On Bicomplex (p,q)-Fibonacci Quaternions

Abstract

Here, we describe the bicomplex $\left(p,q\right)$-Fibonacci numbers and the bicomplex $\left(p,q\right)$-Fibonacci quaternions based on these numbers to show that bicomplex numbers are not defined the same as bicomplex quaternions. Then, we give some of their equations, including the Binet formula, generating function, Catalan, Cassini, and d’Ocagne’s identities, and summation formulas for both. We also create a matrix for bicomplex $\left(p,q\right)$-Fibonacci quaternions, and we obtain the determinant of a special matrix that gives the terms of that quaternion. With this study, we get a general form of the second-order bicomplex number sequences and the second-order bicomplex quaternions. In addition, we show that these two concepts, defined as the same in many studies, are different.

1. Introduction

The number sequences are a subject of study that has been popular for centuries. The sequence that has the most significant role in the emergence of each of these sequences is the Fibonacci sequence. This sequence is shown with $F_n$ and described as, for $n\ge 2$, $F_n=F_{n-1}+F_{n-2}$ with $F_0=0$, $F_1=1$ where $n$ is an integer [1]. A generalization of the Fibonacci sequences is the $\left(p,q\right)$-Fibonacci sequence. Suvarnamani and Tatong [2] defined the $\left(p,q\right)$-Fibonacci sequence, ${\left\{F_n\left(p,q\right)\right\}}_{n=0}^{\infty }$, that has initial terms $0$ and $1$, and for $n\ge 0$, holds the following recurrence relation:

$$F_{n+2}\left(p,q\right)={pF}_{n+1}\left(p,q\right)+q{F}_n\left(p,q\right),$$

(1)

where $p$ and $q$ are nonzero real numbers such that $p^2+4q>0$.

The first few $\left(p,q\right)$-Fibonacci numbers are:

$$\begin{gathered} F_0\left(p,q\right)=0, F_1\left(p,q\right)=1,F_2\left(p,q\right)=p, F_3\left(p,q\right)=\left(p^2+q\right),F_4\left(p,q\right)=\left(p^3+2pq\right),F_5\left(p,q\right)= \\ \left(p^4+3p^{2}q+q^2\right). \end{gathered}$$

The characteristic equation of (1) is:

$${\sigma }^2-p\sigma -q=0.$$

(2)

Binet’s formula for the $F_n\left(p,q\right)$ is as follows:

$$F_n\left(p,q\right)=\frac{{\beta }^n-{\theta }^n}{\beta -\theta },$$

(3)

where $\beta$ and $\theta$ are roots of (2) [2].

A popular subject of study by researchers, especially in recent years, is quaternions. In 1843, Hamilton [3] introduced quaternions that extended complex numbers. In addition, a set of quaternions is defined by Hamilton as follows:

$$H=\{{h=1h}_0+e_1{h}_1+e_2{h}_2+e_3{h}_3: h_0, h_1,h_2, h_3 \in R\}$$

where $R$ is the set of real numbers,

$$\begin{gathered} {e_1}^2={e_2}^2={e_3}^2=-1, e_1{e}_2=-e_2{e}_1=e_3, e_2{e}_3=-e_3{e}_2=e_1, e_3{e}_1= \\ -e_1{e}_3=e_2. \end{gathered}$$

(4)

The quaternions can be thought of as four-dimensional vectors, just as complex numbers can be considered two-dimensional vectors [3] because the quaternions are extensions of complex numbers into a four-dimensional space.

In addition, new quaternions can be defined by combining quaternions and different number sequences. For example, nth $\left(p,q\right)$-Fibonacci quaternions [4] are defined as follows:

$$Q{F}_n\left(p,q\right)=F_n\left(p,q\right)+e_1{F}_{n+1}\left(p,q\right)+e_2{F}_{n+2}\left(p,q\right)+{e_{3}F}_{n+3}\left(p,q\right),$$

(5)

where the $F_n\left(p,q\right)$ is the $nth$ $\left(p,q\right)$-Fibonacci number. Also, the imaginary quaternion units $e_1,e_2,$ and $e_3$ have the rules in (4).

There are many more works on quaternions in the literature. Some of these are as follows:

In [5], Horadam defined Fibonacci quaternions. In [6], the authors described generalized Fibonacci quaternions and Narayana quaternions; in [7], split Fibonacci quaternions; and in [8], Halıcı defined complex Fibonacci quaternions. Moreover, she gave some matrix applications. In [9,10], the authors explained new generalizations of Fibonacci quaternions, and in [11], Cerda-Morales introduced a generalization for Tribonacci quaternions. In [12], the authors gave some properties of $\left(p,q\right)$-Fibonacci and $\left(p,q\right)$-Lucas quaternions, and in [13], Yağmur described hyperbolic $\left(p,q\right)$-Fibonacci quaternions and cited some properties.

Another famous number sequence is the bicomplex number sequence. In 1892, Segre [14] defined bicomplex numbers by four base elements ($1,i,j,ij$), where

$$i^2=j^2=-1 \mathrm{and} ij=ji.$$

(6)

In that case, any bicomplex number $b_c$ can be written as follows:

$${b_c=b_c}_0+i{b_c}_1+j{b_c}_2+ij{b_c}_3={b_c}_0+i{b_c}_1+j{(b_c}_2+i{b_c}_3)$$

where ${b_c}_0,{b_c}_1,{b_c}_2,{b_c}_3\in R$. Let ${b_c=b_c}_0+i{b_c}_1+j{b_c}_2+ij{b_c}_3$ and ${{b_c}^{\prime }={b_c}^{\prime }}_0+i{{b_c}^{\prime }}_1+j{{b_c}^{\prime }}_2+ij{{b_c}^{\prime }}_3$ be two bicomplex numbers. Then, the addition, subtraction, and multiplication of the bicomplex numbers are written in the following form:

$$\begin{gathered} b_c+{b_c}^{\prime }{=(b_c}_0+{{b_c}^{\prime }}_0)+{(b_c}_1+{{b_c}^{\prime }}_1)i+{(b_c}_2+{{b_c}^{\prime }}_2)j+{(b_c}_3+{{b_c}^{\prime }}_3)ij, \\ {b}_c-{b_c}^{\prime }{=(b_c}_0-{{b_c}^{\prime }}_0)+{(b_c}_1-{{b_c}^{\prime }}_1)i+{(b_c}_2-{{b_c}^{\prime }}_2)j+{(b_c}_3-{{b_c}^{\prime }}_3)ij, \\ {b}_c\times {b_c}^{\prime }{=(b_c}_0{{b_c}^{\prime }}_0-{b_c}_1{{b_c}^{\prime }}_1-{b_c}_2{{b_c}^{\prime }}_2+{b_c}_3{{b_c}^{\prime }}_3) \\ +{(b_c}_0{{b_c}^{\prime }}_1+{b_c}_1{{b_c}^{\prime }}_0-{b_c}_2{{b_c}^{\prime }}_3+{b_c}_3{{b_c}^{\prime }}_2)i \\ +({b_c}_0{{b_c}^{\prime }}_2+{b_c}_2{{b_c}^{\prime }}_0-{b_c}_1{{b_c}^{\prime }}_3+{b_c}_3{{b_c}^{\prime }}_1)j \\ +({b_c}_0{{b_c}^{\prime }}_3+{b_c}_3{{b_c}^{\prime }}_0-{b_c}_1{{b_c}^{\prime }}_2+{b_c}_2{{b_c}^{\prime }}_1)ij, \end{gathered}$$

respectively.

Moreover, there are three different conjugations of the bicomplex numbers, as follows:

$$\overline{{b_c}_i}={b_c}_0-i{b_c}_1+j{b_c}_2-ij{b_c}_3,$$

$$\overline{{b_c}_j}={b_c}_0+i{b_c}_1-j{b_c}_2-ij{b_c}_3,$$

$$\overline{{b_c}_{ij}}={b_c}_0-i{b_c}_1-j{b_c}_2+ij{b_c}_3.$$

Many studies have been done on bicomplex numbers. For example, in [15], the authors introduced the algebraic properties of bicomplex numbers. In [16], the authors obtained the elementary functions of bicomplex numbers. In [17], the authors explained the bicomplex Fibonacci and Lucas numbers and gave their properties. In [18], Halıcı described bicomplex Fibonacci numbers; in [19], Yağmur introduced bicomplex $k$-Fibonacci numbers and gave some generalizations of them.

There are also studies in which bicomplex numbers and quaternion sequences are used together. In [20], Torunbalcı defined bicomplex Fibonacci quaternions. In [21], the authors introduced bicomplex generalized $k$-Horadam quaternions, and in [22], Gül described bicomplex Horadam quaternions and obtained some equations regarding them. In [23], the authors introduced new bicomplex generalized Tribonacci quaternions, obtained the Binet formula, generating function, and summation formulas, and gave some matrix applications.

Inspired by the articles above, in this study, we first aim to obtain a new generalization of second-order bicomplex numbers and bicomplex quaternion sequences and show that these two concepts are different, despite being defined as the same in many studies. Also, we provide some of their equations, including the Binet formula, generating function, Catalan, Cassini, d’Ocagne’s identities, and summation formulas for bicomplex $\left(p,q\right)$-Fibonacci numbers and bicomplex $\left(p,q\right)$-Fibonacci quaternions. Then, we show that the bicomplex $\left(p,q\right)$-Fibonacci quaternion can also be expressed with a new matrix of type 4 × 4, whose elements consist of bicomplex $\left(p,q\right)$-Fibonacci numbers. Finally, we aim to create a special matrix for bicomplex $\left(p,q\right)$-Fibonacci quaternions to obtain some equations about the matrix and the determinant of a special matrix that gives the terms of that quaternion.

2. Bicomplex $\left(\mathit{p},\mathit{q}\right)$-Fibonacci Numbers

Here, we obtain a new generalization of the second-order bicomplex numbers. To do this, we describe the bicomplex $\left(p,q\right)$-Fibonacci numbers. We give some equations and summation formulas about the bicomplex $\left(p,q\right)$-Fibonacci number sequence. In addition, we obtain the generating function, Binet’s formula, and Catalan, Cassini, and d’Ocagne’s identities for these number sequences.

Definition 1.

The bicomplex $\left(p,q\right)$-Fibonacci numbers are introduced by

$${BF}_u\left(p,q\right)=F_u+i{F}_{u+1}+j{F}_{u+2}+ij F_{u+3},$$

(7)

where $F_u$ is the $uth \left(p,q\right)$-Fibonacci number and $i,j$ are bicomplex units that provide (6).

The first few terms of the bicomplex $\left(p,q\right)$-Fibonacci sequence are the following:

$$\begin{array}{c}{BF}_0\left(p,q\right)=i+pj+\left(p^2+q\right)ij,\\ {BF}_1\left(p,q\right)=1+pi+\left(p^2+q\right)j+\left(p^3+2pq\right)ij,\\ {BF}_2\left(p,q\right)=p+\left(p^2+q\right)i+\left(p^3+2pq\right)j+\left(p^4+{3p}^{2}q+q^2\right)ij,\\ B{F}_3\left(p,q\right)=\left(p^2+q\right)+\left(p^3+2pq\right)i+\left(p^4+{3p}^{2}q+q^2\right)j+\left(p^5+{4p}^{3}q+{3pq}^2\right)ij,\\ B{F}_4\left(p,q\right)=\left(p^3+2pq\right)+{(p}^4+{3p}^{2}q+q^2)i+\left(p^5+{4p}^{3}q+{3pq}^2\right)j+\left(p^6+{5p}^{4}q+{6p^{2}q}^2+q^3\right)ij.\end{array}$$

For $u\ge 2$, it is given the following identity with a simple calculation:

$${BF}_u\left(p,q\right)={pBF}_{u-1}\left(p,q\right)+q{BF}_{u-2}\left(p,q\right).$$

(8)

Thus, the characteristic equation of (8) is:

$${\phi }^2-p\phi -q=0.$$

(9)

For any non-negative integers $u,v$, let $B{F}_u\left(p,q\right)=F_u+i{F}_{u+1}+j{F}_{u+2}+ij{F}_{u+3}$ and $B{F}_v\left(p,q\right)=F_v+i{F}_{v+1}+j{F}_{v+2}+ij{F}_{v+3}$ be two bicomplex $\left(p,q\right)$-Fibonacci numbers. The addition, subtraction, and multiplication for them are written as follows:

$$B{F}_u\left(p,q\right)\pm B{F}_v\left(p,q\right)=(F_u\pm F_v)+(F_{u+1}\pm F_{v+1})i+(F_{u+2}\pm F_{v+2})j+(F_{u+3}\pm F_{v+3})ij,$$

$$\begin{gathered} B{F}_u\left(p,q\right)\times B{F}_v\left(p,q\right)=(F_u{F}_v-F_{u+1}{F}_{v+1}-F_{u+2}{F}_{v+2}+F_{u+3}{F}_{v+3}) \\ +\left( F_u{F}_{v+1}+F_{u+1}{F}_v-F_{u+2}{F}_{v+3}+F_{u+3}{F}_{v+2}\right)i \\ +(F_u{F}_{v+2}+F_{u+2}{F}_v-F_{u+1}{F}_{v+3}+F_{u+3}{F}_{v+1})j \\ +(F_u{F}_{v+3}+F_{u+3}{F}_v-F_{u+1}{F}_{v+2}+F_{u+2}{F}_{v+1})ij. \end{gathered}$$

The multiplication of a bicomplex $\left(p,q\right)$-Fibonacci number by the real scalar $\mathsf{\mu }$ is described as the following:

$$\mu {BF}_u\left(p,q\right)=\mu F_u+i\mu F_{u+1}+j\mu F_{u+2}+ij \mu F_{u+3}.$$

Furthermore, bicomplex $\left(p,q\right)$-Fibonacci numbers have three different conjugations, which can be written as follows:

$$\overline{{\left({BF}_u\left(p,q\right)\right)}_i}=F_u-i{F}_{u+1}+j{F}_{u+2}-ij{F}_{u+3},$$

(10)

$$\overline{{\left({BF}_u\left(p,q\right)\right)}_j}=F_u+i{F}_{u+1}-j{F}_{u+2}-ij{F}_{u+3},$$

(11)

$$\overline{{\left({BF}_u\left(p,q\right)\right)}_{ij}}=F_u-i{F}_{u+1}-j{F}_{u+2}+ij{F}_{u+3}.$$

(12)

Theorem 1.

Let $B{F}_u$ and $B{F}_v$ be two bicomplex $\left(p,q\right)$-Fibonacci numbers. There hold:

$$\overline{{\left(B{F}_u\right)\left(B{F}_v\right)}_i}=\overline{{\left(B{F}_u\right)}_i} \overline{{\left(B{F}_v\right)}_i}= \overline{{\left(B{F}_v\right)}_i} \overline{{\left(B{F}_u\right)}_i}$$

$$\overline{{\left(B{F}_u\right)\left(B{F}_v\right)}_j}=\overline{{\left(B{F}_u\right)}_j} \overline{{\left(B{F}_v\right)}_j}= \overline{{\left(B{F}_v\right)}_j} \overline{{\left(B{F}_u\right)}_j}$$

$$\overline{{\left(B{F}_u\right)\left(B{F}_v\right)}_{ij}}=\overline{{\left(B{F}_u\right)}_{ij}} \overline{{\left(B{F}_v\right)}_{ij}}= \overline{{\left(B{F}_v\right)}_{ij}} \overline{{\left(B{F}_u\right)}_{ij}}$$

Proof. 

By using (10)–(12), these identities can be obtained with simple mathematical calculations. $\square$

Theorem 2.

Binet’s formula for the bicomplex $\left(p,q\right)$-Fibonacci numbers $\left\{{BF}_u\left(p,q\right)\right\}$ is given in the following equation for $u\ge 0$ ($u$ is any integer),

$${BF}_u=\frac{\underset{\_}{ \tau }{\tau }^u-\underset{\_}{ \omega }{\omega }^u}{\tau -\omega }$$

(13)

where $\tau$ and $\omega$ are roots of (9).

$$\underset{\_}{ \tau }=1+i \tau +j{\tau }^2+ij{\tau }^3$$

$$\underset{\_}{ \omega }=1+i \omega +j{\omega }^2+ij{\omega }^3$$

Proof. 

By using (7) and (3), we have the following equation:

$${BF}_u\left(p,q\right)=\left(\frac{{\tau }^u-{\omega }^u}{\tau -\omega }\right)+i\left(\frac{{\tau }^{u+1}-{\omega }^{u+1}}{\tau -\omega }\right)+j\left(\frac{{\tau }^{u+2}-{\omega }^{u+2}}{\tau -\omega }\right)+ij\left(\frac{{\tau }^{u+3}-{\omega }^{u+3}}{\tau -\omega }\right).$$

Thus, Binet’s formula for the bicomplex $\left(p,q\right)$-Fibonacci numbers is easily given with some simple computation. □

In the remainder of the study, ${BF}_u$ will be considered the $uth$ bicomplex $\left(p,q\right)$-Fibonacci number.

Theorem 3.

The generating function of the bicomplex $\left(p,q\right)$-Fibonacci numbers   ${\{BF}_u\}$ is

$$G_{BF}\left(t\right)=\frac{{BF}_0+\left({BF}_1-p{BF}_0\right)t}{\left(1-pt-q{t}^2\right)}=\frac{{BF}_0+\left(1+qj+\left(p^3+p^2-q\right)ij\right)t}{\left(1-pt-q{t}^2\right)}.$$

Proof. 

To find the generating function of ${\{BF}_u\}$, we will first use the following equation:

$$G_{BF}\left(t\right)={\sum }_{u=0}^{\infty }{BF}_u{t}^u.$$

In that case,

$$G_{BF}\left(t\right)={BF}_0+{BF}_{1}t+{BF}_2{t}^2+\cdots +{BF}_m{t}^m+\cdots .$$

Thus,

$$-pt{G}_{BF}\left(t\right)=-p{BF}_{0}t-p{BF}_1{t}^2-p{BF}_2{t}^3-\cdots -{pBF}_m{t}^{m+1}+\cdots .$$

$${-qt}^2{G}_{BF}\left(t\right)=-q{BF}_0{t}^2-q{BF}_1{t}^3-q{BF}_2{t}^4-\cdots -q{BF}_m{t}^{m+2}+\cdots .$$

We obtain that

$$\begin{gathered} \left(1-pt-q{t}^2\right)G_{BF}\left(t\right)={BF}_0+\left({BF}_1-p{BF}_0\right)t+({BF}_2-p{BF}_1-q{BF}_0)t^2 \\ +\cdots +{(BF}_{m+1}-{pBLe}_m{-q{BF}_{m-1})t}^{m+1}+\cdots . \end{gathered}$$

Using (8) and initial conditions, we have

$$G_{BF}\left(t\right)=\frac{{BF}_0+\left({BF}_1-p{BF}_0\right)t}{\left(1-pt-q{t}^2\right)}=\frac{{BF}_0+\left(1+qj+\left(p^3+p^2-q\right)ij\right)t}{\left(1-pt-q{t}^2\right)}.$$

□

Theorem 4.

The exponential generating function of the bicomplex  $\left(p,q\right)$-Fibonacci numbers ${\{BF}_u\}$ is

$$E_{BF}\left(t\right)=\frac{\underset{\_}{ \tau }{e}^{\tau u}-\underset{\_}{ \omega }{e}^{\omega u}}{\tau -\omega }.$$

Proof. 

To find the exponential generating function of ${\{BF}_u\}$, we first use the following equation:

$$E_{BF}\left(t\right)={\sum }_{u=0}^{\mathrm{\infty }}{BF}_u\frac{t^u}{u!}.$$

(14)

Using (14) and $e^t=\sum _{u=0}^{\infty }\frac{t^u}{u!}$, the exponential generating function of ${BF}_u$ is obtained:

$$E_{BF}\left(t\right)={\sum }_{u=0}^{\mathrm{\infty }}\frac{\underset{\_}{\tau }{\tau }^u-\underset{\_}{\omega }{\omega }^u}{\tau -\omega }\frac{t^u}{u!}=\frac{\underset{\_}{\tau }{e}^{\tau u}-\underset{\_}{\omega }{e}^{\omega u}}{\tau -\omega }.$$

□

Theorem 5.

For $u\ge v$, the Catalan identity for bicomplex $\left(p,q\right)$-Fibonacci numbers is as follows:

$${{{BF}_{u-v}{BF}_{u+v}-BF}_u}^2=\left(\frac{-\underset{\_}{ \tau } \underset{\_}{ \omega }{\left(-q\right)}^{u-v}{\left({\tau }^v-{\omega }^v\right)}^2}{p^2+4q}\right)$$

where $u$ and $v$ are non-negative integers.

Proof. 

$$\begin{array}{cc}{{{BF}_{u-v}{BF}_{u+v}-BF}_u}^2& =\left(\frac{\underset{\_}{\tau }{\tau }^{u-v}-\underset{\_}{\omega }{\omega }^{u-v}}{\tau -\omega }\right)\left(\frac{\underset{\_}{\tau }{\tau }^{u+v}-\underset{\_}{\omega }{\omega }^{u+v}}{\tau -\omega }\right)-\left(\frac{\underset{\_}{\tau }{\tau }^u-\underset{\_}{\omega }{\omega }^u}{\tau -\omega }\right)\left(\frac{\underset{\_}{\tau }{\tau }^u-\underset{\_}{\omega }{\omega }^u}{\tau -\omega }\right)\\ & =\left(\frac{\underset{\_}{ \tau } \underset{\_}{ \omega }{\left(-q\right)}^{u-v}\left({\tau }^v-{\omega }^v\right){\omega }^v-\underset{\_}{ \omega } \underset{\_}{ \tau } {\left(-q\right)}^{u-v}({\tau }^v-{\omega }^v){\tau }^v}{{\left(\tau -\omega \right)}^2}\right).\end{array}$$

Because $\tau$ and $\omega$ are roots of (9), $\tau \omega =(-q)$, $\underset{\_}{\tau }=1+i\tau +j{\tau }^2+ij{\tau }^3$, and $\underset{\_}{\omega }=1+i\omega +j{\omega }^2+ij{\omega }^3$, we obtain

$$\begin{gathered} \underset{\_}{\tau }\underset{\_}{\omega }=\left(1+q-q^2-q^3\right)+i\left(p-p{q}^2\right)+j\left(p^2+2q+p^{2}q+2q^2\right)+ij\left(p^3+2qp\right) \\ =\underset{\_}{ \omega } \underset{\_}{ \tau }. \end{gathered}$$

With this equality, we obtain

$$\begin{array}{cc}{{{BF}_{u-v}{BF}_{u+v}-BF}_u}^2\hfill & =\frac{\underset{\_}{ \tau } \underset{\_}{ \omega }{\left(-q\right)}^{u-v}\left({\tau }^v-{\omega }^v\right){\omega }^v-\underset{\_}{ \omega } \underset{\_}{ \tau } {\left(-q\right)}^{u-v}\left({\tau }^v-{\omega }^v\right){\tau }^v}{{\left(\tau -\omega \right)}^2}\\ & =\left(\frac{-\underset{\_}{\tau }\underset{\_}{\omega }{\left(-q\right)}^{u-v}{\left({\tau }^v-{\omega }^v\right)}^2}{{\left(\tau -\omega \right)}^2}\right)=\left(\frac{-\underset{\_}{\tau }\underset{\_}{\omega }{\left(-q\right)}^{u-v}{\left({\tau }^v-{\omega }^v\right)}^2}{p^2+4q}\right).\end{array}$$

□

If $v=1$ in the Catalan identity, the Cassini identity is obtained as follows:

Corollary 1.

For $u\ge 1$, the Cassini identity for bicomplex $\left(p,q\right)$-Fibonacci numbers is as follows:

$${BF}_{u-1}{BF}_{u+1}{-{BF}_u}^2=-\underset{\_}{ \tau } \underset{\_}{ \omega }{\left(-q\right)}^{u-1}$$

where $u$ is a non-negative integer.

Theorem 6.

D’Ocagne’s identity for bicomplex $\left(p,q\right)$-Fibonacci numbers is as follows:

$${BF}_u{BF}_{v+1}-{BF}_{u+1}{BF}_v=\frac{\underset{\_}{ \tau } \underset{\_}{ \omega }{\left(-q\right)}^v\left({\tau }^{u-v}-{\omega }^{u-v}\right)}{\sqrt{p^2+4q}}.$$

Proof. 

$$\begin{array}{cc}{BF}_u{BF}_{v+1}-{BF}_{u+1}{BF}_v& =\left(\frac{\underset{\_}{ \tau }{\tau }^u-\underset{\_}{ \omega }{\omega }^u}{\tau -\omega }\right)\left(\frac{\underset{\_}{ \tau }{\tau }^{1+v}-\underset{\_}{ \omega }{\omega }^{1+v}}{\tau -\omega }\right)-\left(\frac{\underset{\_}{ \tau }{\tau }^{u+1}-\underset{\_}{ \omega }{\omega }^{u+1}}{\tau -\omega }\right)\left(\frac{\underset{\_}{ \tau }{\tau }^v-\underset{\_}{ \omega }{\omega }^v}{\tau -\omega }\right)\\ & =\left(\frac{\underset{\_}{ \tau } \underset{\_}{ \omega }{\tau }^u{\omega }^v(\tau -\omega )-\underset{\_}{ \omega } \underset{\_}{ \tau } {\omega }^u{\tau }^v(\tau -\omega )}{{\left(\tau -\omega \right)}^2}\right).\end{array}$$

As in the proof of Theorem 5, $\underset{\_}{\tau }\underset{\_}{\omega }=\underset{\_}{\omega }\underset{\_}{\tau }$, $u\ge v$, and $\tau \omega =-q$. Thus,

$$\begin{array}{cc}{BF}_u{BF}_{v+1}-{BF}_{u+1}{BF}_v& =\left(\frac{\underset{\_}{ \tau } \underset{\_}{ \omega }{\tau }^u{\omega }^v(\tau -\omega )-\underset{\_}{ \omega } \underset{\_}{ \tau } {\omega }^u{\tau }^v(\tau -\omega )}{{\left(\tau -\omega \right)}^2}\right)\\ & =\frac{\underset{\_}{ \tau } \underset{\_}{ \omega }{\left(-q\right)}^v\left({\tau }^{u-v}-{\omega }^{u-v}\right)}{(\tau -\omega )}=\frac{\underset{\_}{ \tau } \underset{\_}{ \omega }{\left(-q\right)}^v\left({\tau }^{u-v}-{\omega }^{u-v}\right)}{\sqrt{p^2+4q}}.\end{array}$$

□

Now, we give some identities about summations of terms in the bicomplex $\left(p,q\right)$-Fibonacci numbers.

Theorem 7.

For $k,l$ as natural numbers, the summation formula of bicomplex $\left(p,q\right)$-Fibonacci numbers is

$${\sum }_{k=1}^l{BF}_k=\left\{\begin{array}{c}\frac{{BF}_{l+1}+q{BF}_l-{BF}_1-q{BF}_0}{p+q-1} p+q\ne 1 \\ \frac{{qBF}_l+{BF}_1+\left(l-1\right)(1+i+j+ij)}{1+q } p+q=1\end{array}\right..$$

Proof. 

Firstly, we assume that $p+q\ne 1.$ In this situation,

$${\sum }_{k=1}^l{BF}_k={\sum }_{k=1}^l{F}_k+i{\sum }_{k=1}^l{F}_{k+1}+j{\sum }_{k=1}^l{F}_{k+2}+ij{\sum }_{k=1}^l{F}_{k+3}.$$

In addition, we know from (13) in [24] that

$${\sum }_{k=1}^l{F}_k=\frac{F_{l+1}+q{F}_l-F_1-q{F}_0}{p+q-1}.$$

With simple calculations, we obtain

$${\sum }_{k=1}^l{F}_{k+1}=p{\sum }_{k=1}^l{F}_k+q{\sum }_{k=1}^l{F}_{k-1}=\frac{F_{l+2}+q{F}_{l+1}-F_2-q{F}_1}{p+q-1},$$

$${\sum }_{k=1}^l{F}_{k+2}=\frac{F_{l+3}+q{F}_{l+2}-F_3-q{F}_2}{p+q-1},$$

$${\sum }_{k=1}^l{F}_{k+3}=\frac{F_{l+4}+q{F}_{l+3}-F_4-q{F}_3}{p+q-1}.$$

Thus,

$${\sum }_{k=1}^l{BF}_k=\frac{F_{l+1}+q{F}_l-F_1-q{F}_0}{p+q-1}+i\left(\frac{F_{l+2}+q{F}_{l+1}-F_2-q{F}_1}{p+q-1}\right)+j\left(\frac{F_{l+3}+q{F}_{l+2}-F_3-q{F}_2}{p+q-1}\right)+ij\left(\frac{F_{l+4}+q{F}_{l+3}-F_4-q{F}_3}{p+q-1}\right)=\frac{{BF}_{l+1}+q{BF}_l-{BF}_1-q{BF}_0}{p+q-1}.$$

Now, let $p+q=1$.

We obtain from (13) in [24] that $\sum _{k=1}^l{F}_k=\frac{q{F}_l+(l-1){+F}_1}{1+q}$. Moreover, we have

$${\sum }_{k=1}^l{F}_{k+1}=p{\sum }_{k=1}^l{F}_k+q{\sum }_{k=1}^l{F}_{k-1}=\frac{q{F}_{l+1}+\left(l-1\right)+F_2}{1+q},$$

so we can write

$${\sum }_{k=1}^l{F}_{k+2}=\frac{q{F}_{l+2}+\left(l-1\right)+F_3}{1+q},$$

$\sum _{k=1}^l{F}_{k+3}=\frac{q{F}_{l+3}+\left(l-1\right)+F_4}{1+q}$. Thus,

$${\sum }_{k=1}^l{BF}_k=\frac{q{F}_l+(l-1){+F}_1}{1+q}+i\left(\frac{q{F}_{l+1}+\left(l-1\right)+F_2}{1+q}\right)+j\left(\frac{q{F}_{l+2}+\left(l-1\right)+F_3}{1+q}\right)+ij\left(\frac{q{F}_{l+3}+\left(l-1\right)+F_4}{1+q}\right)=\frac{{qBF}_l+\left(l-1\right)\left(1+i+j+ij\right)+{BF}_1}{1+q}.$$

□

Theorem 8.

For $u,v\ge 0$,

$${BF}_{uv}={\sum }_{k=0}^v\left(\begin{array}{c}v\\ k\end{array}\right){\left(q\right)}^{v-k}{F}_u^k{F}_{u-1}^{v-k}{BF}_k.$$

Proof. 

According to the lemma in [25] for $\left(p,q\right)$-Fibonacci numbers, we know that

$$F_{mn+r}={\sum }_{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right){\left(q\right)}^{n-j}{F}_m^j{F}_{m-1}^{n-j}{F}_{j+r}.$$

(15)

Using (7) and (15),

$$\begin{array}{l}{BF}_{uv}=F_{uv}+i{F}_{uv+1}+j{F}_{uv+2}+ij{F}_{uv+3}\\ ={\displaystyle \sum _{k=0}^v}\left(\begin{array}{c}v\\ k\end{array}\right){\left(q\right)}^{v-k}{F}_u^k{F}_{u-1}^{v-k}{F}_k+i{\displaystyle \sum _{k=0}^v}\left(\begin{array}{c}v\\ k\end{array}\right){\left(q\right)}^{v-k}{F}_u^k{F}_{u-1}^{v-k}{F}_{k+1}\\ +j\sum _{k=0}^v\left(\begin{array}{c}v\\ k\end{array}\right){\left(q\right)}^{v-k}{F}_u^k{F}_{u-1}^{v-k}{F}_{k+2}+ij\sum _{k=0}^v\left(\begin{array}{c}v\\ k\end{array}\right){\left(q\right)}^{v-k}{F}_u^k{F}_{u-1}^{v-k}{F}_{k+3}=\\ \sum _{k=0}^v\left(\begin{array}{c}v\\ k\end{array}\right){\left(q\right)}^{v-k}{F}_u^k{F}_{u-1}^{v-k}{(F_k+iF}_{k+1}+j{F}_{k+2}+ij{F}_{k+3})=\sum _{k=0}^v\left(\begin{array}{c}v\\ k\end{array}\right){\left(q\right)}^{v-k}{F}_u^k{F}_{u-1}^k{BF}_k\end{array}$$

.

□

3. Bicomplex $\left(\mathit{p},\mathit{q}\right)$-Fibonacci Quaternions

Here, we obtain a new generalization of the second-order bicomplex quaternions. To do this, we describe the bicomplex $\left(p,q\right)$-Fibonacci quaternions. We obtain some equations and summation formulas about the bicomplex $\left(p,q\right)$-Fibonacci quaternion sequence. In addition, we give the generating function, Binet’s formula, and Catalan, Cassini, and d’Ocagne’s identities for these quaternions.

Definition 2.

The bicomplex $\left(p,q\right)$-Fibonacci quaternions are defined by

$${BCQF}_u\left(p,q\right)=Q{F}_u\left(p,q\right)+i Q{F}_{u+1}\left(p,q\right)+jQ{F}_{u+2}\left(p,q\right)+ij Q{F}_{u+3}\left(p,q\right),$$

(16)

where $Q{F}_u\left(p,q\right)=F_u{e}_0+F_{u+1}{e}_1$+$F_{u+2}{e}_2+F_{u+3}{e}_3$ is the $uth$ $\left(p,q\right)$-Fibonacci quaternion, $i^2=j^2=-1$, $ij=ji$.

Thus, the bicomplex $\left(p,q\right)$-Fibonacci quaternion with four bicomplex components can be written as

$$\begin{gathered} {BCQF}_u\left(p,q\right)=(F_u+i{F}_{u+1}+j{F}_{u+2}+ij F_{u+3})+{(F}_{u+1}+i{F}_{u+2}+j{F}_{u+3}+ij F_{u+4}) e_1+ \\ (F_{u+2}+i{F}_{u+3}+j{F}_{u+4}+ij F_{u+5}) e_2+(F_{u+3}+i{F}_{u+4}+j{F}_{u+5}+ij F_{u+6}) e_3. \end{gathered}$$

By using (7), we obtain the following equation:

$${BCQF}_u\left(p,q\right)={BF}_u+{BF}_{u+1}{e}_1+{BF}_{u+2}{e}_2+{BF}_{u+3}{e}_3.$$

(17)

Thus, the first few terms of the bicomplex $\left(p,q\right)$-Fibonacci quaternions are

$$\begin{array}{cc}{BCQF}_0\left(p,q\right)=i& +pj+\left(p^2+q\right)ij+1+pi+\left(p^2+q\right)j+\left(p^3+2pq\right)ij e_1+(p\\ & +\left(p^2+q\right)i+\left(p^3+2pq\right)j+\left(p^4+{3p}^{2}q+q^2\right)ij) e_2+(\left(p^2+q\right)\\ & +\left(p^3+2pq\right)i+\left(p^4+{3p}^{2}q+q^2\right)j+\left(p^5+{4p}^{3}q+{3pq}^2\right)ij) e_3,\end{array}$$

$$\begin{gathered} {BCQF}_1\left(p,q\right)=(1+pi+\left(p^2+q\right)j+\left(p^3+2pq\right)ij)+(p+\left(p^2+q\right)i+\left(p^3+2pq\right)j+ \\ \left(p^4+{3p}^{2}q+q^2\right)ij) e_1+(\left(p^2+q\right)+\left(p^3+2pq\right)i+\left(p^4+{3p}^{2}q+q^2\right)j+ \\ \left(p^5+{4p}^{3}q+{3pq}^2\right)ij) e_2+(\left(p^3+2pq\right)+{(p}^4+{3p}^{2}q+q^2)i+\left(p^5+{4p}^{3}q+{3pq}^2\right)j+ \\ \left(p^6+{5p}^{4}q+{6p^{2}q}^2+q^3\right)ij)e_3, \end{gathered}$$

$$\begin{gathered} {BCQF}_2\left(p,q\right)=(p+\left(p^2+q\right)i+\left(p^3+2pq\right)j+\left(p^4+{3p}^{2}q+q^2\right)ij)+(\left(p^2+q\right)+ \\ \left(p^3+2pq\right)i+\left(p^4+{3p}^{2}q+q^2\right)j+\left(p^5+{4p}^{3}q+{3pq}^2\right)ij) e_1+(\left(p^3+2pq\right)+{(p}^4+ \\ {3p}^{2}q+q^2)i+\left(p^5+{4p}^{3}q+{3pq}^2\right)j+\left(p^6+{5p}^{4}q+{6p^{2}q}^2+q^3\right)ij) e_2+ \\ (\left(p^4+{3p}^{2}q+q^2\right)+\left(p^5+{4p}^{3}q+{3pq}^2\right)i+\left(p^6+{5p}^{4}q+{6p^{2}q}^2+q^3\right)j+ \\ \left(p^7+{6p}^{5}q+{10p^{3}q}^2+{4pq}^3\right)ij)e_3. \end{gathered}$$

Therefore, any bicomplex $\left(p,q\right)$-Fibonacci quaternion consists of a scalar part and a vectorial part expressed as follows:

$$S_{{BCQF}_u}={K_0=BF}_u\left(p,q\right)=F_u+i{F}_{u+1}+j{F}_{u+2}+ij F_{u+3},$$

$$V_{{BCQF}_u}={K=BF}_{u+1}\left(p,q\right) e_1+{BF}_{u+2}\left(p,q\right)e_2+{BF}_{u+3}\left(p,q\right) e_3.$$

Here, the set of bicomplex $\left(p,q\right)$-Fibonacci quaternions will be denoted by $H_{{BCQF}_{\left(p,q\right)}}$. And in the remainder of the study, ${BCQF}_u$ and ${QF}_u$ will be considered the $uth$ bicomplex $\left(p,q\right)$-Fibonacci and $\left(p,q\right)$-Fibonacci quaternion, respectively.

Let ${BCQF}_u=K_0+K\mathrm{a}\mathrm{n}\mathrm{d}{BCQF}_v=L_0+L$ be two bicomplex $\left(p,q\right)$-Fibonacci quaternions. The addition and subtraction of them are

$${BCQF}_u\mp {BCQF}_v={(BF}_u\mp {BF}_v)+{(BF}_{u+1}\mp {BF}_{v+1})) e_1+{(BF}_{u+2}\mp {BF}_{v+2}) e_2+{(BF}_{u+3}\mp {BF}_{v+3}) e_3.$$

The multiplication of a bicomplex $\left(p,q\right)$-Fibonacci quaternion by the real scalar $\lambda$ is described as follows:

$${\lambda BCQF}_u=\lambda Q{F}_u+i \lambda Q{F}_{u+1}+j\lambda Q{F}_{u+2}+ij \lambda Q{F}_{u+3}.$$

The product of any two bicomplex $\left(p,q\right)$-Fibonacci quaternions ${BCQF}_u$ and ${BCQF}_v$, such as

$${BCQF}_u=K_0+K \mathrm{and} {BCQF}_v= L_0+L,\mathrm{is}$$

$$\begin{array}{ll}{BCQF}_u{BCQF}_v& ={(K}_0+K)\left(L_0+L\right)=K_0{L}_0+K_{0}L+L_{0}K+KL\\ & =K_0{L}_0+K_{0}L+L_{0}K-K°L+K\times L,\end{array}$$

where $K°L$ and $K\times L$ represent the dot product (scalar product) and the cross product (vector product) of $K$ and $L$, respectively. The conjugate operation in $H_{{BCF}_{\left(p,q\right)}}$ is

$${{BCQF}_u}^{*}=S_{{BCQF}_u}-V_{{BCQF}_u}= {BF}_u\left(p,q\right)-{BF}_{u+1}\left(p,q\right) e_1-{BF}_{u+2}\left(p,q\right)e_2-{BF}_{u+3}\left(p,q\right) e_3,$$

whereas the bicomplex conjugates are

$$\overline{{\left({BCQF}_u\right)}_i}=\overline{{\left(S_{{BCQF}_u}\right)}_i}+\overline{{\left(V_{{BCQF}_u}\right)}_{i }}=\overline{{\left({BF}_u\left(p,q\right)\right)}_i}+\overline{{\left({BF}_{u+1}\left(p,q\right)\right)}_i}{e}_1+\overline{{\left({BF}_{u+2}\left(p,q\right)\right)}_i}{e}_2+\overline{{\left({BF}_{u+3}\left(p,q\right)\right)}_i}{e}_3$$

$$\overline{{\left({BCQF}_u\right)}_j}=\overline{{\left(S_{{BCQF}_u}\right)}_j}+\overline{{\left(V_{{BCQF}_u}\right)}_j}=\overline{{\left({BF}_u\left(p,q\right)\right)}_j}+\overline{{\left({BF}_{u+1}\left(p,q\right)\right)}_j}{e}_1+\overline{{\left({BF}_{u+2}\left(p,q\right)\right)}_j}{e}_2+\overline{{\left({BF}_{u+3}\left(p,q\right)\right)}_j}{e}_3$$

$$\overline{{\left({BCQF}_u\right)}_{ij}}=\overline{{\left(S_{{BCQF}_u}\right)}_{ij}}+\overline{{\left(V_{{BCQF}_u}\right)}_{ij}}=\overline{{\left({BF}_u\left(p,q\right)\right)}_{ij}}+\overline{{\left({BF}_{u+1}\left(p,q\right)\right)}_{ij}}{e}_1+\overline{{\left({BF}_{u+2}\left(p,q\right)\right)}_{ij}}{e}_2+\overline{{\left({BF}_{u+3}\left(p,q\right)\right)}_{ij}}{e}_3.$$

The features of quaternion algebra are adapted to bicomplex quaternions as well as to complex quaternions. In this situation, some key properties of bicomplex quaternions change. Because the norm of a real quaternion $h=(h_0,h_1,h_2,h_3)$ is defined by $‖h‖={h_0}^2+{h_1}^2+{h_2}^2+{h_3}^2$, the norm is positive, definite, and real. But when we consider the complex quaternion, the norm is described according to the inner product of the complex quaternion with itself. That is, for a complex quaternion $ch=({ch}_0,{ch}_1,c{h}_2,{ch}_3)$, the norm of $ch$ can be written as $‖ch‖={c{h}_0}^2+{{ch}_1}^2+{{ch}_2}^2+c{h_3}^2$.

Since the components of $ch$ are complex numbers, the norm of $ch$ has a complex value. In [8], a complex Fibonacci quaternion ($QCF$) is defined, and the norm of a complex Fibonacci quaternion is given as follows:

$$‖QCF‖={C_n}^2+{C_{n+1}}^2+{C_{n+2}}^2+{C_{n+3}}^2$$

where $C_n$ is a complex Fibonacci number.

In addition, we describe the norm of any bicomplex quaternion in terms of the inner product of a bicomplex quaternion with itself, as in the definition of a complex quaternion. Then, for any bicomplex quaternion $bch=(b{ch}_0,b{ch}_1,bc{h}_2,{bch}_3)$, the norm of $\mathrm{b}\mathrm{c}\mathrm{h}$ can be written as $‖bch‖={bc{h}_0}^2+{{bch}_1}^2+{b{ch}_2}^2+bc{h_3}^2$. In this situation, the norm of a bicomplex $\left(p,q\right)$-Fibonacci quaternion can be given as follows:

$$‖BCQF‖=B{F_u}^2+{{BF}_{u+1}}^2+{{BF}_{u+2}}^2+{{BF}_{u+3}}^2.$$

Also, we show that there are four different conjugates of the bicomplex $\left(p,q\right)$-Fibonacci quaternion, whereas there are three different conjugates of the bicomplex $\left(p,q\right)$-Fibonacci numbers. Furthermore, we get the following inequalities about four different conjugations of the bicomplex $\left(p,q\right)$-Fibonacci quaternion:

Theorem 9.

Let ${BCQF}_u$ and ${BCQF}_v$ be two bicomplex $\left(p,q\right)$-Fibonacci quaternions. In that case, we obtain the following inequalities about the four conjugates of them:

$${{\left(\right(BCQF}_u\left){(BCQF}_v\right)) }^{*}\ne {{(BCQF}_u) }^{*}{ {(BCQF}_v) }^{*}$$

$$\overline{{{(BCQF}_u\left){(BCQF}_v\right)}_i} \ne \overline{{{(BCQF}_u)}_i} \overline{{{(BCQF}_v)}_i}$$

$$\overline{{{(BCQF}_u\left){(BCQF}_v\right)}_j} \ne \overline{{{(BCQF}_u)}_j} \overline{{{(BCQF}_v)}_j}$$

$$\overline{{{(BCQF}_u\left){(BCQF}_v\right)}_{ij}} \ne \overline{{{(BCQF}_u)}_{ij}} \overline{{{(BCQF}_v)}_{ij}}$$

Proof. 

Using conjugate operations in $H_{BCQF}$ and (16), the above identities can be easily proved. □

The following equation for the elements of $BCQF$ is easily obtained using (16):

$$BC{QF}_{u+2}=pBC{QF}_{u+1}+qBC{QF}_u.$$

(18)

Thus, the characteristic equation of (18) is

$${\gamma }^2-p\gamma -q=0.$$

(19)

Theorem 10.

Binet’s formula for the bicomplex $\left(p,q\right)$-Fibonacci quaternions $\left\{{BCQF}_u\right\}$ is given by the following equation for $u\ge 0$:

$${BCQF}_u=\frac{\underset{\_}{ \delta }{\acute{\delta }\delta }^u-\underset{\_}{ \rho }{\acute{\rho }\rho }^u}{\delta -\rho }$$

where $\delta$ and $\rho$ are roots of (19) and

$$\underset{\_}{ \delta }=1+i \delta +j{\delta }^2+ij{\delta }^3$$

$$\underset{\_}{ \rho }=1+i \rho +j{\rho }^2+ij{\rho }^3$$

$$\acute{\delta }=1+\delta e_1+ {\delta }^2{e}_2+{\delta }^3 e_{3,}$$

$$\acute{\rho }=1+\rho {\mathrm{e}}_1+{\rho }^2{\mathrm{e}}_2+{\rho }^3{\mathrm{e}}_{3,}$$

Proof. 

By using (13) and (17), we have the following equation:

$${BCQF}_u=\frac{\underset{\_}{ \delta }{\delta }^u-\underset{\_}{ \rho }{\rho }^u}{\delta -\rho }+\left(\frac{\underset{\_}{ \delta }{\delta }^{u+1}-\underset{\_}{ \rho }{\rho }^{u+1}}{\delta -\rho }\right)e_1+\left(\frac{\underset{\_}{ \delta }{\delta }^{u+2}-\underset{\_}{ \rho }{\rho }^{u+2}}{\delta -\rho }\right)e_2+(\frac{\underset{\_}{ \delta }{\delta }^{u+3}-\underset{\_}{ \rho }{\rho }^{u+3}}{\delta -\rho }{) e}_3.$$

Thus, Binet’s formula for the bicomplex $\left(p,q\right)$-Fibonacci quaternion is easily found with some simple computation. □

Theorem 11.

The generating function of the bicomplex $\left(p,q\right)$-Fibonacci quaternions ${\{BCQF}_u\}$ is determined by

$$G_{BCQF}\left(t\right)=\frac{{BCQF}_0+\left({BCQF}_1-p{BCQF}_0\right)t}{\begin{array}{c}\left(1-pt-q{t}^2\right)\end{array}}.$$

Proof. 

To obtain the generating function of ${\left\{{BCQF}_u\right\}}_{u=0}^{\infty }$, we use the power series representation of ${\{BCQF}_u\}$.

$$G_{BCQF}\left(t\right)={\sum }_{u=0}^{\infty }{BCQF}_u{t}^u.$$

That is,

$$G_{BCQF}\left(t\right)={BCQF}_0+{BCQF}_{1}t+{BCQF}_2{t}^2+\cdots +{BCQF}_m{t}^m+\cdots .$$

Thus,

$$-pt{G}_{BCQF}\left(t\right)=-p{BCQF}_{0}t-p{BCQF}_1{t}^2-p{BCQF}_2{t}^3-\cdots -{pBCQF}_m{t}^{m+1}+\cdots .$$

$${-qt}^2{G}_{BCQF}\left(t\right)=-q{BCQF}_0{t}^2-q{BCQF}_1{t}^3-q{BCQF}_2{t}^4-\cdots -q{BCQF}_m{t}^{m+2}+\cdots .$$

We obtain that

$$\begin{gathered} \left(1-pt-q{t}^2\right)G_{BCQF}\left(t\right)={BCQF}_0+\left({BCQF}_1-p{BCQF}_0\right)t+({BCQF}_2-p{BCQF}_1-q{BCQF}_0)t^2 \\ +\cdots +{(BCQF}_{m+1}-{pBCQF}_m{-q{BCQF}_{m-1})t}^{m+1}+\cdots . \end{gathered}$$

Using (18) and initial conditions, we have

$$G_{BCQF}\left(t\right)=\frac{{BCQF}_0+\left({BCQF}_1-p{BCQF}_0\right)t}{\left(1-pt-q{t}^2\right)}.$$

□

Theorem 12.

The exponential generating function of the bicomplex $\left(p,q\right)$-Fibonacci quaternions ${\{BCQF}_u\}$ is

$$E_{BCQF}\left(t\right)=\frac{\underset{\_}{ \delta }{ \acute{\delta } e}^{\delta u}-\underset{\_}{ \rho }{ \acute{\rho } e}^{\rho u}}{\delta -\rho }.$$

Proof. 

To obtain the exponential generating function of ${{\{BCQF}_u\}}_{u=0}^{\infty }$, we use the power series representation of ${\{BCQF}_u\}$.

$$E_{BCQF}\left(t\right)={\sum }_{u=0}^{\mathrm{\infty }}{BCQF}_u\frac{t^u}{u!},$$

(20)

Using (20) and $e^t={\sum }_{u=0}^{\mathrm{\infty }}\frac{t^u}{u!}$, we have

$$E_{BCQF}\left(t\right)={\sum }_{u=0}^{\infty }\frac{\underset{\_}{ \delta }{\acute{\delta }\delta }^u-\underset{\_}{ \rho }{\acute{ \rho }\rho }^u}{\delta -\rho }\frac{t^u}{u!}=\frac{\underset{\_}{ \delta }{ \acute{\delta } e}^{\delta u}-\underset{\_}{ \rho }{ \acute{\rho } e}^{\rho u}}{\delta -\rho }.$$

□

Theorem 13.

For $u\ge v$, the Catalan identity for bicomplex $\left(p,q\right)$-Fibonacci quaternions is as follows:

$${{{BCQF}_{u-v}{BCQF}_{u+v}-BCQF}_u}^2=\left(\frac{\underset{\_}{ \delta } \underset{\_}{ \rho }{ \left(-q\right)}^{u-v}\left({\delta }^v-{\rho }^v\right){(\acute{\delta \stackrel{´}{ \rho }}\rho }^v-\acute{\rho }{\acute{\delta }\delta }^v}{(p^2+4q)}\right)$$

where $u$ and $v$ are positive integers.

Proof. 

$$\begin{array}{ll}{{{BCQF}_{u-v}{BCQF}_{u+v} –BCQF}_u}^2& =\left(\frac{\underset{\_}{ \delta }{\delta }^{u-v}\acute{\delta }-\underset{\_}{ \rho }{\rho }^{u-v}\acute{\rho }}{\delta -\rho }\right)\left(\frac{\underset{\_}{ \delta }{\delta }^{u+v}\acute{\delta }-\underset{\_}{ \rho }{\rho }^{u+v}\acute{\rho }}{\delta -\rho }\right)-\left(\frac{\underset{\_}{ \delta }{\delta }^u\acute{\delta }-\underset{\_}{ \rho }{\rho }^u\acute{\rho }}{\delta -\rho }\right)\left(\frac{\underset{\_}{ \delta }{\delta }^u\acute{\delta }-\underset{\_}{ \rho }{\rho }^u\acute{\rho }}{\delta -\rho }\right)\\ & =\left(\frac{\underset{\_}{ \delta } \underset{\_}{ \rho }{ \acute{\delta \stackrel{´}{\rho }}\left(-q\right)}^{u-v}\left({\delta }^v-{\rho }^v\right){\rho }^v-\underset{\_}{ \rho } \underset{\_}{ \delta } \acute{\rho }{\acute{\delta }\left(-q\right)}^{u-v}\left({\delta }^v-{\rho }^v\right){\delta }^v}{{\left(\delta -\rho \right)}^2}\right)\end{array}$$

Because $\delta$ and $\rho$ are roots of (19), $\delta \rho =\left(-q\right),$ $\underset{\_}{\delta }=1+i\delta +j{\delta }^2+ij{\delta }^3$ and $\underset{\_}{\rho }=1+i\rho +j{\rho }^2+ij{\rho }^3$, we obtain

$$\begin{array}{ll}\underset{\_}{ \delta } \underset{\_}{ \rho }& =\left(1+q-q^2-q^3\right)+i\left(p-p{q}^2\right)+j\left(p^2+2q+p^{2}q+2q^2\right)+ij\left(p^3+2qp\right)\\ & =\underset{\_}{ \rho } \underset{\_}{ \delta }.\end{array}$$

Thus, we obtain

$$\begin{array}{cc}{{{BCQF}_{u-v}{BCQF}_{u+v} –BCQF}_u}^2& =\left(\frac{\underset{\_}{ \delta } \underset{\_}{ \rho }{ \left(-q\right)}^{u-v}\left({\delta }^v-{\rho }^v\right){(\acute{\delta \stackrel{´}{\rho }}\rho }^v-\acute{\rho }{\acute{\delta }\delta }^v}{{\left(\delta -\rho \right)}^2}\right)\\ & =\left(\frac{\underset{\_}{ \delta } \underset{\_}{ \rho }{ \left(-q\right)}^{u-v}\left({\delta }^v-{\rho }^v\right){(\acute{\delta \stackrel{´}{\rho }}\rho }^v-\acute{\rho }{\acute{\delta }\delta }^v}{(p^2+4q)}\right).\end{array}$$

□

If $v=1$ in the Catalan identity, the Cassini identity is obtained as follows:

Corollary 2.

For $u\ge 1$, the Cassini identity for bicomplex $\left(p,q\right)$-Fibonacci quaternions is as follows:

$${BCQF}_{u-1}{BCQF}_{u+1}{-{BCQF}_u}^2=\left(\frac{\underset{\_}{ \delta } \underset{\_}{ \omega }{ \left(-q\right)}^{u-1}(\acute{\delta \stackrel{´}{\rho \rho }}-\acute{\rho \stackrel{´}{\delta \delta }}}{\sqrt{(p^2+4q)}}\right)$$

where $u$ is an integer.

Theorem 14.

D’ocagne’s identity of bicomplex $\left(p,q\right)$-Fibonacci quaternions for $u\ge v$ is as follows:

$${BCQF}_u{BCQF}_{v+1}-{BCQF}_{u+1}{BCQF}_v=\frac{\underset{\_}{ \delta } \underset{\_}{ \rho }{ \acute{\delta }\acute{\rho }\left(-q\right)}^v\left({\acute{\delta }\acute{\rho }\delta }^{u-v}-{\acute{\rho } \acute{\delta }\rho }^{u-v}\right)}{\sqrt{p^2+4q}}.$$

Proof. 

$$\begin{array}{cc}{BCQF}_u{BCQF}_{v+1}& -{BCQF}_{u+1}{BCQF}_v=\\ & =\left(\frac{\underset{\_}{ \delta }{ \acute{\delta }\delta }^u-\underset{\_}{ \rho } \acute{\rho }{\rho }^u}{\delta -\rho }\right)\left(\frac{\underset{\_}{ \delta }{ \acute{\delta }\delta }^{1+v}-\underset{\_}{ \rho }{ \acute{\rho }\rho }^{1+v}}{\delta -\rho }\right)\\ & -\left(\frac{\underset{\_}{ \delta }{ \acute{\delta }\delta }^{u+1}-\underset{\_}{ \rho }{ \acute{\rho }\rho }^{u+1}}{\delta -\rho }\right)\left(\frac{\underset{\_}{ \delta }{ \acute{\delta }\delta }^v-\underset{\_}{ \rho }{ \acute{\rho }\rho }^v}{\delta -\rho }\right)\\ & =\left(\frac{\underset{\_}{ \delta } \underset{\_}{ \rho }\acute{\delta }\acute{\rho }{\left(\delta \rho \right)}^v{\delta }^{u-v}\left(\delta -\rho \right)-\underset{\_}{ \rho } \underset{\_}{ \delta } \acute{\rho } \acute{\delta } {\left(\rho \delta \right)}^v{\rho }^{u-v}(\delta -\rho )}{{\left(\delta -\rho \right)}^2}\right)\end{array}$$

$\underset{\_}{\delta }\underset{\_}{\rho }=\underset{\_}{\rho }\underset{\_}{\delta }$, $u\ge v$ and $\delta \rho =-q=\frac{\underset{\_}{\delta }\underset{\_}{\rho }{\acute{\delta }\acute{\rho }\left(-q\right)}^v\left({\acute{\delta }\acute{\rho }\delta }^{u-v}-{\acute{\rho }\acute{\delta }\rho }^{u-v}\right)}{(\delta -\rho )}=\frac{\underset{\_}{\delta }\underset{\_}{\rho }{\acute{\delta }\acute{\rho }\left(-q\right)}^v\left({\acute{\delta }\acute{\rho }\delta }^{u-v}-{\acute{\rho }\acute{\delta }\rho }^{u-v}\right)}{\sqrt{p^2+4q}}$. □

Now, we give some identities about summations of terms in the bicomplex $\left(p,q\right)$-Fibonacci quaternions.

Theorem 15.

For $k,l$ as natural numbers, the summation formula of bicomplex $\left(p,q\right)$-Fibonacci quaternions is

$${\sum }_{k=1}^l{BCQF}_k=\left\{\begin{array}{c}\frac{{BCQF}_{l+1}+q{BCQF}_l-{BCQF}_1-q{BCQF}_0}{p+q-1} p+q\ne 1 \\ \frac{{qBCQF}_l+{BCQF}_1+\left(l-1\right)(1+e_1+e_2+e_3)\left(1+i+j+ij\right)}{1+q} p+q=1\end{array}\right..$$

Proof. 

Firstly, we assume that $p+q\ne 1.$ In this situation, $\sum _{k=1}^l{BCQF}_k=\sum _{k=1}^l{QF}_k+i\sum _{k=1}^l{QF}_{k+1}+j\sum _{k=1}^l{QF}_{k+2}+ij\sum _{k=1}^l{QF}_{k+3}$. In addition, we obtain that by using (13) in [24], $\sum _{k=1}^l{QF}_k=\frac{{QF}_{l+1}+q{QF}_l-{QF}_1-qQ{F}_0}{\begin{array}{c}p+q-1\end{array}}$. With simple calculations, we get

$${\sum }_{k=1}^l{QF}_{k+1}=p{\sum }_{k=1}^l{QF}_k+q{\sum }_{k=1}^l{QF}_{k-1}=\frac{Q{F}_{l+2}+q{QF}_{l+1}-Q{F}_2-q{QF}_1}{p+q-1},$$

$${\sum }_{k=1}^l{QF}_{k+2}=\frac{Q{F}_{l+3}+qQ{F}_{l+2}-{QF}_3-qQ{F}_2}{p+q-1},$$

$${\sum }_{k=1}^l{QF}_{k+3}=\frac{{QF}_{l+4}+qQ{F}_{l+3}-{QF}_4-qQ{F}_3}{p+q-1}.$$

Thus,

$$\begin{gathered} \sum _{k=1}^l{BCQF}_k=\frac{{QF}_{l+1}+q{QF}_l-{QF}_1-qQ{F}_0}{p+q-1}+i\left(\frac{Q{F}_{l+2}+q{QF}_{l+1}-Q{F}_2-q{QF}_1}{p+q-1}\right)+ \\ j\left(\frac{Q{F}_{l+3}+qQ{F}_{l+2}-{QF}_3-qQ{F}_2}{p+q-1}\right)+ij\left(\frac{{QF}_{l+4}+qQ{F}_{l+3}-{QF}_4-qQ{F}_3}{p+q-1}\right)=\frac{{BCQF}_{l+1}+q{BCQF}_l-{BCQF}_1-q{BCQF}_0}{p+q-1}. \end{gathered}$$

Now, we assume that $p+q=1$. We obtain that by using (13) in [24], ${\sum }_{k=1}^l{QF}_k=\frac{qQ{F}_l+(l-1)(1+e_1+e_2+e_3){+QF}_1}{1+q}$. Moreover, we have

$${\sum }_{k=1}^l{QF}_{k+1}=p{\sum }_{k=1}^l{QF}_k+q{\sum }_{k=1}^l{QF}_{k-1}=\frac{qQ{F}_{l+1}+(l-1)(1+e_1+e_2+e_3){+QF}_2}{1+q},$$

$${\sum }_{k=1}^l{QF}_{k+2}=\frac{qQ{F}_{l+2}+(l-1)(1+e_1+e_2+e_3){+QF}_3}{1+q},$$

$${\sum }_{k=1}^l{QF}_{k+3}=\frac{qQ{F}_{l+3}+(l-1)(1+e_1+e_2+e_3){+QF}_4}{1+q}.$$

So we can write

$${\sum }_{k=1}^l{BCQF}_k={\sum }_{k=1}^l{QF}_k+i{\sum }_{k=1}^l{QF}_{k+1}+j{\sum }_{k=1}^l{QF}_{k+2}+ij{\sum }_{k=1}^l{QF}_{k+3}$$

$$\begin{gathered} {\sum }_{k=1}^l{BCQF}_k=\frac{qQ{F}_l+(l-1)(1+e_1+e_2+e_3){+QF}_1}{1+q}+i\left(\frac{qQ{F}_{l+1}+(l-1)(1+e_1+e_2+e_3){+QF}_2}{1+q}\right)+j\left(\frac{qQ{F}_{l+2}+(l-1)(1+e_1+e_2+e_3){+QF}_3}{1+q}\right)+ij\left(\frac{qQ{F}_{l+3}+\left(l-1\right)\left(1+e_1+e_2+e_3\right){+QF}_4}{1+q}\right). \\ =\frac{{qBCQF}_l+\left(l-1\right)(1+e_1+e_2+e_3)\left(1+i+j+ij\right)+{BCQF}_1}{1+q}. \end{gathered}$$

□

Theorem 16.

For $u,v\ge 0$,

$${BCQF}_{uv}={\sum }_{k=0}^v\left(\begin{array}{c}v\\ k\end{array}\right){\left(q\right)}^{v-k}{F}_u^k{F}_{u-1}^k{BCQF}_k.$$

Proof. 

Using (15), ${BF}_{uv}=\sum _{k=0}^v\left(\begin{array}{c}v\\ k\end{array}\right){\left(q\right)}^{v-k}{F}_u^k{F}_{u-1}^{v-k}{BF}_k$. And, using (17),

$$\begin{array}{cc}{BCQF}_{uv}=& {BF}_{uv}+B{F}_{uv+1}{e}_1+B{F}_{uv+2}{e}_2+B{F}_{uv+3}{e}_3\\ & =\sum _{k=0}^v\left(\begin{array}{c}v\\ k\end{array}\right){\left(q\right)}^{v-k}{F}_u^k{F}_{u-1}^{v-k}{BF}_k+\sum _{k=0}^v\left(\begin{array}{c}v\\ k\end{array}\right){\left(q\right)}^{v-k}{F}_u^k{F}_{u-1}^{v-k}{BF}_{k+1}{e}_1\\ & +\sum _{k=0}^v\left(\begin{array}{c}v\\ k\end{array}\right){\left(q\right)}^{v-k}{F}_u^k{F}_{u-1}^{v-k}{BF}_{k+2}{e}_2+\sum _{k=0}^v\left(\begin{array}{c}v\\ k\end{array}\right){\left(q\right)}^{v-k}{F}_u^k{F}_{u-1}^{v-k}{BF}_{k+3}{e}_3\\ & ={\displaystyle \sum _{k=0}^v}\left(\begin{array}{c}v\\ k\end{array}\right){\left(q\right)}^{v-k}{F}_u^k{F}_{u-1}^{v-k}{(B{F}_k+BF}_{k+1}{e}_1+B{F}_{k+2}{e}_2+B{F}_{k+3}{e}_3)\\ & ={\displaystyle \sum _{k=0}^v}\left(\begin{array}{c}v\\ k\end{array}\right){\left(q\right)}^{v-k}{F}_u^k{F}_{u-1}^k{BCQF}_k.\end{array}$$

□

4. Matrix Representation of Bicomplex $\left(\mathit{p},\mathit{q}\right)$-Fibonacci Quaternions and an Application in This Representation for Them

Firstly, we will use the matrix that generates ${\{F}_u(p,q)\}$, which we define to obtain the N-matrix, which is similar to the definition of the S-matrix defined in [11]. We know that

$$\left(\begin{array}{c}{F}_{u+1}\\ F_u\end{array}\right)={\left(\begin{array}{cc}p& q\\ 1& 0\end{array}\right)}^u\left(\begin{array}{c}{F}_1\\ F_0\end{array}\right).$$

(21)

By using (21), the N-matrix is defined as

$$N^u={\left(\begin{array}{cc}p& q\\ 1& 0\end{array}\right)}^u=\left(\begin{array}{cc}{F}_{u+1}& q{F}_u\\ F_u& q{F}_{\begin{array}{c}u-1\end{array}}\end{array}\right),$$

(22)

where $F_{-2}=\frac{-p}{{\begin{array}{c}q\end{array}}^2}$ and $F_{-1}=\frac{1}{q}$.

Here, we will define the ${BCQF}_N$-matrix that we call the bicomplex $\left(p,q\right)$-Fibonacci quaternion matrix as follows:

$${BCQF}_N=\left(\begin{array}{cc}{BCQF}_3& q{BCQF}_2\\ BCQ{F}_2& qBCQ{F}_{\begin{array}{c}1\end{array}}\end{array}\right).$$

Now, we can give the following theorem about the ${BCQF}_N$-matrix:

Theorem 17.

If $B{CQF}_u$ is the $uth$ bicomplex $\left(p,q\right)$-Fibonacci quaternion, then, for $u\ge 0$,

$${BCQF}_N. {\left(\begin{array}{cc}p& q\\ 1& 0\end{array}\right)}^u=\left(\begin{array}{cc}{BCQF}_{u+3}& q{BCQF}_{u+2}\\ {BCQF}_{u+2}& q{BCQF}_{\begin{array}{c}u+1\end{array}}\end{array}\right).$$

(23)

Proof. 

To do this, we apply induction on $u$. If $u=0$, it is clear that (23) holds. Now, if we suppose that (23) holds for $u=v$, then ${BCQF}_N.{\left(\begin{array}{cc}p& q\\ 1& 0\end{array}\right)}^v=\left(\begin{array}{cc}{BCQF}_{v+3}& q{BCQF}_{v+2}\\ BCQ{F}_{v+2}& qBCQ{F}_{v+1}\end{array}\right)$.

Using Equation (18), for $v\ge 0$, ${BCQF}_{v+2}=p{BCQF}_{v+2}+{qBCQF}_v$. Then, by induction,

$$\begin{array}{lll}{BCQF}_N.N^{v+1}& ={(BCQF}_N.N^v)N& \\ & =\left(\begin{array}{cc}{BCQF}_{v+3}& q{BCQF}_{v+2}\\ BCQ{F}_{v+2}& qBCQ{F}_{\begin{array}{c}v+1\end{array}}\end{array}\right).\left(\begin{array}{cc}p& q\\ 1& 0\end{array}\right)& =\left(\begin{array}{cc}pB{CQF}_{v+3}+qB{CQF}_{v+2}& q{BCQF}_{v+3}\\ pB{CQF}_{v+2}+qB{CQF}_{v+1}& qBCQ{F}_{v+2}\end{array}\right)\\ & & =\left(\begin{array}{cc}{BCQF}_{v+4}& q{BCQF}_{v+3}\\ BCQ{F}_{v+3}& qBCQ{F}_{\begin{array}{c}v+2\end{array}}\end{array}\right).\end{array}$$

Thus, Equation (23) holds for all $u\ge 0$.$\square$

Corollary 3.

For $u\ge 0$,

$${BCQF}_{u+2}={BCQF}_2{F}_{u+1}+\left(qB{CQF}_1\right)F_u.$$

Proof. 

The proof can be easily seen by the coefficient (2, 1) of the matrix ${BCQF}_N.N^u$ and (22). □

Theorem 18.

For $u\ge 1$, $u$ is an integer, and   $v\in \left\{\mathrm{0,1}\right\}$. Then

$$\left(\begin{array}{cc}{BCQF}_{2u+v}& {BCQF}_{2(u-1)+v}\\ {BCQF}_{2(u+1)+v}& {BCQF}_{2u+v}\end{array}\right)=\left(\begin{array}{cc}{BCQF}_{2+v}& {BCQF}_v\\ {BCQF}_{4+v}& {BCQF}_{2+v}\end{array}\right){\left(\begin{array}{cc}{p}^2+2q& 1\\ -q^2& 0\end{array}\right)}^{u-1}.$$

Proof. 

We prove the theorem by induction on $u$. If $u=1$, then the result is clear. Now we assume that, for any integer $d$, such as $1\le d\le u$,

$$\left(\begin{array}{cc}{BCQF}_{2d+v}& {BCQF}_{2(d-1)+v}\\ {BCQF}_{2(d+1)+v}& {BCQF}_{2d+v}\end{array}\right)=\left(\begin{array}{cc}{BCQF}_{2+v}& {BCQF}_v\\ {BCQF}_{4+v}& {BCQF}_{2+v}\end{array}\right){\left(\begin{array}{cc}{p}^2+2q& 1\\ -q^2& 0\end{array}\right)}^{d-1}.$$

Then for $u=d+1$, we obtain

$$\begin{gathered} \left(\begin{array}{cc}{BCQF}_{2+v}& {BF}_v\\ {BCQF}_{4+v}& {BF}_{2+v}\end{array}\right){\left(\begin{array}{cc}{p}^2+2q& 1\\ -q^2& 0\end{array}\right)}^d=\left(\begin{array}{cc}{BF}_{2+v}& {BF}_v\\ {BF}_{4+v}& {BF}_{2+v}\end{array}\right){\left(\begin{array}{cc}{p}^2+2q& 1\\ -q^2& 0\end{array}\right)}^{d-1}\left(\begin{array}{cc}{p}^2+2q& 1\\ -q^2& 0\end{array}\right) \\ =\left(\begin{array}{cc}{BF}_{2d+v}& {BF}_{2(d-1)+v}\\ {BF}_{2(d+1)+v}& {BF}_{2d+v}\end{array}\right)\left(\begin{array}{cc}{p}^2+2q& 1\\ -q^2& 0\end{array}\right)=\left(\begin{array}{cc}{BF}_{2(d+1)+v}& {BF}_{2d+v}\\ {BF}_{2(d+2)+v}& {BF}_{2(d+1)+v}\end{array}\right). \end{gathered}$$

where $v\in \left\{\mathrm{0,1}\right\}$. Therefore, the proof is completed. □

In [8], Halıcı obtained the complex Fibonacci quaternions shown by the 8 × 8 real matrices. Motivated by this paper, we get the matrix form of a bicomplex $\left(p,q\right)$-Fibonacci quaternion ${BCQF}_u$ with the aid of 4 × 4 matrix representations and produce a new 8 × 8 type bicomplex quaternion matrix similarly. We can define the following matrices:

$$E_0=\left(\begin{array}{cc}\alpha & 0\\ 0& \alpha \end{array}\right),E_1=\left(\begin{array}{cc}i\Gamma & 0\\ 0& -i\Gamma \end{array}\right),{ E}_2=\left(\begin{array}{cc}0& \alpha \\ -\alpha & 0\end{array}\right), E_3=\left(\begin{array}{cc}0& j\zeta \\ j\zeta & 0\end{array}\right)$$

where

$$\alpha \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right), \Gamma =\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right), \zeta =\left(\begin{array}{cc}0& -j\\ j& 0\end{array}\right),\mathrm{and} i^2=-1, j^2=-1,{\left(ij\right)}^2=1,ij=ji.$$

Using the matrices $\alpha ,$ $\Gamma$, and $\zeta$, we obtain ${ E_0}^2=I_4$ and ${ E_1}^2={ E_2}^2={ E_3}^2=-I_4$, where $I_4$ is the 4 × 4 identity matrix. Furthermore, it satisfies the following equations:

$$E_1{E}_2=-E_2{E}_1=E_3,E_3{E}_1=-E_1{E}_3=E_2, E_2{E}_3=-E_3{E}_2=E_1{,E}_1{E}_2{E}_3=-I_4.$$

(24)

The bicomplex $\left(p,q\right)$-Fibonacci quaternion $BCQF$ is also expressed by the 4 × 4 matrix with these new matrices. By using the bicomplex $\left(p,q\right)$-Fibonacci number, we can write

$$\begin{gathered} {BCQF}_u={BF}_u{E}_0+{BF}_{u+1} E_1+{BF}_{u+2}{E}_2+{BF}_{u+3} E_3 BCQF= \\ \left(\begin{array}{c}\begin{array}{cc}\begin{array}{c}\begin{array}{c}{BF}_u\\ {-BF}_{u+1}\end{array}\\ {-BF}_{u+2}\end{array}& \begin{array}{c}\begin{array}{c}{BF}_{u+1} \\ {BF}_u\end{array}\\ {BF}_{u+3}\end{array}\\ -{BF}_{u+3}& -{BF}_{u+2}\end{array}\begin{array}{cc}\begin{array}{c}\begin{array}{c}{BF}_{u+2}\\ {-BF}_{u+3}\end{array}\\ {BF}_u\end{array}& \begin{array}{c}\begin{array}{c}{BF}_{u+3}\\ {BF}_{u+2}\end{array}\\ {-BF}_{u+1}\end{array}\\ { BF}_{u+1}& {BF}_u\end{array}\end{array}\right). \end{gathered}$$

Theorem 19.

For $u\ge 0$, the $uth$ term of the bicomplex $\left(p,q\right)$-Fibonacci quaternion sequence with the determinant of a special matrix can be obtained as follows:

$${{BCQF}_u=\left|\begin{array}{lllllllll}{BCQF}_0& -1& 0& 0& 0& \dots & 0& 0& 0\\ {BCQF}_1& 0& -1& 0& 0& \dots & 0& 0& 0\\ {BCQF}_2& 0& 0& -1& 0& \dots & 0& 0& 0\\ 0& 0& q& p& -1& \dots & 0& 0& 0\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ 0& 0& 0& 0& 0& \ddots & -1& 0& 0\\ 0& 0& 0& 0& 0& \ddots & p& -1& 0\\ 0& 0& 0& 0& 0& \ddots & q& p& -1\\ 0& 0& 0& 0& 0& \ddots & 0& q& p\end{array}\right|}_{\begin{array}{c}\left(u+1\right)\times \left(u+1\right)\end{array}}$$

Proof. 

For the proof, we use the induction method on $u$. It is clear that equality holds for $u=0,$ and $1$. Now suppose that equality is true for $2\le k\le u$. Then, we can verify it for $u+1$ as follows:

$$\begin{gathered} {{BCQF}_{u+1}=\left|\begin{array}{llllllllll}{BCQF}_0& -1& 0& 0& 0& \cdots & 0& 0& 0& 0\\ {BCQF}_1& 0& -1& 0& 0& \cdots & 0& 0& 0& 0\\ {BCQF}_2& 0& 0& -1& 0& \cdots & 0& 0& 0& 0\\ 0& 0& q& p& -1& \cdots & 0& 0& 0& 0\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ 0& 0& 0& 0& 1& \ddots & -1& 0& 0& 0\\ 0& 0& 0& 0& 0& \ddots & p& -1& 0& 0\\ 0& 0& 0& 0& 0& \ddots & q& p& -1& 0\\ 0& 0& 0& 0& 0& \ddots & 0& q& p& -1\\ 0& 0& 0& 0& 0& \ddots & 0& 0& q& p\end{array}\right|}_{\begin{array}{c}\left(u+2\right)\times \left(u+2\right)\end{array}} \\ =p{\left(-1\right)}^{2u+4}{\left|\begin{array}{lllllllll}{BCQF}_0& -1& 0& 0& 0& \cdots & 0& 0& 0\\ {BCQF}_1& 0& -1& 0& 0& \cdots & 0& 0& 0\\ {BCQF}_2& 0& 0& -1& 0& \cdots & 0& 0& 0\\ 0& 0& q& p& -1& \cdots & 0& 0& 0\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ 0& 0& 0& 0& 0& \ddots & -1& 0& 0\\ 0& 0& 0& 0& 0& \ddots & p& -1& 0\\ 0& 0& 0& 0& 0& \ddots & q& p& -1\\ 0& 0& 0& 0& 0& \ddots & 0& q& p\end{array}\right|}_{\begin{array}{c}\left(u+1\right)\times \left(u+1\right)\end{array}} \\ +(-1){(-1)}^{2u+3}{\left|\begin{array}{lllllllll}{BCQF}_0& -1& 0& 0& 0& \cdots & 0& 0& 0\\ {BCQF}_1& 0& -1& 0& 0& \cdots & 0& 0& 0\\ {BCQF}_2& 0& 0& -1& 0& \cdots & 0& 0& 0\\ 0& 0& q& p& -1& 0& 0& 0& 0\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ 0& 0& 0& 0& 0& \ddots & -1& 0& 0\\ 0& 0& 0& 0& 0& \ddots & p& -1& 0\\ 0& 0& 0& 0& 0& \ddots & q& p& -1\\ 0& 0& 0& 0& 0& \ddots & 0& 0& q\end{array}\right|}_{\begin{array}{c}\left(u+1\right)\times \left(u+1\right)\end{array}} \\ =p{BCQF}_u+q{\left(-1\right)}^{2u+2}{\left|\begin{array}{llllllll}{BCQF}_0& -1& 0& 0& 0& \cdots & 0& 0\\ {BCQF}_1& 0& -1& 0& 0& \cdots & 0& 0\\ {BCQF}_2& 0& 0& -1& 0& \cdots & 0& 0\\ 0& 0& q& p& -1& \cdots & 0& 0\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ 0& 0& 0& 0& 0& \ddots & -1& 0\\ 0& 0& 0& 0& 0& \ddots & p& -1\\ 0& 0& 0& 0& 0& \cdots & q& p\end{array}\right|}_{\begin{array}{c}u\times u\end{array}} \\ =p{BCQF}_u+q{BCQF}_{u-1} \end{gathered}$$

Thus, the proof is completed.$\square$

5. Conclusions

Here, we describe the bicomplex $\left(p,q\right)$-Fibonacci numbers and the bicomplex $\left(p,q\right)$-Fibonacci quaternions based on these numbers to show that bicomplex numbers are not defined as bicomplex quaternions. Furthermore, we obtain some of their equations, including the Binet formula, generating function, Catalan, Cassini, and d’Ocagne’s identities, and summation formulas for both.

In addition, we describe a matrix that we call an N-matrix of type 4 × 4 for bicomplex $\left(p,q\right)$-Fibonacci quaternions whose terms are bicomplex $\left(p,q\right)$-Fibonacci numbers. Then, we show that the bicomplex $\left(p,q\right)$-Fibonacci quaternions can be expressed as 8 × 8 real matrices. With the help of the new four matrices we define in 4 × 4 type, we obtain {${E_0, E}_1,E_2, E_3\}$, which is used as the basic elements of real quaternions $\{{1, e}_1,e_2, e_3\}$. Also, we show that the bicomplex $\left(p,q\right)$-Fibonacci quaternion can be expressed with a new matrix of type 4 × 4, whose elements consist of bicomplex $\left(p,q\right)$-Fibonacci numbers. Finally, we create a matrix for bicomplex $\left(p,q\right)$-Fibonacci quaternions, and we obtain the determinant of a special matrix that gives the terms of that quaternion.

Consequently, in this study, we present a general form of the second-order bicomplex number sequences and the second-order bicomplex quaternions. We show that these two concepts, defined as the same in many studies, are different.