delta }^{2}q\right)}^s{\mathbb{H}}_{a-1}^{\left(s\right)}\left(\delta ,q\right).$$

(7)

Proof. 

Making use of the Binet-like formula of the ${\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)$, direct computation gives

$$\begin{array}{ccc}{\mathbb{H}}_{a+1}^{\left(s\right)}\left(\delta ,q\right)\hfill & =& {\displaystyle \frac{{\delta }^{sa}\left(\mathfrak{h}-q^{s\left(a+1\right)}\mathfrak{g}\right)}{1-q^s}}\hfill \\ & =& {\displaystyle \frac{{\delta }^{s\left(a-1\right)}{\delta }^s\left(\mathfrak{h}-q^{sa}\mathfrak{g}+q^{sa}\mathfrak{g}-q^{s\left(a+1\right)}\mathfrak{g}\right)}{1-q^s}}\hfill \\ & =& {\delta }^s{\displaystyle \frac{{\delta }^{s\left(a-1\right)}\left(\mathfrak{h}-q^{sa}\mathfrak{g}\right)}{1-q^s}}+{\delta }^{s\left(a-1\right)}{\delta }^s{\displaystyle \frac{q^{sa}\mathfrak{g}-q^{s\left(a+1\right)}\mathfrak{g}}{1-q^s}}\hfill \\ & =& {\delta }^s{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)+{\delta }^{sa}{\displaystyle \frac{q^s{q}^{s\left(a-1\right)}\mathfrak{g}-q^{s\left(a+1\right)}\mathfrak{g}}{1-q^s}}\hfill \\ & =& {\delta }^s{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)+{\delta }^{sa}{\displaystyle \frac{-q^s\mathfrak{h}+q^s{q}^{s\left(a-1\right)}\mathfrak{g}+q^s\mathfrak{h}-q^{s\left(a+1\right)}\mathfrak{g}}{1-q^s}}\hfill \\ & =& {\delta }^s{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)-{\delta }^{sa}{\displaystyle \frac{q^s\mathfrak{h}-q^s{q}^{s\left(a-1\right)}\mathfrak{g}-q^s\mathfrak{h}+q^{s\left(a+1\right)}\mathfrak{g}}{1-q^s}}\hfill \\ & =& {\delta }^s{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)-{\delta }^{sa}\left({\displaystyle \frac{q^s\left(\mathfrak{h}-q^{s\left(a-1\right)}\mathfrak{g}\right)}{1-q^s}}-{\displaystyle \frac{q^s\left(\mathfrak{h}-q^{sa}\mathfrak{g}\right)}{1-q^s}}\right)\hfill \\ & =& {\delta }^s{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)-{\delta }^{sa}{q}^s\left({\displaystyle \frac{\left(\mathfrak{h}-q^{s\left(a-1\right)}\mathfrak{g}\right)}{1-q^s}}-{\displaystyle \frac{\mathfrak{h}-q^{sa}\mathfrak{g}}{1-q^s}}\right)\hfill \\ & =& {\delta }^s{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)-{\delta }^{2s}{q}^s{\displaystyle \frac{{\delta }^{s\left(a-2\right)}\left(\mathfrak{h}-q^{s\left(a-1\right)}\mathfrak{g}\right)}{1-q^s}}+{\delta }^s{q}^s{\displaystyle \frac{{\delta }^{s\left(a-1\right)}\left(\mathfrak{h}-q^{sa}\mathfrak{g}\right)}{1-q^s}}\hfill \\ & =& {\delta }^s{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)-{\delta }^{2s}{q}^s{\mathbb{H}}_{a-1}^{\left(s\right)}\left(\delta ,q\right)+{\delta }^s{q}^s{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)\hfill \\ & =& {\delta }^s\left(1+q^s\right){\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)-{\left({\delta }^{2}q\right)}^s{\mathbb{H}}_{a-1}^{\left(s\right)}\left(\delta ,q\right).\hfill \end{array}$$

Thus, the proof is completed. □

Example 2.

From definition of the ${\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)$, we obtain the hybrid numbers as follows:

$$\begin{array}{c}{\mathbb{H}}_1^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right)=1+3\mathit{i}+8\mathsf{\epsilon }+21\mathit{h}\hfill \\ {\mathbb{H}}_2^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right)=3+8\mathit{i}+21\mathsf{\epsilon }+55\mathit{h}\hfill \\ {\mathbb{H}}_3^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right)=8+21\mathit{i}+55\mathsf{\epsilon }+144\mathit{h}\hfill \\ {\mathbb{H}}_4^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right)=21+55\mathit{i}+144\mathsf{\epsilon }+377\mathit{h}\hfill \\ {\mathbb{H}}_5^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right)=55+144\mathit{i}+377\mathsf{\epsilon }+987\mathit{h}\hfill \\ {\mathbb{H}}_6^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right)=144+377\mathit{i}+987\mathsf{\epsilon }+2584\mathit{h}\hfill \\ {\mathbb{H}}_7^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right)=377+987\mathit{i}+2584\mathsf{\epsilon }+6765\mathit{h}\hfill \\ {\mathbb{H}}_8^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right)=987+2584\mathit{i}+6765\mathsf{\epsilon }+17711\mathit{h}\hfill \\ {\mathbb{H}}_9^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right)=2584+6765\mathit{i}+17711\mathsf{\epsilon }+46368\mathit{h}\hfill \\ {\mathbb{H}}_{10}^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right)=6765+17711\mathit{i}+46368\mathsf{\epsilon }+121393\mathit{h}.\hfill \end{array}$$

(8)

Setting $\delta ={\displaystyle \frac{1+\sqrt{5}}{2}}$, $q=-{\displaystyle \frac{1}{{\delta }^2}}={\displaystyle \frac{-3+\sqrt{5}}{2}}$, $s=2$, and $a=2$ in (7) and using the above hybrid numbers, we verify the following equation:

$$\begin{array}{ccc}& & {\left({\displaystyle \frac{1+\sqrt{5}}{2}}\right)}^2\left(1+{\left({\displaystyle \frac{-3+\sqrt{5}}{2}}\right)}^2\right)\left(3+8\mathit{i}+21\mathsf{\epsilon }+55\mathit{h}\right)-{\left(-1\right)}^2\left(1+3\mathit{i}+8\mathsf{\epsilon }+21\mathit{h}\right)\hfill \\ & =& 8+21\mathit{i}+55\mathsf{\epsilon }+144\mathit{h}\hfill \\ & =& {\mathbb{H}}_3^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right).\hfill \end{array}$$

The Binet-like formula, Vajda-like identity, Catalan-like identity, Cassini-like identity, and d’Ocagne-like identity are recognized as having significant importance in the study of special numbers. The computation of these identities provides valuable information about the sequences under investigation. For further details, readers are directed to [1,19]. Among the most notable identities encountered in the examination of number sequences is Vajda’s identity, which has been the subject of scholarly investigation since 1901 [20]. We will now give the Vajda’s identity for ${\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)$.

Theorem 8.

For c, d, $s\in \mathbb{Z}$, $\delta \in \mathbb{R}$, and $q\in \mathbb{R}\setminus \left\{1\right\}$, the Vajda’s identity for ${\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)$ is as follows:

$${\mathbb{H}}_{a+c}^{\left(s\right)}\left(\delta ,q\right){\mathbb{H}}_{a+d}^{\left(s\right)}\left(\delta ,q\right)-{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right){\mathbb{H}}_{a+c+d}^{\left(s\right)}\left(\delta ,q\right)={\displaystyle \frac{{\delta }^{s\left(2a+c+d\right)}\left(q^{sc}-1\right)q^{sa}}{{\delta }^{2s}{\left(1-q^s\right)}^2}}\left(q^{sd}\mathfrak{hg}-\mathfrak{gh}\right).$$

(9)

Proof. 

Using the Binet-like formula and direct computation yields

$$\begin{array}{ccc}& & {\mathbb{H}}_{a+c}^{\left(s\right)}\left(\delta ,q\right){\mathbb{H}}_{a+d}^{\left(s\right)}\left(\delta ,q\right)-{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right){\mathbb{H}}_{a+c+d}^{\left(s\right)}\left(\delta ,q\right)\hfill \\ & =& {\displaystyle \frac{{\delta }^{s\left(a+c-1\right)}\left(\mathfrak{h}-q^{s\left(a+c\right)}\mathfrak{g}\right)}{1-q^s}}{\displaystyle \frac{{\delta }^{s\left(a+d-1\right)}\left(\mathfrak{h}-q^{s\left(a+d\right)}\mathfrak{g}\right)}{1-q^s}}\hfill \\ & & -{\displaystyle \frac{{\delta }^{s\left(a-1\right)}\left(\mathfrak{h}-q^{sa}\mathfrak{g}\right)}{1-q^s}}{\displaystyle \frac{{\delta }^{s\left(a+c+d-1\right)}\left(\mathfrak{h}-q^{s\left(a+c+d\right)}\mathfrak{g}\right)}{1-q^s}}\hfill \\ & & \\ & =& {\displaystyle \frac{1}{{\left(1-q^s\right)}^2}}\left(\begin{array}{c}{\delta }^{s\left(2a+c+d-2\right)}\left(\mathfrak{h}-q^{s\left(a+c\right)}\mathfrak{g}\right)\left(\mathfrak{h}-q^{s\left(a+d\right)}\mathfrak{g}\right)\\ -{\delta }^{s\left(2a+c+d-2\right)}\left(\mathfrak{h}-q^{sa}\mathfrak{g}\right)\left(\mathfrak{h}-q^{s\left(a+c+d\right)}\mathfrak{g}\right)\end{array}\right)\hfill \\ & & \\ & =& {\displaystyle \frac{{\delta }^{s\left(2a+c+d-2\right)}}{{\left(1-q^s\right)}^2}}\left(\begin{array}{c}{\mathfrak{h}}^2-q^{s\left(a+d\right)}\mathfrak{hg}-q^{s\left(a+c\right)}\mathfrak{gh}+q^{s\left(2a+c+d\right)}{\mathfrak{g}}^2\\ -{\mathfrak{h}}^2+q^{s\left(a+c+d\right)}\mathfrak{hg}+q^{sa}\mathfrak{gh}-q^{s\left(2a+c+d\right)}{\mathfrak{g}}^2\end{array}\right)\hfill \\ & & \\ & =& {\displaystyle \frac{{\delta }^{s\left(2a+c+d-2\right)}}{{\left(1-q^s\right)}^2}}\left(\left(q^{s\left(a+c+d\right)}-q^{s\left(a+d\right)}\right)\mathfrak{hg}+\left(q^{sa}-q^{s\left(a+c\right)}\right)\mathfrak{gh}\right)\hfill \\ & =& {\displaystyle \frac{{\delta }^{s\left(2a+c+d-2\right)}}{{\left(1-q^s\right)}^2}}\left(q^{s\left(a+d\right)}\left(q^{sc}-1\right)\mathfrak{hg}+q^{sa}\left(1-q^{sc}\right)\mathfrak{gh}\right)\hfill \\ & =& {\displaystyle \frac{{\delta }^{s\left(2a+c+d-2\right)}\left(q^{sc}-1\right)q^{sa}}{{\left(1-q^s\right)}^2}}\left(q^{sd}\mathfrak{hg}-\mathfrak{gh}\right).\hfill \end{array}$$

Therefore, we obtain Formula (9). □

Example 3.

Letting $a=2$, $c=3$, and $d=4$ in (9) and using Table 1 and Equations (2) and (8), we obtain

$$\begin{array}{ccc}& & {\mathbb{H}}_5^{\left(2\right)}\left(\delta ,q\right){\mathbb{H}}_6^{\left(2\right)}\left(\delta ,q\right)-{\mathbb{H}}_2^{\left(2\right)}\left(\delta ,q\right){\mathbb{H}}_9^{\left(2\right)}\left(\delta ,q\right)\hfill \\ & =& \left(55+144\mathit{i}+377\mathsf{\epsilon }+987\mathit{h}\right)\left(144+377\mathit{i}+987\mathsf{\epsilon }+2584\mathit{h}\right)\hfill \\ & & -\left(3+8\mathit{i}+21\mathsf{\epsilon }+55\mathit{h}\right)\left(2584+6765\mathit{i}+17711\mathsf{\epsilon }+46368\mathit{h}\right)\hfill \\ & =& 672+1632\mathit{i}+1928\mathsf{\epsilon }+2648\mathit{h}.\hfill \end{array}$$

On the other hand, for the right-hand side, by using the (5), we have

$$\begin{array}{ccc}\mathfrak{hg}\hfill & =& 4+\left(3-3\sqrt{5}\right)\mathit{i}+\left(7-2\sqrt{5}\right)\mathsf{\epsilon }+\left(18+\sqrt{5}\right)\mathit{h}\hfill \\ \mathfrak{gh}\hfill & =& 4+\left(3+3\sqrt{5}\right)\mathit{i}+\left(7+2\sqrt{5}\right)\mathsf{\epsilon }+\left(18-\sqrt{5}\right)\mathit{h}.\hfill \end{array}$$

Then, we obtain

$$\begin{array}{cc}{\displaystyle \frac{{\delta }^{22}\left(q^6-1\right)q^4}{{\delta }^4{\left(1-q^2\right)}^2}}\left(q^8\mathfrak{hg}-\mathfrak{gh}\right)\hfill & ={\displaystyle \frac{-420-188\sqrt{5}}{5}}\left({\displaystyle \frac{2207-987\sqrt{5}}{2}}\mathfrak{hg}-\mathfrak{gh}\right)\hfill \\ \hfill & =672+1632\mathit{i}+1928\mathsf{\epsilon }+2648\mathit{h}.\hfill \end{array}$$

Hence, the equality is satisfied.

Corollary 2.

The following formulas hold:

(i) 

(Catalan Identity):

$${\mathbb{H}}_{a+c}^{\left(s\right)}\left(\delta ,q\right){\mathbb{H}}_{a-c}^{\left(s\right)}\left(\delta ,q\right)-{\left({\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)\right)}^2={\displaystyle \frac{{\left({\delta }^{2}q\right)}^{sa}\left(q^{sc}-1\right)}{{\delta }^{2s}{\left(1-q^s\right)}^2}}\left(q^{-sc}\mathfrak{hg}-\mathfrak{gh}\right).$$

(10)

(ii) 

(Cassini’s Identity):

$${\mathbb{H}}_{a+1}^{\left(s\right)}\left(\delta ,q\right){\mathbb{H}}_{a-1}^{\left(s\right)}\left(\delta ,q\right)-{\left({\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)\right)}^2={\displaystyle \frac{{\left({\delta }^{2}q\right)}^{sa}\left(q^s-1\right)}{{\delta }^{2s}{\left(1-q^s\right)}^2}}\left(q^{-s}\mathfrak{hg}-\mathfrak{gh}\right).$$

(11)

(iii) 

(d’Ocagne Identity):

$${\mathbb{H}}_b^{\left(s\right)}\left(\delta ,q\right){\mathbb{H}}_{a+1}^{\left(s\right)}\left(\delta ,q\right)-{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right){\mathbb{H}}_{b+1}^{\left(s\right)}\left(\delta ,q\right)={\displaystyle \frac{{\delta }^{s\left(a+b\right)}\left(q^{s\left(b-a\right)}-1\right)q^{sa}}{{\delta }^s{\left(1-q^s\right)}^2}}\left(q^s\mathfrak{hg}-\mathfrak{gh}\right).$$

(12)

Proof. 

(i)

Formula (10) can be shown by taking $d=-c$ in (9).

(ii)

Formula (11) can be verified by taking $c=-d=1$ in (9).

(iii)

Formula (12) can be proved by taking $b\in \mathbb{Z}$, $d=1$ and $a+c=b$ in (9).

□

Example 4.

Letting $a=2$ and $c=3$ in (10), we have

$$\begin{array}{ccc}{\mathbb{H}}_5^{\left(s\right)}\left(\delta ,q\right){\mathbb{H}}_{-1}^{\left(s\right)}\left(\delta ,q\right)-{\left({\mathbb{H}}_2^{\left(s\right)}\left(\delta ,q\right)\right)}^2\hfill & =& \left(55+144\mathit{i}+377\mathsf{\epsilon }+987\mathit{h}\right)\left(-1+0\mathit{i}+1\mathsf{\epsilon }+3\mathit{h}\right)\hfill \\ & & -\left(3306+48\mathit{i}+126\mathsf{\epsilon }+330\mathit{h}\right)\hfill \\ & =& -256+240\mathit{i}-160\mathsf{\epsilon }-1296\mathit{h},\hfill \end{array}$$

where $\delta ={\displaystyle \frac{1+\sqrt{5}}{2}}$, $q=-{\displaystyle \frac{1}{{\delta }^2}}={\displaystyle \frac{-3+\sqrt{5}}{2}}$.

Example 5.

Letting $a=2$ in (10), we have

$$\begin{array}{ccc}& & {\mathbb{H}}_3^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right){\mathbb{H}}_1^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right)-{\left({\mathbb{H}}_2^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right)\right)}^2\hfill \\ & =& -4+6\mathit{i}-\mathsf{\epsilon }-21\mathit{h}.\hfill \end{array}$$

Example 6.

Letting $a=2$ and $b=5$ in (12), we have

$$\begin{array}{cc}\hfill & {\mathbb{H}}_5^{\left(s\right)}\left(\delta ,q\right){\mathbb{H}}_3^{\left(s\right)}\left(\delta ,q\right)-{\mathbb{H}}_2^{\left(s\right)}\left(\delta ,q\right){\mathbb{H}}_6^{\left(s\right)}\left(\delta ,q\right)\hfill \\ \hfill & =\left(55+144\mathit{i}+377\mathsf{\epsilon }+987\mathit{h}\right)\left(21+55\mathit{i}+144\mathsf{\epsilon }+377\mathit{h}\right)\hfill \\ \hfill & -\left(3+8\mathit{i}+21\mathsf{\epsilon }+55\mathit{h}\right)\left(144+377\mathit{i}+987\mathsf{\epsilon }+2584\mathit{h}\right)\hfill \\ \hfill & =32+96\mathit{i}+104\mathsf{\epsilon }+120\mathit{h},\hfill \end{array}$$

where $\delta ={\displaystyle \frac{1+\sqrt{5}}{2}}$, $q=-{\displaystyle \frac{1}{{\delta }^2}}={\displaystyle \frac{-3+\sqrt{5}}{2}}$.

Theorem 9.

For a, b, $s\in \mathbb{Z}$, $\delta \in \mathbb{R}$, and $q\in \mathbb{R}\setminus \left\{1\right\}$, the Honsberger Identity for ${\mathbb{H}}_{a+c}^{\left(s\right)}\left(\delta ,q\right)$ is expressed by

$$\begin{array}{cc}\hfill & {\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right){\mathbb{H}}_b^{\left(s\right)}\left(\delta ,q\right)+{\mathbb{H}}_{a+1}^{\left(s\right)}\left(\delta ,q\right){\mathbb{H}}_{b+1}^{\left(s\right)}\left(\delta ,q\right)\hfill \\ \hfill & ={\displaystyle \frac{{\delta }^{s\left(a+b\right)}}{{\left(1-q^s\right)}^2}}\left(\begin{array}{c}\left(1+{\delta }^{-2s}\right){\mathfrak{h}}^2+q^{\left(a+b\right)}\left(q^{2s}+{\delta }^{-2s}\right){\mathfrak{g}}^2\\ -\left(q^s+{\delta }^{-2s}\right)\left(q^{sb}\mathfrak{hg}+q^{sa}\mathfrak{gh}\right)\end{array}\right).\hfill \end{array}$$

Proof. 

Applying the Binet-like formula given in (4), we obtain

$$\begin{array}{ccc}& & {\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right){\mathbb{H}}_b^{\left(s\right)}\left(\delta ,q\right)+{\mathbb{H}}_{a+1}^{\left(s\right)}\left(\delta ,q\right){\mathbb{H}}_{b+1}^{\left(s\right)}\left(\delta ,q\right)\hfill \\ & =& {\displaystyle \frac{{\delta }^{s\left(a-1\right)}\left(\mathfrak{h}-q^{sa}\mathfrak{g}\right)}{1-q^s}}{\displaystyle \frac{{\delta }^{s\left(b-1\right)}\left(\mathfrak{h}-q^{sb}\mathfrak{g}\right)}{1-q^s}}\hfill \\ & & +{\displaystyle \frac{{\delta }^{sa}\left(\mathfrak{h}-q^{s\left(a+1\right)}\mathfrak{g}\right)}{1-q^s}}{\displaystyle \frac{{\delta }^{sb}\left(\mathfrak{h}-q^{s\left(b+1\right)}\mathfrak{g}\right)}{1-q^s}}\hfill \\ & =& {\displaystyle \frac{1}{{\left(1-q^s\right)}^2}}\left(\begin{array}{c}{\delta }^{s\left(a-1\right)}\left(\mathfrak{h}-q^{sa}\mathfrak{g}\right){\delta }^{s\left(b-1\right)}\left(\mathfrak{h}-q^{sb}\mathfrak{g}\right)\\ +{\delta }^{sa}\left(\mathfrak{h}-q^{s\left(a+1\right)}\mathfrak{g}\right){\delta }^{sb}\left(\mathfrak{h}-q^{s\left(b+1\right)}\mathfrak{g}\right)\end{array}\right)\hfill \\ & =& {\displaystyle \frac{1}{{\left(1-q^s\right)}^2}}\left(\begin{array}{c}{\delta }^{s\left(a+b-2\right)}\left({\mathfrak{h}}^2-\mathfrak{h}{q}^{sb}\mathfrak{g}-q^{sa}\mathfrak{gh}+q^{sa}\mathfrak{g}{q}^{sb}{\mathfrak{g}}^2\right)\\ +{\delta }^{s\left(a+b\right)}\left({\mathfrak{h}}^2-\mathfrak{h}{q}^{s\left(b+1\right)}\mathfrak{g}-q^{s\left(a+1\right)}\mathfrak{gh}+q^{s\left(a+1\right)}\mathfrak{g}{q}^{s\left(b+1\right)}\mathfrak{g}\right)\end{array}\right)\hfill \\ & & \\ & =& {\displaystyle \frac{{\delta }^{s\left(a+b\right)}}{{\left(1-q^s\right)}^2}}\left(\begin{array}{c}{\delta }^{-2s}{\mathfrak{h}}^2-{\delta }^{-2s}{q}^{sb}\mathfrak{hg}-{\delta }^{-2s}{q}^{sa}\mathfrak{gh}+{\delta }^{-2s}{q}^{\left(a+b\right)}{\mathfrak{g}}^2\\ +{\mathfrak{h}}^2-q^{s\left(b+1\right)}\mathfrak{hg}-q^{s\left(a+1\right)}\mathfrak{gh}+q^{s\left(a+b+2\right)}{\mathfrak{g}}^2\end{array}\right)\hfill \\ & & \\ & =& {\displaystyle \frac{{\delta }^{s\left(a+b\right)}}{{\left(1-q^s\right)}^2}}\left(\begin{array}{c}\left(1+{\delta }^{-2s}\right){\mathfrak{h}}^2-q^{sb}\left(q^s+{\delta }^{-2s}\right)\mathfrak{hg}\\ -q^{sa}\left(q^s+{\delta }^{-2s}\right)\mathfrak{gh}+q^{\left(a+b\right)}\left(q^{2s}+{\delta }^{-2s}\right){\mathfrak{g}}^2\end{array}\right)\hfill \\ & & \\ & =& {\displaystyle \frac{{\delta }^{s\left(a+b\right)}}{{\left(1-q^s\right)}^2}}\left(\begin{array}{c}\left(1+{\delta }^{-2s}\right){\mathfrak{h}}^2+q^{\left(a+b\right)}\left(q^{2s}+{\delta }^{-2s}\right){\mathfrak{g}}^2\\ -\left(q^s+{\delta }^{-2s}\right)\left(q^{sb}\mathfrak{hg}+q^{sa}\mathfrak{gh}\right)\end{array}\right).\hfill \end{array}$$

The proof is completed. □

Theorem 10.

For a, b, $s\in \mathbb{Z}$, $\delta \in \mathbb{R}$, and $q\in \mathbb{R}\setminus \left\{1\right\}$, following equation is true:

$$\sum _{a=1}^b{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)={\displaystyle \frac{1}{1-q^s}}\left({\displaystyle \frac{1-{\delta }^{sb}}{1-{\delta }^s}}\mathfrak{h}-q^s{\displaystyle \frac{1-{\left(\delta q\right)}^{sb}}{1-{\left(\delta q\right)}^s}}\mathfrak{g}\right).$$

(13)

Proof. 

By using the Equation (4), we obtain

$$\begin{array}{ccc}\sum _{a=1}^b{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)\hfill & =& \sum _{a=1}^b{\displaystyle \frac{{\delta }^{s\left(a-1\right)}\left(\mathfrak{h}-q^{sa}\mathfrak{g}\right)}{1-q^s}}\hfill \\ & & \\ & =& {\displaystyle \frac{1}{1-q^s}}\left(\begin{array}{c}\left(\mathfrak{h}-q^s\mathfrak{g}\right)+{\delta }^s\left(\mathfrak{h}-q^{2s}\mathfrak{g}\right)+{\delta }^{2s}\left(\mathfrak{h}-q^{3s}\mathfrak{g}\right)\\ +\dots +{\delta }^{s\left(b-2\right)}\left(\mathfrak{h}-q^{s\left(b-1\right)}\mathfrak{g}\right)+{\delta }^{s\left(b-1\right)}\left(\mathfrak{h}-q^{sb}\mathfrak{g}\right)\end{array}\right)\hfill \\ & & \\ & =& {\displaystyle \frac{1}{1-q^s}}\left(\begin{array}{c}\left(1+{\delta }^s+{\delta }^{2s}+\dots +{\delta }^{s\left(b-2\right)}+{\delta }^{s\left(b-1\right)}\right)\mathfrak{h}\\ -q^s\left(1+{\left(\delta q\right)}^s+{\left(\delta q\right)}^{2s}+\dots +{\left(\delta q\right)}^{s\left(b-2\right)}+{\left(\delta q\right)}^{s\left(b-1\right)}\right)\mathfrak{g}\end{array}\right)\hfill \\ & & \\ & =& {\displaystyle \frac{1}{1-q^s}}\left({\displaystyle \frac{1-{\delta }^{sb}}{1-{\delta }^s}}\mathfrak{h}-q^s{\displaystyle \frac{1-{\left(\delta q\right)}^{sb}}{1-{\left(\delta q\right)}^s}}\mathfrak{g}\right).\hfill \end{array}$$

The proof is completed. □

Example 7.

Using the hybrid numbers in (8), we have

$$\sum _{a=1}^5{\mathbb{H}}_a^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right)=88+231\mathit{i}+605\mathsf{\epsilon }+1584\mathit{h}.$$

(14)

On the other hand, from the right-hand side of Equation (13), we have

$${\displaystyle \frac{1}{1-q^2}}\left({\displaystyle \frac{1-{\delta }^{10}}{1-{\delta }^2}}\mathfrak{h}-q^s{\displaystyle \frac{1-{\left(\delta q\right)}^{10}}{1-{\left(\delta q\right)}^2}}\mathfrak{g}\right)=88+231\mathit{i}+605\mathsf{\epsilon }+1584\mathit{h}.$$

(15)

Therefore, our results are verified.

Theorem 11.

For a, b, $s\in \mathbb{Z}$, $\delta \in \mathbb{R}$, and $q\in \mathbb{R}\setminus \left\{1\right\}$, the following equation is satisfied:

$$\sum _{a=1}^b{\mathbb{H}}_{2a}^{\left(s\right)}\left(\delta ,q\right)={\displaystyle \frac{{\delta }^s}{1-q^s}}\left(\mathfrak{h}{\displaystyle \frac{\left(1-{\delta }^{2sb}\right)}{1-{\delta }^{2s}}}-\mathfrak{g}{\displaystyle \frac{q^{2s}\left(1-{\left(\delta q\right)}^{2sb}\right)}{1-{\left(\delta q\right)}^{2s}}}\right).$$

(16)

Proof. 

By using Equation (4), we have

$$\begin{array}{ccc}\sum _{a=1}^b{\mathbb{H}}_{2a}^{\left(s\right)}\left(\delta ,q\right)\hfill & =& \sum _{a=1}^b{\displaystyle \frac{{\delta }^{s\left(2a-1\right)}\left(\mathfrak{h}-q^{2sa}\mathfrak{g}\right)}{1-q^s}}\hfill \\ & =& {\displaystyle \frac{{\delta }^{-s}}{1-q^s}}\left(\mathfrak{h}\sum _{a=1}^b{\delta }^{2sa}-\mathfrak{g}\sum _{a=1}^b{\left(\delta q\right)}^{2sa}\right)\hfill \\ & =& {\displaystyle \frac{{\delta }^{-s}}{1-q^s}}\left(\mathfrak{h}{\displaystyle \frac{{\delta }^{2s}\left(1-{\delta }^{2sb}\right)}{1-{\delta }^{2s}}}-\mathfrak{g}{\displaystyle \frac{{\left(\delta q\right)}^{2s}\left(1-{\left(\delta q\right)}^{2sb}\right)}{1-{\left(\delta q\right)}^{2s}}}\right)\hfill \\ & =& {\displaystyle \frac{{\delta }^s}{1-q^s}}\left(\mathfrak{h}{\displaystyle \frac{1-{\delta }^{2sb}}{1-{\delta }^{2s}}}-\mathfrak{g}{\displaystyle \frac{q^{2s}\left(1-{\left(\delta q\right)}^{2sb}\right)}{1-{\left(\delta q\right)}^{2s}}}\right).\hfill \end{array}$$

The proof is completed. □

Example 8.

Using the hybrid numbers in (8), we have

$$\begin{array}{cc}\hfill & \sum _{a=1}^5{\mathbb{H}}_{2a}^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right)\hfill \\ \hfill & ={\mathbb{H}}_2^{\left(2\right)}\left(\delta ,q\right)+{\mathbb{H}}_4^{\left(2\right)}\left(\delta ,q\right)+{\mathbb{H}}_6^{\left(2\right)}\left(\delta ,q\right)+{\mathbb{H}}_8^{\left(2\right)}\left(\delta ,q\right)+{\mathbb{H}}_{10}^{\left(2\right)}\left(\delta ,q\right)\hfill \\ \hfill & =7920+20735\mathit{i}+54285\mathsf{\epsilon }+142120\mathit{h}.\hfill \end{array}$$

On the other hand, from the right side of Equation (16), we have

$${\displaystyle \frac{{\delta }^2}{1-q^2}}\left(\mathfrak{h}{\displaystyle \frac{\left(1-{\delta }^{20}\right)}{1-{\delta }^4}}-\mathfrak{g}{\displaystyle \frac{q^4\left(1-{\left(\delta q\right)}^{20}\right)}{1-{\left(\delta q\right)}^4}}\right)=7920+20735\mathit{i}+54285\mathsf{\epsilon }+142120\mathit{h}.$$

So, our result is verified.

Theorem 12.

For b, $s\in \mathbb{Z}$, $\delta \in \mathbb{R}$, and $q\in \mathbb{R}\setminus \left\{1\right\}$, the following equation holds:

$$\sum _{a=1}^b{\mathbb{H}}_{2a-1}^{\left(s\right)}\left(\delta ,q\right)={\displaystyle \frac{1}{1-q^s}}\left(\mathfrak{h}{\displaystyle \frac{1-{\delta }^{2sb}}{1-{\delta }^{2s}}}-q^s\mathfrak{g}{\displaystyle \frac{1-{\left(\delta q\right)}^{2sb}}{1-{\left(\delta q\right)}^{2s}}}\right).$$

(17)

Proof. 

Applying Equation (4), we obtain

$$\begin{array}{ccc}\sum _{a=1}^b{\mathbb{H}}_{2a-1}^{\left(s\right)}\left(\delta ,q\right)\hfill & =& \sum _{a=1}^b{\displaystyle \frac{{\delta }^{s\left(2a-2\right)}\left(\mathfrak{h}-q^{s\left(2a-1\right)}\mathfrak{g}\right)}{1-q^s}}\hfill \\ & =& {\displaystyle \frac{{\delta }^{-2s}}{1-q^s}}\left(\mathfrak{h}\sum _{a=1}^b{\delta }^{2sa}-q^{-s}\mathfrak{g}\sum _{a=1}^b{\left(\delta q\right)}^{2sa}\right)\hfill \\ & =& {\displaystyle \frac{{\delta }^{-2s}}{1-q^s}}\left(\mathfrak{h}{\displaystyle \frac{{\delta }^{2s}\left(1-{\delta }^{2sb}\right)}{1-{\delta }^{2s}}}-q^{-s}\mathfrak{g}{\displaystyle \frac{{\left(\delta q\right)}^{2s}\left(1-{\left(\delta q\right)}^{2sb}\right)}{1-{\left(\delta q\right)}^{2s}}}\right)\hfill \\ & =& {\displaystyle \frac{1}{1-q^s}}\left(\mathfrak{h}{\displaystyle \frac{1-{\delta }^{2sb}}{1-{\delta }^{2s}}}-q^s\mathfrak{g}{\displaystyle \frac{1-{\left(\delta q\right)}^{2sb}}{1-{\left(\delta q\right)}^{2s}}}\right).\hfill \\ & \end{array}$$

The proof is completed. □

Example 9.

Using the hybrid numbers in (8), we have

$$\begin{array}{cc}\hfill & \sum _{a=1}^5{\mathbb{H}}_{2a-1}^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right)\hfill \\ \hfill & ={\mathbb{H}}_1^{\left(2\right)}\left(\delta ,q\right)+{\mathbb{H}}_3^{\left(2\right)}\left(\delta ,q\right)+{\mathbb{H}}_5^{\left(2\right)}\left(\delta ,q\right)+{\mathbb{H}}_7^{\left(2\right)}\left(\delta ,q\right)+{\mathbb{H}}_9^{\left(2\right)}\left(\delta ,q\right)\hfill \\ \hfill & =3025+7920\mathit{i}+20735\mathsf{\epsilon }+54285\mathit{h},\hfill \end{array}$$

and from the right-hand side of Equation (17), we have

$${\displaystyle \frac{1}{1-q^2}}\left(\mathfrak{h}{\displaystyle \frac{1-{\delta }^{20}}{1-{\delta }^4}}-q^2\mathfrak{g}{\displaystyle \frac{1-{\left(\delta q\right)}^{20}}{1-{\left(\delta q\right)}^4}}\right)=3025+7920\mathit{i}+20735\mathsf{\epsilon }+54285\mathit{h}.$$

Therefore, our results are verified.

3. Some Matrix Representations for ${\mathbb{H}}_{\mathit{n}}^{\left(\mathit{s}\right)}\left(\mathit{\delta },\mathit{q}\right)$

This section undertakes an examination of various properties of higher-order generalized Fibonacci hybrid numbers with quantum integer components, utilizing the matrices defined below:

$${\mathbf{W}}_{a,\delta ,q}^{\left(s\right)}=\left(\begin{array}{cc}{\mathbb{H}}_{a+1}^{\left(s\right)}\left(\delta ,q\right)& {\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)\\ {\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)& {\mathbb{H}}_{a-1}^{\left(s\right)}\left(\delta ,q\right)\end{array}\right)$$

(18)

and

$${\mathbf{X}}_{\delta ,q}^{\left(s\right)}=\left(\begin{array}{cc}{\delta }^s\left(1+q^s\right)& -{\left({\delta }^{2}q\right)}^s\\ 1& 0\end{array}\right).$$

Lemma 1.

For $a\in \mathbb{N}$, we have

$${\left({\mathbf{X}}_{\delta ,q}^{\left(s\right)}\right)}^a=\left(\begin{array}{cc}{\delta }^{sa}{\left[a+1\right]}_{q^s}& -{\delta }^{s\left(a+1\right)}{q}^s{\left[a\right]}_{q^s}\\ {\delta }^{s\left(a-1\right)}{\left[a\right]}_{q^s}& -{\delta }^{sa}{q}^s{\left[a-1\right]}_{q^s}\end{array}\right).$$

(19)

Proof. 

The proof will be conducted using mathematical induction on a. It is evident that Equation (19) is true for $a=1$. Suppose that Equation (19) is true for k. By using the definition of a q-integer, we have

$$\begin{array}{ccc}{\left({\mathbf{X}}_{\delta ,q}^{\left(s\right)}\right)}^{k+1}\hfill & =& \left(\begin{array}{cc}{\delta }^s\left(1+q^s\right)& -{\left({\delta }^{2}q\right)}^s\\ 1& 0\end{array}\right){\left(\begin{array}{cc}{\delta }^s\left(1+q^s\right)& -{\left({\delta }^{2}q\right)}^s\\ 1& 0\end{array}\right)}^k\hfill \\ & =& \left(\begin{array}{cc}{\delta }^s\left(1+q^s\right)& -{\left({\delta }^{2}q\right)}^s\\ 1& 0\end{array}\right)\left(\begin{array}{cc}{\delta }^{sk}{\left[k+1\right]}_{q^s}& -{\delta }^{s\left(k+1\right)}{q}^s{\left[k\right]}_{q^s}\\ {\delta }^{s\left(k-1\right)}{\left[k\right]}_{q^s}& -{\delta }^{sk}{q}^s{\left[k-1\right]}_{q^s}\end{array}\right)\hfill \\ & =& \left(\begin{array}{cc}{\delta }^{s\left(k+1\right)}\left(\left(1+q^s\right){\left[k+1\right]}_{q^s}-q^s{\left[k\right]}_{q^s}\right)& -{\delta }^{s\left(k+2\right)}{q}^s\left(\left(1+q^s\right){\left[k\right]}_{q^s}-q^s{\left[k-1\right]}_{q^s}\right)\\ {\delta }^{sk}{\left[k+1\right]}_{q^s}& -{\delta }^{s\left(k+1\right)}{q}^s{\left[k\right]}_{q^s}\end{array}\right)\hfill \\ & =& \left(\begin{array}{cc}{\delta }^{s\left(k+1\right)}{\left[k+2\right]}_{q^s}& -{\delta }^{s\left(k+2\right)}{q}^s{\left[k+1\right]}_{q^s}\\ {\delta }^{sk}{\left[k+1\right]}_{q^s}& -{\delta }^{s\left(k+1\right)}{q}^s{\left[k\right]}_{q^s}\end{array}\right).\hfill \end{array}$$

The proof is completed. □

Lemma 2.

For $a\ge 1,$ we have

$${\mathbf{W}}_{a,\delta ,q}^{\left(s\right)}={\mathbf{X}}_{\delta ,q}^{\left(s\right)}{\mathbf{W}}_{a-1,\delta ,q}^{\left(s\right)}.$$

(20)

Proof. 

Using Equation (7) and matrix multiplication, we have

$$\begin{array}{ccc}{\mathbf{X}}_{\delta ,q}^{\left(s\right)}{\mathbf{W}}_{a-1\delta ,q}^{\left(s\right)}\hfill & =& \left(\begin{array}{cc}{\delta }^s\left(1+q^s\right)& -{\left({\delta }^{2}q\right)}^s\\ 1& 0\end{array}\right)\left(\begin{array}{cc}{\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)& {\mathbb{H}}_{a-1}^{\left(s\right)}\left(\delta ,q\right)\\ {\mathbb{H}}_{a-1}^{\left(s\right)}\left(\delta ,q\right)& {\mathbb{H}}_{a-2}^{\left(s\right)}\left(\delta ,q\right)\end{array}\right)\hfill \\ & =& \left(\begin{array}{cc}{\mathbb{H}}_{a+1}^{\left(s\right)}\left(\delta ,q\right)& {\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)\\ {\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)& {\mathbb{H}}_{a-1}^{\left(s\right)}\left(\delta ,q\right)\end{array}\right)={\mathbf{W}}_{a,\delta ,q}^{\left(s\right)}.\hfill \end{array}$$

Hence, the conclusion is proved. □

Example 10.

Substituting $\delta ={\displaystyle \frac{1+\sqrt{5}}{2}}$, $q=-{\displaystyle \frac{1}{{\delta }^2}}={\displaystyle \frac{-3+\sqrt{5}}{2}}$, $s=2$, and $a=4$ in (18) and using (8), we have

$${\mathbf{W}}_{4,\delta ,q}^{\left(2\right)}=\left(\begin{array}{cc}55+144\mathit{i}+377\mathsf{\epsilon }+987\mathit{h}& 21+55\mathit{i}+144\mathsf{\epsilon }+377\mathit{h}\\ 21+55\mathit{i}+144\mathsf{\epsilon }+377\mathit{h}& 8+21\mathit{i}+55\mathsf{\epsilon }+144\mathit{h}\end{array}\right).$$

Similarly, for $a=4,$ we have

$${\mathbf{W}}_{3,\delta ,q}^{\left(2\right)}=\left(\begin{array}{cc}21+55\mathit{i}+144\mathsf{\epsilon }+377\mathit{h}& 8+21\mathit{i}+55\mathsf{\epsilon }+144\mathit{h}\\ 8+21\mathit{i}+55\mathsf{\epsilon }+144\mathit{h}& 3+8\mathit{i}+21\mathsf{\epsilon }+55\mathit{h}\end{array}\right).$$

On the other hand, for the right-hand side of Equation (20), we have

$$\begin{array}{ccc}{\mathbf{X}}_{\delta ,q}^{\left(2\right)}{\mathbf{W}}_{3,\delta ,q}^{\left(2\right)}\hfill & =& \left(\begin{array}{cc}3& -1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}{\mathbb{H}}_4^{\left(2\right)}\left(\delta ,q\right)& {\mathbb{H}}_3^{\left(2\right)}\left(\delta ,q\right)\\ {\mathbb{H}}_3^{\left(2\right)}\left(\delta ,q\right)& {\mathbb{H}}_2^{\left(2\right)}\left(\delta ,q\right)\end{array}\right)\hfill \\ & =& \left(\begin{array}{cc}3{\mathbb{H}}_4^{\left(2\right)}\left(\delta ,q\right)-{\mathbb{H}}_3^{\left(2\right)}\left(\delta ,q\right)& 3{\mathbb{H}}_3^{\left(2\right)}\left(\delta ,q\right)-{\mathbb{H}}_2^{\left(2\right)}\left(\delta ,q\right)\\ {\mathbb{H}}_4^{\left(2\right)}\left(\delta ,q\right)& {\mathbb{H}}_3^{\left(2\right)}\left(\delta ,q\right)\end{array}\right)\hfill \\ & =& \left(\begin{array}{cc}55+144\mathit{i}+377\mathsf{\epsilon }+987\mathit{h}& 21+55\mathit{i}+144\mathsf{\epsilon }+377\mathit{h}\\ 21+55\mathit{i}+144\mathsf{\epsilon }+377\mathit{h}& 8+21\mathit{i}+55\mathsf{\epsilon }+144\mathit{h}\end{array}\right).\hfill \end{array}$$

Therefore, our results are verified.

Lemma 3.

For $a\ge 1$, we have

$${\mathbf{W}}_{a,\delta ,q}^{\left(s\right)}={\left({\mathbf{X}}_{\delta ,q}^{\left(s\right)}\right)}^a{\mathbf{W}}_{0,\delta ,q}^{\left(s\right)}.$$

(21)

Proof. 

The proof will be conducted using mathematical induction on $a.$ For $a=1$, thanks to the recurrence relation of ${\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)$, we obtain

$$\begin{array}{ccc}{\left({\mathbf{X}}_{\delta ,q}^{\left(s\right)}\right)}^1{\mathbf{W}}_{0,\delta ,q}^{\left(s\right)}\hfill & =& \left(\begin{array}{cc}{\delta }^s\left(1+q^s\right)& -{\left({\delta }^{2}q\right)}^s\\ 1& 0\end{array}\right)\left(\begin{array}{cc}{\mathbb{H}}_1^{\left(s\right)}\left(\delta ,q\right)& {\mathbb{H}}_0^{\left(s\right)}\left(\delta ,q\right)\\ {\mathbb{H}}_0^{\left(s\right)}\left(\delta ,q\right)& {\mathbb{H}}_{-1}^{\left(s\right)}\left(\delta ,q\right)\end{array}\right)\hfill \\ & =& \left(\begin{array}{cc}{\mathbb{H}}_2^{\left(s\right)}\left(\delta ,q\right)& {\mathbb{H}}_1^{\left(s\right)}\left(\delta ,q\right)\\ {\mathbb{H}}_1^{\left(s\right)}\left(\delta ,q\right)& {\mathbb{H}}_0^{\left(s\right)}\left(\delta ,q\right)\end{array}\right)\hfill \\ & =& {\mathbf{W}}_{1,\delta ,q}^{\left(s\right)}.\hfill \end{array}$$

Suppose that it is true for a. Then, for $a+1$, using Equation (7), we have

$$\begin{array}{ccc}{\left({\mathbf{X}}_{\delta ,q}^{\left(s\right)}\right)}^{a+1}{\mathbf{W}}_{0,\delta ,q}^{\left(s\right)}\hfill & =& {\mathbf{X}}_{\delta ,q}^{\left(s\right)}{\left({\mathbf{X}}_{\delta ,q}^{\left(s\right)}\right)}^a{\mathbf{W}}_{0,\delta ,q}^{\left(s\right)}\hfill \\ & =& {\mathbf{X}}_{\delta ,q}^{\left(s\right)}{\mathbf{W}}_{a,\delta ,q}^{\left(s\right)}\hfill \\ & =& \left(\begin{array}{cc}{\delta }^s\left(1+q^s\right)& -{\left({\delta }^{2}q\right)}^s\\ 1& 0\end{array}\right)\left(\begin{array}{cc}{\mathbb{H}}_{a+1}^{\left(s\right)}\left(\delta ,q\right)& {\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)\\ {\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)& {\mathbb{H}}_{a-1}^{\left(s\right)}\left(\delta ,q\right)\end{array}\right)\hfill \\ & =& \left(\begin{array}{cc}{\mathbb{H}}_{a+2}^{\left(s\right)}\left(\delta ,q\right)& {\mathbb{H}}_{a+1}^{\left(s\right)}\left(\delta ,q\right)\\ {\mathbb{H}}_{a+1}^{\left(s\right)}\left(\delta ,q\right)& {\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)\end{array}\right)\hfill \\ & =& {\mathbf{W}}_{a+1,\delta ,q}^{\left(s\right)}.\hfill \end{array}$$

So, the proof is completed. □

Theorem 13.

For $a\in \mathbb{N}$, the following equalities are satisfied:

$${\mathbb{H}}_{a+1}^{\left(s\right)}\left(\delta ,q\right){\mathbb{H}}_{a-1}^{\left(s\right)}\left(\delta ,q\right)-{\left({\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)\right)}^2={\displaystyle \frac{{\left({\delta }^{2}q\right)}^{sa}{\delta }^{-2s}}{{\left(1-q^s\right)}^2}}\left(\left(1-q^{-s}\right)\mathfrak{hg}+\left(1-q^s\right)\mathfrak{gh}\right),$$

(22)

$${\mathbb{H}}_{a-1}^{\left(s\right)}\left(\delta ,q\right){\mathbb{H}}_{a+1}^{\left(s\right)}\left(\delta ,q\right)-{\left({\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)\right)}^2={\displaystyle \frac{{\left({\delta }^{2}q\right)}^{sa}{\delta }^{-2s}}{{\left(1-q^s\right)}^2}}\left(\left(1-q^s\right)\mathfrak{hg}+\left(1-q^{-s}\right)\mathfrak{gh}\right).$$

(23)

Proof. 

For the proofs, we use “multiplication from above to down below” and “multiplication from down below to above” rules and Equation (21). Note that

$$\left|\begin{array}{cc}{\mathbb{H}}_{a+1}^{\left(s\right)}\left(\delta ,q\right)& {\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)\\ {\mathbb{H}}_a^{\left(s\right)}\left(\delta ,q\right)& {\mathbb{H}}_{a-1}^{\left(s\right)}\left(\delta ,q\right)\end{array}\right|={\left|\begin{array}{cc}{\delta }^s\left(1+q^s\right)& -{\left({\delta }^{2}q\right)}^s\\ 1& 0\end{array}\right|}^a\left|\begin{array}{cc}{\mathbb{H}}_1^{\left(s\right)}\left(\delta ,q\right)& {\mathbb{H}}_0^{\left(s\right)}\left(\delta ,q\right)\\ {\mathbb{H}}_0^{\left(s\right)}\left(\delta ,q\right)& {\mathbb{H}}_{-1}^{\left(s\right)}\left(\delta ,q\right)\end{array}\right|.$$

We now show (22) holds. We have

$$\begin{array}{ccc}& & {\left({\delta }^{2}q\right)}^{sa}\left({\mathbb{H}}_1^{\left(s\right)}\left(\delta ,q\right){\mathbb{H}}_{-1}^{\left(s\right)}\left(\delta ,q\right)-{\left({\mathbb{H}}_0^{\left(s\right)}\left(\delta ,q\right)\right)}^2\right)\hfill \\ & =& {\displaystyle \frac{{\left({\delta }^{2}q\right)}^{sa}}{{\left(1-q^s\right)}^2}}\left({\delta }^0\left(\mathfrak{h}-q^s\mathfrak{g}\right){\delta }^{-2s}\left(\mathfrak{h}-q^{-s}\mathfrak{g}\right)-{\left({\delta }^{-s}\left(\mathfrak{h}-q^0\mathfrak{g}\right)\right)}^2\right)\hfill \\ & =& {\displaystyle \frac{{\left({\delta }^{2}q\right)}^{sa}}{{\left(1-q^s\right)}^2}}\left({\delta }^{-2s}\left(\mathfrak{h}-q^s\mathfrak{g}\right)\left(\mathfrak{h}-q^{-s}\mathfrak{g}\right)-{\delta }^{-2s}{\left(\mathfrak{h}-\mathfrak{g}\right)}^2\right)\hfill \\ & =& {\displaystyle \frac{{\left({\delta }^{2}q\right)}^{sa}{\delta }^{-2s}}{{\left(1-q^s\right)}^2}}\left(\left(\mathfrak{h}-q^s\mathfrak{g}\right)\left(\mathfrak{h}-q^{-s}\mathfrak{g}\right)-{\left(\mathfrak{h}-\mathfrak{g}\right)}^2\right)\hfill \\ & =& {\displaystyle \frac{{\left({\delta }^{2}q\right)}^{sa}{\delta }^{-2s}}{{\left(1-q^s\right)}^2}}\left(\begin{array}{c}\left({\mathfrak{h}}^2-q^{-s}\mathfrak{hg}-q^s\mathfrak{gh}+{\mathfrak{g}}^2\right)\\ -\left({\mathfrak{h}}^2-\mathfrak{hg}-\mathfrak{gh}+{\mathfrak{g}}^2\right)\end{array}\right)\hfill \\ & =& {\displaystyle \frac{{\left({\delta }^{2}q\right)}^{sa}{\delta }^{-2s}}{{\left(1-q^s\right)}^2}}\left(\begin{array}{c}{\mathfrak{h}}^2-q^{-s}\mathfrak{hg}-q^s\mathfrak{gh}+{\mathfrak{g}}^2\\ -{\mathfrak{h}}^2+\mathfrak{hg}+\mathfrak{gh}-{\mathfrak{g}}^2\end{array}\right)\hfill \\ & =& {\displaystyle \frac{{\left({\delta }^{2}q\right)}^{sa}{\delta }^{-2s}}{{\left(1-q^s\right)}^2}}\left(\left(1-q^{-s}\right)\mathfrak{hg}+\left(1-q^s\right)\mathfrak{gh}\right).\hfill \end{array}$$

For the proof of (23), we have

$$\begin{array}{ccc}& & {\left({\delta }^{2}q\right)}^{sa}\left({\mathbb{H}}_{-1}^{\left(s\right)}\left(\delta ,q\right){\mathbb{H}}_1^{\left(s\right)}\left(\delta ,q\right)-{\left({\mathbb{H}}_0^{\left(s\right)}\left(\delta ,q\right)\right)}^2\right)\hfill \\ & =& {\displaystyle \frac{{\left({\delta }^{2}q\right)}^{sa}}{{\left(1-q^s\right)}^2}}\left({\delta }^{-2s}\left(\mathfrak{h}-q^{-s}\mathfrak{g}\right){\delta }^0\left(\mathfrak{h}-q^s\mathfrak{g}\right)-{\left({\delta }^{-s}\left(\mathfrak{h}-q^0\mathfrak{g}\right)\right)}^2\right)\hfill \\ & =& {\displaystyle \frac{{\left({\delta }^{2}q\right)}^{sa}}{{\left(1-q^s\right)}^2}}\left({\delta }^{-2s}\left(\mathfrak{h}-q^{-s}\mathfrak{g}\right)\left(\mathfrak{h}-q^s\mathfrak{g}\right)-{\delta }^{-2s}{\left(\mathfrak{h}-\mathfrak{g}\right)}^2\right)\hfill \\ & =& {\displaystyle \frac{{\left({\delta }^{2}q\right)}^{sa}{\delta }^{-2s}}{{\left(1-q^s\right)}^2}}\left(\left(\mathfrak{h}-q^{-s}\mathfrak{g}\right)\left(\mathfrak{h}-q^s\mathfrak{g}\right)-{\left(\mathfrak{h}-\mathfrak{g}\right)}^2\right)\hfill \\ & =& {\displaystyle \frac{{\left({\delta }^{2}q\right)}^{sa}{\delta }^{-2s}}{{\left(1-q^s\right)}^2}}\left(\begin{array}{c}\left({\mathfrak{h}}^2-q^s\mathfrak{hg}-q^{-s}\mathfrak{gh}+{\mathfrak{g}}^2\right)\\ -\left({\mathfrak{h}}^2-\mathfrak{hg}-\mathfrak{gh}+{\mathfrak{g}}^2\right)\end{array}\right)\hfill \\ & =& {\displaystyle \frac{{\left({\delta }^{2}q\right)}^{sa}{\delta }^{-2s}}{{\left(1-q^s\right)}^2}}\left(\begin{array}{c}{\mathfrak{h}}^2-q^s\mathfrak{hg}-q^{-s}\mathfrak{gh}+{\mathfrak{g}}^2\\ -{\mathfrak{h}}^2+\mathfrak{hg}+\mathfrak{gh}-{\mathfrak{g}}^2\end{array}\right)\hfill \\ & =& {\displaystyle \frac{{\left({\delta }^{2}q\right)}^{sa}{\delta }^{-2s}}{{\left(1-q^s\right)}^2}}\left(\left(1-q^s\right)\mathfrak{hg}+\left(1-q^{-s}\right)\mathfrak{gh}\right).\hfill \end{array}$$

Hence, the conclusion is proved. □

Remark 1.

It is worth noting that it is quite challenging when we manually determine the product of two hybrid numbers, especially when dealing with large or irrational coefficients.

The below Maple code is written and presented to find the multiplication of $h=h_1+h_2\mathbf{i}+h_3\mathsf{ϵ}+h_4\mathbf{h}$ and $k=k_1+k_2\mathbf{i}+k_3\mathsf{ϵ}+k_4\mathbf{h}$. Here, the desired product is obtained by typing the coefficients of h and k, respectively, in procedure b. For example, the number (1) in Figure 1 is the multiplication of ${\mathbb{H}}_3^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right)$ and ${\mathbb{H}}_1^{\left(2\right)}\left({\displaystyle \frac{1+\sqrt{5}}{2}},{\displaystyle \frac{-3+\sqrt{5}}{2}}\right)$. Similarly, number(2)equals $\mathfrak{gh}$.

Figure 1. Maple 2021 code for multiplication of hybrid numbers.

4. Conclusions

In this paper, several new identities for higher-order generalized Fibonacci hybrid numbers with q-integer components are derived through the application of specific types of matrices. This study explores an alternative generalization of Fibonacci hybrid numbers, which encompasses both existing families of Fibonacci-type hybrid numbers documented in the literature and introduces new families based on the parameters s, $\delta$, and q. Our findings extend and enhance several existing results in the relevant literature [21], thereby facilitating the development of novel families of Fibonacci-type hybrid numbers and their associated proof techniques in subsequent research.