New Properties and Matrix Representations on Higher-Order Generalized Fibonacci Quaternions with q-Integer Components

Abstract

In this paper, by using q-integers and higher-order generalized Fibonacci numbers, we define the higher-order generalized Fibonacci quaternions with q-integer components. We give some special cases of these newly established quaternions. This article examines q-calculus and quaternions together. We obtain a Binet-like formula, some new identities, a generating function, a recurrence relation, an exponential generating function, and sum properties of quaternions with quantum integer coefficients. In addition, we obtain some new identities for these types of quaternions by using three new special matrices.

1. Introduction and Preliminaries

The famous Irish mathematician William Rowan Hamilton was the inventor of quaternions, defined as a non-commutative number system that extends complex numbers in 1843 [1]. He defined the set of quaternions as a four-dimensional real vector space and established a multiplicative operation on it. Recall that a quaternion p is defined in the form

$$p=p_0+p_1\mathbf{i}+p_2\mathbf{j}+p_3\mathbf{k},$$

(1)

where $p_0,p_1,p_2,p_3$ are real numbers, and $\mathbf{1},$$\mathbf{i},$$\mathbf{j},$$\mathbf{k}$ are the standard orthonormal basis in ${\mathbb{R}}^4$ which satisfy the quaternion multiplication rules as follows:

$${\mathbf{i}}^2={\mathbf{j}}^2={\mathbf{k}}^2=\mathbf{ijk}=-1,$$

(2)

$$\mathbf{ij}=\mathbf{k}=-\mathbf{ji},\mathbf{jk}=\mathbf{i}=-\mathbf{kj},\mathbf{ki}=\mathbf{j}=-\mathbf{ik}.$$

(3)

The norm of a quaternion p as in Equation (1) is defined by

$${∥p∥}^2=p_0^2+p_1^2+p_2^2+p_3^2.$$

If we write $p=p_0+u$, where $u=p_1\mathbf{i}+p_2\mathbf{j}+p_3\mathbf{k}$, then the conjugate of the quaternion p is denoted by $\overline{p}=p_0-p_1\mathbf{i}-p_2\mathbf{j}-p_3\mathbf{k}.$ Quaternions have proven to be highly beneficial for both theoretical and practical applications in research. A significant number of research articles on quaternions are frequently published in journals of mathematical physics and quantum mechanics, with quaternion analysis being considered part of mainstream physics. In engineering, quaternions are widely utilized in control systems, and in the field of computer science, they play a critical role in computer graphics. For a more comprehensive understanding of the advances in these areas, the monographs and papers [2,3,4] are valuable resources for interested readers.

The utilization of quaternions by mathematicians extends to the domain of defining quaternions whose coefficients are characterized by special integer sequences or special polynomials, and subsequently examining the algebraic properties of these quaternion types. Horadam [5] introduced the Fibonacci quaternions as

$$\mathbb{Q}{F}_n=F_n+F_{n+1}\mathbf{i}+F_{n+2}\mathbf{j}+F_{n+3}\mathbf{k},$$

(4)

where $F_n$ is the n-th Fibonacci number defined by

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

(5)

for $n\ge 2,$ with the initial values $F_0=0$, $F_1=1.$ The Binet formula for the Fibonacci sequence is

$$F_n={\displaystyle \frac{{\alpha }^n-{\beta }^n}{\sqrt{5}}},$$

where $\alpha ={\displaystyle \frac{1+\sqrt{5}}{2}}$ and $\beta ={\displaystyle \frac{1-\sqrt{5}}{2}}.$ Also, the Binet formula of the $\mathbb{Q}{F}_n$ is as follows:

$$\mathbb{Q}{F}_n={\displaystyle \frac{\underline{\alpha }{\alpha }^n-\underline{\beta }{\beta }^n}{\sqrt{5}}},$$

(6)

where

$$\begin{array}{ccc}\hfill \underline{\alpha }& =& 1+\alpha \mathbf{i}+{\alpha }^2\mathbf{j}+{\alpha }^3\mathbf{k},\hfill \\ \hfill \underline{\beta }& =& 1+\beta \mathbf{i}+{\beta }^2\mathbf{j}+\beta \mathbf{k}.\hfill \end{array}$$

Previous research on this construction has been conducted by numerous mathematicians; see, for example, Refs. [6,7,8,9,10,11,12,13,14,15,16,17]. Additionally, Pashaev and Özvatan studied the Fibonacci divisor, also known as higher-order Fibonacci numbers (see [18,19] for details). These higher-order Fibonacci numbers, also known as the Fibonacci divisor or conjugate to $F_s$, are defined for any integer $s\ge 1$ as follows:

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

(7)

Considering the divisibility of $F_{ns}$ by $F_s$, it can be inferred that the ratio ${\displaystyle \frac{F_{ns}}{Fs}}$ is an integer. Consequently, all higher-order Fibonacci numbers, specifically $F_n^{\left(s\right)}$, are integer. When $s=1,$ the higher-order Fibonacci number $F_n^{\left(1\right)}$ becomes an ordinary Fibonacci number. The initial few values of the higher-order Fibonacci numbers $F_n^{\left(s\right)}$ are

• For $s=1$, $F_n^{\left(1\right)}=F_n=1,1,2,3,\dots$ ;
• For $s=2$, $F_n^{\left(2\right)}=F_{2n}=1,3,8,21,\dots$ ;
• For $s=3$, $F_n^{\left(3\right)}={\displaystyle \frac{1}{2}}{F}_{3n}=1,4,17,72,\dots$ ;
• For $s=4$, $F_n^{\left(4\right)}={\displaystyle \frac{1}{3}}{F}_{4n}=1,7,48,329,\dots$ ;
• For $s=5$, $F_n^{\left(5\right)}={\displaystyle \frac{1}{5}}{F}_{5n}=1,11,122,1353,\dots$ .

Quantum calculus (q-calculus for short) was originally proposed by Jackson [20] and is one of the most important extensions of ordinary calculus. The q-integer is a generalization of traditional calculus $(q\to 1)$, and is defined as the following:

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

for $n\in \mathbb{N}$ and $q\in \mathbb{R}\setminus \left\{1\right\}$. Certain properties of q-numbers, including the q-addition formula, the q-subtraction formula, and the q-product formula are as follows:

$${\left[-n\right]}_q=-q^{-n}{\left[n\right]}_q,$$

$${\left[n+m\right]}_q={\left[m\right]}_q+q^m{\left[n\right]}_q,$$

$${\left[n-m\right]}_q={\left[n\right]}_q-q^{n-m}{\left[m\right]}_q,$$

and

$${\left[nm\right]}_q={\left[n\right]}_{q^m}{\left[m\right]}_q.$$

where ${\left[n\right]}_{q^m}={\displaystyle \frac{1-q^{mn}}{1-q^m}}.$ By using the definition of the q-integer, the following identity holds:

$${\left[n+2\right]}_{q^s}=\left(1+q^s\right){\left[n+1\right]}_{q^s}-q^s{\left[n\right]}_{q^s}.$$

(8)

For more details, interested readers are encouraged to consult the remarkable monograph [21]. Pashaev revealed crucial aspects of the higher-order Fibonacci numbers and concurrently exhibited their utility in numerous physical scenarios. For example, in [19], by using the quantum calculus, the infinite hierarchy of Golden quantum oscillators with integer spectrum determined by Fibonacci divisors, the hierarchy of Golden coherent states, and related Fock–Bargman representations in space of complex analytic functions were derived.

In [22], the authors defined quaternions with quantum coefficients, including Fibonacci and Lucas quaternions. The authors also provided some combinatorial properties and applications related to time evolution and rotation. The objective of this paper is to extend the quaternion family introduced by Akkus and Kizilaslan by considering a parameter s. Furthermore, we aim to define quaternion families that may or may not have been previously published, by utilizing higher-order generalized Fibonacci numbers and q-integers for arbitrary integer values of s. The following is the outline of our paper: In the second section, we present the main definition. Subsequently, we obtain various properties of the new quaternion family, such as Binet-like formulas, generating functions, recurrence relations, linearization properties, sum formulas, and more. In the third part of our paper, we obtain new properties for these quaternions using the interrelations of the three matrices of the special type we have defined. The fourth section of the paper summarizes the obtained results and outlines future research directions.

2. The Higher-Order Generalized Fibonacci Quaternions with $\mathbf{q}$-Integer Components

In this section, we introduce the concept of higher-order generalized Fibonacci quaternions with q-integer components.

Definition 1.

Let $n,s\in \mathbb{Z}$ (the set of integers), $p\in \mathbb{R}$, and $q\in \mathbb{R}\setminus \left\{1\right\}$. The higher-order generalized Fibonacci quaternions with q-integer components are defined by

$$\begin{array}{c}\hfill {\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)=p^{s\left(n-1\right)}{\displaystyle \frac{{\left[ns\right]}_q}{{\left[s\right]}_q}}+p^{sn}{\displaystyle \frac{{\left[\left(n+1\right)s\right]}_q}{{\left[s\right]}_q}}\mathit{i}+p^{s\left(n+1\right)}{\displaystyle \frac{{\left[\left(n+2\right)s\right]}_q}{{\left[s\right]}_q}}\mathit{j}+p^{s\left(n+2\right)}{\displaystyle \frac{{\left[\left(n+3\right)s\right]}_q}{{\left[s\right]}_q}}\mathit{k}\end{array}$$

(9)

or simply,

$${\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)=p^{s\left(n-1\right)}{\left[n\right]}_{q^s}+p^{sn}{\left[n+1\right]}_{q^s}\mathit{i}+p^{s\left(n+1\right)}{\left[n+2\right]}_{q^s}\mathit{j}+p^{s\left(n+2\right)}{\left[n+3\right]}_{q^s}\mathit{k}.$$

(10)

In fact, ${\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)$ contains many important quaternions as special cases:

1.

For $s=1$, Equation (10) becomes the q-Fibonacci-type quaternion [22].

2.

For $p={\displaystyle \frac{k+\sqrt{k^2+4}}{2}}$, $q=-{\displaystyle \frac{1}{p^2}}$, Equation (10) becomes the higher-order k-Fibonacci quaternion.

3.

For $p={\displaystyle \frac{1+\sqrt{5}}{2}}$, $q=-{\displaystyle \frac{1}{p^2}}$, Equation (10) becomes the higher-order Fibonacci quaternion [14].

4.

For $p={\displaystyle \frac{1+\sqrt{5}}{2}}$, $q=-{\displaystyle \frac{1}{p^2}}$, $s=1$, Equation (10) becomes the Fibonacci quaternion [5].

5.

For $p={\displaystyle \frac{k+\sqrt{k^2+8}}{2}}$, $q=-{\displaystyle \frac{2}{p^2}}$, Equation (10) becomes the higher-order k-Jacobsthal quaternion.

6.

For $p=2$, $q=-{\displaystyle \frac{1}{2}}$, Equation (10) becomes the higher-order Jacobsthal quaternion [23].

7.

For $p=2$, $q=-{\displaystyle \frac{1}{2}}$, $s=1$, Equation (10) becomes the Jacobsthal quaternion.

8.

For $p=1+\sqrt{1+k}$, $q=-{\displaystyle \frac{k}{p^2}}$, Equation (10) becomes the higher-order k-Pell quaternion.

9.

For $p=1+\sqrt{2}$, $q=-{\displaystyle \frac{1}{p^2}}$, Equation (10) becomes the higher-order Pell quaternion [24].

10.

For $p=1+\sqrt{2}$, $q=-{\displaystyle \frac{1}{p^2}}$, $s=1$, Equation (10) becomes the Pell quaternion [25].

We note that the following identities hold:

$${\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)+\overline{{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)}=2p^{s\left(n-1\right)}{\left[n\right]}_{q^s}$$

and

$$\begin{array}{c}\hfill {\left({\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)\right)}^2=-{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)\overline{{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)}+2p^{s\left(n-1\right)}{\left[n\right]}_{q^s}{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right).\end{array}$$

Theorem 1.

For $n,s\in \mathbb{Z}$, $p\in \mathbb{R}$, and $q\in \mathbb{R}\setminus \left\{1\right\}$. The Binet-like formula of ${\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)$ can be expressed by

$${\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)={\displaystyle \frac{p^{s\left(n-1\right)}\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}{1-q^s}},$$

(11)

where

$$\mathrm{\Theta }=1+p^s\mathit{i}+p^{2s}\mathit{j}+p^{3s}\mathit{k}$$

and

$$\mathrm{{\rm Y}}=1+{\left(pq\right)}^s\mathit{i}+{\left(pq\right)}^{2s}\mathit{j}+{\left(pq\right)}^{3s}\mathit{k}.$$

Proof. 

By using Equation (10), we obtain

$$\begin{array}{ccc}\hfill {\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)& =& p^{s\left(n-1\right)}{\left[n\right]}_{q^s}+p^{sn}{\left[n+1\right]}_{q^s}\mathit{i}+p^{s\left(n+1\right)}{\left[n+2\right]}_{q^s}\mathit{j}+p^{s\left(n+2\right)}{\left[n+3\right]}_{q^s}\mathit{k}\hfill \\ & =& {\displaystyle \frac{p^{s\left(n-1\right)}}{1-q^s}}\left(1-q^{sn}+p^s\left(1-q^{s\left(n+1\right)}\right)\mathit{i}+p^{2s}\left(1-q^{s\left(n+2\right)}\right)\mathit{j}+p^{3s}\left(1-q^{s\left(n+3\right)}\right)\mathit{k}\right)\hfill \\ & =& {\displaystyle \frac{p^{s\left(n-1\right)}}{1-q^s}}\left(1+p^s\mathit{i}+p^{2s}\mathit{j}+p^{3s}\mathit{k}-q^{sn}-p^s{q}^{s\left(n+1\right)}\mathit{i}-p^{2s}{q}^{s\left(n+2\right)}\mathit{j}-p^{3s}{q}^{s\left(n+3\right)}\mathit{k}\right)\hfill \\ & =& {\displaystyle \frac{p^{s\left(n-1\right)}}{1-q^s}}\left(1+p^s\mathit{i}+p^{2s}\mathit{j}+p^{3s}\mathit{k}-q^{sn}\left(1+{\left(pq\right)}^s\mathit{i}+{\left(pq\right)}^{2s}\mathit{j}+{\left(pq\right)}^{3s}\mathit{k}\right)\right)\hfill \\ & =& {\displaystyle \frac{p^{s\left(n-1\right)}\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}{1-q^s}}.\hfill \end{array}$$

The proof is completed. □

Applying the Binet-like formula of ${\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)$ (i.e., Theorem 1), we obtain the following result.

Proposition 1.

The following new equalities for ${\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)$ hold true:

(1) 

${\mathbb{G}}_{-n}^{\left(-s\right)}\left(p,q\right)=-{\left(p^{2}q\right)}^s{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right),$

(2) 

${\mathbb{G}}_n^{\left(-s\right)}\left(p,q\right)=-p^{-2s\left(n-1\right)}{q}^s{\displaystyle \frac{\left(\mathrm{\Theta }-q^{-sn}\mathrm{{\rm Y}}\right)}{\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}}{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right),$

(3) 

${\mathbb{G}}_{-n}^{\left(s\right)}\left(p,q\right)=p^{-2sn}{\displaystyle \frac{\left(\mathrm{\Theta }-q^{-sn}\mathrm{{\rm Y}}\right)}{\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}}{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right).$

Proof. 

Applying Theorem 1, we have

$$\begin{array}{ccc}\hfill -p^{2s}{q}^s{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)& =& -p^{2s}{q}^s{\displaystyle \frac{p^{s\left(n-1\right)}\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}{1-q^s}}\hfill \\ & =& -p^{2s}{q}^s{\displaystyle \frac{p^{s\left(n-1\right)}\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}{-q^s\left(1-q^{-s}\right)}}\hfill \\ & =& {\displaystyle \frac{p^{s\left(n+1\right)}\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}{\left(1-q^{-s}\right)}}\hfill \\ & =& {\displaystyle \frac{p^{-s\left(-n-1\right)}\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}{\left(1-q^{-s}\right)}}\hfill \\ & =& {\mathbb{G}}_{-n}^{\left(-s\right)}\left(p,q\right).\hfill \end{array}$$

Hence, (1) is proved. Applying Theorem 1, we have

$$\begin{array}{ccc}\hfill {\displaystyle \frac{{\mathbb{G}}_n^{\left(-s\right)}\left(p,q\right)}{{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)}}& =& {\displaystyle \frac{\frac{p^{-s\left(n-1\right)}\left(\mathrm{\Theta }-q^{-sn}\mathrm{{\rm Y}}\right)}{1-q^{-s}}}{\frac{p^{s\left(n-1\right)}\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}{1-q^s}}}\hfill \\ & =& {\displaystyle \frac{p^{-2s\left(n-1\right)}\left(\mathrm{\Theta }-q^{-sn}\mathrm{{\rm Y}}\right)\left(1-q^s\right)}{\left(1-q^{-s}\right)\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}}\hfill \\ & =& -p^{-2s\left(n-1\right)}{q}^s{\displaystyle \frac{\left(\mathrm{\Theta }-q^{-sn}\mathrm{{\rm Y}}\right)}{\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}},\hfill \end{array}$$

which implies the assertion (2). Finally, let us verify Equation (3). Applying Theorem 1, we have

$${\displaystyle \frac{{\mathbb{G}}_{-n}^{\left(s\right)}\left(p,q\right)}{{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)}}={\displaystyle \frac{\frac{p^{s\left(-n-1\right)}\left(\mathrm{\Theta }-q^{-sn}\mathrm{{\rm Y}}\right)}{1-q^s}}{\frac{p^{s\left(n-1\right)}\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}{1-q^s}}}={\displaystyle \frac{p^{-2sn}\left(\mathrm{\Theta }-q^{-sn}\mathrm{{\rm Y}}\right)}{\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}}.$$

The proof is completed. □

Theorem 2.

For $m,s\in \mathbb{Z}$, $n\in \mathbb{N}\cup \left\{0\right\}$, $p\in \mathbb{R}$ and $q\in \mathbb{R}\setminus \left\{1\right\}$, the generating function of ${\mathbb{G}}_{m+n}^{\left(s\right)}\left(p,q\right)$ is

$$\sum _{n=0}^{\infty }{\mathbb{G}}_{m+n}^{\left(s\right)}\left(p,q\right)t^n={\displaystyle \frac{p^{s\left(m-1\right)}}{1-q^s}}\left({\displaystyle \frac{\mathrm{\Theta }-q^{sm}\mathrm{{\rm Y}}-{\left(pq\right)}^s\left(\mathrm{\Theta }-q^{s(m-1)}\mathrm{{\rm Y}}\right)t}{1-p^s\left(1+q^s\right)t+{\left(p^{2}q\right)}^s{t}^2}}\right).$$

(12)

Proof. 

Employing Theorem 1, we obtain

$$\begin{array}{ccc}\hfill \sum _{n=0}^{\infty }{\mathbb{G}}_{m+n}^{\left(s\right)}\left(p,q\right)t^n& =& \sum _{n=0}^{\infty }{\displaystyle \frac{p^{s\left(m+n-1\right)}\left(\mathrm{\Theta }-q^{s\left(m+n\right)}\mathrm{{\rm Y}}\right)}{1-q^s}}{t}^n\hfill \\ & =& {\displaystyle \frac{p^{s\left(m-1\right)}}{1-q^s}}\left(\mathrm{\Theta }\sum _{n=0}^{\infty }{\left(p^{s}t\right)}^n-q^{sm}\mathrm{{\rm Y}}\sum _{n=0}^{\infty }{\left({\left(pq\right)}^{s}t\right)}^n\right)\hfill \\ & =& {\displaystyle \frac{p^{s\left(m-1\right)}}{1-q^s}}\left(\mathrm{\Theta }{\displaystyle \frac{1}{1-p^{s}t}}-q^{sm}\mathrm{{\rm Y}}{\displaystyle \frac{1}{1-{\left(pq\right)}^{s}t}}\right)\hfill \\ & =& {\displaystyle \frac{p^{s\left(m-1\right)}}{1-q^s}}\left({\displaystyle \frac{\mathrm{\Theta }-\mathrm{\Theta }{\left(pq\right)}^{s}t-q^{sm}\mathrm{{\rm Y}}+q^{sm}\mathrm{{\rm Y}}{p}^{s}t}{\left(1-p^{s}t\right)\left(1-{\left(pq\right)}^{s}t\right)}}\right)\hfill \\ & =& {\displaystyle \frac{p^{s\left(m-1\right)}}{1-q^s}}\left({\displaystyle \frac{\mathrm{\Theta }-q^{sm}\mathrm{{\rm Y}}-{\left(pq\right)}^s\left(\mathrm{\Theta }-q^{s(m-1)}\mathrm{{\rm Y}}\right)t}{1-p^s\left(1+q^s\right)t+{\left(p^{2}q\right)}^s{t}^2}}\right).\hfill \end{array}$$

The required proof is completed. □

When taking $m=0$ in Equation (12), we can derive the following conclusion immediately.

Corollary 1.

For $n\in \mathbb{N}\cup \left\{0\right\}$, $s\in \mathbb{Z}$, $p\in \mathbb{R}$ and $q\in \mathbb{R}\setminus \left\{1\right\}$, the generating function of ${\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)$ is

$$\sum _{n=0}^{\infty }{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)t^n={\displaystyle \frac{1}{p^s\left(1-q^s\right)}}\left({\displaystyle \frac{\mathrm{\Theta }-\mathrm{{\rm Y}}-p^s\left(q^s\mathrm{\Theta }-\mathrm{{\rm Y}}\right)t}{1-p^s\left(1+q^s\right)t+{\left(p^{2}q\right)}^s{t}^2}}\right).$$

Theorem 3.

For $n\in \mathbb{N}\cup \left\{0\right\}$, $s\in \mathbb{Z}$, $p\in \mathbb{R}$ and $q\in \mathbb{R}\setminus \left\{1\right\}$, the exponential generating function of ${\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)$ is

$$\sum _{n=0}^{\infty }{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right){\displaystyle \frac{t^n}{n!}}={\displaystyle \frac{\mathrm{\Theta }{e}^{p^{s}t}-\mathrm{{\rm Y}}{e}^{{\left(pq\right)}^{s}t}}{p^s\left(1-q^s\right)}}.$$

Proof. 

Thanks to Equation (11), we obtain

$$\begin{array}{ccc}\hfill \sum _{n=0}^{\infty }{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right){\displaystyle \frac{t^n}{n!}}& =& \sum _{n=0}^{\infty }{\displaystyle \frac{p^{s\left(n-1\right)}\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}{1-q^s}}{\displaystyle \frac{t^n}{n!}}\hfill \\ & =& {\displaystyle \frac{1}{p^s\left(1-q^s\right)}}\left(\mathrm{\Theta }\sum _{n=0}^{\infty }{p}^{sn}{\displaystyle \frac{t^n}{n!}}-\mathrm{{\rm Y}}\sum _{n=0}^{\infty }{p}^{sn}{q}^{sn}{\displaystyle \frac{t^n}{n!}}\right)\hfill \\ & =& {\displaystyle \frac{\mathrm{\Theta }{e}^{p^{s}t}-\mathrm{{\rm Y}}{e}^{{\left(pq\right)}^{s}t}}{p^s\left(1-q^s\right)}}.\hfill \end{array}$$

The proof is completed. □

Theorem 4.

For $n,s\in \mathbb{Z}$, $p\in \mathbb{R}$ and $q\in \mathbb{R}\setminus \left\{1\right\}$, the recurrence relation for ${\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)$ is as follows:

$${\mathbb{G}}_{n+1}^{\left(s\right)}\left(p,q\right)=p^s\left(1+q^s\right){\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)-{\left(p^{2}q\right)}^s{\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right).$$

Proof. 

Utilizing Theorem 1 leads to

$$\begin{array}{ccc}\hfill {\mathbb{G}}_{n+1}^{\left(s\right)}\left(p,q\right)& =& {\displaystyle \frac{p^{sn}\left(\mathrm{\Theta }-q^{s\left(n+1\right)}\mathrm{{\rm Y}}\right)}{1-q^s}}\hfill \\ & =& {\displaystyle \frac{p^{s\left(n-1\right)}{p}^s\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}+q^{sn}\mathrm{{\rm Y}}-q^{s\left(n+1\right)}\mathrm{{\rm Y}}\right)}{1-q^s}}\hfill \\ & =& p^s{\displaystyle \frac{p^{s\left(n-1\right)}\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}{1-q^s}}+p^{s\left(n-1\right)}{p}^s{\displaystyle \frac{q^{sn}\mathrm{{\rm Y}}-q^{s\left(n+1\right)}\mathrm{{\rm Y}}}{1-q^s}}\hfill \\ & =& p^s{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)+p^{s\left(n-2\right)}{p}^{2s}{\displaystyle \frac{q^s\mathrm{\Theta }-q^{s\left(n+1\right)}\mathrm{{\rm Y}}-q^s\mathrm{\Theta }+q^s{q}^{s\left(n-1\right)}\mathrm{{\rm Y}}}{1-q^s}}\hfill \\ & =& p^s\left({\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)+p^{s\left(n-2\right)}{p}^s{q}^s{\displaystyle \frac{\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}-\mathrm{\Theta }+q^{s\left(n-1\right)}\mathrm{{\rm Y}}}{1-q^s}}\right)\hfill \\ & =& p^s\left({\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)+q^s{\displaystyle \frac{p^{s\left(n-1\right)}\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}{1-q^s}}-p^s{q}^s{\displaystyle \frac{p^{s\left(n-2\right)}\left(\mathrm{\Theta }-q^{s\left(n-1\right)}\mathrm{{\rm Y}}\right)}{1-q^s}}\right)\hfill \\ & =& p^s\left(1+q^s\right){\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)-{\left(p^{2}q\right)}^s{\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right).\hfill \end{array}$$

The proof of Theorem 4 is thus completed. □

We now establish the following identities for ${\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)$.

Theorem 5.

For $k,n,r,s\in \mathbb{Z}$, $p\in \mathbb{R}$ and $q\in \mathbb{R}\setminus \left\{1\right\}$, we have

$${\mathbb{G}}_{n+r}^{\left(s\right)}\left(p,q\right){\mathbb{G}}_{n+k}^{\left(s\right)}\left(p,q\right)-{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right){\mathbb{G}}_{n+r+k}^{\left(s\right)}\left(p,q\right)={\displaystyle \frac{p^{s\left(2n+r+k\right)}\left(q^{sr}-1\right)q^{sn}}{p^{2s}{\left(1-q^s\right)}^2}}\left(q^{sk}\mathrm{\Theta }\mathrm{{\rm Y}}-\mathrm{{\rm Y}}\mathrm{\Theta }\right).$$

(13)

Proof. 

By using Theorem 1, we obtain

$$\begin{array}{ccc}& & {\mathbb{G}}_{n+r}^{\left(s\right)}\left(p,q\right){\mathbb{G}}_{n+k}^{\left(s\right)}\left(p,q\right)-{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right){\mathbb{G}}_{n+r+k}^{\left(s\right)}\left(p,q\right)\hfill \\ & =& {\displaystyle \frac{p^{s\left(n+r-1\right)}\left(\mathrm{\Theta }-q^{s\left(n+r\right)}\mathrm{{\rm Y}}\right)}{1-q^s}}{\displaystyle \frac{p^{s\left(n+k-1\right)}\left(\mathrm{\Theta }-q^{s\left(n+k\right)}\mathrm{{\rm Y}}\right)}{1-q^s}}\hfill \\ & & -{\displaystyle \frac{p^{s\left(n-1\right)}\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}{1-q^s}}{\displaystyle \frac{p^{s\left(n+r+k-1\right)}\left(\mathrm{\Theta }-q^{s\left(n+r+k\right)}\mathrm{{\rm Y}}\right)}{1-q^s}}\hfill \\ & =& {\displaystyle \frac{1}{{\left(1-q^s\right)}^2}}\left(\begin{array}{c}{p}^{s\left(2n+r+k-2\right)}\left(\mathrm{\Theta }-q^{s\left(n+r\right)}\mathrm{{\rm Y}}\right)\left(\mathrm{\Theta }-q^{s\left(n+k\right)}\mathrm{{\rm Y}}\right)\\ -p^{s\left(2n+r+k-2\right)}\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)\left(\mathrm{\Theta }-q^{s\left(n+r+k\right)}\mathrm{{\rm Y}}\right)\end{array}\right)\hfill \\ & =& {\displaystyle \frac{p^{s\left(2n+r+k-2\right)}}{{\left(1-q^s\right)}^2}}\left(\begin{array}{c}{\mathrm{\Theta }}^2-\mathrm{\Theta }{q}^{s\left(n+k\right)}\mathrm{{\rm Y}}-q^{s\left(n+r\right)}\mathrm{{\rm Y}}\mathrm{\Theta }+q^{s\left(n+r\right)}\mathrm{{\rm Y}}{q}^{s\left(n+k\right)}\mathrm{{\rm Y}}\\ -{\mathrm{\Theta }}^2+\mathrm{\Theta }{q}^{s\left(n+r+k\right)}\mathrm{{\rm Y}}+q^{sn}\mathrm{{\rm Y}}\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}{q}^{s\left(n+r+k\right)}\mathrm{{\rm Y}}\end{array}\right)\hfill \\ & =& {\displaystyle \frac{p^{s\left(2n+r+k-2\right)}}{{\left(1-q^s\right)}^2}}\left(\left(q^{s\left(n+r+k\right)}-q^{s\left(n+k\right)}\right)\mathrm{\Theta }\mathrm{{\rm Y}}+\left(q^{sn}-q^{s\left(n+r\right)}\right)\mathrm{{\rm Y}}\mathrm{\Theta }\right)\hfill \\ & =& {\displaystyle \frac{p^{s\left(2n+r+k-2\right)}\left(q^{sr}-1\right)q^{sn}}{{\left(1-q^s\right)}^2}}\left(q^{sk}\mathrm{\Theta }\mathrm{{\rm Y}}-\mathrm{{\rm Y}}\mathrm{\Theta }\right).\hfill \end{array}$$

The proof is completed. □

Remark 1.

As applications of Theorem 5, we have the following results:

(1) 

When taking $k\to -r$ in Equation (13), we obtain

$${\mathbb{G}}_{n+r}^{\left(s\right)}\left(p,q\right){\mathbb{G}}_{n-r}^{\left(s\right)}\left(p,q\right)-{\left({\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)\right)}^2={\displaystyle \frac{p^{2sn}\left(q^{sr}-1\right)q^{sn}}{p^{2s}{\left(1-q^s\right)}^2}}\left(q^{-sr}\mathrm{\Theta }\mathrm{{\rm Y}}-\mathrm{{\rm Y}}\mathrm{\Theta }\right).$$

(14)

(2) 

When taking $k\to -1$ and $r\to 1$ in Equation (13), we have

$${\mathbb{G}}_{n+1}^{\left(s\right)}\left(p,q\right){\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right)-{\left({\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)\right)}^2={\displaystyle \frac{p^{2sn}\left(q^s-1\right)q^{sn}}{p^{2s}{\left(1-q^s\right)}^2}}\left(q^{-s}\mathrm{\Theta }\mathrm{{\rm Y}}-\mathrm{{\rm Y}}\mathrm{\Theta }\right).$$

(15)

(3) 

When taking $k\to 1$ and $n+r\to m$ in Equation (13), we obtain

$${\mathbb{G}}_m^{\left(s\right)}\left(p,q\right){\mathbb{G}}_{n+1}^{\left(s\right)}\left(p,q\right)-{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right){\mathbb{G}}_{m+1}^{\left(s\right)}\left(p,q\right)={\displaystyle \frac{p^{s\left(n+m+1\right)}\left(q^{s\left(m-n\right)}-1\right)q^{sn}}{p^{2s}{\left(1-q^s\right)}^2}}\left(q^s\mathrm{\Theta }\mathrm{{\rm Y}}-\mathrm{{\rm Y}}\mathrm{\Theta }\right).$$

(16)

Theorem 6.

For $m,n,s\in \mathbb{Z}$, $p\in \mathbb{R}$ and $q\in \mathbb{R}\setminus \left\{1\right\}$, we have

$${\mathbb{G}}_n^{\left(s\right)}\left(p,q\right){\mathbb{G}}_m^{\left(s\right)}\left(p,q\right)+{\mathbb{G}}_{n+1}^{\left(s\right)}\left(p,q\right){\mathbb{G}}_{m+1}^{\left(s\right)}\left(p,q\right)={\displaystyle \frac{p^{s\left(n+m\right)}}{{\left(1-q^s\right)}^2}}\left(\begin{array}{c}\left(1+p^{-2s}\right){\mathrm{\Theta }}^2+q^{\left(n+m\right)}\left(p^{-2s}+q^{2s}\right){\mathrm{{\rm Y}}}^2\\ -\left(q^s+p^{-2s}\right)\left(q^{sm}\mathrm{\Theta }\mathrm{{\rm Y}}+q^{sn}\mathrm{{\rm Y}}\mathrm{\Theta }\right)\end{array}\right).$$

Proof. 

By virtue of Theorem 1, we conclude

$$\begin{array}{ccc}& & {\mathbb{G}}_n^{\left(s\right)}\left(p,q\right){\mathbb{G}}_m^{\left(s\right)}\left(p,q\right)+{\mathbb{G}}_{n+1}^{\left(s\right)}\left(p,q\right){\mathbb{G}}_{m+1}^{\left(s\right)}\left(p,q\right)\hfill \\ & =& {\displaystyle \frac{p^{s\left(n-1\right)}\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}{1-q^s}}{\displaystyle \frac{p^{s\left(m-1\right)}\left(\mathrm{\Theta }-q^{sm}\mathrm{{\rm Y}}\right)}{1-q^s}}\hfill \\ & & +{\displaystyle \frac{p^{sn}\left(\mathrm{\Theta }-q^{s\left(n+1\right)}\mathrm{{\rm Y}}\right)}{1-q^s}}{\displaystyle \frac{p^{sm}\left(\mathrm{\Theta }-q^{s\left(m+1\right)}\mathrm{{\rm Y}}\right)}{1-q^s}}\hfill \\ & & \\ & =& {\displaystyle \frac{1}{{\left(1-q^s\right)}^2}}\left(\begin{array}{c}{p}^{s\left(n+m-2\right)}\left({\mathrm{\Theta }}^2-\mathrm{\Theta }{q}^{sm}\mathrm{{\rm Y}}-q^{sn}\mathrm{{\rm Y}}\mathrm{\Theta }+q^{sn}\mathrm{{\rm Y}}{q}^{sm}\mathrm{{\rm Y}}\right)\\ +p^{s\left(n+m\right)}\left({\mathrm{\Theta }}^2-\mathrm{\Theta }{q}^{s\left(m+1\right)}\mathrm{{\rm Y}}-q^{s\left(n+1\right)}\mathrm{{\rm Y}}\mathrm{\Theta }+q^{s\left(n+1\right)}\mathrm{{\rm Y}}{q}^{s\left(m+1\right)}\mathrm{{\rm Y}}\right)\end{array}\right)\hfill \\ & & \\ & =& {\displaystyle \frac{p^{s\left(n+m\right)}}{{\left(1-q^s\right)}^2}}\left(\begin{array}{c}\left(1+p^{-2s}\right){\mathrm{\Theta }}^2-q^{sm}\left(q^s+p^{-2s}\right)\mathrm{\Theta }\mathrm{{\rm Y}}\\ -q^{sn}\left(q^s+p^{-2s}\right)\mathrm{{\rm Y}}\mathrm{\Theta }+q^{\left(n+m\right)}\left(p^{-2s}+q^{2s}\right){\mathrm{{\rm Y}}}^2\end{array}\right)\hfill \\ & & \\ & =& {\displaystyle \frac{p^{s\left(n+m\right)}}{{\left(1-q^s\right)}^2}}\left(\begin{array}{c}\left(1+p^{-2s}\right){\mathrm{\Theta }}^2+q^{\left(n+m\right)}\left(p^{-2s}+q^{2s}\right){\mathrm{{\rm Y}}}^2\\ -\left(q^s+p^{-2s}\right)\left(q^{sm}\mathrm{\Theta }\mathrm{{\rm Y}}+q^{sn}\mathrm{{\rm Y}}\mathrm{\Theta }\right)\end{array}\right).\hfill \end{array}$$

The proof is completed. □

Remark 2.

In Theorems 5 and 6 and also in Remark 1, we can obtain some identities for special cases of p, q, and s.

Example 1 

([26]). When taking $p={\displaystyle \frac{k+\sqrt{k^2+4}}{2}}$, $q=-{\displaystyle \frac{1}{p^2}}$ and $s=1$ in Equation (14), we obtain

$$\mathbb{Q}{F}_{k,n-r}\mathbb{Q}{F}_{k,n+r}-{\left(\mathbb{Q}{F}_{k,n}\right)}^2={\left(-1\right)}^{n-r+1}\left(2F_{k,r}\mathbb{Q}{F}_{k,r}-\left(k^2+2\right)F_{k,2r}\mathbf{k}\right),$$

where $n\ge r\ge 1.$

Example 2 

([27]). When taking $p={\displaystyle \frac{1+\sqrt{5}}{2}}$, $q=-{\displaystyle \frac{1}{p^2}}$, and $s=1$ in Equation (15), we obtain

$$\mathbb{Q}{F}_{n-1}\mathbb{Q}{F}_{n+1}-{\left(\mathbb{Q}{F}_n\right)}^2={\left(-1\right)}^n\left(2\mathbb{Q}{F}_1-3\mathbf{k}\right).$$

Theorem 7.

Let $n,s\in \mathbb{Z}$, $p\in \mathbb{R}$ and $q\in \mathbb{R}\setminus \left\{1\right\}$. If $p^{2}q=t$, then the linearization of ${\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)$ can be represented as

$$p^s{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)+\left(-t^s\right){\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right)=p^{sn}\mathrm{\Theta }.$$

Proof. 

Applying Theorem 1 yields

$$\begin{array}{ccc}\hfill {\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)-{\left(pq\right)}^s{\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right)& =& {\displaystyle \frac{p^{s\left(n-1\right)}\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}{1-q^s}}-{\left(pq\right)}^s{\displaystyle \frac{p^{s\left(n-2\right)}\left(\mathrm{\Theta }-q^{s\left(n-1\right)}\mathrm{{\rm Y}}\right)}{1-q^s}}\hfill \\ & =& {\displaystyle \frac{p^{s\left(n-1\right)}\mathrm{\Theta }-p^{s\left(n-1\right)}{q}^{sn}\mathrm{{\rm Y}}-p^{s\left(n-1\right)}{q}^s\mathrm{\Theta }+p^{s\left(n-1\right)}{q}^{sn}\mathrm{{\rm Y}}}{1-q^s}}\hfill \\ & =& p^{s\left(n-1\right)}\mathrm{\Theta }.\hfill \end{array}$$

Multiplying both sides of the above equation by $p^s$, we obtain the linearization of ${\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)$ as

$$p^s{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)-{\left(p^{2}q\right)}^s{\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right)=p^{sn}\mathrm{\Theta },$$

or equivalence

$$p^s{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)-t^s{\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right)=p^{sn}\mathrm{\Theta }.$$

The proof is completed. □

Here, we give an example illustrating Theorem 7.

Example 3.

Let $p={\displaystyle \frac{k+\sqrt{k^2+4}}{2}}$.

(1) 

For $q:=-{\displaystyle \frac{1}{p^2}}$, we have $p^{2}q=-1$. Making use of Theorem 7 yields

$$p^s{\mathbb{G}}_n^{\left(s\right)}\left(p,-p^{-2}\right)+{\left(-1\right)}^{s+1}{\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,-p^{-2}\right)=p^{sn}\mathrm{\Theta }.$$

(2) 

For $q:=-{\displaystyle \frac{2}{p^2}}$, we have $p^{2}q=-2$. Making use of Theorem 7 yields

$$p^s{\mathbb{G}}_n^{\left(s\right)}\left(p,-2p^{-2}\right)-{\left(-2\right)}^s{\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,-2p^{-2}\right)=p^{sn}\mathrm{\Theta }.$$

Theorem 8.

The following summation formulas for ${\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)$ hold true:

$$\begin{array}{ccc}& & \sum _{n=1}^m{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)={\displaystyle \frac{1}{1-q^s}}\left(\mathrm{\Theta }{\displaystyle \frac{1-p^m}{1-p}}-q^s\mathrm{{\rm Y}}{\displaystyle \frac{1-{\left(pq\right)}^m}{1-pq}}\right),\hfill \end{array}$$

(17)

$$\begin{array}{ccc}& & \sum _{n=1}^m{\mathbb{G}}_{2n}^{\left(s\right)}\left(p,q\right)={\displaystyle \frac{p^s}{1-q^s}}\left(\mathrm{\Theta }{\displaystyle \frac{1-p^{2sm}}{1-p^{2s}}}-q^{2s}\mathrm{{\rm Y}}{\displaystyle \frac{1-{\left(pq\right)}^{2sm}}{1-{\left(pq\right)}^{2s}}}\right),\hfill \end{array}$$

(18)

$$\begin{array}{ccc}& & \sum _{n=1}^m{\mathbb{G}}_{2n-1}^{\left(s\right)}\left(p,q\right)={\displaystyle \frac{1}{1-q^s}}\left(\mathrm{\Theta }{\displaystyle \frac{1-p^{2sm}}{1-p^{2s}}}-q^s\mathrm{{\rm Y}}{\displaystyle \frac{1-{\left(pq\right)}^{2sm}}{1-{\left(pq\right)}^{2s}}}\right).\hfill \end{array}$$

(19)

Proof. 

Applying Theorem 1, we have

$$\begin{array}{ccc}\hfill \sum _{n=1}^m{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)& =& \sum _{n=1}^m{\displaystyle \frac{p^{s\left(n-1\right)}\left(\mathrm{\Theta }-q^{sn}\mathrm{{\rm Y}}\right)}{1-q^s}}\hfill \\ & & \\ & =& {\displaystyle \frac{1}{1-q^s}}\left(\begin{array}{c}\left(\mathrm{\Theta }-q^s\mathrm{{\rm Y}}\right)+p^s\left(\mathrm{\Theta }-q^{2s}\mathrm{{\rm Y}}\right)+p^{2s}\left(\mathrm{\Theta }-q^{3s}\mathrm{{\rm Y}}\right)\\ +\dots +p^{s\left(m-2\right)}\left(\mathrm{\Theta }-q^{s\left(m-1\right)}\mathrm{{\rm Y}}\right)+p^{s\left(m-1\right)}\left(\mathrm{\Theta }-q^{sm}\mathrm{{\rm Y}}\right)\end{array}\right)\hfill \\ & & \\ & =& {\displaystyle \frac{1}{1-q^s}}\left(\begin{array}{c}\mathrm{\Theta }\left(1+p^s+p^{2s}+\dots +p^{s\left(m-2\right)}+p^{s\left(m-1\right)}\right)\\ -q^s\mathrm{{\rm Y}}\left(1+{\left(pq\right)}^s+{\left(pq\right)}^{2s}+\dots +{\left(pq\right)}^{s\left(m-2\right)}+{\left(pq\right)}^{s\left(m-1\right)}\right)\end{array}\right)\hfill \\ & & \\ & =& {\displaystyle \frac{1}{1-q^s}}\left(\mathrm{\Theta }{\displaystyle \frac{1-p^m}{1-p}}-q^s\mathrm{{\rm Y}}{\displaystyle \frac{1-{\left(pq\right)}^m}{1-pq}}\right).\hfill \end{array}$$

This shows Equation (17). Equations (18) and (19) can be proved similarly. □

Theorem 9.

Let $k,l,m\in \mathbb{N}$, $p\in \mathbb{R}$, $q\in \mathbb{R}\setminus \left\{1\right\}$ and $\chi =p^s\left(1-q^s\right)$.

(1) 

If m is even, then

$$\begin{array}{c}\hfill \sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{\mathbb{G}}_{2l+k}^{\left(s\right)}\left(p,q\right)={\mathbb{G}}_{k+m}^{\left(s\right)}\left(p,q\right){\chi }^m.\end{array}$$

(20)

(2) 

If m is odd, then

$$\begin{array}{c}\hfill \sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{\mathbb{G}}_{2l+k}^{\left(s\right)}\left(p,q\right)=p^{s\left(m+k\right)}{\chi }^{m-1}\left(\mathrm{\Theta }+q^{s\left(k+m\right)}\mathrm{{\rm Y}}\right).\end{array}$$

(21)

Proof. 

Applying Theorem 1, we derive

$$\begin{array}{ccc}\hfill \sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{\mathbb{G}}_{2l+k}^{\left(s\right)}\left(p,q\right)& =& \sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{\displaystyle \frac{p^{s\left(2l+k-1\right)}\left(\mathrm{\Theta }-q^{s\left(2l+k\right)}\mathrm{{\rm Y}}\right)}{1-q^s}}\hfill \\ & =& {\displaystyle \frac{1}{1-q^s}}\left(\begin{array}{c}\mathrm{\Theta }\sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{p}^{s\left(2l+k-1\right)}\\ -\mathrm{{\rm Y}}\sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{p}^{s\left(2l+k-1\right)}{q}^{s\left(2l+k\right)}\end{array}\right)\hfill \\ \\ & =& {\displaystyle \frac{p^{s\left(k-1\right)}}{1-q^s}}\left(\begin{array}{c}\mathrm{\Theta }\sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{\left(p^{2s}\right)}^l\\ -q^{sk}\mathrm{{\rm Y}}\sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{\left(p^{2s}{q}^{2s}\right)}^l\end{array}\right)\hfill \\ \\ & =& {\displaystyle \frac{p^{s\left(k-1\right)}}{1-q^s}}\left(\mathrm{\Theta }{\left(p^{2s}-p^{2s}{q}^s\right)}^m-q^{sk}\mathrm{{\rm Y}}{\left(p^{2s}{q}^{2s}-p^{2s}{q}^s\right)}^m\right)\hfill \\ \\ & =& {\displaystyle \frac{p^{s\left(2m+k-1\right)}}{1-q^s}}\left(\mathrm{\Theta }{\left(1-q^s\right)}^m-q^{s\left(k+m\right)}\mathrm{{\rm Y}}{\left(q^s-1\right)}^m\right).\hfill \end{array}$$

(1) If m is even, then

$$\begin{array}{ccc}\hfill \sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{\mathbb{G}}_{2l+k}^{\left(s\right)}\left(p,q\right)& =& {\displaystyle \frac{p^{s\left(2m+k-1\right)}}{1-q^s}}\left(\mathrm{\Theta }{\left(1-q^s\right)}^m-q^{s\left(k+m\right)}\mathrm{{\rm Y}}{\left(1-q^s\right)}^m\right)\hfill \\ & =& {\displaystyle \frac{p^{s\left(2m+k-1\right)}{\left(1-q^s\right)}^m}{1-q^s}}\left(\mathrm{\Theta }-q^{s\left(k+m\right)}\mathrm{{\rm Y}}\right)\hfill \\ & =& {\displaystyle \frac{p^{s\left(m+k-1\right)}\left(\mathrm{\Theta }-q^{s\left(k+m\right)}\mathrm{{\rm Y}}\right)}{1-q^s}}{p}^{sm}{\left(1-q^s\right)}^m\hfill \\ & =& {\mathbb{G}}_{k+m}^{\left(s\right)}\left(p,q\right){\chi }^m.\hfill \end{array}$$

(2) If m is odd, then

$$\begin{array}{ccc}\hfill \sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{\mathbb{G}}_{2l+k}^{\left(s\right)}\left(p,q\right)& =& {\displaystyle \frac{p^{s\left(2m+k-1\right)}}{1-q^s}}\left(\mathrm{\Theta }{\left(1-q^s\right)}^m+q^{s\left(k+m\right)}\mathrm{{\rm Y}}{\left(1-q^s\right)}^m\right)\hfill \\ & =& p^{s\left(2m+k-1\right)}{\left(1-q^s\right)}^{m-1}\left(\mathrm{\Theta }+q^{s\left(k+m\right)}\mathrm{{\rm Y}}\right)\hfill \\ & =& p^{s\left(m+k\right)}{\chi }^{m-1}\left(\mathrm{\Theta }+q^{s\left(k+m\right)}\mathrm{{\rm Y}}\right).\hfill \end{array}$$

The proof is thus completed. □

Theorem 10.

Let $k,l,m\in \mathbb{N}$, $p\in \mathbb{R}$ and $q\in \mathbb{R}\setminus \left\{1\right\}$. Then, we have

$$\begin{array}{c}\hfill \sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(-1\right)}^l{\left(-p^{2s}{q}^s\right)}^{m-l}{\mathbb{G}}_{2l+k}^{\left(s\right)}\left(p,q\right)={\mathbb{G}}_{k+m}^{\left(s\right)}\left(p,q\right){\left(-p^s\left(1+q^s\right)\right)}^m\end{array}$$

(22)

and

$$\begin{array}{c}\hfill \sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(p^s\left(1+q^s\right)\right)}^l{\left(-p^{2s}{q}^s\right)}^{m-l}{\mathbb{G}}_l^{\left(s\right)}\left(p,q\right)={\mathbb{G}}_{2m}^{\left(s\right)}\left(p,q\right).\end{array}$$

(23)

Proof. 

Thanks to Theorem 1, we deduce

$$\begin{array}{ccc}\hfill \sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(-1\right)}^l{\left(-p^{2s}{q}^s\right)}^{m-l}{\mathbb{G}}_{2l+k}^{\left(s\right)}\left(p,q\right)& =& \sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(-1\right)}^l{\left(-p^{2s}{q}^s\right)}^{m-l}{\displaystyle \frac{p^{s\left(2l+k-1\right)}\left(\mathrm{\Theta }-q^{s\left(2l+k\right)}\mathrm{{\rm Y}}\right)}{1-q^s}}\hfill \\ & =& {\displaystyle \frac{p^{s\left(k-1\right)}}{1-q^s}}\left(\begin{array}{c}\mathrm{\Theta }\sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{\left(-p^{2s}\right)}^l\\ -q^{sk}\mathrm{{\rm Y}}\sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{\left(-p^{2s}{q}^{2s}\right)}^l\end{array}\right)\hfill \\ & =& {\displaystyle \frac{p^{s\left(k-1\right)}}{1-q^s}}\left(\mathrm{\Theta }{\left(-p^{2s}-p^{2s}{q}^s\right)}^m-q^{sk}\mathrm{{\rm Y}}{\left(-p^{2s}{q}^{2s}-p^{2s}{q}^s\right)}^m\right)\hfill \\ & =& {\displaystyle \frac{p^{s\left(k-1\right)}}{1-q^s}}\left(\mathrm{\Theta }{\left(-p^{2s}\left(1+q^s\right)\right)}^m-q^{sk}\mathrm{{\rm Y}}{\left(-p^{2s}{q}^s\left(q^s+1\right)\right)}^m\right)\hfill \\ & =& {\displaystyle \frac{p^{s\left(k-1\right)}}{1-q^s}}\left(\mathrm{\Theta }-q^{s\left(k+m\right)}\mathrm{{\rm Y}}\right)p^{sm}{\left(-p^s\left(1+q^s\right)\right)}^m\hfill \\ & =& {\displaystyle \frac{p^{s\left(k+m-1\right)}}{1-q^s}}\left(\mathrm{\Theta }-q^{s\left(k+m\right)}\mathrm{{\rm Y}}\right){\left(-p^s\left(1+q^s\right)\right)}^m\hfill \\ & =& {\mathbb{G}}_{k+m}^{\left(s\right)}\left(p,q\right){\left(-p^s\left(1+q^s\right)\right)}^m.\hfill \end{array}$$

This shows Equation (22). Similarly, by Theorem 1 again, we obtain

$$\begin{array}{ccc}\hfill \sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(p^s\left(1+q^s\right)\right)}^l{\left(-p^{2s}{q}^s\right)}^{m-l}{\mathbb{G}}_l^{\left(s\right)}\left(p,q\right)& =& \sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(p^s\left(1+q^s\right)\right)}^l{\left(-p^{2s}{q}^s\right)}^{m-l}{\displaystyle \frac{p^{s\left(l-1\right)}\left(\mathrm{\Theta }-q^{sl}\mathrm{{\rm Y}}\right)}{1-q^s}}\hfill \\ & =& {\displaystyle \frac{1}{1-q^s}}\left(\begin{array}{c}\mathrm{\Theta }\sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(p^s\left(1+q^s\right)\right)}^l{\left(-p^{2s}{q}^s\right)}^{m-l}{p}^{s\left(l-1\right)}\\ -\mathrm{{\rm Y}}\sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(p^s\left(1+q^s\right)\right)}^l{\left(-p^{2s}{q}^s\right)}^{m-l}{p}^{s\left(l-1\right)}{q}^{sl}\end{array}\right)\hfill \\ & =& {\displaystyle \frac{p^{-s}}{1-q^s}}\left(\begin{array}{c}\mathrm{\Theta }\sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(p^{2s}\left(1+q^s\right)\right)}^l{\left(-p^{2s}{q}^s\right)}^{m-l}\\ -\mathrm{{\rm Y}}\sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(p^{2s}{q}^s\left(1+q^s\right)\right)}^l{\left(-p^{2s}{q}^s\right)}^{m-l}\end{array}\right)\hfill \\ & =& {\displaystyle \frac{p^{-s}}{1-q^s}}\left(\mathrm{\Theta }{\left(p^{2s}\left(1+q^s\right)-p^{2s}{q}^s\right)}^m-\mathrm{{\rm Y}}{\left(p^{2s}{q}^s\left(1+q^s\right)-p^{2s}{q}^s\right)}^m\right)\hfill \\ & =& {\displaystyle \frac{p^{s\left(2m-1\right)}}{1-q^s}}\left(\mathrm{\Theta }-\mathrm{{\rm Y}}{q}^{2sm}\right)\hfill \\ & =& {\mathbb{G}}_{2m}^{\left(s\right)}\left(p,q\right).\hfill \end{array}$$

Hence, Equation (23) holds. The proof is completed. □

Theorem 11.

Let $k,l,m\in \mathbb{N}$, $p\in \mathbb{R}$, $q\in \mathbb{R}\setminus \left\{1\right\}$ and $\chi =p^s\left(1-q^s\right)$.

(1) 

If m is even, then

$$\begin{array}{c}\hfill \sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{\mathbb{G}}_{4l}^{\left(s\right)}\left(p,q\right)={\displaystyle \frac{{\mathbb{G}}_m^{\left(s\right)}\left(p,q\right)}{2}}\left({\chi }^m+{\left(p^s\left(1+q^s\right)\right)}^m\right).\end{array}$$

(24)

(2) 

If m is odd, then

$$\begin{array}{c}\hfill \sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{\mathbb{G}}_{4l}^{\left(s\right)}\left(p,q\right)={\displaystyle \frac{1}{2}}\left(p^{sm}{\chi }^{m-1}\left(\mathrm{\Theta }+q^{sm}\mathrm{{\rm Y}}\right)-{\mathbb{G}}_m^{\left(s\right)}\left(p,q\right){\left(p^s\left(1+q^s\right)\right)}^m\right).\end{array}$$

(25)

Proof. 

By using Theorem 1, we obtain

$$\begin{array}{ccc}\hfill \sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{\mathbb{G}}_{4l}^{\left(s\right)}\left(p,q\right)& =& {\displaystyle \frac{1}{2}}\sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right)\left(1+{\left(-1\right)}^l\right){\left(-p^{2s}{q}^s\right)}^{m-l}{\mathbb{G}}_{2l}^{\left(s\right)}\left(p,q\right)\hfill \\ & =& {\displaystyle \frac{1}{2}}\left(\begin{array}{c}\sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{\mathbb{G}}_{2l}^{\left(s\right)}\left(p,q\right)\\ +\sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{l}}\right){\left(-1\right)}^l{\left(-p^{2s}{q}^s\right)}^{m-l}{\mathbb{G}}_{2l}^{\left(s\right)}\left(p,q\right)\end{array}\right).\hfill \end{array}$$

(1)

If m is even, then, from (20) and (22), we obtain

$$\begin{array}{ccc}\hfill \sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{\mathbb{G}}_{4l}^{\left(s\right)}\left(p,q\right)& =& {\displaystyle \frac{1}{2}}\left({\mathbb{G}}_m^{\left(s\right)}\left(p,q\right){\chi }^m+{\mathbb{G}}_m^{\left(s\right)}\left(p,q\right){\left(p^s\left(1+q^s\right)\right)}^m\right)\hfill \end{array}$$

(26)

$$\begin{array}{ccc}& =& {\displaystyle \frac{{\mathbb{G}}_m^{\left(s\right)}\left(p,q\right)}{2}}\left({\chi }^m+{\left(p^s\left(1+q^s\right)\right)}^m\right).\hfill \end{array}$$

(27)

(2)

If m is odd, then a direct calculation provides

$$\sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{\mathbb{G}}_{4l}^{\left(s\right)}\left(p,q\right)={\displaystyle \frac{1}{2}}\left(p^{sm}{\chi }^{m-1}\left(\mathrm{\Theta }+q^{sm}\mathrm{{\rm Y}}\right)-{\mathbb{G}}_m^{\left(s\right)}\left(p,q\right){\left(p^s\left(1+q^s\right)\right)}^m\right).$$

Now, we arrive at the result. □

Following a similar argument to that in the proof of Theorem 11, we can obtain the following result.

Theorem 12.

Let $k,l,m\in \mathbb{N}$, $p\in \mathbb{R}$, $q\in \mathbb{R}\setminus \left\{1\right\}$, and $\chi =p^s\left(1-q^s\right)$.

(1) 

If m is even, then

$$\begin{array}{c}\hfill \sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{\mathbb{G}}_{4l+1}^{\left(s\right)}\left(p,q\right)={\displaystyle \frac{{\mathbb{G}}_{m+1}^{\left(s\right)}\left(p,q\right)}{2}}\left({\chi }^m+{\left(p^s\left(1+q^s\right)\right)}^m\right).\end{array}$$

(2) 

If m is odd, then

$$\begin{array}{c}\hfill \sum _{l=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2l}}\right){\left(-p^{2s}{q}^s\right)}^{m-l}{\mathbb{G}}_{4l+1}^{\left(s\right)}\left(p,q\right)={\displaystyle \frac{1}{2}}\left(p^{s\left(m+1\right)}{\chi }^{m-1}\left(\mathrm{\Theta }+q^{s\left(1+m\right)}\mathrm{{\rm Y}}\right)-{\mathbb{G}}_{m+1}^{\left(s\right)}\left(p,q\right){\left(p^s\left(1+q^s\right)\right)}^m\right).\end{array}$$

3. New Matrix Representations and Formulas for ${\mathbb{G}}_{\mathit{n}}^{\left(\mathit{s}\right)}\left(\mathit{p},\mathit{q}\right)$

In this section, we will examine various properties of the higher-order generalized q-Fibonacci quaternions with quantum integer components by using the matrices defined below:

$$\begin{array}{ccc}\hfill {\mathbf{A}}_{n,p,q}^{\left(s\right)}& =& \left(\begin{array}{cc}{\mathbb{G}}_{n+1}^{\left(s\right)}\left(p,q\right)& {\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)\\ {\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)& {\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right)\end{array}\right),\hfill \end{array}$$

(28)

$$\begin{array}{ccc}\hfill {\mathbf{B}}_{p,q}^{\left(s\right)}& =& \left(\begin{array}{cc}{p}^s\left(1+q^s\right)& -{\left(p^{2}q\right)}^s\\ 1& 0\end{array}\right),\hfill \end{array}$$

(29)

$$\begin{array}{ccc}\hfill {\mathbf{C}}_{n,p,q}^{\left(s\right)}& =& \left(\begin{array}{cc}{p}^{sn}{\left[n+1\right]}_{q^s}& p^{s\left(n-1\right)}{\left[n\right]}_{q^s}\\ p^{s\left(n-1\right)}{\left[n\right]}_{q^s}& p^{s\left(n-2\right)}{\left[n-1\right]}_{q^s}\end{array}\right).\hfill \end{array}$$

(30)

More precisely, we will study on ${\mathbf{A}}_{n,p,q}^{\left(s\right)}$, ${\mathbf{B}}_{p,q}^{\left(s\right)}$, ${\mathbf{C}}_{n,p,q}^{\left(s\right)}$ matrices whose entries are higher-order generalized Fibonacci quaternions with quantum integer components ${\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)$, coefficients of recurrence relation of ${\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)$ and coefficients of the ${\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)$, respectively.

The following matrix formulas are crucial in this section.

Theorem 13.

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

$${\left({\mathbf{B}}_{p,q}^{\left(s\right)}\right)}^n=\left(\begin{array}{cc}{p}^{sn}{\left[n+1\right]}_{q^s}& -p^{sn}{q}^s{\left[n\right]}_{q^s}\\ p^{s(n-1)}{\left[n\right]}_{q^s}& -p^{s\left(n-1\right)}{q}^s{\left[n-1\right]}_{q^s}\end{array}\right).$$

(31)

Proof. 

The proof will be by mathematical induction on n. Clearly, Equation (31) is true for $n=1$. Suppose that Equation (31) is true for $n=k$. By using Equation (8) and applying the induction hypothesis, we have

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

Therefore, (31) is also true for $n=k+1$. This completes the induction, and hence our conclusion is proved. □

Theorem 14.

Let $n\in \mathbb{N}$. Then, the following formulas hold:

$$\begin{array}{cc}\hfill {\mathbf{A}}_{n,p,q}^{\left(s\right)}& ={\mathbf{B}}_{p,q}^{\left(s\right)}{\mathbf{A}}_{n-1,p,q}^{\left(s\right)}\hfill \end{array}$$

(32)

and

$$\begin{array}{cc}\hfill {\mathbf{C}}_{n,p,q}^{\left(s\right)}& ={\mathbf{B}}_{p,q}^{\left(s\right)}{\mathbf{C}}_{n-1,p,q}^{\left(s\right)}.\hfill \end{array}$$

(33)

Moreover, we have

$$\begin{array}{cc}\hfill {\mathbf{A}}_{n,p,q}^{\left(s\right)}& ={\left({\mathbf{B}}_{p,q}^{\left(s\right)}\right)}^n{\mathbf{A}}_{0,p,q}^{\left(s\right)}\hfill \end{array}$$

(34)

and

$$\begin{array}{cc}\hfill {\mathbf{C}}_{n,p,q}^{\left(s\right)}& ={\left({\mathbf{B}}_{p,q}^{\left(s\right)}\right)}^n{\mathbf{C}}_{0,p,q}^{\left(s\right)}.\hfill \end{array}$$

(35)

Proof. 

From Theorem 4, by using matrix multiplication, we obtain

$$\begin{array}{ccc}\hfill {\mathbf{B}}_{p,q}^{\left(s\right)}{\mathbf{A}}_{n-1,p,q}^{\left(s\right)}& =& \left(\begin{array}{cc}{p}^s\left(1+q^s\right)& -{\left(p^{2}q\right)}^s\\ 1& 0\end{array}\right)\left(\begin{array}{cc}{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)& {\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right)\\ {\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right)& {\mathbb{G}}_{n-2}^{\left(s\right)}\left(p,q\right)\end{array}\right)\hfill \\ & =& \left(\begin{array}{cc}\begin{array}{c}{p}^s\left(1+q^s\right){\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)-{\left(p^{2}q\right)}^s{\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right)\end{array}& \begin{array}{c}{p}^s\left(1+q^s\right){\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right)\\ -{\left(p^{2}q\right)}^s{\mathbb{G}}_{n-2}^{\left(s\right)}\left(p,q\right)\end{array}\\ {\mathbb{G}}_n^{\left(s\right)}& {\mathbb{G}}_{n-1}^{\left(s\right)}\end{array}\right)\hfill \\ & =& \left(\begin{array}{cc}{\mathbb{G}}_{n+1}^{\left(s\right)}\left(p,q\right)& {\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)\\ {\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)& {\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right)\end{array}\right)={\mathbf{A}}_{n,p,q}^{\left(s\right)}.\hfill \end{array}$$

This shows Equation (32). Following a similar argument to that above, we can prove (33). Moreover, Formulas (34) and (35) can be obtained by repeatedly substituting Formulas (32) and (33) for a finite number of times, respectively. □

As applications of Theorems 13 and 14, we establish the following new formulas for ${\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)$.

Theorem 15.

Let $n\in \mathbb{N}$. Then, the following formulas hold:

$${\mathbb{G}}_{n+1}^{\left(s\right)}\left(p,q\right){\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right)-{\left({\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)\right)}^2={\displaystyle \frac{{\left(p^{2}q\right)}^{sn}}{p^{2s}{\left(1-q^s\right)}^2}}\left(\left(1-p^{4s}\right)\left({\mathrm{\Theta }}^2+{\mathrm{{\rm Y}}}^2\right)-\left(q^{-s}-p^{4s}\right)\mathrm{\Theta }\mathrm{{\rm Y}}-\left(q^s-p^{4s}\right)\mathrm{{\rm Y}}\mathrm{\Theta }\right),$$

(36)

$${\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right){\mathbb{G}}_{n+1}^{\left(s\right)}\left(p,q\right)-{\left({\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)\right)}^2={\displaystyle \frac{{\left(p^{2}q\right)}^{sn}}{p^{2s}{\left(1-q^s\right)}^2}}\left(\left(1-p^{4s}\right)\left({\mathrm{\Theta }}^2+{\mathrm{{\rm Y}}}^2\right)-\left(q^s-p^{4s}\right)\mathrm{\Theta }\mathrm{{\rm Y}}-\left(q^{-s}-p^{4s}\right)\mathrm{{\rm Y}}\mathrm{\Theta }\right).$$

(37)

Proof. 

In our proofs, we use “multiplication from above to down below” and “multiplication from down below to above” rules, respectively, for determinant of Equation (34). Note that

$$\begin{array}{ccc}\hfill \left|\begin{array}{cc}{\mathbb{G}}_{n+1}^{\left(s\right)}\left(p,q\right)& {\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)\\ {\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)& {\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right)\end{array}\right|& =& {\left|\begin{array}{cc}{p}^s\left(1+q^s\right)& -{\left(p^{2}q\right)}^s\\ 1& 0\end{array}\right|}^n\left|\begin{array}{cc}{\mathbb{G}}_1^{\left(s\right)}\left(p,q\right)& {\mathbb{G}}_0^{\left(s\right)}\left(p,q\right)\\ {\mathbb{G}}_0^{\left(s\right)}\left(p,q\right)& {\mathbb{G}}_{-1}^{\left(s\right)}\left(p,q\right)\end{array}\right|\hfill \end{array}$$

(38)

We now show (36) holds. Since

$$\begin{array}{ccc}& & {\left(p^{2}q\right)}^{sn}\left({\mathbb{G}}_1^{\left(s\right)}\left(p,q\right){\mathbb{G}}_{-1}^{\left(s\right)}\left(p,q\right)-{\left({\mathbb{G}}_0^{\left(s\right)}\left(p,q\right)\right)}^2\right)\hfill \\ & =& {\displaystyle \frac{{\left(p^{2}q\right)}^{sn}}{{\left(1-q^s\right)}^2}}\left(p^0\left(\mathrm{\Theta }-q^s\mathrm{{\rm Y}}\right)p^{-2s}\left(\mathrm{\Theta }-q^{-s}\mathrm{{\rm Y}}\right)-{\left(p^{-s}\left(\mathrm{\Theta }-q^0\mathrm{{\rm Y}}\right)\right)}^2\right)\hfill \\ & =& {\displaystyle \frac{{\left(p^{2}q\right)}^{sn}}{{\left(1-q^s\right)}^2}}\left(p^{-2s}\left(\mathrm{\Theta }-q^s\mathrm{{\rm Y}}\right)\left(\mathrm{\Theta }-q^{-s}\mathrm{{\rm Y}}\right)-p^{2s}{\left(\mathrm{\Theta }-\mathrm{{\rm Y}}\right)}^2\right)\hfill \\ & =& {\displaystyle \frac{{\left(p^{2}q\right)}^{sn}}{p^{2s}{\left(1-q^s\right)}^2}}\left({\mathrm{\Theta }}^2-q^{-s}\mathrm{\Theta }\mathrm{{\rm Y}}-q^s\mathrm{{\rm Y}}\mathrm{\Theta }+{\mathrm{{\rm Y}}}^2-p^{4s}{\mathrm{\Theta }}^2+p^{4s}\mathrm{\Theta }\mathrm{{\rm Y}}+p^{4s}\mathrm{{\rm Y}}\mathrm{\Theta }-p^{4s}{\mathrm{{\rm Y}}}^2\right)\hfill \\ & =& {\displaystyle \frac{{\left(p^{2}q\right)}^{sn}}{p^{2s}{\left(1-q^s\right)}^2}}\left({\mathrm{\Theta }}^2-p^{4s}{\mathrm{\Theta }}^2+{\mathrm{{\rm Y}}}^2-p^{4s}{\mathrm{{\rm Y}}}^2-q^{-s}\mathrm{\Theta }\mathrm{{\rm Y}}+p^{4s}\mathrm{\Theta }\mathrm{{\rm Y}}-q^s\mathrm{{\rm Y}}\mathrm{\Theta }+p^{4s}\mathrm{{\rm Y}}\mathrm{\Theta }\right)\hfill \\ & =& {\displaystyle \frac{{\left(p^{2}q\right)}^{sn}}{p^{2s}{\left(1-q^s\right)}^2}}\left(\left(1-p^{4s}\right){\mathrm{\Theta }}^2+\left(1-p^{4s}\right){\mathrm{{\rm Y}}}^2-\left(q^{-s}-p^{4s}\right)\mathrm{\Theta }\mathrm{{\rm Y}}-\left(q^s-p^{4s}\right)\mathrm{{\rm Y}}\mathrm{\Theta }\right)\hfill \\ & =& {\displaystyle \frac{{\left(p^{2}q\right)}^{sn}}{p^{2s}{\left(1-q^s\right)}^2}}\left(\left(1-p^{4s}\right)\left({\mathrm{\Theta }}^2+{\mathrm{{\rm Y}}}^2\right)-\left(q^{-s}-p^{4s}\right)\mathrm{\Theta }\mathrm{{\rm Y}}-\left(q^s-p^{4s}\right)\mathrm{{\rm Y}}\mathrm{\Theta }\right),\hfill \end{array}$$

combining this with Equation (38), we can prove Equation (36). Similarly, by using the rule “multiplication from down below to above”, Equation (37) can be verified. The proof is completed. □

Theorem 16.

Let $m,n,r,t\in \mathbb{N}$. If $m+n=r+t$, then

$$p^{s\left(m-1\right)}{\left[m\right]}_{q^s}{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)-p^{s\left(r-1\right)}{\left[r\right]}_{q^s}{\mathbb{G}}_t^{\left(s\right)}\left(p,q\right)=p^{s\left(m-1\right)}{q}^s{\left[m-1\right]}_{q^s}{\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right)-p^{s\left(r-1\right)}{q}^s{\left[r-1\right]}_{q^s}{\mathbb{G}}_{t-1}^{\left(s\right)}\left(p,q\right).$$

Proof. 

Based on Equations (32) and (34), we obtain

$$\begin{array}{cc}\hfill {\left({\mathbf{B}}_{p,q}^{\left(s\right)}\right)}^m{\mathbf{A}}_{n,p,q}^{\left(s\right)}& ={\left({\mathbf{B}}_{p,q}^{\left(s\right)}\right)}^m{\left({\mathbf{B}}_{p,q}^{\left(s\right)}\right)}^n{\mathbf{A}}_{0,p,q}^{\left(s\right)}\hfill \\ \hfill & ={\left({\mathbf{B}}_{p,q}^{\left(s\right)}\right)}^{m+n}{\mathbf{A}}_{0,p,q}^{\left(s\right)}\hfill \\ \hfill & ={\left({\mathbf{B}}_{p,q}^{\left(s\right)}\right)}^{r+t}{\mathbf{A}}_{0,p,q}^{\left(s\right)}\hfill \\ \hfill & ={\left({\mathbf{B}}_{p,q}^{\left(s\right)}\right)}^r{\mathbf{A}}_{t,p,q}^{\left(s\right)}.\hfill \end{array}$$

By using the matrix multiplication, we conclude that ${\left({\mathbf{B}}_{p,q}^{\left(s\right)}\right)}^m{\mathbf{A}}_{n,p,q}^{\left(s\right)}$ equals

$$\left(\begin{array}{cc}{p}^{sm}\left({\left[m+1\right]}_{q^s}{\mathbb{G}}_{n+1}^{\left(s\right)}\left(p,q\right)-q^s{\left[m\right]}_{q^s}{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)\right)& p^{sm}\left({\left[m+1\right]}_{q^s}{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)-q^s{\left[m\right]}_{q^s}{\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right)\right)\\ p^{s\left(m-1\right)}\left({\left[m\right]}_{q^s}{\mathbb{G}}_{n+1}^{\left(s\right)}\left(p,q\right)-q^s{\left[m-1\right]}_{q^s}{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)\right)& p^{s\left(m-1\right)}\left({\left[m\right]}_{q^s}{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)-q^s{\left[m-1\right]}_{q^s}{\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right)\right)\end{array}\right).$$

Similarly, ${\left({\mathbf{B}}_{p,q}^{\left(s\right)}\right)}^r{\mathbf{A}}_{t,p,q}^{\left(s\right)}$ equals

$$\left(\begin{array}{cc}{p}^{sr}\left({\left[r+1\right]}_{q^s}{\mathbb{G}}_{t+1}^{\left(s\right)}\left(p,q\right)-q^s{\left[r\right]}_{q^s}{\mathbb{G}}_t^{\left(s\right)}\left(p,q\right)\right)& p^{sr}\left({\left[r+1\right]}_{q^s}{\mathbb{G}}_t^{\left(s\right)}\left(p,q\right)-q^s{\left[r\right]}_{q^s}{\mathbb{G}}_{t-1}^{\left(s\right)}\left(p,q\right)\right)\\ p^{s\left(r-1\right)}\left({\left[r\right]}_{q^s}{\mathbb{G}}_{t+1}^{\left(s\right)}\left(p,q\right)-q^s{\left[r-1\right]}_{q^s}{\mathbb{G}}_t^{\left(s\right)}\left(p,q\right)\right)& p^{s\left(r-1\right)}\left({\left[r\right]}_{q^s}{\mathbb{G}}_t^{\left(s\right)}\left(p,q\right)-q^s{\left[r-1\right]}_{q^s}{\mathbb{G}}_{t-1}^{\left(s\right)}\left(p,q\right)\right)\end{array}\right).$$

Since ${\left({\mathbf{B}}_{p,q}^{\left(s\right)}\right)}^m{\mathbf{A}}_{n,p,q}^{\left(s\right)}={\left({\mathbf{B}}_{p,q}^{\left(s\right)}\right)}^r{\mathbf{A}}_{t,p,q}^{\left(s\right)}$, we acquire

$$p^{s\left(m-1\right)}\left({\left[m\right]}_{q^s}{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)-q^s{\left[m-1\right]}_{q^s}{\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right)\right)=p^{s\left(r-1\right)}\left({\left[r\right]}_{q^s}{\mathbb{G}}_t^{\left(s\right)}\left(p,q\right)-q^s{\left[r-1\right]}_{q^s}{\mathbb{G}}_{t-1}^{\left(s\right)}\left(p,q\right)\right),$$

which implies

$$p^{s\left(m-1\right)}{\left[m\right]}_{q^s}{\mathbb{G}}_n^{\left(s\right)}\left(p,q\right)-p^{s\left(r-1\right)}{\left[r\right]}_{q^s}{\mathbb{G}}_t^{\left(s\right)}\left(p,q\right)=p^{s\left(m-1\right)}{q}^s{\left[m-1\right]}_{q^s}{\mathbb{G}}_{n-1}^{\left(s\right)}\left(p,q\right)-p^{s\left(r-1\right)}{q}^s{\left[r-1\right]}_{q^s}{\mathbb{G}}_{t-1}^{\left(s\right)}\left(p,q\right).$$

The proof is completed. □

Theorem 17.

For $n,m\in \mathbb{N}$, we have

$${\mathbb{G}}_{n+m}^{\left(s\right)}\left(p,q\right)=p^{sn}{\left[n+1\right]}_{q^s}{\mathbb{G}}_m^{\left(s\right)}\left(p,q\right)-p^{s\left(n+1\right)}{q}^s{\left[n\right]}_{q^s}{\mathbb{G}}_{m-1}^{\left(s\right)}\left(p,q\right).$$

Proof. 

Thanks to Equations (34) and (35), we obtain

$${\mathbf{A}}_{n+m,p,q}^{\left(s\right)}={\left({\mathbf{B}}_{p,q}^{\left(s\right)}\right)}^{n+m}{\mathbf{A}}_{0,p,q}^{\left(s\right)}={\left({\mathbf{B}}_{p,q}^{\left(s\right)}\right)}^n{\left({\mathbf{B}}_{p,q}^{\left(s\right)}\right)}^m{\mathbf{A}}_{0,p,q}^{\left(s\right)}={\left({\mathbf{B}}_{p,q}^{\left(s\right)}\right)}^n{\mathbf{A}}_{m,p,q}^{\left(s\right)}$$

and

$${\left({\mathbf{B}}_{p,q}^{\left(s\right)}\right)}^n={\mathbf{C}}_{n,p,q}^{\left(s\right)}{\left({\mathbf{C}}_{0,p,q}^{\left(s\right)}\right)}^{-1}.$$

By (28), we have

$${\mathbf{A}}_{n+m,p,q}^{\left(s\right)}=\left(\begin{array}{cc}{\mathbb{G}}_{n+m+1}^{\left(s\right)}\left(p,q\right)& {\mathbb{G}}_{n+m}^{\left(s\right)}\left(p,q\right)\\ {\mathbb{G}}_{n+m}^{\left(s\right)}\left(p,q\right)& {\mathbb{G}}_{n+m-1}^{\left(s\right)}\left(p,q\right)\end{array}\right).$$

(39)

On the other hand, we acquire

$$\begin{array}{cc}\hfill {\mathbf{A}}_{n+m,p,q}^{\left(s\right)}& ={\mathbf{C}}_{n,p,q}^{\left(s\right)}{\left({\mathbf{C}}_{0,p,q}^{\left(s\right)}\right)}^{-1}{\mathbf{A}}_{m,p,q}^{\left(s\right)}\hfill \\ \hfill & ={\mathbf{C}}_{n,p,q}^{\left(s\right)}{\left(\begin{array}{cc}{\left[1\right]}_{q^s}& p^{-s}{\left[0\right]}_{q^s}\\ p^{-s}{\left[0\right]}_{q^s}& p^{-2s}{\left[-1\right]}_{q^s}\end{array}\right)}^{-1}\left(\begin{array}{cc}{\mathbb{G}}_{m+1}^{\left(s\right)}\left(p,q\right)& {\mathbb{G}}_m^{\left(s\right)}\left(p,q\right)\\ {\mathbb{G}}_m^{\left(s\right)}\left(p,q\right)& {\mathbb{G}}_{m-1}^{\left(s\right)}\left(p,q\right)\end{array}\right)\hfill \\ \hfill & ={\mathbf{C}}_{n,p,q}^{\left(s\right)}\left(\begin{array}{cc}1& 0\\ 0& -p^{2s}{q}^s\end{array}\right)\left(\begin{array}{cc}{\mathbb{G}}_{m+1}^{\left(s\right)}\left(p,q\right)& {\mathbb{G}}_m^{\left(s\right)}\left(p,q\right)\\ {\mathbb{G}}_m^{\left(s\right)}\left(p,q\right)& {\mathbb{G}}_{m-1}^{\left(s\right)}\left(p,q\right)\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}{p}^{sn}{\left[n+1\right]}_{q^s}& p^{s\left(n-1\right)}{\left[n\right]}_{q^s}\\ p^{s\left(n-1\right)}{\left[n\right]}_{q^s}& p^{s\left(n-2\right)}{\left[n-1\right]}_{q^s}\end{array}\right)\left(\begin{array}{cc}{\mathbb{G}}_{m+1}^{\left(s\right)}\left(p,q\right)& {\mathbb{G}}_m^{\left(s\right)}\left(p,q\right)\\ -p^{2s}{q}^s{\mathbb{G}}_m^{\left(s\right)}\left(p,q\right)& -p^{2s}{q}^s{\mathbb{G}}_{m-1}^{\left(s\right)}\left(p,q\right)\end{array}\right).\hfill \end{array}$$

Comparing the above two formulas, we can conclude

$${\mathbb{G}}_{n+m}^{\left(s\right)}\left(p,q\right)=p^{sn}{\left[n+1\right]}_{q^s}{\mathbb{G}}_m^{\left(s\right)}\left(p,q\right)-p^{s\left(n+1\right)}{q}^s{\left[n\right]}_{q^s}{\mathbb{G}}_{m-1}^{\left(s\right)}\left(p,q\right).$$

The proof is completed. □

4. Conclusions

In this paper, we have examined a different generalization of Fibonacci quaternions, which not only provides some already existing families of Fibonacci-type quaternions in the literature but also some new families of Fibonacci-type quaternions according to the values of parameters s, p, and q. In the last section, we have achieved some identities for higher-order generalized q-Fibonacci quaternions with integer components by using some matrices of special type. Therefore, our new results generalize and improve several results available in the corresponding literature and will assist us in obtaining novel new families of Fibonacci-type quaternions as well as their proof techniques in future research.