Powers of Hybrid Numbers: A Unified Algebraic and Combinatorial Framework

1. Introduction

The study of hybrid numbers and hybrid rings has recently gained attention through the pioneering work of Özdemir [10]. His construction introduces a noncommutative multiplication law that extends classical number systems and sequence structures. The hybrid ring combines elements of commutative algebra with noncommutative behaviour, producing a rich framework for generalisations of recurrence relations and matrix theory.

Hybrid rings arise naturally when extending sequences, such as Fibonacci-type or Leonardo-type numbers, into algebraic structures with multiple components. Özdemir’s approach defines a multiplication that is associative but not commutative, thereby opening new directions in noncommutative algebra. This multiplication is equivalent to his original formulation, yet it admits alternative representations that facilitate applications in linear algebra and number theory.

The motivation for studying these rings lies in their potential to unify diverse algebraic systems under a single framework. This paper aims to refine the definitions, establish foundational properties, and illustrate applications with explicit examples. Specifically, this paper resolves the computational complexity of hybrid exponentiation by establishing four mathematically equivalent frameworks: a recursive sequence approach, a Lucas-equivalence formulation, a structural root expansion, and a purely real combinatorial method.

1.1. Hybrid Numbers

Since the late nineteenth century, two-dimensional number systems have garnered continuous attention from the mathematical community. By that period, complex numbers were rigorously formalized, whereas hyperbolic (split-complex) and dual numbers were introduced as notable generalizations. Consequently, these algebraic structures emerged as pivotal subjects of research, yielding a fertile framework for algebraic inquiry and motivating subsequent generalizations throughout the twentieth century.

The paradigm of hybrid numbers constitutes a unified algebraic architecture that subsumes complex, hyperbolic, and dual numbers. Within this comprehensive framework, each classical number system manifests as a distinct special case, even as the hybrid structure facilitates the formulation of mixed and composite states. This concept was initially proposed by Özdemir [10] and subsequently expanded upon in [11]. The fundamental set of hybrid numbers is formally designated by

$$\mathbb{K}=\{a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}:a,b,c,d\in \mathbb{R},{\mathbf{i}}^2=-1,{\mathbf{\epsilon }}^2=0,{\mathbf{h}}^2=1,\mathbf{ih}=-\mathbf{hi}=\mathbf{\epsilon }+\mathbf{i}\}.$$

Each element of $\mathbb{K}$ can be represented as a quadruple, with the fundamental hybrid units given by

$$1\leftrightarrow (1,0,0,0),i\leftrightarrow (0,1,0,0),\epsilon \leftrightarrow (0,0,1,0),h\leftrightarrow (0,0,0,1).$$

Equality between two hybrid numbers is satisfied if and only if their respective four components are strictly identical. The operation of addition is executed component-wise, wherein the zero element serves as the additive identity, and the corresponding additive inverse is derived through the negation of every component. The multiplicative operation, designated as the Hybridian product, is governed by the axiomatic rules detailed in Table 1.

The evaluation of the hybrid product is accomplished by executing a rightward distribution of terms while strictly maintaining the non-commutative multiplication order of the basis units. Subsequently, the resulting unit products are reduced by applying the fundamental defining relations ${\mathit{i}}^2=-1,$${\mathit{\epsilon }}^2=0,$${\mathbf{h}}^2=1,$i$\mathbf{h}=-\mathbf{h}$i=$\epsilon$+i.

For an arbitrary hybrid number $\mathbf{Z}=a+b$i$+c$$\epsilon$$+d\mathbf{h}$,, a number of core definitions are established (refer to Özdemir [10]):

• The scalar part is defined as $S\left(\mathbf{Z}\right)=a$, vector part as $V\left(\mathbf{Z}\right)=b$i$+c$$\epsilon$$+d\mathbf{h}$, conjugate as $\overline{\mathbf{Z}}=a-b$i$-c$$\epsilon$$-d\mathbf{h}$., and the characteristic number is given by

  $$\begin{array}{ccc}\hfill C\left(\mathbf{Z}\right)& =& \mathbf{Z}\overline{\mathbf{Z}}\hfill \\ & =& a^2+{(b-c)}^2-c^2-d^2\hfill \\ & =& a^2+b^2-2bc-d^2.\hfill \end{array}$$
• The type number (vector discriminant) is formulated as

  $$\Delta \left(\mathbf{Z}\right)={\left(V\left(\mathbf{Z}\right)\right)}^2=-{(b-c)}^2+c^2+d^2=-b^2+2bc+d^2.$$
  So $\Delta \left(\mathbf{Z}\right)=a^2-C\left(\mathbf{Z}\right),$ i.e., ${\left(V\left(\mathbf{Z}\right)\right)}^2=a^2-C\left(\mathbf{Z}\right).$
• Classification criteria:

  $$\mathbf{Z}\mathrm{is}\mathrm{elliptic}\mathrm{if}\Delta \left(\mathbf{Z}\right)<0,\mathbf{Z}\mathrm{is}\mathrm{hyperbolic}\mathrm{if}\Delta \left(\mathbf{Z}\right)>0,\mathbf{Z}\mathrm{is}\mathrm{parabolic}\mathrm{if}\Delta \left(\mathbf{Z}\right)=0.$$
• The norm is articulated as$\parallel \mathbf{Z}\parallel =\sqrt{\left|C\right(\mathbf{Z}\left)\right|},$ and the multiplicative inverse is

  $${\mathbf{Z}}^{-1}={\displaystyle \frac{\overline{\mathbf{Z}}}{C\left(\mathbf{Z}\right)}},$$

  strictly under the condition that $\parallel \mathbf{Z}\parallel \ne 0.$

In summary, the set of hybrid numbers, $\mathbb{K}$, possesses the ensuing algebraic properties:

• The multiplication operation in $\mathbb{K}$ is associative and not commutative.
• The characteristic number preserves multiplicativity, such that $\mathcal{C}($${\mathit{Z}}_1$${\mathit{Z}}_2)=\mathcal{C}($${\mathit{Z}}_1\left)\mathcal{C}\right($${\mathit{Z}}_2)$ for ${\mathit{Z}}_1,$${\mathit{Z}}_2\in \mathbb{K}.$
• The additive identity is denoted by $0=0+0$i$+0$$\epsilon$$+0\mathbf{h}\in \mathbb{K}$ whereas the multiplicative identity is represented by $1=1+0$i$+0$$\epsilon$$+0\mathbf{h}\in \mathbb{K}$.
• $(\mathbb{K},+)$ is an Abelian group.
• $\mathbb{K}$ is a non-commutative ring with respect to the addition and multiplication operations.

The typological classification of hybrid numbers naturally arises from the evaluation of the determinant and trace associated with their matrix representations. Comprehensive elaborations on this subject can be found in the literature [3,10,11].

Given that the inherent multiplication table poses memorization challenges and non-commutativity introduces computational complexity, adopting a matrix representation provides a highly efficacious alternative. To this end, Özdemir [10] articulated the subsequent isomorphic relationship.

Theorem 1

(Özdemir [10]). The hybrid number ring $\mathbb{K}$ shares an algebraic isomorphism with the ring of real $2\times 2$ matrices, $M_{2\times 2}\left(\mathbb{R}\right)$, defined through the mapping

$${\phi }_{\mathbf{Z}}(a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h})=\left(\begin{array}{cc}a+c& b-c+d\\ c-b+d& a-c\end{array}\right).$$

The resulting matrix ${\phi }_{\mathbf{Z}}\left(\mathbf{Z}\right)$ is formally recognized as the hybrid matrix corresponding to $\mathbf{Z}.$ ${\phi }_{\mathbf{Z}}\left(\mathbf{Z}\right)$ is also denoted as $M\left(Z\right)$. Furthermore, the inverse transformation is given as

$${\phi }_{\mathbf{Z}}^{-1}\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\frac{a+d}{2}+\frac{a+b-c-d}{2}\mathbf{i}+\frac{a-d}{2}\mathbf{\epsilon }+\frac{b+c}{2}\mathbf{h}.$$

Consequently, the foundational hybrid basis units are mapped to the ensuing matrix representations:

$${\phi }_{\mathbf{Z}}\left(1\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),{\phi }_{\mathbf{Z}}\left(i\right)=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right),{\phi }_{\mathbf{Z}}\left(\epsilon \right)=\left(\begin{array}{cc}1& -1\\ 1& -1\end{array}\right),{\phi }_{\mathbf{Z}}\left(h\right)=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right).$$

By leveraging these matrix equivalents, the multiplication of hybrid numbers is systematically reduced to standard matrix multiplication, thereby significantly streamlining advanced algebraic computations.

Although the foundational definition of the hybrid ring restricts its internal components strictly to real scalars, the algebraic methodologies developed in this paper possess a much broader mathematical applicability. To fully support the exponentiation of polynomial sequences and complex variables, we formalize the following structural extension.

Remark 1

(Functional and Complex Extension of Hybrid Components). While the fundamental set of hybrid numbers $\mathbb{K}$ is formally defined with scalar components $a,b,c,d\in \mathbb{R}$, throughout this overarching framework we assume that these components can equally be arbitrary real-valued or complex-valued functions of an independent variable x (i.e., $a\left(x\right),b\left(x\right),c\left(x\right),d\left(x\right)$). Because the ring of such functions is commutative and interacts with the non-commutative hybrid basis units exactly as $\mathbb{R}$ does, all algebraic reductions, geometric characteristic equations, and combinatorial exponentiation identities established in this work remain strictly true and mathematically valid.

Furthermore, when these functions evaluate to complex values, the algebraic structure seamlessly adapts based on the topological boundary of the imaginary units: it either expands into the 8-dimensional tensor space $\mathbb{C}\otimes \mathbb{K}$ when the complex unit I and hybrid unit $\mathbf{i}$ are distinct ($I\ne \mathbf{i}$), or it undergoes dimensional collapse when they are explicitly equated ($I=\mathbf{i}$).

In this extended context, the hybrid number operates effectively as a "hybrid function," enabling the seamless exponentiation of sequence polynomials and complex characteristic roots across all topological boundary conditions.

1.2. Standard Multiplication and the Left Regular Matrix Representation

To establish a rigorous computational framework for evaluating powers and sequences, we must first formalize the non-commutative multiplication of two distinct hybrid numbers.

Definition 2.

Let ${\mathbf{Z}}_1=a_1+b_1\mathbf{i}+c_1\mathbf{\epsilon }+d_1\mathbf{h}$ and ${\mathbf{Z}}_2=a_2+b_2\mathbf{i}+c_2\mathbf{\epsilon }+d_2\mathbf{h}$ be two hybrid numbers. Derived directly from the fundamental basis multiplication rules of the hybrid algebra, their non-commutative product ${\mathbf{Z}}_1{\mathbf{Z}}_2$ (also denoted as ${\mathbf{Z}}_1·{\mathbf{Z}}_2$ ) is explicitly defined as:

$$\begin{array}{cc}\hfill {\mathbf{Z}}_1{\mathbf{Z}}_2& =(a_1{a}_2-b_1{b}_2+b_1{c}_2+c_1{b}_2+d_1{d}_2)\hfill \end{array}$$

$$\begin{array}{cc}\hfill & +(a_1{b}_2+b_1{a}_2+b_1{d}_2-d_1{b}_2)\mathbf{i}\hfill \end{array}$$

(1.1)

$$\begin{array}{cc}\hfill & +(a_1{c}_2+c_1{a}_2+b_1{d}_2-d_1{b}_2+d_1{c}_2-c_1{d}_2)\mathbf{\epsilon }\hfill \end{array}$$

(1.2)

$$\begin{array}{cc}\hfill & +(a_1{d}_2+d_1{a}_2-b_1{c}_2+c_1{b}_2)\mathbf{h}.\hfill \end{array}$$

(1.3)

While this explicit algebraic formula provides the exact components of the product, manually calculating it is tedious and highly susceptible to sign and permutation errors. To circumvent these inherent algebraic vulnerabilities, we utilize the left regular matrix representation of the hybrid algebra.

This approach establishes a robust isomorphism between the non-commutative ring of hybrid numbers, denoted $\mathbb{K}$, and a specific subalgebra of $4\times 4$ real matrices, ${\mathcal{M}}_4\left(\mathbb{R}\right)$. By mapping hybrid numbers to these transformation matrices, we translate error-prone symbolic multiplication into standard, associative linear algebra operations.

Let $\mathbf{Z}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$ be a hybrid number. We establish a bijective mapping between $\mathbf{Z}$ and a standard real column vector ${\mathbf{v}}_{\mathbf{Z}}\in {\mathbb{R}}^4$:

$$\mathbf{Z}\mapsto {\mathbf{v}}_{\mathbf{Z}}=\left(\begin{array}{c}a\\ b\\ c\\ d\end{array}\right)={(a,b,c,d)}^T$$

By treating the units $\{1,\mathbf{i},\mathbf{\epsilon },\mathbf{h}\}$ as the standard basis, we define four fundamental $4\times 4$ transformation matrices. Each matrix corresponds to the linear operation of left-multiplication by a specific base unit across the entire basis, derived directly from the hybrid multiplication table (Öztürk and Özdemir [12]):

$$M_1=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right),M_i=\left(\begin{array}{cccc}0& -1& 1& 0\\ 1& 0& 0& 1\\ 0& 0& 0& 1\\ 0& 0& -1& 0\end{array}\right),$$

$$M_{\epsilon }=\left(\begin{array}{cccc}0& 1& 0& 0\\ 0& 0& 0& 0\\ 1& 0& 0& -1\\ 0& 1& 0& 0\end{array}\right),M_h=\left(\begin{array}{cccc}0& 0& 0& 1\\ 0& -1& 0& 0\\ 0& -1& 1& 0\\ 1& 0& 0& 0\end{array}\right).$$

Due to the linearity of the algebra, the left-multiplication of any arbitrary hybrid number $\mathbf{Z}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$ can be represented by a unique matrix operator $M_{\mathbf{Z}}$. This operator is precisely the linear combination of the fundamental basis matrices:

$$M_{\mathbf{Z}}=a{M}_1+b{M}_{\mathbf{i}}+c{M}_{\mathbf{\epsilon }}+d{M}_{\mathbf{h}}$$

Expanding this linear combination yields the explicit general transformation matrix for any hybrid number. For $\mathbf{Z}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$, its corresponding left-multiplication matrix $L\left(\mathbf{Z}\right)$ is constructed as follows (Öztürk and Özdemir [12]):

$$L\left(\mathbf{Z}\right)=M_{\mathbf{Z}}=\left(\begin{array}{cccc}a& c-b& b& d\\ b& a-d& 0& b\\ c& -d& a+d& b-c\\ d& c& -b& a\end{array}\right).$$

(1.4)

Because the mapping $\mathbf{Z}\mapsto M_{\mathbf{Z}}$ is a true algebra isomorphism, the multiplication of two hybrid numbers X and Y behaves identically to the multiplication of their corresponding real matrices. We define two primary computational methods for evaluating the product $XY$:

1.

Matrix-Vector Product: To find the explicit coefficients of the resulting hybrid number, apply the transformation matrix of the left operand, $M_X$, to the column vector of the right operand, ${\mathbf{v}}_Y$:

$${\mathbf{v}}_{XY}=M_X{\mathbf{v}}_Y$$

2.

Matrix-Matrix Product: To generate the transformation matrix of the combined product (which is especially useful for chained multiplications of three or more elements), multiply the respective operator matrices:

$$M_{XY}=M_X{M}_Y$$

Note: The first column of the resulting combined matrix$M_{XY}$inherently yields the coefficient vector${\mathbf{v}}_{XY}$of the desired product.

Remark 2

(Demonstration of the Matrix Isomorphism). To explicitly verify that this matrix formulation seamlessly recovers Definition 2, consider the matrix-vector product $L\left({\mathbf{Z}}_1\right){\mathbf{v}}_2$:

$$L\left({\mathbf{Z}}_1\right){\mathbf{v}}_2=\left(\begin{array}{cccc}{a}_1& c_1-b_1& b_1& d_1\\ b_1& a_1-d_1& 0& b_1\\ c_1& -d_1& a_1+d_1& b_1-c_1\\ d_1& c_1& -b_1& a_1\end{array}\right)\left(\begin{array}{c}{a}_2\\ b_2\\ c_2\\ d_2\end{array}\right)$$

Performing this standard matrix multiplication yields a $4\times 1$ column vector:

$${\mathbf{v}}_{result}=\left(\begin{array}{c}{a}_1{a}_2+(c_1-b_1)b_2+b_1{c}_2+d_1{d}_2\\ b_1{a}_2+(a_1-d_1)b_2+0·c_2+b_1{d}_2\\ c_1{a}_2-d_1{b}_2+(a_1+d_1)c_2+(b_1-c_1)d_2\\ d_1{a}_2+c_1{b}_2-b_1{c}_2+a_1{d}_2\end{array}\right)$$

By distributing the terms within each row, we directly isolate the coefficients for the basis elements $1,\mathbf{i},\mathbf{\epsilon }$, and $\mathbf{h}$:

$$\begin{array}{cc}\hfill \mathrm{Row}1\left(\mathit{Scalar}\right)\text{:}& a_1{a}_2-b_1{b}_2+c_1{b}_2+b_1{c}_2+d_1{d}_2\hfill \\ \hfill \mathrm{Row}2\left(\mathbf{i}\right):& a_1{b}_2+b_1{a}_2-d_1{b}_2+b_1{d}_2\hfill \\ \hfill \mathrm{Row}3\left(\mathbf{\epsilon }\right):& c_1{a}_2-d_1{b}_2+a_1{c}_2+d_1{c}_2+b_1{d}_2-c_1{d}_2\hfill \\ \hfill \mathrm{Row}4\left(\mathbf{h}\right):& d_1{a}_2+c_1{b}_2-b_1{c}_2+a_1{d}_2\hfill \end{array}$$

Translating this vector back into algebraic form perfectly reproduces Equation (1). By standardizing multiplication through this matrix representation, we computationally guarantee associativity and entirely eliminate the risk of manual errors.

Similarly, we can compute ${\mathbf{Z}}_2{\mathbf{Z}}_1.$ While swapping the indices is straightforward in theory, rearranging the terms to clearly see the sign changes takes time and is easy to slip up on. By writing it out explicitly, one can clearly show that the scalar part is entirely commutative, while specific cross-terms in the $\mathbf{i}$, $\mathbf{\epsilon }$, and $\mathbf{h}$ components experience sign reversals due to the non-commutative multiplication rules.

Here is the exact expansion for ${\mathbf{Z}}_2{\mathbf{Z}}_1$, arranged with the indices in standard $(1,2)$ order so the differences from ${\mathbf{Z}}_1{\mathbf{Z}}_2$ are immediately visible:

$$\begin{array}{cc}\hfill {\mathbf{Z}}_2{\mathbf{Z}}_1& =(a_1{a}_2-b_1{b}_2+b_1{c}_2+c_1{b}_2+d_1{d}_2)\hfill \\ \hfill & +(a_1{b}_2+b_1{a}_2-b_1{d}_2+d_1{b}_2)\mathbf{i}\hfill \\ \hfill & +(a_1{c}_2+c_1{a}_2-b_1{d}_2+d_1{b}_2-d_1{c}_2+c_1{d}_2)\mathbf{\epsilon }\hfill \\ \hfill & +(a_1{d}_2+d_1{a}_2+b_1{c}_2-c_1{b}_2)\mathbf{h}.\hfill \end{array}$$

(1.5)

Remark 3.

If we take $\mathbf{Z}={\mathbf{Z}}_1={\mathbf{Z}}_2=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$ in Definition 2, we obtain the closed-form expansion for the square of a single hybrid number:

$$\begin{array}{ccc}\hfill \mathbf{ZZ}& =& (aa-bb+bc+cb+dd)+(ab+ba+bd-db)\mathbf{i}\hfill \\ & & +(ac+ca+bd-db+dc-cd)\mathbf{\epsilon }+(ad+da-bc+cb)\mathbf{h}\hfill \end{array}$$

equivalently

$$\begin{array}{ccc}\hfill {\mathbf{Z}}^2& =& (d^2+a^2-b^2+2cb)+2ab\mathbf{i}+2ac\mathbf{\epsilon }+2da\mathbf{h}\hfill \\ & =& (d^2+a^2-b^2+2cb)+2a(b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h})\hfill \end{array}$$

(1.6)

1.3. Second-Order (Horadam) Polynomial Sequences and Binet Formulas

Before evaluating the powers and mixed products of hybrid recurrence sequences, it is necessary to establish the algebraic foundations of the underlying polynomial sequences. In this subsection, we introduce the Horadam polynomials, defining their recurrence relations, characteristic equations, and explicit Binet-style formulas.

These Binet formulas—and specifically the roots of their characteristic equations—will serve as the foundational scalar inputs for generating the higher-dimensional hybrid sequences in the subsequent sections.

The Horadam polynomials represent a vast generalization of second-order recurrence polynomials, encompassing the Fibonacci, Lucas, Pell, and Jacobsthal polynomials as special cases.

The generalized Fibonacci polynomials (or Horadam polynomials or generalized $\left(r\right(x),s(x\left)\right)$-polynomials or $\left(r\right(x),s(x\left)\right)$ Horadam polynomials)

$${\left\{W_n(W_0,W_1;r\left(x\right),s\left(x\right))\right\}}_{n\ge 0}$$

(or ${\left\{W_n\left(x\right)\right\}}_{n\ge 0}$ or shortly ${\left\{W_n\right\}}_{n\ge 0}$) is defined as follows:

$$W_n\left(x\right)=r\left(x\right)W_{n-1}\left(x\right)+s\left(x\right)W_{n-2}\left(x\right),W_0\left(x\right)=a\left(x\right),W_1\left(x\right)=b\left(x\right),n\ge 2$$

(1.7)

where $W_0\left(x\right),W_1\left(x\right)$are arbitrary complex (or real) polynomials with real coefficients and $r\left(x\right)$ and $s\left(x\right)$ are polynomials with real coefficients with $r\left(x\right)\ne 0,$$s\left(x\right)\ne 0.$

The sequence ${\left\{W_n\right\}}_{n\ge 0}$ can be extended to negative subscripts by defining

$$W_{-n}\left(x\right)=-{\displaystyle \frac{r\left(x\right)}{s\left(x\right)}}{W}_{-(n-1)}\left(x\right)+{\displaystyle \frac{1}{s\left(x\right)}}{W}_{-(n-2)}\left(x\right)$$

for $n=1,2,3,...$ when $s\left(x\right)\ne 0.$ Therefore, recurrence (7) holds for all integers $n.$ Note that $W_{-n}\left(x\right)$ need not to be a polynomial in the ordinary sense. For more information on Horadam polynomials see Soykan [15].

Binet’s formula of generalized Fibonacci (Horadam) polynomials can be calculated using its characteristic equation (the quadratic equation, polynomial) which is given as

$$y^2-r\left(x\right)y-s\left(x\right)=0.$$

(1.8)

The roots of characteristic equation are

$$\alpha \left(x\right):=\alpha ={\displaystyle \frac{r\left(x\right)+\sqrt{r^2\left(x\right)+4s\left(x\right)}}{2}},\beta \left(x\right):=\beta ={\displaystyle \frac{r\left(x\right)-\sqrt{r^2\left(x\right)+4s\left(x\right)}}{2}},$$

(1.9)

and the followings hold

$$\begin{array}{ccc}\hfill \alpha +\beta & =& r\left(x\right),\hfill \\ \hfill \alpha \beta & =& -s\left(x\right).\hfill \end{array}$$

If the roots $\alpha$ and $\beta$ of characteristic equation (1.8) are distinct, i.e., $\alpha \ne \beta$ then $r^2\left(x\right)+4s\left(x\right)\ne 0$ and if the roots $\alpha$ and $\beta$ of characteristic equation (1.8) are equal, i.e., $\alpha =\beta$ then (1.8) can be written as

$$y^2-r\left(x\right)y-s\left(x\right)={(y-\alpha )}^2=y^2-2\alpha y+{\alpha }^2=0$$

and, in this case,

$$\begin{array}{ccc}\hfill \alpha & =& {\displaystyle \frac{r\left(x\right)}{2}},\hfill \\ \hfill r\left(x\right)& =& 2\alpha ,\hfill \\ \hfill s\left(x\right)& =& -{\alpha }^2=-{\displaystyle \frac{r^2\left(x\right)}{4}},\hfill \\ \hfill r^2\left(x\right)+4s\left(x\right)& =& 0.\hfill \end{array}$$

Now, we define two special cases of the polynomials $W_n\left(x\right)$. $\left(r\right(x),s(x\left)\right)$-Fibonacci polynomials ${\left\{G_n(0,1;r\left(x\right),s\left(x\right))\right\}}_{n\ge 0}$ (or shortly $G_n\left(x\right)$) and $\left(r\right(x),s(x\left)\right)$-Lucas polynomials ${\left\{H_n(2,r\left(x\right);r\left(x\right),s\left(x\right))\right\}}_{n\ge 0}$ (or shortly $H_n\left(x\right)$) are defined, respectively, by the second-order recurrence relations

$$\begin{array}{ccc}\hfill G_{n+2}\left(x\right)& =& r\left(x\right)G_{n+1}+s\left(x\right)G_n\left(x\right),G_0\left(x\right)=0,G_1\left(x\right)=1,\hfill \end{array}$$

(1.10)

$$\begin{array}{ccc}\hfill H_{n+2}\left(x\right)& =& r\left(x\right)H_{n+1}+s\left(x\right)H_n\left(x\right),H_0\left(x\right)=2,H_1\left(x\right)=r\left(x\right).\hfill \end{array}$$

(1.11)

The (sequences of polynomials) ${\left\{G_n\left(x\right)\right\}}_{n\ge 0}$ and ${\left\{H_n\left(x\right)\right\}}_{n\ge 0}$ can be extended to negative subscripts by defining

$$\begin{array}{ccc}\hfill G_{-n}\left(x\right)& =& -{\displaystyle \frac{r\left(x\right)}{s\left(x\right)}}{G}_{-(n-1)}\left(x\right)+{\displaystyle \frac{1}{s\left(x\right)}}{G}_{-(n-2)}\left(x\right),\hfill \\ \hfill H_{-n}\left(x\right)& =& -{\displaystyle \frac{r\left(x\right)}{s\left(x\right)}}{H}_{-(n-1)}\left(x\right)+{\displaystyle \frac{1}{s\left(x\right)}}{H}_{-(n-2)}\left(x\right),\hfill \end{array}$$

for $n=1,2,3,...$ respectively. Therefore, recurrences (1.10) and (1.11) hold for all integers $n.$

NOTE: For the sake of simplicity throughout the rest of the paper we use

$$W_n,r,s,W_0,W_1,\alpha ,\beta ,G_n,H_n,G_0,G_1,H_0,H_1$$

instead of

$$W_n\left(x\right),r\left(x\right),s\left(x\right),W_0\left(x\right),W_1\left(x\right),\alpha \left(x\right),\beta \left(x\right),G_n\left(x\right),H_n\left(x\right),G_0\left(x\right),G_1\left(x\right),H_0\left(x\right),H_1\left(x\right),$$

respectively. For example, we write

$$W_n=r{W}_{n-1}+s{W}_{n-2},W_0=a,W_1=b,n\ge 2$$

for the equation (1.7).

Using the roots $\alpha ,\beta$ and recurrence relation (1.7), Binet’s formula can be given as follows:

Theorem 3

([15], Theorem 2). The general term of the generalized Fibonacci (Horadam) polynomials $W_n$ can be presented by the following Binet’s formula:

$$\begin{array}{ccc}\hfill W_n& =& \left\{\begin{array}{ccc}{\displaystyle \frac{W_1-\beta W_0}{\alpha -\beta }}{\alpha }^n-{\displaystyle \frac{W_1-\alpha W_0}{\alpha -\beta }}{\beta }^n& ,& \mathit{if}\alpha \ne \beta (\mathit{Distinct}\mathit{Roots}\mathit{Case})\\ (n{W}_1-\alpha \left(n-1\right)W_0){\alpha }^{n-1}& ,& \mathit{if}\alpha =\beta (\mathit{Single}\mathit{Root}\mathit{Case})\end{array}\right.\hfill \\ & =& \left\{\begin{array}{ccc}{\displaystyle \frac{W_1-\beta W_0}{\alpha -\beta }}{\alpha }^n-{\displaystyle \frac{W_1-\alpha W_0}{\alpha -\beta }}{\beta }^n& ,& \mathit{if}\alpha \ne \beta (\mathit{Distinct}\mathit{Roots}\mathit{Case})\\ (n{W}_1-{\displaystyle \frac{r}{2}}\left(n-1\right)W_0){\left({\displaystyle \frac{r}{2}}\right)}^{n-1}& ,& \mathit{if}\alpha =\beta (\mathit{Single}\mathit{Root}\mathit{Case})\end{array}\right..\hfill \end{array}$$

(1.12)

For all integers $n,$$(r,s)$-Fibonacci and $(r,s)$-Lucas polynomials can be expressed using Binet’s formulas as

Corollary 1

([15], p.31).

$$G_n=\left\{\begin{array}{ccc}{\displaystyle \frac{{\alpha }^n-{\beta }^n}{\alpha -\beta }}& ,& \mathit{if}\alpha \ne \beta (\mathit{Distinct}\mathit{Roots}\mathit{Case})\\ n{\alpha }^{n-1}& ,& \mathit{if}\alpha =\beta (\mathit{Sin}\mathit{gle}\mathit{Root}\mathit{Case})\end{array}\right.,$$

$$H_n=\left\{\begin{array}{ccc}{\alpha }^n+{\beta }^n& ,& \mathit{if}\alpha \ne \beta (\mathit{Distinct}\mathit{Roots}\mathit{Case})\\ 2{\alpha }^n& ,& \mathit{if}\alpha =\beta (\mathit{Sin}\mathit{gle}\mathit{Root}\mathit{Case})\end{array}\right..$$

The explicit expressions of $G_n$ and $H_n$ can be given as follows.

Corollary 2

([15], Corollary 40). For any $n\ge 0,$ if $\alpha \ne \beta$ then

(a) 

$$2^{n-1}{G}_n={\displaystyle \sum _{m=0}^n}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2m+1}}\right)r^{n-2m-1}{(r^2+4s)}^m.$$

(1.13)

(b) 

$$2^{n-1}{H}_n={\displaystyle \sum _{m=0}^n}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2m}}\right)r^{n-2m}{\left(r^2+4s\right)}^m.$$

(1.14)

2. The Characteristic Equation

In this section, we transition from the general non-commutative multiplication of distinct operands to the specific algebraic properties of self-multiplication. When we square a single hybrid number, the complex non-commutative cross-terms simplify elegantly. We demonstrate that every hybrid number fundamentally satisfies a unique quadratic characteristic equation based strictly on its real trace and hybrid norm. By exploiting this property, we uncover a direct correspondence between hybrid numbers and second-order linear recurrence sequences. This establishes a mathematically unified framework for evaluating arbitrary powers of $\mathbf{Z}$.

Theorem 4.

Let $\mathbf{Z}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$ be a general hybrid number. $\mathbf{Z}$ satisfies the quadratic characteristic equation:

$${\mathbf{Z}}^2=u\mathbf{Z}+v$$

(2.1)

where the scalar trace is $u=2a$ and the scalar norm parameter is $v=-C\left(\mathbf{Z}\right)$.

In other words, the square of a single hybrid number bypasses the complex non-commutative cross-terms and reduces to a remarkable linear combination. This characteristic equation demonstrates that the self-multiplication of $\mathbf{Z}$ is governed entirely by its real trace ($u=2a$) and its hybrid norm ($v=-C\left(\mathbf{Z}\right)$), which we explicitly derive in the proof below.

Explicitly, the roots of the characteristic equation (2.1) depend on the vector discriminant $\Delta =a^2-C\left(\mathbf{Z}\right)$:

• If the roots are distinct (${\lambda }_1\ne {\lambda }_2$, equivalently $\Delta \ne 0$), they are given by:

  $$\begin{array}{cc}\hfill {\lambda }_1& =a+\sqrt{\Delta }=a+\sqrt{a^2-C\left(\mathbf{Z}\right)},\hfill \end{array}$$

  (2.2)

  $$\begin{array}{cc}\hfill {\lambda }_2& =a-\sqrt{\Delta }=a-\sqrt{a^2-C\left(\mathbf{Z}\right)}.\hfill \end{array}$$

  (2.3)
• If the roots of the characteristic equation (2.1) are equal (equivalently $\Delta =0$), they degenerate to:

  $${\lambda }_1={\lambda }_2=\lambda =a.$$

  (2.4)

Proof. 

Every hybrid number $\mathbf{Z}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$ possesses a conjugate defined as $\overline{\mathbf{Z}}=a-b\mathbf{i}-c\mathbf{\epsilon }-d\mathbf{h}$. For Özdemir’s non-commutative hybrid numbers, the hybrid characteristic number (or norm) is defined by the product $C\left(\mathbf{Z}\right)=\mathbf{Z}\overline{\mathbf{Z}}=\overline{\mathbf{Z}}\mathbf{Z}=a^2+b^2-2bc-d^2$.

By rearranging the explicit component expansion of ${\mathbf{Z}}^2$ (as established in Equation (6)), we can express ${\mathbf{Z}}^2$ as a linear combination of $\mathbf{Z}$ and purely real scalars. We isolate the vector components by strategically adding and subtracting $2a^2$:

$$\begin{array}{cc}\hfill {\mathbf{Z}}^2& =(d^2+a^2-b^2+2bc)+2a(b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h})\hfill \\ \hfill & =(d^2+a^2-b^2+2bc)-2a^2+2a^2+2a(b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h})\hfill \\ \hfill & =(d^2-a^2-b^2+2bc)+2a(a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h})\hfill \\ \hfill & =-(a^2+b^2-2bc-d^2)+2a\mathbf{Z}\hfill \\ \hfill & =-C\left(\mathbf{Z}\right)+2a\mathbf{Z}\hfill \end{array}$$

Rearranging this identity directly yields the quadratic characteristic equation based on its real scalar part and its hybrid character:

$${\mathbf{Z}}^2-2a\mathbf{Z}+C\left(\mathbf{Z}\right)=0$$

which perfectly matches the form ${\mathbf{Z}}^2-u\mathbf{Z}-v=0$ where $u=2a$ and $v=-C\left(\mathbf{Z}\right)$.

We now evaluate the algebraic roots of this characteristic equation:

$${\lambda }_{1,2}={\displaystyle \frac{u\pm \sqrt{u^2+4v}}{2}}.$$

Substituting $u=2a$ and $v=-C\left(\mathbf{Z}\right)$ yields $u^2+4v=4a^2-4C\left(\mathbf{Z}\right)$. By defining the vector discriminant $\Delta =a^2-C\left(\mathbf{Z}\right)$, the expression under the square root becomes $4\Delta$. Substituting this back provides the exact roots:

$${\lambda }_{1,2}={\displaystyle \frac{2a\pm \sqrt{4\Delta }}{2}}=a\pm \sqrt{\Delta }.$$

(2.5)

Finally, we examine the degenerate case where the roots are equal (i.e., ${\lambda }_1={\lambda }_2=\lambda$). Under this condition, the characteristic polynomial becomes a perfect square:

$${(\mathbf{Z}-\lambda )}^2={\mathbf{Z}}^2-2\lambda \mathbf{Z}+{\lambda }^2=0.$$

By comparing coefficients with our originally established characteristic equation ${\mathbf{Z}}^2-2a\mathbf{Z}+C\left(\mathbf{Z}\right)=0$, we immediately obtain $\lambda =a$ and ${\lambda }^2=C\left(\mathbf{Z}\right)$. This implies $a^2=C\left(\mathbf{Z}\right)$, which strictly requires the vector discriminant to vanish:

$$\Delta =a^2-C\left(\mathbf{Z}\right)=0.$$

This formally verifies all stated properties of the characteristic roots, completing the proof. □

Since establishing Theorem 4, we observe that hybrid numbers structurally satisfy a recursive quadratic equation. Motivated by the nature of this characteristic equation, we can naturally associate a linear recurrence sequence with any hybrid number. This sequence captures the scaling coefficients required to evaluate higher powers of $\mathbf{Z}$.

Definition 5.

Let $\mathbf{Z}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$ be a hybrid number. The fundamental sequence $G_n$ associated with $\mathbf{Z}$ is defined by the linear recurrence relation:

$$G_n=u{G}_{n-1}+v{G}_{n-2}=2a{G}_{n-1}-C\left(\mathbf{Z}\right)G_{n-2}$$

(2.6)

with the initial conditions $G_0=0$ and $G_1=1$.

This provides a new perspective: hybrid numbers are not only algebraic entities but also generators of recurrence sequences. We now introduce a complementary companion sequence related to $G_n$, which we also require for expanding the powers of hybrid numbers.

Definition 6.

Let $\mathbf{Z}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$ be a hybrid number. The fundamental-Lucas sequence $H_n$ associated with $\mathbf{Z}$ is defined by the exact same second-order linear recurrence relation as $G_n$:

$$H_n=u{H}_{n-1}+v{H}_{n-2}=2a{H}_{n-1}-C\left(\mathbf{Z}\right)H_{n-2}$$

(2.7)

with the initial conditions $H_0=2$ and $H_1=u=2a$.

To conceptually unify these two companion sequences and streamline our terminology throughout the remainder of this work, we establish the following convention.

Remark 4.

Because both the fundamental sequence $G_n$ and the fundamental-Lucas sequence $H_n$ are derived directly from the invariants of the characteristic equation ${\mathbf{Z}}^2-u\mathbf{Z}-v=0$ and share the identical recurrence relation, we collectively refer to them as the characteristic sequences of the hybrid number $\mathbf{Z}$.

The characteristic sequences $G_n$ and $H_n$ established in Definition 5 and Definition 6 can be naturally extended to negative indices via the relations $G_{k-2}={\displaystyle \frac{1}{v}}(G_k-u{G}_{k-1})$ and $H_{k-2}={\displaystyle \frac{1}{v}}(H_k-u{H}_{k-1})$ provided that $C\left(\mathbf{Z}\right)\ne 0$ (so that $v\ne 0$). In this extended domain, the resulting structure is technically a two-sided sequence; however, as is customary in the literature, we will simply continue to refer to them as the fundamental and fundamental-Lucas sequences associated with the hybrid number.

For all integers $n,$ the characteristic sequences can be expressed using Binet’s formulas as

Corollary 3.

Binet’s formulas of the fundamental sequence $G_n$ and the fundamental-Lucas sequence $H_n$ are

$$G_n=\left\{\begin{array}{ccc}{\displaystyle \frac{{\lambda }_1^n-{\lambda }_2^n}{{\lambda }_1-{\lambda }_2}}& ,& \mathit{if}\Delta \ne 0({\lambda }_1\ne {\lambda }_2)(\mathit{Distinct}\mathit{Roots}\mathit{Case})\\ n{\lambda }^{n-1}(=n{a}^{n-1})& ,& \mathit{if}\Delta =0({\lambda }_1={\lambda }_2=\lambda =a)(\mathit{Single}\mathit{Root}\mathit{Case})\end{array}\right.,$$

and

$$H_n=\left\{\begin{array}{ccc}{\lambda }_1^n+{\lambda }_2^n& ,& \mathit{if}\Delta \ne 0({\lambda }_1\ne {\lambda }_2)(\mathit{Distinct}\mathit{Roots}\mathit{Case})\\ 2{\lambda }^n(=2a^n)& ,& \mathit{if}\Delta =0({\lambda }_1={\lambda }_2=\lambda =a)(\mathit{Single}\mathit{Root}\mathit{Case})\end{array}\right..$$

equivalently

$$G_n=\left\{\begin{array}{ccc}{\displaystyle \frac{{(a+\sqrt{\Delta })}^n-{(a-\sqrt{\Delta })}^n}{2\sqrt{\Delta }}}& ,& \mathit{if}\Delta \ne 0({\lambda }_1\ne {\lambda }_2)(\mathit{Distinct}\mathit{Roots}\mathit{Case})\\ n{a}^{n-1}& ,& \mathit{if}\Delta =0({\lambda }_1={\lambda }_2=\lambda =a)(\mathit{Single}\mathit{Root}\mathit{Case})\end{array}\right.,$$

and

$$H_n=\left\{\begin{array}{ccc}{(a+\sqrt{\Delta })}^n+{(a-\sqrt{\Delta })}^n& ,& \mathit{if}\Delta \ne 0({\lambda }_1\ne {\lambda }_2)(\mathit{Distinct}\mathit{Roots}\mathit{Case})\\ 2a^n& ,& \mathit{if}\Delta =0({\lambda }_1={\lambda }_2=\lambda =a)(\mathit{Single}\mathit{Root}\mathit{Case})\end{array}\right..$$

Proof. 

Apply Corollary 1 with $r=u=2a$ and $s=v=-C\left(\mathbf{Z}\right)=\Delta -a^2$ and the roots of the characteristic equation ${\mathbf{Z}}^2-u\mathbf{Z}-v=0$ given in (2.5). □

This mapping from a hybrid number $\mathbf{Z}\in \mathbb{K}$ to specific second-order linear recurrence sequences naturally raises the question of uniqueness: Does a specific linear recurrence uniquely identify a hybrid number, or do multiple hybrid numbers share the same characteristic sequences? We formally address the nature of this correspondence in the following lemma.

Lemma 1.

Let $\mathcal{S}$ denote the set of all second-order linear recurrence sequences uniquely determined by the parameters $(u,v)\in {\mathbb{R}}^2$. The correspondence $\mathrm{\Phi }:\mathbb{K}\to \mathcal{S}$ that maps a hybrid number $\mathbf{Z}$ to its characteristic sequences ($G_n$ and $H_n$, where $u=2a$ and $v=-C\left(\mathbf{Z}\right)$) is surjective (onto) but is highly non-injective (not one-to-one). An infinite surface of hybrid numbers maps to the exact same pair of characteristic sequences.

Proof. 

To test for surjectivity, we must show that for any given sequence generated by arbitrary real parameters $(u,v)$, there exists at least one hybrid number $\mathbf{Z}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$ mapping to it. From $u=2a$, we immediately fix $a={\displaystyle \frac{u}{2}}$. From $v=-C\left(\mathbf{Z}\right)$, we require $a^2+b^2-2bc-d^2=-v$. Substituting a yields:

$$b^2-2bc-d^2=-\left(v+{\displaystyle \frac{u^2}{4}}\right)$$

We can trivially satisfy this equation by setting $c=0$, which reduces the condition to $b^2-d^2=-(v+u^2/4)$. Because the expression $x^2-y^2$ ranges over all real numbers, there will always exist real solutions for $(b,d)$ regardless of the values of u and v. Thus, the mapping is surjective.

To prove it is non-injective, we observe the remaining degree of freedom. Once a is fixed by the trace u, any combination of $(b,c,d)$ satisfying $b^2-2bc-d^2=-v-a^2$ will generate the exact same sequences. For example, if the required constant is $K=1$, the hybrid components $(b,c,d)=(1,0,0)$ and $(b,c,d)=(\sqrt{2},0,1)$ both satisfy $b^2-2bc-d^2=1$. Consequently, an uncountably infinite set of hybrid numbers collapses to the same characteristic sequences, confirming the mapping is not one-to-one. □

Multiplying both sides of (2.1) by ${\mathbf{Z}}^{n-2}$ ($n\ge 2$) immediately reveals that the powers of any hybrid number form a second-order linear recurrence relation:

$${\mathbf{Z}}^n=2a{\mathbf{Z}}^{n-1}-C\left(\mathbf{Z}\right){\mathbf{Z}}^{n-2}.$$

(2.8)

To illustrate how this characteristic equation (2.1) maps seamlessly onto well-known recurrence sequences, let us consider a specific hybrid number with unit coefficients.

Example 7.

Suppose that $a=1,b=1,c=1$, and crucially $d=1$, so that $C\left(\mathbf{Z}\right)=a^2+b^2-2bc-d^2=1^2+1^2-2\left(1\right)\left(1\right)-1^2=-1.$

Because $u=2a=2$ and $v=-C\left(\mathbf{Z}\right)=1$, the characteristic sequences associated with $\mathbf{Z}=1+\mathbf{i}+\mathbf{\epsilon }+\mathbf{h}$ share the second-order recurrence relation:

$$W_n=2W_{n-1}+W_{n-2}$$

(2.9)

Applying the initial conditions $G_0=0$ and $G_1=1$, the fundamental sequence exactly generates the well-known Pell numbers ($G_n=P_n$). Simultaneously, applying the Lucas-type initial conditions $H_0=2$ and $H_1=u=2$, the fundamental-Lucas sequence exactly generates the Pell-Lucas numbers ($H_n=Q_n$).

However, because the mapping is not one-to-one, this is not the only hybrid number associated with these sequences. Any hybrid number with $a=1$ and components satisfying $b^2-2bc-d^2=-2$ will generate both the Pell and Pell-Lucas numbers. Thus, the purely real number $\mathbf{Z}=1$ predictably does not generate them (since $C\left(1\right)=1\ne -1$), but a number like $\mathbf{Z}=1+\sqrt{2}\mathbf{h}$ (where $b=0,c=0,d=\sqrt{2}$) will exactly reproduce both sequences simultaneously.

To concretely illustrate the surjective yet non-injective nature of the mapping $\mathrm{\Phi }:\mathbb{K}\to \mathcal{S}$, we construct explicit families of hybrid numbers that generate several renowned recurrence sequences. By prescribing the recurrence coefficients u and v to match these classical sequences, we immediately determine the fixed real trace a and the algebraic surface equation $b^2-2bc-d^2=-v-a^2$ that governs the remaining hybrid components.

Example 8.

By setting the parameters $(u,v)$ to target classical recurrence sequences, we can identify the constraints on the hybrid components $(b,c,d)$. As established in Lemma 1, these components are never unique.

(a) 

The characteristic sequences sharing the recurrence relation

$$W_n=W_{n-1}+W_{n-2}$$

(2.10)

require $u=1$ and $v=1$. Consequently, $a={\displaystyle \frac{1}{2}}$ and $C\left(\mathbf{Z}\right)=-1$. The components must satisfy $b^2-2bc-d^2=-{\displaystyle \frac{5}{4}}$. For example, if we choose $b=1$ and $c=-2$, we find $d=\pm {\displaystyle \frac{5}{2}}$. Thus, the hybrid numbers $\mathbf{Z}={\displaystyle \frac{1}{2}}+\mathbf{i}-2\mathbf{\epsilon }\pm {\displaystyle \frac{5}{2}}\mathbf{h}$ perfectly generate the Fibonacci sequence ($G_n=F_n$). Furthermore, applying $H_0=2$ and $H_1=1$, these exact same hybrid numbers simultaneously generate the classical Lucas sequence ($H_n=L_n$).

(b) 

The recurrence $W_n=W_{n-1}+2W_{n-2}$ requires $u=1$ and $v=2$. Consequently, $a={\displaystyle \frac{1}{2}}$ and $C\left(\mathbf{Z}\right)=-2$. The non-unique components must satisfy the surface equation $b^2-2bc-d^2=-{\displaystyle \frac{9}{4}}$. By choosing $b=0$ and $c=0$, we yield $d=\pm {\displaystyle \frac{3}{2}}$. Thus, the hybrid numbers $\mathbf{Z}={\displaystyle \frac{1}{2}}\pm {\displaystyle \frac{3}{2}}\mathbf{h}$ successfully generate the Jacobsthal sequence ($G_n=J_n$). Correspondingly, with initial conditions $H_0=2$ and $H_1=1$, these exact same hybrid numbers simultaneously generate the Jacobsthal-Lucas numbers ($H_n=j_n$).

(c) 

The recurrence $W_n=3W_{n-1}-2W_{n-2}$ requires $u=3$ and $v=-2$. Consequently, $a={\displaystyle \frac{3}{2}}$ and $C\left(\mathbf{Z}\right)=2$. The matching components must satisfy $b^2-2bc-d^2=-{\displaystyle \frac{1}{4}}$. By choosing $b=0$ and $c=0$, we find $d=\pm {\displaystyle \frac{1}{2}}$. Therefore, the hybrid numbers $\mathbf{Z}={\displaystyle \frac{3}{2}}\pm {\displaystyle \frac{1}{2}}\mathbf{h}$ generate the Mersenne sequence ($G_n=M_n$). Simultaneously, applying $H_0=2$ and $H_1=3$, these identical hybrid numbers generate the Mersenne-Lucas numbers.

(d) 

The recurrence $W_n=6W_{n-1}-W_{n-2}$ requires $u=6$ and $v=-1$. Consequently, $a=3$ and $C\left(\mathbf{Z}\right)=1$. The corresponding components must satisfy $b^2-2bc-d^2=-8$. For a non-trivial recurrence solution, setting $c=0$ and $b=1$ gives $d=\pm 3$. Thus, the hybrid numbers $\mathbf{Z}=3+\mathbf{i}\pm 3\mathbf{h}$ generate the balancing numbers ($G_n=B_n$). Concurrently, applying $H_0=2$ and $H_1=6$, these same hybrid numbers generate the modified Lucas-balancing numbers.

(e) 

The sequence $W_n=W_{n-1}-{\displaystyle \frac{1}{4}}{W}_{n-2}$ represents a degenerate linear recurrence with a double characteristic root, requiring $u=1$ and $v=-{\displaystyle \frac{1}{4}}$. Consequently, $a={\displaystyle \frac{1}{2}}$ and $C\left(\mathbf{Z}\right)={\displaystyle \frac{1}{4}}$. Any hybrid number mapping to this sequence must satisfy the cone equation $b^2-2bc-d^2=0$. A trivial solution is the purely real number $\mathbf{Z}={\displaystyle \frac{1}{2}}$. A non-trivial hybrid solution is obtained by setting $d=0$ and $b=2$, yielding $c=1$. Thus, $\mathbf{Z}={\displaystyle \frac{1}{2}}+2\mathbf{i}+\mathbf{\epsilon }$ also generates the modified Oresme sequence. By applying $H_0=2$ and $H_1=1$, these exact same hybrid numbers also simultaneously generate the Oresme-Lucas numbers.

3. Anti-Involutions, Character Identities, and Algebraic Symmetry

In this section, we transition from the computational mechanics of hybrid multiplication to the deeper structural symmetries of the hybrid ring. We formally investigate the behavior of the hybrid conjugate under non-commutative multiplication and establish the rigorous algebraic proofs for the character identities of hybrid exponents. By isolating these foundational properties, we provide the necessary mathematical scaffolding to support the general exponentiation theorems presented later in this work.

3.1. The Anti-Involutive Property of the Hybrid Conjugate

At first glance, the behavior of the conjugate under hybrid multiplication appears counterintuitive. However, it highlights one of the most beautiful and profound structural necessities of non-commutative algebras. It is precisely because hybrid multiplication is generally non-commutative ($Z_1{Z}_2\ne Z_2{Z}_1$) that the conjugate must reverse the order of the operands. If the conjugate preserved the operational order in a non-commutative space, the underlying algebraic structure would collapse. The following lemma provides the rigorous foundation for this order-reversing phenomenon.

Lemma 2

(Anti-involution of the Hybrid Conjugate). For any two hybrid numbers $Z_1$ and $Z_2$, the conjugate of their product is equal to the product of their conjugates in reverse order:

$$\overline{Z_1{Z}_2}=\overline{Z_2}·\overline{Z_1}.$$

(3.1)

Proof. 

We prove this property by leveraging the ring isomorphism between the set of hybrid numbers $\mathbb{K}$ and the set of $2\times 2$ real matrices $M_2\left(\mathbb{R}\right)$. Let $Z=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$ be an arbitrary hybrid number. Its standard matrix representation, $M\left(Z\right)$, is defined as:

$$M\left(Z\right)=\left(\begin{array}{cc}a+c& b-c+d\\ c-b+d& a-c\end{array}\right).$$

(3.2)

By definition, the conjugate of Z is $\overline{Z}=a-b\mathbf{i}-c\mathbf{\epsilon }-d\mathbf{h}$. We can construct the matrix representation of the conjugate, $M\left(\overline{Z}\right)$, by substituting the coefficients $(a,-b,-c,-d)$ into the matrix mapping:

$$\begin{array}{cc}\hfill M\left(\overline{Z}\right)& =\left(\begin{array}{cc}a+(-c)& (-b)-(-c)+(-d)\\ (-c)-(-b)+(-d)& a-(-c)\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}a-c& -b+c-d\\ -c+b-d& a+c\end{array}\right).\hfill \end{array}$$

In linear algebra, the adjugate of a $2\times 2$ matrix $A=\left(\begin{array}{cc}x& y\\ z& w\end{array}\right)$ is given by $\mathrm{adj}\left(A\right)=\left(\begin{array}{cc}w& -y\\ -z& x\end{array}\right)$. Applying this definition to our original matrix $M\left(Z\right)$:

$$\begin{array}{cc}\hfill \mathrm{adj}\left(M\right(Z\left)\right)& =\left(\begin{array}{cc}a-c& -(b-c+d)\\ -(c-b+d)& a+c\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}a-c& -b+c-d\\ -c+b-d& a+c\end{array}\right).\hfill \end{array}$$

Comparing the two results, we establish the fundamental relation that the matrix representation of a hybrid conjugate is exactly the adjugate of its matrix representation:

$$M\left(\overline{Z}\right)=\mathrm{adj}\left(M\left(Z\right)\right).$$

(3.3)

A well-known theorem of matrix algebra states that the adjugate of a product of matrices reverses the order of multiplication: $\mathrm{adj}\left(AB\right)=\mathrm{adj}\left(B\right)\mathrm{adj}\left(A\right)$. Using this theorem alongside our established isomorphism, we evaluate the conjugate of the product $Z_1{Z}_2$:

$$\begin{array}{cc}\hfill M\left(\overline{Z_1{Z}_2}\right)& =\mathrm{adj}\left(M\left(Z_1{Z}_2\right)\right)\hfill \\ \hfill & =\mathrm{adj}\left(M\left(Z_1\right)M\left(Z_2\right)\right)(\mathrm{by}\mathrm{the}\mathrm{property}\mathrm{of}\mathrm{ring}\mathrm{isomorphism})\hfill \\ \hfill & =\mathrm{adj}\left(M\left(Z_2\right)\right)\mathrm{adj}\left(M\left(Z_1\right)\right)(\mathrm{by}\mathrm{the}\mathrm{matrix}\mathrm{adjugate}\mathrm{product}\mathrm{rule})\hfill \\ \hfill & =M\left(\overline{Z_2}\right)M\left(\overline{Z_1}\right)\hfill \\ \hfill & =M(\overline{Z_2}·\overline{Z_1}).\hfill \end{array}$$

Since the mapping between hybrid numbers and $2\times 2$ matrices is a bijection, $M\left(\overline{Z_1{Z}_2}\right)=M(\overline{Z_2}·\overline{Z_1})$ implies that the hybrid numbers themselves must be equal. Therefore, we conclude:

$$\overline{Z_1{Z}_2}=\overline{Z_2}·\overline{Z_1}.$$

(3.4)

This completes the proof. □

This order-reversing behavior is a ubiquitous structural requirement in linear algebra. While it may appear unusual in the context of numbers, it mirrors the exact behavior of standard matrix operators. Even though matrix multiplication is non-commutative ($AB\ne BA$), both the transpose and the inverse of a matrix product flawlessly reverse the operand order: ${\left(AB\right)}^T=B^T{A}^T$ and ${\left(AB\right)}^{-1}=B^{-1}{A}^{-1}$.

In advanced abstract algebra, an operation is classified as an involution if applying it twice returns the original element (e.g., $\overline{\overline{Z}}=Z$). When an involution simultaneously reverses the order of multiplication, it is formally termed an anti-involution. The conjugate of a hybrid number, the transpose of a matrix, and the conjugate of a quaternion are all quintessential examples of anti-involutions. Thus, the identity $\overline{Z_1{Z}_2}=\overline{Z_2}·\overline{Z_1}$ is not a mathematical anomaly, but rather the mechanism by which perfect algebraic symmetry is maintained in a non-commutative space.

Because a hybrid number trivially commutes with its own powers, this anti-involutive property simplifies elegantly when applied to exponentiation. We formalize this in the following theorem.

Theorem 9.

For any hybrid number Z and any integer n (provided $C\left(Z\right)\ne 0$ when $n<0$), the identity

$$\overline{Z^n}={\left(\overline{Z}\right)}^n$$

holds.

Proof. 

We base our proof on the fundamental property of the conjugate of a product of hybrid numbers. For any two hybrid numbers $Z_1$ and $Z_2$, the conjugate of their product reverses the order: $\overline{Z_1{Z}_2}=\overline{Z_2}\overline{Z_1}$ (Lemma 2).

We divide the proof into three cases based on the integer n.

Case 1: Positive Integers ($n>0$)

We proceed by mathematical induction.

Base Case ($n=1$):$\overline{Z^1}={\left(\overline{Z}\right)}^1$, which is trivially true.

Inductive Step: Assume the proposition holds for some integer $k\ge 1$, such that $\overline{Z^k}={\left(\overline{Z}\right)}^k$. We examine the case for $k+1$:

$$\begin{array}{cc}\hfill \overline{Z^{k+1}}& =\overline{Z^k·Z}\hfill \\ \hfill & =\overline{Z}·\overline{Z^k}(\mathrm{by}\mathrm{the}\mathrm{product}\mathrm{conjugation}\mathrm{property})\hfill \\ \hfill & =\overline{Z}·{\left(\overline{Z}\right)}^k(\mathrm{by}\mathrm{the}\mathrm{inductive}\mathrm{hypothesis})\hfill \\ \hfill & ={\left(\overline{Z}\right)}^{k+1}.\hfill \end{array}$$

Thus, by induction, the identity holds for all positive integers.

Case 2: Zero ($n=0$)

By definition of the zero exponent, $Z^0=1$ (the real identity element $1+0\mathbf{i}+0\mathbf{\epsilon }+0\mathbf{h}$). Taking the conjugate yields $\overline{Z^0}=\overline{1}=1$. Similarly, ${\left(\overline{Z}\right)}^0=1$. Therefore, $\overline{Z^0}={\left(\overline{Z}\right)}^0$ holds perfectly.

Case 3: Negative Integers ($n<0$)

For negative powers to be defined, Z must be invertible, which requires $C\left(Z\right)\ne 0$. First, we establish the identity for $n=-1$. By the definition of the inverse, we have:

$$Z{Z}^{-1}=1.$$

Taking the conjugate of both sides gives:

$$\begin{array}{cc}\hfill \overline{Z{Z}^{-1}}& =\overline{1},\hfill \\ \hfill \overline{Z^{-1}}·\overline{Z}& =1.\hfill \end{array}$$

Multiplying both sides on the right by ${\left(\overline{Z}\right)}^{-1}$ (which exists since $C\left(\overline{Z}\right)=C\left(Z\right)\ne 0$) yields:

$$\overline{Z^{-1}}={\left(\overline{Z}\right)}^{-1}.$$

Now, let $n=-m$ where $m>0$. Using the properties established in Case 1 and the inverse identity above, we can express the n-th power without relying on a backward induction loop:

$$\begin{array}{cc}\hfill \overline{Z^n}& =\overline{Z^{-m}}\hfill \\ \hfill & =\overline{{\left(Z^{-1}\right)}^m}\hfill \\ \hfill & ={\left(\overline{Z^{-1}}\right)}^m(\sin \mathrm{ce}m\mathrm{is}\mathrm{positive},\mathrm{by}\mathrm{Case}1)\hfill \\ \hfill & ={\left({\left(\overline{Z}\right)}^{-1}\right)}^m(\mathrm{by}\mathrm{the}\mathrm{inverse}\mathrm{conjugate}\mathrm{identity})\hfill \\ \hfill & ={\left(\overline{Z}\right)}^{-m}\hfill \\ \hfill & ={\left(\overline{Z}\right)}^n.\hfill \end{array}$$

This confirms that the identity holds rigorously for all negative integers, completing the proof. □

3.2. Proof of the Character Identity $C\left(Z^k\right)={\left(C\left(Z\right)\right)}^k$

While the multiplicative property of the hybrid character, $C\left(Z_1{Z}_2\right)=C\left(Z_1\right)C\left(Z_2\right)$, is a widely recognized consequence of its matrix isomorphism (see Özdemir [10], Çakır and Özdemir [2]), explicit formal proofs for the generalized exponentiation identity $C\left(Z^k\right)={\left(C\left(Z\right)\right)}^k$ are notably absent from the current literature. In particular, the necessary mathematical extensions to non-positive integers ($k\le 0$) and the analytical edge-case of degenerate characteristic roots have been entirely omitted from standard texts. To resolve this gap in the literature, we present complete, rigorous proofs of this identity utilizing two distinct approaches: mathematical induction and structural root analysis.

To establish the foundation for the inductive proof, we first formalize the multiplicativity of the character.

Lemma 3

(Multiplicativity of the Character). Let $Z_1$ and $Z_2$ be two hybrid numbers. Then $C\left(Z_1{Z}_2\right)=C\left(Z_1\right)C\left(Z_2\right)$.

Proof. 

The ring of hybrid numbers is isomorphic to the ring of $2\times 2$ real matrices. The standard matrix representation $M\left(Z\right)$ of a hybrid number $Z=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$ is given by:

$$M\left(Z\right)=\left(\begin{array}{cc}a+c& b-c+d\\ c-b+d& a-c\end{array}\right).$$

(3.5)

The determinant of this matrix precisely corresponds to the character of Z:

$$\begin{array}{cc}\hfill \det \left(M\right(Z\left)\right)& =(a+c)(a-c)-(b-c+d)(c-b+d)\hfill \\ \hfill & =a^2-c^2-\left[d^2-{(b-c)}^2\right]\hfill \\ \hfill & =a^2+{(b-c)}^2-c^2-d^2\hfill \\ \hfill & =C\left(Z\right).\hfill \end{array}$$

Because the determinant of a product of square matrices equals the product of their determinants ($\det \left(AB\right)=\det \left(A\right)\det \left(B\right)$), the multiplicativity follows immediately:

$$C\left(Z_1{Z}_2\right)=\det \left(M\left(Z_1{Z}_2\right)\right)=\det \left(M\left(Z_1\right)M\left(Z_2\right)\right)=\det \left(M\left(Z_1\right)\right)\det \left(M\left(Z_2\right)\right)=C\left(Z_1\right)C\left(Z_2\right).$$

(3.6)

□

Building upon this multiplicative lemma, we now establish the universal character identity for all integer powers of a hybrid number.

Theorem 10.

For any hybrid number Z and any integer k (with the condition that $C\left(Z\right)\ne 0$ if $k\le 0$), the character satisfies

$$C\left(Z^k\right)={\left(C\left(Z\right)\right)}^k.$$

Proof. Method 1: Proof via Matrix Isomorphism and Induction

We divide the proof into two parts based on the sign of the integer k.

Forward Induction for Positive Integers ($k>0$):

Base Case ($k=1$):$C\left(Z^1\right)={\left(C\left(Z\right)\right)}^1$, which holds trivially.

Inductive Step: Assume $C\left(Z^n\right)={\left(C\left(Z\right)\right)}^n$ holds for some integer $n\ge 1$. We evaluate the case for $n+1$:

$$\begin{array}{cc}\hfill C\left(Z^{n+1}\right)& =C\left(Z^{n}Z\right)\hfill \\ \hfill & =C\left(Z^n\right)C\left(Z\right)(\mathrm{by}\mathrm{Lemma})\hfill \\ \hfill & ={\left(C\left(Z\right)\right)}^{n}C\left(Z\right)(\mathrm{by}\mathrm{the}\mathrm{inductive}\mathrm{hypothesis})\hfill \\ \hfill & ={\left(C\left(Z\right)\right)}^{n+1}.\hfill \end{array}$$

Thus, the proposition holds for all $k>0$.

Backward Induction for Non-Positive Integers ($k\le 0$):

For $Z^k$ to be mathematically defined for non-positive integers, the hybrid number Z must be invertible. In the hybrid number ring, this requires that Z is not lightlike, i.e., $C\left(Z\right)\ne 0$. We proceed under this strictly necessary condition.

Base Case ($k=0$): By definition, $Z^0=1$ (the identity hybrid number $1+0\mathbf{i}+0\mathbf{\epsilon }+0\mathbf{h}$). The character is $C\left(Z^0\right)=C\left(1\right)=1^2-0=1$. Since ${\left(C\left(Z\right)\right)}^0=1$, the equality holds.

Inductive Step: Assume the proposition holds for some non-positive integer $m\le 0$, meaning $C\left(Z^m\right)={\left(C\left(Z\right)\right)}^m$. We must demonstrate it holds for the next downward integer, $m-1$.

Consider the standard exponent relation:

$$Z^{m-1}Z=Z^m.$$

(3.7)

Because Z is invertible, $Z^{m-1}$ is a well-defined hybrid number. Taking the character of both sides and applying Lemma 1:

$$C\left(Z^{m-1}\right)C\left(Z\right)=C\left(Z^m\right).$$

(3.8)

Applying the inductive hypothesis $C\left(Z^m\right)={\left(C\left(Z\right)\right)}^m$ to the right side yields:

$$C\left(Z^{m-1}\right)C\left(Z\right)={\left(C\left(Z\right)\right)}^m.$$

(3.9)

Since Z is invertible, we know $C\left(Z\right)\ne 0$. This guarantees we can safely divide both sides by $C\left(Z\right)$ without any logical contradictions:

$$C\left(Z^{m-1}\right)={\displaystyle \frac{{\left(C\left(Z\right)\right)}^m}{C\left(Z\right)}}={\left(C\left(Z\right)\right)}^{m-1}.$$

(3.10)

This completes the downward induction for all non-positive integers.

Method 2: Alternative Proof via Characteristic Roots

A deeper structural proof utilizes the characteristic equation of the hybrid number. Let $Z=a+V$, where $V=b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$ is the vector part, satisfying $V^2=a^2-C\left(Z\right)$.

The characteristic equation for Z is given by:

$${\lambda }^2-2a\lambda +C\left(Z\right)=0.$$

(3.11)

The characteristic roots are ${\lambda }_{1,2}=a\pm \sqrt{a^2-C\left(Z\right)}$. Notably, the product of the roots is ${\lambda }_1{\lambda }_2=C\left(Z\right)$ and the sum is ${\lambda }_1+{\lambda }_2=2a$.

We evaluate the identity by considering the two cases for the discriminant of the characteristic equation.

Case 1: The roots are distinct (${\lambda }_1\ne {\lambda }_2$)

Using Sylvester’s matrix formula (see Remark 5 below), (or by solving the linear recurrence for $Z^k$), we can express the k-th power of Z as:

$$Z^k={\displaystyle \frac{{\lambda }_1^k-{\lambda }_2^k}{{\lambda }_1-{\lambda }_2}}Z-{\lambda }_1{\lambda }_2{\displaystyle \frac{{\lambda }_1^{k-1}-{\lambda }_2^{k-1}}{{\lambda }_1-{\lambda }_2}}I.$$

(3.12)

Substituting $Z=a+V$ and grouping the scalar and vector components gives:

$$Z^k=\left(a{\displaystyle \frac{{\lambda }_1^k-{\lambda }_2^k}{{\lambda }_1-{\lambda }_2}}-{\lambda }_1{\lambda }_2{\displaystyle \frac{{\lambda }_1^{k-1}-{\lambda }_2^{k-1}}{{\lambda }_1-{\lambda }_2}}\right)+{\displaystyle \frac{{\lambda }_1^k-{\lambda }_2^k}{{\lambda }_1-{\lambda }_2}}V.$$

(3.13)

Using the identity $a={\displaystyle \frac{{\lambda }_1+{\lambda }_2}{2}}$, the scalar part simplifies exactly to ${\displaystyle \frac{{\lambda }_1^k+{\lambda }_2^k}{2}}$, yielding:

$$Z^k={\displaystyle \frac{{\lambda }_1^k+{\lambda }_2^k}{2}}+{\displaystyle \frac{{\lambda }_1^k-{\lambda }_2^k}{{\lambda }_1-{\lambda }_2}}V.$$

(3.14)

We know that by definition, $C\left(Z\right)=a^2-V^2$ for $Z=a+V.$ So the character of $Z^k$ is defined as the scalar part squared minus the square of the vector coefficient times $V^2$:

$$C\left(Z^k\right)={\left({\displaystyle \frac{{\lambda }_1^k+{\lambda }_2^k}{2}}\right)}^2-{\left({\displaystyle \frac{{\lambda }_1^k-{\lambda }_2^k}{{\lambda }_1-{\lambda }_2}}\right)}^2{V}^2.$$

(3.15)

Substitute $V^2=a^2-C\left(Z\right)={\left({\displaystyle \frac{{\lambda }_1-{\lambda }_2}{2}}\right)}^2={\displaystyle \frac{{({\lambda }_1-{\lambda }_2)}^2}{4}}$:

$$\begin{array}{cc}\hfill C\left(Z^k\right)& ={\displaystyle \frac{{({\lambda }_1^k+{\lambda }_2^k)}^2}{4}}-{\displaystyle \frac{{({\lambda }_1^k-{\lambda }_2^k)}^2}{4}}\hfill \\ \hfill & ={\displaystyle \frac{4{\lambda }_1^k{\lambda }_2^k}{4}}\hfill \\ \hfill & ={\left({\lambda }_1{\lambda }_2\right)}^k.\hfill \end{array}$$

Since ${\lambda }_1{\lambda }_2=C\left(Z\right)$, we arrive at $C\left(Z^k\right)={\left(C\left(Z\right)\right)}^k$.

Case 2: The roots are equal (${\lambda }_1={\lambda }_2$)

When the characteristic roots are degenerate, we have ${\lambda }_1={\lambda }_2=\lambda$. This occurs precisely when the discriminant is zero, meaning $a^2-C\left(Z\right)=0$. Consequently, $C\left(Z\right)=a^2$ and $V^2=0$.

Because $V^2=0$, the hybrid number $Z=a+V$ acts as a dual-like number. For any integer k, the binomial expansion terminates completely after just two terms. (By the generalized binomial theorem, this applies to negative integers as well, provided $a\ne 0$, which is guaranteed since Z must be invertible for negative powers):

$$Z^k={(a+V)}^k=a^k+k{a}^{k-1}V.$$

(3.16)

Taking the character $C\left(Z^k\right)=Z^k\overline{Z^k}$:

$$\begin{array}{cc}\hfill C\left(Z^k\right)& =(a^k+k{a}^{k-1}V)(a^k-k{a}^{k-1}V)\hfill \\ \hfill & =a^{2k}-k^2{a}^{2k-2}{V}^2.\hfill \end{array}$$

Because $V^2=0$, the vector term vanishes entirely:

$$C\left(Z^k\right)=a^{2k}={\left(a^2\right)}^k={\left(C\left(Z\right)\right)}^k.$$

(3.17)

This formally confirms that the character identity maintains its structural integrity even in the degenerate case where the characteristic roots are equal. □

To provide context for the algebraic decomposition utilized in Theorem 10, we briefly review the foundational matrix theorem that enables this structural root analysis.

Remark 5

(Sylvester’s Formula). The structural derivation of the n-th power of a hybrid number via its characteristic roots fundamentally relies on Sylvester’s matrix theorem (often referred to simply as Sylvester’s formula). Because the ring of hybrid numbers is isomorphic to the ring of $2\times 2$ real matrices, analytic matrix functions–including matrix exponentiation–can be directly applied to hybrid numbers. For any $2\times 2$ matrix A with distinct eigenvalues ${\lambda }_1$ and ${\lambda }_2$, Sylvester’s formula evaluates a function $f\left(A\right)$ using the projection matrices associated with its eigenvalues:

$$f\left(A\right)=f\left({\lambda }_1\right){\displaystyle \frac{A-{\lambda }_{2}I}{{\lambda }_1-{\lambda }_2}}+f\left({\lambda }_2\right){\displaystyle \frac{A-{\lambda }_{1}I}{{\lambda }_2-{\lambda }_1}}.$$

By setting $f\left(A\right)=A^k$, replacing A with our hybrid number $\mathbf{Z}$, and using the characteristic roots ${\lambda }_{1,2}$ with $f\left({\lambda }_1\right)={\lambda }_1^k$ and $f\left({\lambda }_2\right)={\lambda }_2^k$, the formula perfectly isolates the scalar and vector components into powers of $Z$as

$$Z^k={\lambda }_1^k{\displaystyle \frac{Z-{\lambda }_{2}I}{{\lambda }_1-{\lambda }_2}}+{\lambda }_2^k{\displaystyle \frac{Z-{\lambda }_{1}I}{{\lambda }_2-{\lambda }_1}}={\displaystyle \frac{{\lambda }_1^k-{\lambda }_2^k}{{\lambda }_1-{\lambda }_2}}Z-{\lambda }_1{\lambda }_2{\displaystyle \frac{{\lambda }_1^{k-1}-{\lambda }_2^{k-1}}{{\lambda }_1-{\lambda }_2}}I.$$

Comprehensive treatments and rigorous proofs of Sylvester’s matrix formula can be found in foundational matrix analysis texts, such as Higham (2008) [6] and Horn and Johnson (2012) [7].

4. Powers of Hybrid Numbers

Having established the fundamental algebraic properties and characteristic equation of a single hybrid number, we now turn our attention to the generalized exponentiation of $\mathbf{Z}$. In this section, we develop four distinct, yet mathematically unified, methodological frameworks to evaluate arbitrary integer powers of a hybrid number. By systematically exploring recursive sequence identities, Lucas-type equivalences structural root formulas and combinatorial expansions, we construct a comprehensive computational toolkit for hybrid algebra.

4.1. Method 1: The Recursive Framework via the Fundamental Sequence

With the fundamental sequence properly defined, we are now positioned to formulate a general theorem. This theorem explicitly expresses any integer power of a hybrid number as a linear combination of the number itself and the real scalar v, scaled exactly by the terms of the fundamental sequence.

Theorem 11.

Let

$$\mathbf{Z}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$$

be a hybrid number satisfying the characteristic equation

$${\mathbf{Z}}^2-u\mathbf{Z}-v=0,$$

where $u=2a$ and $v=-C\left(\mathbf{Z}\right)$. Let $G_n$ be the fundamental sequence defined by the recurrence relation

$$G_k=u{G}_{k-1}+v{G}_{k-2}$$

with initial conditions $G_0=0$ and $G_1=1$.

For any integer $n\ge 1$, the n-th power of $\mathbf{Z}$ is given by the identity:

$$\begin{array}{ccc}\hfill {\mathbf{Z}}^n& =& G_n\mathbf{Z}+v{G}_{n-1}\hfill \\ & =& g_1+g_2\mathbf{i}+g_3\mathbf{\epsilon }+g_4\mathbf{h}\hfill \end{array}$$

where $g_1=a{G}_n+v{G}_{n-1}$, $g_2=b{G}_n$, $g_3=c{G}_n$, and $g_4=d{G}_n.$

Furthermore, if $\mathbf{Z}$ is invertible (i.e., $C\left(\mathbf{Z}\right)\ne 0$, implying $v\ne 0$), this identity (i.e., ${\mathbf{Z}}^n=G_n\mathbf{Z}+v{G}_{n-1}$) holds for all integers $n\in \mathbb{Z}$, where the sequence $G_n$ is extended to negative indices via the backward relation $G_{n-2}={\displaystyle \frac{1}{v}}(G_n-u{G}_{n-1})$.

Proof. 

We present two methods for proving this theorem. The first method provides a direct algebraic proof utilizing polynomial evaluation and the properties of the characteristic roots. The second method employs mathematical induction, explicitly providing a comprehensive proof for all integer cases $n\in \mathbb{Z}$ by separating the logic into forward induction for positive powers and a distinct backward induction for non-positive powers.

First Method: Direct Method.

Consider the characteristic polynomial associated with the hybrid number:

$$P\left(t\right)=t^2-ut-v.$$

(4.1)

Let ${\lambda }_1$ and ${\lambda }_2$ be the roots of the equation $P\left(t\right)=0$. From Vieta’s formulas, the sum and product of the roots are given by:

$$\begin{array}{cc}\hfill {\lambda }_1+{\lambda }_2& =u,\hfill \end{array}$$

(4.2)

$$\begin{array}{cc}\hfill {\lambda }_1{\lambda }_2& =-v.\hfill \end{array}$$

(4.3)

By the division algorithm for polynomials, dividing $t^n$ by $P\left(t\right)$ yields a quotient $Q\left(t\right)$ and a linear remainder $At+B$:

$$t^n=Q\left(t\right)(t^2-ut-v)+At+B.$$

(4.4)

To determine the coefficients A and B, we must consider two cases depending on whether the roots of the characteristic polynomial are distinct or equal.

Case 1: The roots are distinct (${\lambda }_1\ne {\lambda }_2$).

To determine the coefficients A and B, we evaluate equation (4.4) at the roots $t={\lambda }_1$ and $t={\lambda }_2$. Since $P\left({\lambda }_1\right)=0$ and $P\left({\lambda }_2\right)=0$, the $Q\left(t\right)$ term vanishes, resulting in a system of two linear equations:

$$\begin{array}{cc}\hfill {\lambda }_1^n& =A{\lambda }_1+B\hfill \end{array}$$

(4.5)

$$\begin{array}{cc}\hfill {\lambda }_2^n& =A{\lambda }_2+B.\hfill \end{array}$$

(4.6)

Subtracting equation (4.6) from (4.5) eliminates B:

$$\begin{array}{cc}\hfill {\lambda }_1^n-{\lambda }_2^n& =A({\lambda }_1-{\lambda }_2),\hfill \\ \hfill A& ={\displaystyle \frac{{\lambda }_1^n-{\lambda }_2^n}{{\lambda }_1-{\lambda }_2}}.\hfill \end{array}$$

We recognize this expression as the Binet formula for the sequence $G_n=u{G}_{n-1}+v{G}_{n-2}$ with initial conditions $G_0=0$ and $G_1=1$. Therefore, we have $A=G_n$.

Next, to solve for B, we multiply equation (4.5) by ${\lambda }_2$, equation (4.6) by ${\lambda }_1$, and subtract the latter from the former:

$$\begin{array}{cc}\hfill {\lambda }_1^n{\lambda }_2-{\lambda }_2^n{\lambda }_1& =B({\lambda }_2-{\lambda }_1),\hfill \\ \hfill B({\lambda }_2-{\lambda }_1)& ={\lambda }_1{\lambda }_2({\lambda }_1^{n-1}-{\lambda }_2^{n-1}),\hfill \\ \hfill B& ={\displaystyle \frac{{\lambda }_1{\lambda }_2({\lambda }_1^{n-1}-{\lambda }_2^{n-1})}{{\lambda }_2-{\lambda }_1}},\hfill \\ \hfill B& =-{\lambda }_1{\lambda }_2\left({\displaystyle \frac{{\lambda }_1^{n-1}-{\lambda }_2^{n-1}}{{\lambda }_1-{\lambda }_2}}\right).\hfill \end{array}$$

Substituting the product of the roots ${\lambda }_1{\lambda }_2=-v$ and recognizing the Binet formula for $G_{n-1}$, we obtain:

$$B=-(-v)G_{n-1}=v{G}_{n-1}.$$

Case 2: The roots are equal (${\lambda }_1={\lambda }_2=\lambda$).

In this case, the characteristic polynomial is a perfect square, $P\left(t\right)={(t-\lambda )}^2$, where $u=2\lambda$ and $v=-{\lambda }^2$. Equation (4.4) becomes:

$$t^n=Q\left(t\right){(t-\lambda )}^2+At+B.$$

(4.7)

Evaluating this at $t=\lambda$ gives our first equation:

$${\lambda }^n=A\lambda +B.$$

(4.8)

Because $\lambda$ is a double root, we can obtain a second equation by taking the derivative of equation (4.7) with respect to t:

$$n{t}^{n-1}=Q^{\prime }\left(t\right){(t-\lambda )}^2+2Q\left(t\right)(t-\lambda )+A.$$

Evaluating this derivative at $t=\lambda$ isolates A:

$$A=n{\lambda }^{n-1}.$$

Substituting A back into equation (4.8) allows us to solve for B:

$$B={\lambda }^n-A\lambda ={\lambda }^n-\left(n{\lambda }^{n-1}\right)\lambda ={\lambda }^n-n{\lambda }^n=(1-n){\lambda }^n.$$

Now, we relate these coefficients to the fundamental sequence $G_n$. For a linear recurrence relation with a double root $\lambda$, the Binet formula takes the modified form $G_n=(C_1+C_{2}n){\lambda }^n$. Using the initial conditions $G_0=0$ and $G_1=1$, we find $C_1=0$ and $C_2=1/\lambda$. Thus, the explicit formula for the sequence is:

$$G_n=n{\lambda }^{n-1}.$$

This perfectly matches our derived value for A, so $A=G_n$.

To verify B, we use the identity $v=-{\lambda }^2$ alongside our sequence formula for $G_{n-1}$:

$$v{G}_{n-1}=(-{\lambda }^2)\left((n-1){\lambda }^{n-2}\right)=-(n-1){\lambda }^n=(1-n){\lambda }^n.$$

This perfectly matches our value for B, confirming that $B=v{G}_{n-1}$ holds for the equal roots case as well.

Conclusion and Extension of the First Method.

For $n\ge 0$, substituting A and B back into the polynomial division equation gives:

$$t^n=Q\left(t\right)(t^2-ut-v)+G_{n}t+v{G}_{n-1}.$$

Evaluating this polynomial identity at the hybrid number $t=\mathbf{Z}$ causes the quotient term to vanish, establishing ${\mathbf{Z}}^n=G_n\mathbf{Z}+v{G}_{n-1}$, i.e., because $\mathbf{Z}$ satisfies its own characteristic equation, ${\mathbf{Z}}^2-u\mathbf{Z}-v=0$, the quotient term vanishes completely:

$$\begin{array}{cc}\hfill {\mathbf{Z}}^n& =Q\left(\mathbf{Z}\right)\left(0\right)+G_n\mathbf{Z}+v{G}_{n-1},\hfill \\ \hfill {\mathbf{Z}}^n& =G_n\mathbf{Z}+v{G}_{n-1}.\hfill \end{array}$$

Furthermore, this algebraic method naturally extends to non-positive integers ($n\le 0$). When $C\left(\mathbf{Z}\right)\ne 0$, we have $v\ne 0$. Consequently, the product of the roots ${\lambda }_1{\lambda }_2=-v\ne 0$, ensuring that the roots are non-zero. This guarantees that ${\lambda }^n$ is well-defined for negative powers. By evaluating the functional identity on the spectrum of the invertible operator $\mathbf{Z}$, we obtain the exact same linear systems for both the distinct and equal root cases:

$$\begin{array}{cc}\hfill {\lambda }_1^n& =A{\lambda }_1+B,\hfill \\ \hfill {\lambda }_2^n& =A{\lambda }_2+B.\hfill \end{array}$$

Because the Binet formula analytically extends to negative indices, solving this system yields the identical algebraic solutions $A=G_n$ and $B=v{G}_{n-1}$. Thus, evaluating the identity at $t=\mathbf{Z}$ confirms that ${\mathbf{Z}}^n=G_n\mathbf{Z}+v{G}_{n-1}$ holds universally for all $n\in \mathbb{Z}$.

Second Method: Mathematical Induction.

We proceed by mathematical induction on n.

Forward Induction for Positive Integers ($n\ge 1$):

For the base case $n=1$, the formula yields:

$${\mathbf{Z}}^1=G_1\mathbf{Z}+v{G}_0=1·\mathbf{Z}+v·0=\mathbf{Z},$$

which trivially holds. For $n=2$, the formula yields:

$${\mathbf{Z}}^2=G_2\mathbf{Z}+v{G}_1=u\mathbf{Z}+v,$$

which is exactly the characteristic equation established in Theorem 4.

Now, assume that the identity ${\mathbf{Z}}^n=G_n\mathbf{Z}+v{G}_{n-1}$ holds for some arbitrary positive integer n. We evaluate the $(n+1)$-th power:

$$\begin{array}{cc}\hfill {\mathbf{Z}}^{n+1}& =\mathbf{Z}·{\mathbf{Z}}^n\hfill \\ \hfill & =\mathbf{Z}(G_n\mathbf{Z}+v{G}_{n-1})\hfill \\ \hfill & =G_n{\mathbf{Z}}^2+v{G}_{n-1}\mathbf{Z}.\hfill \end{array}$$

Substituting the characteristic equation ${\mathbf{Z}}^2=u\mathbf{Z}+v$ into this expression gives:

$$\begin{array}{cc}\hfill {\mathbf{Z}}^{n+1}& =G_n(u\mathbf{Z}+v)+v{G}_{n-1}\mathbf{Z}\hfill \\ \hfill & =(u{G}_n+v{G}_{n-1})\mathbf{Z}+v{G}_n.\hfill \end{array}$$

By the definition of the fundamental sequence, $G_{n+1}=u{G}_n+v{G}_{n-1}$. Substituting this recurrence relation back into our equation yields:

$${\mathbf{Z}}^{n+1}=G_{n+1}\mathbf{Z}+v{G}_n,$$

which completes the forward induction step.

Backward Induction for Non-Positive Integers ($n\le 0$):

When $C\left(\mathbf{Z}\right)\ne 0$, we have $v\ne 0$, ensuring both that ${\mathbf{Z}}^{-1}$ exists and that the sequence $G_n$ can be extended to negative indices via the relation $G_{n-2}={\displaystyle \frac{1}{v}}(G_n-u{G}_{n-1})$.

For the base case $n=0$, we evaluate the sequence recurrence at index 1:

$$G_1=u{G}_0+v{G}_{-1}\Rightarrow 1=u\left(0\right)+v{G}_{-1}\Rightarrow v{G}_{-1}=1$$

Applying the theorem’s identity for $n=0$ yields:

$${\mathbf{Z}}^0=G_0\mathbf{Z}+v{G}_{-1}=0·\mathbf{Z}+1=1,$$

which holds perfectly.

Assume the identity holds for some integer $n\le 0$. We must show it holds for $n-1$. Multiplying the assumed identity ${\mathbf{Z}}^n=G_n\mathbf{Z}+v{G}_{n-1}$ by ${\mathbf{Z}}^{-1}$ yields:

$${\mathbf{Z}}^{n-1}=G_n+v{G}_{n-1}{\mathbf{Z}}^{-1}.$$

From the characteristic equation ${\mathbf{Z}}^2-u\mathbf{Z}-v=0$, multiplying by ${\mathbf{Z}}^{-1}$ gives $\mathbf{Z}-u-v{\mathbf{Z}}^{-1}=0$, which rearranges to $v{\mathbf{Z}}^{-1}=\mathbf{Z}-u$. Substituting this into our expression for ${\mathbf{Z}}^{n-1}$:

$$\begin{array}{cc}\hfill {\mathbf{Z}}^{n-1}& =G_n+G_{n-1}(\mathbf{Z}-u)\hfill \\ \hfill & =G_{n-1}\mathbf{Z}+(G_n-u{G}_{n-1}).\hfill \end{array}$$

By the backward recurrence relation of the fundamental sequence, we know that $v{G}_{n-2}=G_n-u{G}_{n-1}$. Substituting this directly yields:

$${\mathbf{Z}}^{n-1}=G_{n-1}\mathbf{Z}+v{G}_{n-2}.$$

This completes the backward induction, proving that the structural identity gracefully extends to all non-positive integer powers.

Final Component Extraction:

Since the identity ${\mathbf{Z}}^n=G_n\mathbf{Z}+v{G}_{n-1}$ holds universally for all $n\in \mathbb{Z}$ (provided $v\ne 0$ for $n\le 0$), we substitute the general definition $\mathbf{Z}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$ into the established formula to extract the explicit components:

$$\begin{array}{cc}\hfill {\mathbf{Z}}^n& =G_n(a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h})+v{G}_{n-1}\hfill \\ \hfill & =(a{G}_n+v{G}_{n-1})+\left(b{G}_n\right)\mathbf{i}+\left(c{G}_n\right)\mathbf{\epsilon }+\left(d{G}_n\right)\mathbf{h}.\hfill \end{array}$$

This exactly matches the claimed coefficients $g_1,g_2,g_3$, and $g_4$, completing the proof. □

To demonstrate the computational efficiency of the characteristic equation and the fundamental sequence formulas we developed, let us explicitly calculate the positive and negative powers of a specific hybrid number.

Example 12.

For the hybrid number $\mathbf{Z}=2+3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h}$, we have $a=2,b=3,c=4,d=7$. Its real trace is $u=2\left(2\right)=4$, and its norm is $C\left(\mathbf{Z}\right)=2^2+3^2-2\left(3\right)\left(4\right)-7^2=4+9-24-49=-60$. Thus, $v=-C\left(\mathbf{Z}\right)=60$.

Using the structural identity ${\mathbf{Z}}^n=G_n\mathbf{Z}+v{G}_{n-1}$ (formally proved in Theorem 11), where $G_n=4G_{n-1}+60G_{n-2}$, we compute the forward sequence values up to $n=7$: $G_0=0,G_1=1,G_2=4,G_3=76,G_4=544,G_5=6736,G_6=59584,G_7=642496$.

Running the sequence backward via $G_{n-2}={\displaystyle \frac{1}{60}}(G_n-4G_{n-1})$, a profound pattern emerges when we express the fractions with a common denominator of $v^n$:

$$\begin{array}{ccc}\hfill G_{-1}& =& {\displaystyle \frac{1}{60}},G_{-2}=-{\displaystyle \frac{4}{{60}^2}},G_{-3}={\displaystyle \frac{76}{{60}^3}},G_{-4}=-{\displaystyle \frac{544}{{60}^4}},\hfill \\ \hfill G_{-5}& =& {\displaystyle \frac{6736}{{60}^5}},G_{-6}=-{\displaystyle \frac{59584}{{60}^6}},G_{-7}={\displaystyle \frac{642496}{{60}^7}},G_{-8}=-{\displaystyle \frac{6145024}{{60}^8}}.\hfill \end{array}$$

Notice the immediate symmetric relation $G_{-n}={\displaystyle \frac{{(-1)}^{n+1}}{v^n}}{G}_n$. Substituting the forward sequence into the fundamental identity yields the positive powers:

$$\begin{array}{ccc}\hfill {\mathbf{Z}}^2& =& 4\mathbf{Z}+60\left(1\right)=68+12\mathbf{i}+16\mathbf{\epsilon }+28\mathbf{h},\hfill \\ \hfill {\mathbf{Z}}^3& =& 76\mathbf{Z}+60\left(4\right)=392+228\mathbf{i}+304\mathbf{\epsilon }+532\mathbf{h},\hfill \\ \hfill {\mathbf{Z}}^4& =& 544\mathbf{Z}+60\left(76\right)=5648+1632\mathbf{i}+2176\mathbf{\epsilon }+3808\mathbf{h},\hfill \\ \hfill {\mathbf{Z}}^5& =& 6736\mathbf{Z}+60\left(544\right)=46112+20208\mathbf{i}+26944\mathbf{\epsilon }+47152\mathbf{h},\hfill \\ \hfill {\mathbf{Z}}^6& =& 59584\mathbf{Z}+60\left(6736\right)=523328+178752\mathbf{i}+238336\mathbf{\epsilon }+417088\mathbf{h},\hfill \\ \hfill {\mathbf{Z}}^7& =& 642496\mathbf{Z}+60\left(59584\right)=4860032+1927488\mathbf{i}+2569984\mathbf{\epsilon }+4497472\mathbf{h},\hfill \end{array}$$

Substituting the backward sequence yields the negative powers. By keeping the denominator unsimplified, a striking conjugate relationship becomes immediately apparent across all powers:

$$\begin{array}{ccc}\hfill {\mathbf{Z}}^{-1}& =& {\displaystyle \frac{1}{60}}\mathbf{Z}+60\left(-{\displaystyle \frac{4}{{60}^2}}\right)={\displaystyle \frac{-2+3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h}}{60}}={\displaystyle \frac{-\overline{{\mathbf{Z}}^1}}{{60}^1}},\hfill \\ \hfill {\mathbf{Z}}^{-2}& =& -{\displaystyle \frac{4}{{60}^2}}\mathbf{Z}+60\left({\displaystyle \frac{76}{{60}^3}}\right)={\displaystyle \frac{68-12\mathbf{i}-16\mathbf{\epsilon }-28\mathbf{h}}{{60}^2}}={\displaystyle \frac{\overline{{\mathbf{Z}}^2}}{{60}^2}},\hfill \\ \hfill {\mathbf{Z}}^{-3}& =& {\displaystyle \frac{76}{{60}^3}}\mathbf{Z}+60\left(-{\displaystyle \frac{544}{{60}^4}}\right)={\displaystyle \frac{-392+228\mathbf{i}+304\mathbf{\epsilon }+532\mathbf{h}}{{60}^3}}={\displaystyle \frac{-\overline{{\mathbf{Z}}^3}}{{60}^3}},\hfill \\ \hfill {\mathbf{Z}}^{-4}& =& -{\displaystyle \frac{544}{{60}^4}}\mathbf{Z}+60\left({\displaystyle \frac{6736}{{60}^5}}\right)={\displaystyle \frac{5648-1632\mathbf{i}-2176\mathbf{\epsilon }-3808\mathbf{h}}{{60}^4}}={\displaystyle \frac{\overline{{\mathbf{Z}}^4}}{{60}^4}},\hfill \\ \hfill {\mathbf{Z}}^{-5}& =& {\displaystyle \frac{6736}{{60}^5}}\mathbf{Z}+60\left(-{\displaystyle \frac{59584}{{60}^6}}\right)={\displaystyle \frac{-46112+20208\mathbf{i}+26944\mathbf{\epsilon }+47152\mathbf{h}}{{60}^5}}={\displaystyle \frac{-\overline{{\mathbf{Z}}^5}}{{60}^5}},\hfill \\ \hfill {\mathbf{Z}}^{-6}& =& -{\displaystyle \frac{59584}{{60}^6}}\mathbf{Z}+60\left({\displaystyle \frac{642496}{{60}^7}}\right)={\displaystyle \frac{523328-178752\mathbf{i}-238336\mathbf{\epsilon }-417088\mathbf{h}}{{60}^6}}={\displaystyle \frac{\overline{{\mathbf{Z}}^6}}{{60}^6}},\hfill \\ \hfill {\mathbf{Z}}^{-7}& =& {\displaystyle \frac{642496}{{60}^7}}\mathbf{Z}+60\left(-{\displaystyle \frac{6145024}{{60}^8}}\right)={\displaystyle \frac{-4860032+1927488\mathbf{i}+2569984\mathbf{\epsilon }+4497472\mathbf{h}}{{60}^7}}={\displaystyle \frac{-\overline{{\mathbf{Z}}^7}}{{60}^7}}.\hfill \end{array}$$

4.2. Method 2: The Lucas-Equivalence Framework

In the previous subsection, we established the recursive framework (Method 1) for evaluating powers of a hybrid number. We now present a highly elegant alternative that entirely decouples the real scalar part from the vector part of the hybrid exponentiation. By utilizing both the fundamental sequence $G_n$ and its companion fundamental-Lucas sequence $H_n$, we establish the foundational theorem from which all subsequent structural and combinatorial representations will be derived.

Theorem 13.

Let $\mathbf{Z}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}=a+V$ be a hybrid number satisfying ${\mathbf{Z}}^2-u\mathbf{Z}-v=0$, where $u=2a$ and $v=-C\left(\mathbf{Z}\right)$. Let $G_n$ and $H_n$ be its associated characteristic sequences. For any non-negative integer n, the n-th power of $\mathbf{Z}$ can be uniquely expressed as the decoupled linear combination:

$${\mathbf{Z}}^n={\displaystyle \frac{1}{2}}{H}_n+G_{n}V$$

(4.9)

Furthermore, if $\mathbf{Z}$ is invertible (i.e., $C\left(\mathbf{Z}\right)\ne 0$), this decoupled identity extends natively to all negative integers. For any positive integer n, we have:

$${\mathbf{Z}}^{-n}={\displaystyle \frac{1}{2}}{H}_{-n}+G_{-n}V$$

(4.10)

Proof. 

We prove the theorem in two parts, utilizing the recursive framework established in Theorem 11.

Part 1: Positive Powers. Theorem 11 dictates that ${\mathbf{Z}}^n=G_n\mathbf{Z}+v{G}_{n-1}$. By substituting the explicit definition of the hybrid number $\mathbf{Z}=a+V$ into this identity, we obtain:

$${\mathbf{Z}}^n=G_n(a+V)+v{G}_{n-1}=(a{G}_n+v{G}_{n-1})+G_{n}V.$$

We must now resolve the scalar component $a{G}_n+v{G}_{n-1}$. From the fundamental definitions of the characteristic sequences, we recall the standard linear recurrence relationship connecting the companion sequence $H_n$ to the fundamental sequence $G_n$ (see, for example, Soykan ([15], Lemma 9) or Soykan [16], Lemma 1.4):

$$H_n=2G_{n+1}-u{G}_n.$$

By shifting the index of the fundamental recurrence $G_{n+1}=u{G}_n+v{G}_{n-1}$, we substitute $G_{n+1}$ into our relation for $H_n$:

$$H_n=2(u{G}_n+v{G}_{n-1})-u{G}_n=u{G}_n+2v{G}_{n-1}.$$

Substituting the trace parameter $u=2a$ yields $H_n=2a{G}_n+2v{G}_{n-1}$, which immediately simplifies to:

$${\displaystyle \frac{1}{2}}{H}_n=a{G}_n+v{G}_{n-1}.$$

Substituting this scalar equivalence back into our exponentiation identity directly yields ${\mathbf{Z}}^n={\displaystyle \frac{1}{2}}{H}_n+G_{n}V$.

Part 2: Negative Powers. For negative indices (where $n>0$), Theorem 11 guarantees that the fundamental relation holds via backwards extension: ${\mathbf{Z}}^{-n}=G_{-n}\mathbf{Z}+v{G}_{-(n+1)}$. Substituting $\mathbf{Z}=a+V$ yields:

$${\mathbf{Z}}^{-n}=G_{-n}(a+V)+v{G}_{-(n+1)}=(a{G}_{-n}+v{G}_{-(n+1)})+G_{-n}V.$$

Because the sequence relation $H_k=2G_{k+1}-u{G}_k$ is universally valid for all integers $k\in \mathbb{Z}$, we evaluate it strictly at $k=-n$:

$$H_{-n}=2G_{-n+1}-u{G}_{-n}.$$

Applying the backwards recurrence $G_{-n+1}=u{G}_{-n}+v{G}_{-(n+1)}$ gives:

$$H_{-n}=2(u{G}_{-n}+v{G}_{-(n+1)})-u{G}_{-n}=u{G}_{-n}+2v{G}_{-(n+1)}.$$

Substituting $u=2a$ yields $H_{-n}=2a{G}_{-n}+2v{G}_{-(n+1)}$, which identically halves to:

$${\displaystyle \frac{1}{2}}{H}_{-n}=a{G}_{-n}+v{G}_{-(n+1)}.$$

Substituting this equivalence natively into the negative power expansion directly yields ${\mathbf{Z}}^{-n}={\displaystyle \frac{1}{2}}{H}_{-n}+G_{-n}V$, completing the proof. □

To concretely demonstrate the computational superiority of the unified Lucas-equivalence framework established in Theorem 13, we now explicitly evaluate the higher-order positive and negative powers of a specific hybrid number. By utilizing the characteristic sequences $G_n$ and $H_n$, we entirely bypass the tedious requirement of recursive vector multiplication, resolving the exponentiation strictly through rapid scalar sequence generation.

Example 14.

We now calculate the powers of the exact same hybrid number, $\mathbf{Z}=2+3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h}$, strictly utilizing the Lucas-Equivalence identity ${\mathbf{Z}}^n={\displaystyle \frac{1}{2}}{H}_n+G_{n}V$. For the hybrid number $\mathbf{Z}=2+3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h}$, the scalar part is $a=2$ and the vector part is $V=3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h}$. The scalar trace is $u=2\left(2\right)=4$, and its hybrid norm evaluates to $C\left(\mathbf{Z}\right)=2^2+3^2-2\left(3\right)\left(4\right)-7^2=-60$. Thus, the recurrence parameter is $v=-C\left(\mathbf{Z}\right)=60$.

Using the structural (fundamental) identity ${\mathbf{Z}}^n={\displaystyle \frac{1}{2}}{H}_n+G_{n}V$ (formally proved in Theorem 13), where $G_n=4G_{n-1}+60G_{n-2}$ and $H_n=4H_{n-1}+60H_{n-2}$ (since the recurrence parameters are $u=4$ and $v=60$), we compute the forward characteristic sequence values up to $n=7$:

$$G_0=0,G_1=1,G_2=4,G_3=76,G_4=544,G_5=6736,G_6=59584,G_7=642496$$

and

$$H_0=2,H_1=4,H_2=136,H_3=784,H_4=11296,H_5=92224,H_6=1046656,H_7=9720064$$

Running the sequences backward via the inverse relation $W_{n-2}={\displaystyle \frac{1}{60}}(W_n-4W_{n-1})$, we obtain the negative indices, keeping the denominators explicitly as powers of 60 to preserve their algebraic symmetry:

$$G_{-1}={\displaystyle \frac{1}{60}},G_{-2}=-{\displaystyle \frac{4}{{60}^2}},G_{-3}={\displaystyle \frac{76}{{60}^3}},G_{-4}=-{\displaystyle \frac{544}{{60}^4}},G_{-5}={\displaystyle \frac{6736}{{60}^5}},G_{-6}=-{\displaystyle \frac{59584}{{60}^6}},G_{-7}={\displaystyle \frac{642496}{{60}^7}},$$

and

$$H_{-1}=-{\displaystyle \frac{4}{60}},H_{-2}={\displaystyle \frac{136}{{60}^2}},H_{-3}=-{\displaystyle \frac{784}{{60}^3}},H_{-4}={\displaystyle \frac{11296}{{60}^4}},H_{-5}=-{\displaystyle \frac{92224}{{60}^5}},H_{-6}={\displaystyle \frac{1046656}{{60}^6}},H_{-7}=-{\displaystyle \frac{9720064}{{60}^7}}.$$

Substituting the forward sequences into the fundamental identity yields the explicit positive powers of $\mathbf{Z}$:

$$\begin{array}{cc}\hfill {\mathbf{Z}}^2& ={\displaystyle \frac{1}{2}}{H}_2+G_{2}V=68+4(3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h})=68+12\mathbf{i}+16\mathbf{\epsilon }+28\mathbf{h},\hfill \\ \hfill {\mathbf{Z}}^3& ={\displaystyle \frac{1}{2}}{H}_3+G_{3}V=392+228\mathbf{i}+304\mathbf{\epsilon }+532\mathbf{h},\hfill \\ \hfill {\mathbf{Z}}^4& ={\displaystyle \frac{1}{2}}{H}_4+G_{4}V=5648+1632\mathbf{i}+2176\mathbf{\epsilon }+3808\mathbf{h},\hfill \\ \hfill {\mathbf{Z}}^5& ={\displaystyle \frac{1}{2}}{H}_5+G_{5}V=46112+20208\mathbf{i}+26944\mathbf{\epsilon }+47152\mathbf{h},\hfill \\ \hfill {\mathbf{Z}}^6& ={\displaystyle \frac{1}{2}}{H}_6+G_{6}V=523328+178752\mathbf{i}+238336\mathbf{\epsilon }+417088\mathbf{h},\hfill \\ \hfill {\mathbf{Z}}^7& ={\displaystyle \frac{1}{2}}{H}_7+G_{7}V=4860032+1927488\mathbf{i}+2569984\mathbf{\epsilon }+4497472\mathbf{h}.\hfill \end{array}$$

Substituting the backward sequences perfectly generates the negative powers, exposing the profound structural property that ${\mathbf{Z}}^{-n}={\left(C\left(\mathbf{Z}\right)\right)}^{-n}\overline{{\mathbf{Z}}^n}$:

$$\begin{array}{cc}\hfill {\mathbf{Z}}^{-1}& ={\displaystyle \frac{1}{2}}{H}_{-1}+G_{-1}V={\displaystyle \frac{1}{60}}(-2+3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h}),\hfill \\ \hfill {\mathbf{Z}}^{-2}& ={\displaystyle \frac{1}{2}}{H}_{-2}+G_{-2}V={\displaystyle \frac{1}{{60}^2}}(68-12\mathbf{i}-16\mathbf{\epsilon }-28\mathbf{h}),\hfill \\ \hfill {\mathbf{Z}}^{-3}& ={\displaystyle \frac{1}{2}}{H}_{-3}+G_{-3}V={\displaystyle \frac{1}{{60}^3}}(-392+228\mathbf{i}+304\mathbf{\epsilon }+532\mathbf{h}),\hfill \\ \hfill {\mathbf{Z}}^{-4}& ={\displaystyle \frac{1}{2}}{H}_{-4}+G_{-4}V={\displaystyle \frac{1}{{60}^4}}(5648-1632\mathbf{i}-2176\mathbf{\epsilon }-3808\mathbf{h}),\hfill \\ \hfill {\mathbf{Z}}^{-5}& ={\displaystyle \frac{1}{2}}{H}_{-5}+G_{-5}V={\displaystyle \frac{1}{{60}^5}}(-46112+20208\mathbf{i}+26944\mathbf{\epsilon }+47152\mathbf{h}),\hfill \\ \hfill {\mathbf{Z}}^{-6}& ={\displaystyle \frac{1}{2}}{H}_{-6}+G_{-6}V={\displaystyle \frac{1}{{60}^6}}(523328-178752\mathbf{i}-238336\mathbf{\epsilon }-417088\mathbf{h}),\hfill \\ \hfill {\mathbf{Z}}^{-7}& ={\displaystyle \frac{1}{2}}{H}_{-7}+G_{-7}V={\displaystyle \frac{1}{{60}^7}}(-4860032+1927488\mathbf{i}+2569984\mathbf{\epsilon }+4497472\mathbf{h}).\hfill \end{array}$$

By establishing this Lucas-Equivalence as the foundational structural identity, the algebraic and combinatorial frameworks developed in the following subsections can now be derived instantaneously.

4.3. Method 3: The Structural Root Framework via Binet’s Formulas

Building upon the Lucas-Equivalence identity, we can now derive the explicit structural root expansion. Because this formulation completely evaluates the n-th power using solely the internal components of $\mathbf{Z}$ and its vector discriminant, it stands as a fundamental theorem of hybrid exponentiation.

Theorem 15

(Structural Root Expansion). Let $\mathbf{Z}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}=a+V$ be a hybrid number with vector discriminant $\Delta =-b^2+2bc+d^2=V^2$. We define the structural scaling sequences ${\mathcal{A}}_n$ and ${\mathcal{B}}_n$ as follows:

Case 1: Distinct Roots ($\Delta \ne 0$).

$${\mathcal{A}}_n={\displaystyle \frac{{(a+\sqrt{\Delta })}^n+{(a-\sqrt{\Delta })}^n}{2}}\mathit{and}{\mathcal{B}}_n={\displaystyle \frac{{(a+\sqrt{\Delta })}^n-{(a-\sqrt{\Delta })}^n}{2\sqrt{\Delta }}}.$$

(4.11)

Case 2: Equal Roots ($\Delta =0$).

$${\mathcal{A}}_n=a^n\mathit{and}{\mathcal{B}}_n=n{a}^{n-1}.$$

(4.12)

For any non-negative integer n, the n-th power of $\mathbf{Z}$ evaluates directly as:

$${\mathbf{Z}}^n={\mathcal{A}}_n+{\mathcal{B}}_{n}V.$$

(4.13)

Furthermore, if $\mathbf{Z}$ is invertible ($C\left(\mathbf{Z}\right)=a^2-\Delta \ne 0$), the negative powers are explicitly generated by dividing the algebraic conjugate by the character:

$${\mathbf{Z}}^{-n}={\displaystyle \frac{{\mathcal{A}}_n-{\mathcal{B}}_{n}V}{{(a^2-\Delta )}^n}}.$$

(4.14)

Proof. Positive Powers: By Theorem 13, ${\mathbf{Z}}^n={\displaystyle \frac{1}{2}}{H}_n+G_{n}V$. The roots of the characteristic equation ${\mathbf{Z}}^2-u\mathbf{Z}-v=0$ are ${\lambda }_{1,2}={\displaystyle \frac{u\pm \sqrt{u^2+4v}}{2}}$. Substituting $u=2a$ and $v=\Delta -a^2$ yields the exact roots ${\lambda }_{1,2}=a\pm \sqrt{\Delta }$.

If $\Delta \ne 0$ (${\lambda }_1\ne {\lambda }_2$), substituting the standard generalized Binet formulas $G_n={\displaystyle \frac{{\lambda }_1^n-{\lambda }_2^n}{{\lambda }_1-{\lambda }_2}}={\mathcal{B}}_n$ and $H_n={\lambda }_1^n+{\lambda }_2^n=2{\mathcal{A}}_n$ directly into the Lucas-equivalence identity immediately yields ${\mathbf{Z}}^n={\mathcal{A}}_n+{\mathcal{B}}_{n}V$.

If $\Delta =0$ (${\lambda }_1={\lambda }_2=a$), standard linear recurrence theory yields the degenerate Binet formulas $G_n=n{\lambda }^{n-1}=n{a}^{n-1}={\mathcal{B}}_n$ and $H_n=2{\lambda }^n=2a^n=2{\mathcal{A}}_n$. Substituting these into the Lucas-equivalence identity directly yields ${\mathbf{Z}}^n={\displaystyle \frac{1}{2}}\left(2{\mathcal{A}}_n\right)+\left({\mathcal{B}}_n\right)V={\mathcal{A}}_n+{\mathcal{B}}_{n}V$.

Negative Powers: For an invertible hybrid number, we utilize the algebraic conjugate property ${\mathbf{Z}}^{-n}={\displaystyle \frac{\overline{{\mathbf{Z}}^n}}{{\left(C\left(\mathbf{Z}\right)\right)}^n}}$. The expansion of ${\mathbf{Z}}^{-n}$ relies on the established multiplicativity of the character (Theorem 10), $C\left({\mathbf{Z}}^n\right)={\left(C\left(\mathbf{Z}\right)\right)}^n$, and the anti-involutive property of the conjugate (Theorem 9), $\overline{{\mathbf{Z}}^n}={\left(\overline{\mathbf{Z}}\right)}^n$. Because the derived scalar components ${\mathcal{A}}_n$ and ${\mathcal{B}}_n$ are purely real in both cases, applying the algebraic conjugate simply negates the vector part V. Substituting $C\left(\mathbf{Z}\right)=a^2-\Delta$ into the denominator exactly yields the fractional formula ${\displaystyle \frac{{\mathcal{A}}_n-{\mathcal{B}}_{n}V}{{(a^2-\Delta )}^n}}$, completing the proof. □

To observe the mechanics of the structural root formula in practice, we revisit our standard test hybrid number.

Example 16.

For the hybrid number $\mathbf{Z}=2+3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h}$, $a=2,b=3,c=4,d=7$ and the character is $C\left(\mathbf{Z}\right)=a^2+b^2-2bc-d^2=-60$. The vector part is $V=b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}=3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h}$ and the vector discriminant is $\Delta =-b^2+2bc+d^2=64$. By Theorem 15, the explicit sequences ${\mathcal{A}}_n$ and ${\mathcal{B}}_n$ evaluate as:

$$\begin{array}{ccc}\hfill {\mathcal{A}}_n& =& {\displaystyle \frac{{(a+\sqrt{\Delta })}^n+{(a-\sqrt{\Delta })}^n}{2}}={\displaystyle \frac{{10}^n+{(-6)}^n}{2}},\hfill \\ \hfill {\mathcal{B}}_n& =& {\displaystyle \frac{{(a+\sqrt{\Delta })}^n-{(a-\sqrt{\Delta })}^n}{2\sqrt{\Delta }}}={\displaystyle \frac{{10}^n-{(-6)}^n}{16}}.\hfill \end{array}$$

For any non-negative integer n, the n-th power is strictly determined by ${\mathbf{Z}}^n={\mathcal{A}}_n+{\mathcal{B}}_{n}V$, and for any positive integer $n\in \mathbb{N}$, the negative power is generated via ${\mathbf{Z}}^{-n}={\displaystyle \frac{{\mathcal{A}}_n-{\mathcal{B}}_{n}V}{{(-60)}^n}}$.

Evaluating these exact analytical sums up to $n=10$ yields:

$$\begin{array}{ccc}\hfill {\mathbf{Z}}^2& =& {\displaystyle \frac{{10}^2+{(-6)}^2}{2}}+{\displaystyle \frac{{10}^2-{(-6)}^2}{16}}(3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h})=68+12\mathbf{i}+16\mathbf{\epsilon }+28\mathbf{h},\hfill \\ \hfill {\mathbf{Z}}^3& =& {\displaystyle \frac{{10}^3+{(-6)}^3}{2}}+{\displaystyle \frac{{10}^3-{(-6)}^3}{16}}(3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h})=392+228\mathbf{i}+304\mathbf{\epsilon }+532\mathbf{h},\hfill \\ \hfill {\mathbf{Z}}^4& =& {\displaystyle \frac{{10}^4+{(-6)}^4}{2}}+{\displaystyle \frac{{10}^4-{(-6)}^4}{16}}(3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h})=5648+1632\mathbf{i}+2176\mathbf{\epsilon }+3808\mathbf{h},\hfill \\ \hfill {\mathbf{Z}}^5& =& {\displaystyle \frac{{10}^5+{(-6)}^5}{2}}+{\displaystyle \frac{{10}^5-{(-6)}^5}{16}}(3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h})=46112+20208\mathbf{i}+26944\mathbf{\epsilon }+47152\mathbf{h},\hfill \\ \hfill {\mathbf{Z}}^6& =& {\displaystyle \frac{{10}^6+{(-6)}^6}{2}}+{\displaystyle \frac{{10}^6-{(-6)}^6}{16}}(3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h})=523328+178752\mathbf{i}+238336\mathbf{\epsilon }+417088\mathbf{h},\hfill \\ \hfill {\mathbf{Z}}^7& =& {\displaystyle \frac{{10}^7+{(-6)}^7}{2}}+{\displaystyle \frac{{10}^7-{(-6)}^7}{16}}(3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h})=4860032+1927488\mathbf{i}+2569984\mathbf{\epsilon }+4497472\mathbf{h},\hfill \\ \hfill {\mathbf{Z}}^8& =& {\displaystyle \frac{{10}^8+{(-6)}^8}{2}}+{\displaystyle \frac{{10}^8-{(-6)}^8}{16}}V=50839808+18435072\mathbf{i}+24580096\mathbf{\epsilon }+43015168\mathbf{h},\hfill \\ \hfill {\mathbf{Z}}^9& =& {\displaystyle \frac{{10}^9+{(-6)}^9}{2}}+{\displaystyle \frac{{10}^9-{(-6)}^9}{16}}V=494961152+189389568\mathbf{i}+252519424\mathbf{\epsilon }+441908992\mathbf{h},\hfill \\ \hfill {\mathbf{Z}}^{10}& =& {\displaystyle \frac{{10}^{10}+{(-6)}^{10}}{2}}+{\displaystyle \frac{{10}^{10}-{(-6)}^{10}}{16}}V=5030233088+1863662592\mathbf{i}+2484883456\mathbf{\epsilon }+4348546048\mathbf{h}.\hfill \end{array}$$

And running the exact fractional formulas natively generates the negative powers:

$$\begin{array}{ccc}\hfill {\mathbf{Z}}^{-1}& =& {\displaystyle \frac{\frac{{10}^1+{(-6)}^1}{2}-\frac{{10}^1-{(-6)}^1}{16}(3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h})}{{(-60)}^1}}={\displaystyle \frac{1}{60}}(-2+3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h}),\hfill \\ \hfill {\mathbf{Z}}^{-2}& =& {\displaystyle \frac{\frac{{10}^2+{(-6)}^2}{2}-\frac{{10}^2-{(-6)}^2}{16}(3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h})}{{(-60)}^2}}={\displaystyle \frac{1}{{60}^2}}(68-12\mathbf{i}-16\mathbf{\epsilon }-28\mathbf{h}),\hfill \\ \hfill {\mathbf{Z}}^{-3}& =& {\displaystyle \frac{\frac{{10}^3+{(-6)}^3}{2}-\frac{{10}^3-{(-6)}^3}{16}(3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h})}{{(-60)}^3}}={\displaystyle \frac{1}{{60}^3}}(-392+228\mathbf{i}+304\mathbf{\epsilon }+532\mathbf{h}),\hfill \\ \hfill {\mathbf{Z}}^{-4}& =& {\displaystyle \frac{\frac{{10}^4+{(-6)}^4}{2}-\frac{{10}^4-{(-6)}^4}{16}(3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h})}{{(-60)}^4}}={\displaystyle \frac{1}{{60}^4}}(5648-1632\mathbf{i}-2176\mathbf{\epsilon }-3808\mathbf{h}),\hfill \\ \hfill {\mathbf{Z}}^{-5}& =& {\displaystyle \frac{\frac{{10}^5+{(-6)}^5}{2}-\frac{{10}^5-{(-6)}^5}{16}(3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h})}{{(-60)}^5}}={\displaystyle \frac{1}{{60}^5}}(-46112+20208\mathbf{i}+26944\mathbf{\epsilon }+47152\mathbf{h}),\hfill \\ \hfill {\mathbf{Z}}^{-6}& =& {\displaystyle \frac{\frac{{10}^6+{(-6)}^6}{2}-\frac{{10}^6-{(-6)}^6}{16}(3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h})}{{(-60)}^6}}={\displaystyle \frac{1}{{60}^6}}(523328-178752\mathbf{i}-238336\mathbf{\epsilon }-417088\mathbf{h}),\hfill \\ \hfill {\mathbf{Z}}^{-7}& =& {\displaystyle \frac{\frac{{10}^7+{(-6)}^7}{2}-\frac{{10}^7-{(-6)}^7}{16}(3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h})}{{(-60)}^7}}={\displaystyle \frac{1}{{60}^7}}(-4860032+1927488\mathbf{i}+2569984\mathbf{\epsilon }+4497472\mathbf{h}),\hfill \\ \hfill {\mathbf{Z}}^{-8}& =& {\displaystyle \frac{\frac{{10}^8+{(-6)}^8}{2}-\frac{{10}^8-{(-6)}^8}{16}V}{{(-60)}^8}}={\displaystyle \frac{1}{{60}^8}}(50839808-18435072\mathbf{i}-24580096\mathbf{\epsilon }-43015168\mathbf{h}),\hfill \\ \hfill {\mathbf{Z}}^{-9}& =& {\displaystyle \frac{\frac{{10}^9+{(-6)}^9}{2}-\frac{{10}^9-{(-6)}^9}{16}V}{{(-60)}^9}}={\displaystyle \frac{1}{{60}^9}}(-494961152+189389568\mathbf{i}+252519424\mathbf{\epsilon }+441908992\mathbf{h}),\hfill \\ \hfill {\mathbf{Z}}^{-10}& =& {\displaystyle \frac{\frac{{10}^{10}+{(-6)}^{10}}{2}-\frac{{10}^{10}-{(-6)}^{10}}{16}V}{{(-60)}^{10}}}={\displaystyle \frac{1}{{60}^{10}}}(5030233088-1863662592\mathbf{i}-2484883456\mathbf{\epsilon }-4348546048\mathbf{h}).\hfill \end{array}$$

4.4. Method 4: The Combinatorial Framework via Horadam Polynomials

While the structural root formulas established in Method 3 provide an elegant analytic method for evaluating arbitrary powers, calculating $\sqrt{\Delta }$ introduces temporary complex arithmetic if the vector discriminant is negative ($\Delta <0$). We now demonstrate that the exact same universal scaling sequences ${\mathcal{A}}_n$ and ${\mathcal{B}}_n$ from Theorem 15 can be generated without any square roots. By utilizing a structured binomial expansion, we evaluate the exponentiation of $\mathbf{Z}$ purely through real combinatorial mathematics

Theorem 17

(Combinatorial Expansion). Let $\mathbf{Z}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}=a+V$ be a hybrid number with vector discriminant $\Delta =-b^2+2bc+d^2=V^2$. We define the combinatorial scaling sequences ${\mathcal{A}}_n$ and ${\mathcal{B}}_n$ as follows:

Case 1: Distinct Roots ($\Delta \ne 0$).

$${\mathcal{A}}_n={\displaystyle \sum _{k=0}^{\lfloor n/2\rfloor }}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)a^{n-2k}{\Delta }^k\mathit{and}{\mathcal{B}}_n={\displaystyle \sum _{k=0}^{\lfloor (n-1)/2\rfloor }}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k+1}}\right)a^{n-1-2k}{\Delta }^k$$

(4.15)

Case 2: Equal Roots ($\Delta =0$).Because the combinatorial summations elegantly collapse to their $k=0$ terms, we directly yield:

$${\mathcal{A}}_n=a^n\mathit{and}{\mathcal{B}}_n=n{a}^{n-1}$$

(4.16)

For any non-negative integer n, the n-th power of $\mathbf{Z}$ evaluates directly as:

$${\mathbf{Z}}^n={\mathcal{A}}_n+{\mathcal{B}}_{n}V$$

(4.17)

Furthermore, if $\mathbf{Z}$ is invertible ($C\left(\mathbf{Z}\right)=a^2-\Delta \ne 0$), the negative powers are explicitly generated by dividing the algebraic conjugate by the character:

$${\mathbf{Z}}^{-n}={\displaystyle \frac{{\mathcal{A}}_n-{\mathcal{B}}_{n}V}{{(a^2-\Delta )}^n}}$$

(4.18)

Proof. Positive Powers: In the study of generalized Fibonacci (Horadam) polynomials, Soykan (2023) established explicit combinatorial expressions for the generalized sequences $G_n$ and $H_n$ defined by the recurrence parameters r and s. According to Corollary 2 of that work, for distinct roots, the exact combinatorial expansions are given as (1.13) and (1.14).

In our hybrid algebra framework, the recurrence parameters map exactly as $r=u=2a$ and $(r^2+4s)=(u^2+4v)=4\Delta$. Substituting these mapped parameters into combinatorial identity (1.14) for $H_n$ and dividing by $2^n$ yields exactly ${\displaystyle \frac{1}{2}}{H}_n={\mathcal{A}}_n$. Applying the identical substitution to (1.13) for $G_n$ and dividing by $2^{n-1}$ identically yields $G_n={\mathcal{B}}_n$.

Furthermore, when $\Delta =0$ (which corresponds perfectly to $r^2+4s=0$), all terms containing ${\Delta }^k$ for $k\ge 1$ in the summation completely vanish, leaving only the $k=0$ components, perfectly recovering the degenerate formulas ${\mathcal{A}}_n=\left({\displaystyle \genfrac{}{}{0pt}{}{n}{0}}\right)a^n=a^n$ and ${\mathcal{B}}_n=\left({\displaystyle \genfrac{}{}{0pt}{}{n}{1}}\right)a^{n-1}=n{a}^{n-1}$.

By substituting these precise combinatorial equivalences back into Theorem 13 (${\mathbf{Z}}^n={\displaystyle \frac{1}{2}}{H}_n+G_{n}V$), the combinatorial framework is instantaneously proven.

Negative Powers: For negative indices, we apply the exact same conjugate property ${\mathbf{Z}}^{-n}={\displaystyle \frac{\overline{{\mathbf{Z}}^n}}{{(a^2-\Delta )}^n}}$. Because the combinatorial sequences ${\mathcal{A}}_n$ and ${\mathcal{B}}_n$ are purely real summations over integer powers (which naturally truncate for $\Delta =0$), the conjugate negates only the vector part V, strictly verifying the exact negative combinatorial expansion across both the distinct and equal root cases. □

Remark 6.

From the explicit derivations constructed in the proofs of Theorem 15 and Theorem 17, we establish the universal scaling equivalence that intrinsically bridges the structural, combinatorial, and Lucas-equivalence frameworks:

$${\mathcal{A}}_n={\displaystyle \frac{1}{2}}{H}_n\mathit{and}{\mathcal{B}}_n=G_n.$$

To demonstrate how these combinatorial sequences generate highly structured patterns during actual computation, we will now explicitly expand the alternating sequence.

Example 18.

By extracting the alternating scalar and vector terms from the combinatorial expansions defined in Theorem 17, a highly structured combinatorial pattern emerges intuitively. Because the real scalar a commutes with the vector V, higher powers of V simply alternate between scalar multiples of Δ and vector multiples of V (e.g., $V^2=\Delta$, $V^3=\Delta V,$$V^4={\Delta }^2,$$V^5={\Delta }^{2}V,$$V^6={\Delta }^3,$$V^7={\Delta }^{3}V$). Consequently, the positive powers of any hybrid number $\mathbf{Z}$ up to $n=10$ elegantly expand via pure real combinatorics as:

$$\begin{array}{ccc}\hfill {\mathbf{Z}}^2& =& (a^2+\Delta )+\left(2a\right)V,\hfill \\ \hfill {\mathbf{Z}}^3& =& (a^3+3a\Delta )+(3a^2+\Delta )V,\hfill \\ \hfill {\mathbf{Z}}^4& =& (a^4+6a^2\Delta +{\Delta }^2)+(4a^3+4a\Delta )V,\hfill \\ \hfill {\mathbf{Z}}^5& =& (a^5+10a^3\Delta +5a{\Delta }^2)+(5a^4+10a^2\Delta +{\Delta }^2)V,\hfill \\ \hfill {\mathbf{Z}}^6& =& (a^6+15a^4\Delta +15a^2{\Delta }^2+{\Delta }^3)+(6a^5+20a^3\Delta +6a{\Delta }^2)V,\hfill \\ \hfill {\mathbf{Z}}^7& =& (a^7+21a^5\Delta +35a^3{\Delta }^2+7a{\Delta }^3)+(7a^6+35a^4\Delta +21a^2{\Delta }^2+{\Delta }^3)V,\hfill \\ \hfill {\mathbf{Z}}^8& =& (a^8+28a^6\Delta +70a^4{\Delta }^2+28a^2{\Delta }^3+{\Delta }^4)+(8a^7+56a^5\Delta +56a^3{\Delta }^2+8a{\Delta }^3)V,\hfill \\ \hfill {\mathbf{Z}}^9& =& (a^9+36a^7\Delta +126a^5{\Delta }^2+84a^3{\Delta }^3+9a{\Delta }^4)+(9a^8+84a^6\Delta +126a^4{\Delta }^2+36a^2{\Delta }^3+{\Delta }^4)V,\hfill \\ \hfill {\mathbf{Z}}^{10}& =& (a^{10}+45a^8\Delta +210a^6{\Delta }^2+210a^4{\Delta }^3+45a^2{\Delta }^4+{\Delta }^5)+(10a^9+120a^7\Delta +252a^5{\Delta }^2+120a^3{\Delta }^3+10a{\Delta }^4)V.\hfill \end{array}$$

For an invertible hybrid number ($C\left(\mathbf{Z}\right)=a^2-\Delta \ne 0$), we seamlessly reuse these combinatorial expansions for negative indices. By negating the vector component and scaling by the character, the negative powers unfold perfectly symmetrically:

$$\begin{array}{ccc}\hfill {\mathbf{Z}}^{-1}& =& {\displaystyle \frac{1}{a^2-\Delta }}(a-V),\hfill \\ \hfill {\mathbf{Z}}^{-2}& =& {\displaystyle \frac{1}{{(a^2-\Delta )}^2}}((a^2+\Delta )-\left(2a\right)V),\hfill \\ \hfill {\mathbf{Z}}^{-3}& =& {\displaystyle \frac{1}{{(a^2-\Delta )}^3}}((a^3+3a\Delta )-(3a^2+\Delta )V),\hfill \\ \hfill {\mathbf{Z}}^{-4}& =& {\displaystyle \frac{1}{{(a^2-\Delta )}^4}}((a^4+6a^2\Delta +{\Delta }^2)-(4a^3+4a\Delta )V),\hfill \\ \hfill {\mathbf{Z}}^{-5}& =& {\displaystyle \frac{1}{{(a^2-\Delta )}^5}}((a^5+10a^3\Delta +5a{\Delta }^2)-(5a^4+10a^2\Delta +{\Delta }^2)V),\hfill \\ \hfill {\mathbf{Z}}^{-6}& =& {\displaystyle \frac{1}{{(a^2-\Delta )}^6}}((a^6+15a^4\Delta +15a^2{\Delta }^2+{\Delta }^3)-(6a^5+20a^3\Delta +6a{\Delta }^2)V),\hfill \\ \hfill {\mathbf{Z}}^{-7}& =& {\displaystyle \frac{1}{{(a^2-\Delta )}^7}}((a^7+21a^5\Delta +35a^3{\Delta }^2+7a{\Delta }^3)-(7a^6+35a^4\Delta +21a^2{\Delta }^2+{\Delta }^3)V),\hfill \\ \hfill {\mathbf{Z}}^{-8}& =& {\displaystyle \frac{1}{{(a^2-\Delta )}^8}}((a^8+28a^6\Delta +70a^4{\Delta }^2+28a^2{\Delta }^3+{\Delta }^4)-(8a^7+56a^5\Delta +56a^3{\Delta }^2+8a{\Delta }^3)V),\hfill \\ \hfill {\mathbf{Z}}^{-9}& =& {\displaystyle \frac{1}{{(a^2-\Delta )}^9}}((a^9+36a^7\Delta +126a^5{\Delta }^2+84a^3{\Delta }^3+9a{\Delta }^4)-(9a^8+84a^6\Delta +126a^4{\Delta }^2+36a^2{\Delta }^3+{\Delta }^4)V),\hfill \\ \hfill {\mathbf{Z}}^{-10}& =& {\displaystyle \frac{1}{{(a^2-\Delta )}^{10}}}((a^{10}+45a^8\Delta +210a^6{\Delta }^2+210a^4{\Delta }^3+45a^2{\Delta }^4+{\Delta }^5)\hfill \\ & & -(10a^9+120a^7\Delta +252a^5{\Delta }^2+120a^3{\Delta }^3+10a{\Delta }^4)V).\hfill \end{array}$$

For the hybrid number $\mathbf{Z}=2+3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h}$, by simply substituting $a=2,V=3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h}$ and $\Delta =64$ directly into these combinatorial polynomials, we flawlessly reproduce the exact coefficient vectors previously derived in Examples 12, 14 and 16.

4.5. Summary and Comprehensive Methodological Comparison

To conclude our investigation of hybrid exponentiation with real components, we synthesize the four distinct methodologies developed in this section into a single, comprehensive algebraic statement. There is a profound mathematical elegance in how these seemingly disparate concepts intricately connect. By observing how a single hybrid number simultaneously behaves as a geometric vector, a root-based algebraic structure, and a generator for classical Fibonacci and Lucas sequences, the intrinsic beauty of this algebra is fully revealed.

Specifically, the Lucas-Equivalence Framework (Method 2) acts as the ultimate unifying bridge between the recursive and structural worlds. Because it relies on $G_n$ and $H_n$, it operates natively as a recursive sequence; yet, by explicitly decoupling the scalar and vector parts, it perfectly satisfies the algebraic conjugate constraints of the structural formulations.

Remark 7

(Comprehensive Methodological Comparison). The mathematical architecture of hybrid exponentiation is now fully supported by four distinct, yet perfectly unified, analytical frameworks. Comparing the mechanics of the four examples evaluated throughout this section (where $\mathbf{Z}=2+3\mathbf{i}+4\mathbf{\epsilon }+7\mathbf{h}$) provides a striking validation of this overarching algebraic theory:

• Method 1 (The Recursive Framework - Theorem 11):Demonstrated in Example 12, this method computes ${\mathbf{Z}}^n=G_n\mathbf{Z}+v{G}_{n-1}$. It is ideal for continuous, sequential step-by-step generation using only the single fundamental sequence $G_n$. However, it requires recursively distributing $G_n$ across the entire hybrid number $\mathbf{Z}$ and continually shifting the real scalar part at every step.
• Method 2 (The Lucas-Equivalence Framework - Theorem 13):Demonstrated in Example 14, this calculates ${\mathbf{Z}}^n={\displaystyle \frac{1}{2}}{H}_n+G_{n}V$. This approach is algebraically superior for sequential mapping because it perfectly decouples the scalar and vector components into the generalized Lucas ($H_n$) and Fibonacci ($G_n$) sequences. This clean separation entirely eliminates the need to recursively recombine real and vector terms.
• Method 3 (The Structural Root Framework - Theorem 15):Demonstrated in Example 16, this method evaluates the power analytically using $\sqrt{\Delta }$. It is theoretically profound, providing instantaneous "random access" to any specific power without computing preceding sequence terms. However, it requires evaluating square roots, which temporarily introduces complex arithmetic if the vector discriminant is negative ($\Delta <0$).
• Method 4 (The Combinatorial Framework - Theorem 17):Demonstrated in Example 18, this approach offers the ultimate discrete algorithmic solution. By expressing the scaling sequences ${\mathcal{A}}_n$ and ${\mathcal{B}}_n$ purely as combinatorial summations over integer powers of Δ, it completely bypasses the need to evaluate $\sqrt{\Delta }$. It computes arbitrary isolated positive and negative powers directly using strictly real algebraic mathematics, making it vastly superior for direct software implementation and cryptography.

Ultimately, all four methodologies fundamentally compute the exact same positive and negative polynomial expansions, as empirical calculations beautifully reveal that the multiplicative property of the hybrid norm extends symmetrically across all integer powers.

The following Master Theorem formally guarantees that regardless of the computational perspective chosen, the exponentiation of any hybrid number with real coefficients evaluates universally and consistently.

Theorem 19

(Unified Master Theorem of Hybrid Exponentiation). Let $\mathbf{Z}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}=a+V$ be an arbitrary hybrid number satisfying the characteristic equation ${\mathbf{Z}}^2-u\mathbf{Z}-v=0,$ where $u=2a$ and $v=-C\left(\mathbf{Z}\right)$. Let its character be $C\left(\mathbf{Z}\right)=a^2-\Delta$, where the vector discriminant is $\Delta =-b^2+2bc+d^2=V^2$.

For any integer $n\ge 1$, the n-th power of $\mathbf{Z}$ can be precisely evaluated using any of the following four algebraically equivalent identities:

(a) 

Method 1 (Fundamental Sequence Recursive Framework):

$${\mathbf{Z}}^n=G_n\mathbf{Z}+v{G}_{n-1}$$

(4.19)

where $G_n=u{G}_{n-1}+v{G}_{n-2}$ with $G_0=0,G_1=1$.

(b) 

Method 2 (Characteristic Sequences Recursive Framework):

$${\mathbf{Z}}^n={\displaystyle \frac{1}{2}}{H}_n+G_{n}V$$

(4.20)

where $H_n=u{H}_{n-1}+v{H}_{n-2}$ with Lucas-type initial conditions $H_0=2,H_1=u$.

(c) 

Method 3 (Structural Root Framework via Binet’s Formulas):

$${\mathbf{Z}}^n=\left({\displaystyle \frac{{(a+\sqrt{\Delta })}^n+{(a-\sqrt{\Delta })}^n}{2}}\right)+\left({\displaystyle \frac{{(a+\sqrt{\Delta })}^n-{(a-\sqrt{\Delta })}^n}{2\sqrt{\Delta }}}\right)V$$

(4.21)

(d) 

Method 4 (Combinatorial Framework):

$${\mathbf{Z}}^n=\left({\displaystyle \sum _{k=0}^{\lfloor n/2\rfloor }}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)a^{n-2k}{\Delta }^k\right)+\left({\displaystyle \sum _{k=0}^{\lfloor (n-1)/2\rfloor }}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k+1}}\right)a^{n-1-2k}{\Delta }^k\right)V$$

(4.22)

Furthermore, if $\mathbf{Z}$ is invertible (i.e., $C\left(\mathbf{Z}\right)\ne 0$), these identities seamlessly extend to evaluate the negative powers (denoted as $-n$ for $n\ge 1$).

• Under Methods 1 and 2 (The Recursive Frameworks):The negative powers are evaluated natively by extending the characteristic sequences ($G_n$ and $H_n$) backward via the relation $W_{k-2}={\displaystyle \frac{1}{v}}(W_k-u{W}_{k-1})$:

  $$\begin{array}{cc}\hfill \mathit{Method}\mathit{1}\text{:}& {\mathbf{Z}}^{-n}=G_{-n}\mathbf{Z}+v{G}_{-(n+1)}\hfill \end{array}$$

  (4.23)

  $$\begin{array}{cc}\hfill \mathit{Method}\mathit{2}\text{:}& {\mathbf{Z}}^{-n}={\displaystyle \frac{1}{2}}{H}_{-n}+G_{-n}V\hfill \end{array}$$

  (4.24)
• Under Methods 3 and 4 (The Structural Frameworks):The negative powers are equivalently generated by dividing the algebraic conjugate of the positive expansion by the n-th power of the character:

  $${\mathbf{Z}}^{-n}={\displaystyle \frac{\overline{{\mathbf{Z}}^n}}{{\left(C\left(\mathbf{Z}\right)\right)}^n}}={\displaystyle \frac{a_n-V_n}{{(a^2-\Delta )}^n}}$$

  (4.25)

  where the explicit scalar component ($a_n$) and vector component ($V_n$) corresponding to each respective framework are given by:

  Method 3:

  $a_n=\left({\displaystyle \frac{{(a+\sqrt{\Delta })}^n+{(a-\sqrt{\Delta })}^n}{2}}\right)$ and $V_n=\left({\displaystyle \frac{{(a+\sqrt{\Delta })}^n-{(a-\sqrt{\Delta })}^n}{2\sqrt{\Delta }}}\right)V.$

  Method 4:

  $a_n={\displaystyle \sum _{k=0}^{\lfloor n/2\rfloor }}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)a^{n-2k}{\Delta }^k$ and $V_n=\left({\displaystyle \sum _{k=0}^{\lfloor (n-1)/2\rfloor }}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k+1}}\right)a^{n-1-2k}{\Delta }^k\right)V.$

5. Mixed Products of Hybrid Powers

We now extend our investigation from the exponentiation of a single hybrid number to the mixed products of arbitrary hybrid powers, specifically evaluating terms of the form $Z_1^k{Z}_2^m$. Due to the non-commutative nature of hybrid multiplication, evaluating such products manually via standard expansion is computationally prohibitive and highly susceptible to cross-term errors. However, by leveraging the distinct computational frameworks established in the previous sections—namely, recursive sequences, structural root analysis, and combinatorial vector decompositions—we can derive exact, closed-form algebraic identities for these mixed non-commutative products.

To establish our first mixed product identity, we leverage the recursive framework. In the following theorem, we take into consideration Theorem 11, which states that for $\mathbf{Z}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$, the integer power satisfies the property:

$${\mathbf{Z}}^k=G_k\mathbf{Z}+v{G}_{k-1}$$

where the fundamental sequence $G_n$ is defined by the recurrence relation $G_k=u{G}_{k-1}+v{G}_{k-2}$ with initial conditions $G_0=0$ and $G_1=1$. By applying this property simultaneously to two distinct hybrid numbers, we elegantly resolve their non-commutative mixed product.

Theorem 20.

Let $Z_1=a_1+b_1\mathbf{i}+c_1\mathbf{\epsilon }+d_1\mathbf{h}$ and $Z_2=a_2+b_2\mathbf{i}+c_2\mathbf{\epsilon }+d_2\mathbf{h}$ be two hybrid numbers. Let $W_k$ and $Y_m$ be their respective fundamental real sequences defined by the recurrence relations:

$$\begin{array}{cc}\hfill W_k& =u_1{W}_{k-1}+v_1{W}_{k-2},(W_0=0,W_1=1),\hfill \\ \hfill Y_m& =u_2{Y}_{m-1}+v_2{Y}_{m-2},(Y_0=0,Y_1=1),\hfill \end{array}$$

where $u_1=2a_1$, $v_1=-C\left(Z_1\right)$ and $u_2=2a_2$, $v_2=-C\left(Z_2\right)$.

Then the mixed product of their powers, $Z_1^k{Z}_2^m$, is given directly by the linear combination:

$$Z_1^k{Z}_2^m=W_k{Y}_m\left(Z_1{Z}_2\right)+v_2{W}_k{Y}_{m-1}{Z}_1+v_1{W}_{k-1}{Y}_m{Z}_2+v_1{v}_2{W}_{k-1}{Y}_{m-1},$$

(5.1)

where the non-commutative product $Z_1{Z}_2$ can be evaluated explicitly using Definition 2 or directly via the $4\times 4$ left-multiplication matrix representation of $Z_1$ applied to the components of $Z_2$.

Proof. 

From the previously established power identities for hybrid numbers (Theorem 11), we can express the k-th power of $Z_1$ and the m-th power of $Z_2$ as:

$$\begin{array}{cc}\hfill Z_1^k& =W_k{Z}_1+v_1{W}_{k-1},\hfill \end{array}$$

(5.2)

$$\begin{array}{cc}\hfill Z_2^m& =Y_m{Z}_2+v_2{Y}_{m-1}.\hfill \end{array}$$

(5.3)

To find the mixed product, we multiply equations (5.2) and (5.3) in strict sequence:

$$Z_1^k{Z}_2^m=(W_k{Z}_1+v_1{W}_{k-1})(Y_m{Z}_2+v_2{Y}_{m-1}).$$

(5.4)

Because the sequence terms ($W_k,W_{k-1},Y_m,Y_{m-1}$) and the parameter coefficients ($v_1,v_2$) are strictly real scalars, they commute freely with the hybrid numbers $Z_1$ and $Z_2$. Distributing the terms while strictly preserving the non-commutative order of the hybrid product $Z_1{Z}_2$ yields:

$$Z_1^k{Z}_2^m=W_k{Y}_m\left(Z_1{Z}_2\right)+W_k{v}_2{Y}_{m-1}{Z}_1+v_1{W}_{k-1}{Y}_m{Z}_2+v_1{v}_2{W}_{k-1}{Y}_{m-1}.$$

(5.5)

Rearranging the real scalars into a standard presentation format yields the final identity:

$$Z_1^k{Z}_2^m=W_k{Y}_m\left(Z_1{Z}_2\right)+v_2{W}_k{Y}_{m-1}{Z}_1+v_1{W}_{k-1}{Y}_m{Z}_2+v_1{v}_2{W}_{k-1}{Y}_{m-1}.$$

This completes the proof. □

While Theorem 20 is highly effective for sequential generation, we can also evaluate the mixed product directly by decomposing the hybrid numbers into their commuting scalar and non-commuting vector parts. Utilizing the universal scaling sequences ${\mathcal{A}}_n$ and ${\mathcal{B}}_n$ derived analytically via the Structural Root Framework (Method 3, Theorem 15) or combinatorially via the Combinatorial Framework (Method 4, Theorem 17), we formulate a powerful closed-form identity for arbitrary positive mixed powers.

Theorem 21.

Let $Z_1=a_1+V_1$ and $Z_2=a_2+V_2$ be two hybrid numbers with vector parts $V_1$ and $V_2$. For any non-negative integers k and m, let ${\mathcal{A}}_{1,k},{\mathcal{B}}_{1,k}$ be the scalar scaling sequences for $Z_1$, and ${\mathcal{A}}_{2,m},{\mathcal{B}}_{2,m}$ be the scalar scaling sequences for $Z_2$ (defined explicitly via Theorem 15 or combinatorially via Theorem 17).

The mixed product $Z_1^k{Z}_2^m$ is explicitly given by:

$$Z_1^k{Z}_2^m={\mathcal{A}}_{1,k}{\mathcal{A}}_{2,m}+{\mathcal{B}}_{1,k}{\mathcal{A}}_{2,m}{V}_1+{\mathcal{A}}_{1,k}{\mathcal{B}}_{2,m}{V}_2+{\mathcal{B}}_{1,k}{\mathcal{B}}_{2,m}\left(V_1{V}_2\right),$$

(5.6)

where the product of the pure vector components $\left(V_1{V}_2\right)$ is determined directly via Definition 2 by setting the scalar terms $a_1=0$ and $a_2=0$.

Proof. 

By Theorem 15 and Theorem 17, the isolated powers of $Z_1$ and $Z_2$ expand perfectly into their scalar and vector scaling sequences:

$$\begin{array}{cc}\hfill Z_1^k& ={\mathcal{A}}_{1,k}+{\mathcal{B}}_{1,k}{V}_1,\hfill \\ \hfill Z_2^m& ={\mathcal{A}}_{2,m}+{\mathcal{B}}_{2,m}{V}_2.\hfill \end{array}$$

Multiplying these two powers in strict order gives:

$$Z_1^k{Z}_2^m=({\mathcal{A}}_{1,k}+{\mathcal{B}}_{1,k}{V}_1)({\mathcal{A}}_{2,m}+{\mathcal{B}}_{2,m}{V}_2).$$

Because ${\mathcal{A}}_{1,k}$, ${\mathcal{B}}_{1,k}$, ${\mathcal{A}}_{2,m}$, and ${\mathcal{B}}_{2,m}$ are strictly real numbers, they commute globally across the equation. The only non-commutative interaction occurs between the distinct hybrid vectors $V_1$ and $V_2$. Expanding the product and factoring the scalars yields:

$$Z_1^k{Z}_2^m={\mathcal{A}}_{1,k}{\mathcal{A}}_{2,m}+{\mathcal{A}}_{1,k}\left({\mathcal{B}}_{2,m}{V}_2\right)+\left({\mathcal{B}}_{1,k}{V}_1\right){\mathcal{A}}_{2,m}+\left({\mathcal{B}}_{1,k}{V}_1\right)\left({\mathcal{B}}_{2,m}{V}_2\right),$$

$$Z_1^k{Z}_2^m={\mathcal{A}}_{1,k}{\mathcal{A}}_{2,m}+{\mathcal{B}}_{1,k}{\mathcal{A}}_{2,m}{V}_1+{\mathcal{A}}_{1,k}{\mathcal{B}}_{2,m}{V}_2+{\mathcal{B}}_{1,k}{\mathcal{B}}_{2,m}\left(V_1{V}_2\right).$$

This explicitly separates the commutative scalar scaling behavior from the non-commutative structural interaction of the vectors, completing the proof. □

Furthermore, this vector decomposition elegantly preserves its algebraic symmetry when evaluated over negative integers. By applying the conjugate property established in the previous frameworks, we can seamlessly extend the mixed product identity to accommodate any combination of negative integer powers.

Corollary 4.

Let $Z_1=a_1+V_1$ and $Z_2=a_2+V_2$ be invertible hybrid numbers such that $C\left(Z_1\right)\ne 0$ and $C\left(Z_2\right)\ne 0$. For any positive integers k and m, the mixed products involving negative powers are given explicitly by:

(a)

Both Negative Powers:

$$Z_1^{-k}{Z}_2^{-m}={\displaystyle \frac{{\mathcal{A}}_{1,k}{\mathcal{A}}_{2,m}-{\mathcal{B}}_{1,k}{\mathcal{A}}_{2,m}{V}_1-{\mathcal{A}}_{1,k}{\mathcal{B}}_{2,m}{V}_2+{\mathcal{B}}_{1,k}{\mathcal{B}}_{2,m}\left(V_1{V}_2\right)}{{\left(C\left(Z_1\right)\right)}^k{\left(C\left(Z_2\right)\right)}^m}}.$$

(5.7)

(b)

First Positive, Second Negative:

$$Z_1^k{Z}_2^{-m}={\displaystyle \frac{{\mathcal{A}}_{1,k}{\mathcal{A}}_{2,m}+{\mathcal{B}}_{1,k}{\mathcal{A}}_{2,m}{V}_1-{\mathcal{A}}_{1,k}{\mathcal{B}}_{2,m}{V}_2-{\mathcal{B}}_{1,k}{\mathcal{B}}_{2,m}\left(V_1{V}_2\right)}{{\left(C\left(Z_2\right)\right)}^m}}.$$

(5.8)

(c)

First Negative, Second Positive:

$$Z_1^{-k}{Z}_2^m={\displaystyle \frac{{\mathcal{A}}_{1,k}{\mathcal{A}}_{2,m}-{\mathcal{B}}_{1,k}{\mathcal{A}}_{2,m}{V}_1+{\mathcal{A}}_{1,k}{\mathcal{B}}_{2,m}{V}_2-{\mathcal{B}}_{1,k}{\mathcal{B}}_{2,m}\left(V_1{V}_2\right)}{{\left(C\left(Z_1\right)\right)}^k}}.$$

(5.9)

Proof. 

We prove each case sequentially by applying the established positive and negative expansion identities.

(a)

By Theorem 17, the exact algebraic expansions for the negative powers are:

$$\begin{array}{cc}\hfill Z_1^{-k}& ={\displaystyle \frac{{\mathcal{A}}_{1,k}-{\mathcal{B}}_{1,k}{V}_1}{{\left(C\left(Z_1\right)\right)}^k}},\hfill \\ \hfill Z_2^{-m}& ={\displaystyle \frac{{\mathcal{A}}_{2,m}-{\mathcal{B}}_{2,m}{V}_2}{{\left(C\left(Z_2\right)\right)}^m}}.\hfill \end{array}$$

We multiply $Z_1^{-k}$ and $Z_2^{-m}$. The denominators are real scalars representing powers of the hybrid character, meaning they commute and can be multiplied securely:

$$Z_1^{-k}{Z}_2^{-m}={\displaystyle \frac{({\mathcal{A}}_{1,k}-{\mathcal{B}}_{1,k}{V}_1)({\mathcal{A}}_{2,m}-{\mathcal{B}}_{2,m}{V}_2)}{{\left(C\left(Z_1\right)\right)}^k{\left(C\left(Z_2\right)\right)}^m}}.$$

Distributing the numerator identically to Theorem 21 requires careful observation of the negative signs. The cross terms acquire a negative sign, while the product of the two negative vector terms yields $(-V_1)(-V_2)=V_1{V}_2$:

$$\mathrm{Numerator}={\mathcal{A}}_{1,k}{\mathcal{A}}_{2,m}-{\mathcal{A}}_{1,k}{\mathcal{B}}_{2,m}{V}_2-{\mathcal{B}}_{1,k}{\mathcal{A}}_{2,m}{V}_1+{\mathcal{B}}_{1,k}{\mathcal{B}}_{2,m}\left(V_1{V}_2\right).$$

Rearranging the terms to mirror the standard presentation structure directly produces identity (a).

(b)

For the mixed product $Z_1^k{Z}_2^{-m}$, we multiply the positive expansion of $Z_1^k$ (Theorem 17) by the negative expansion of $Z_2^{-m}$ (Theorem 17):

$$Z_1^k{Z}_2^{-m}=({\mathcal{A}}_{1,k}+{\mathcal{B}}_{1,k}{V}_1)\left({\displaystyle \frac{{\mathcal{A}}_{2,m}-{\mathcal{B}}_{2,m}{V}_2}{{\left(C\left(Z_2\right)\right)}^m}}\right).$$

Because the denominator is a real scalar, we can factor it out entirely. Expanding the numerator yields:

$$({\mathcal{A}}_{1,k}+{\mathcal{B}}_{1,k}{V}_1)({\mathcal{A}}_{2,m}-{\mathcal{B}}_{2,m}{V}_2)={\mathcal{A}}_{1,k}{\mathcal{A}}_{2,m}-{\mathcal{A}}_{1,k}{\mathcal{B}}_{2,m}{V}_2+{\mathcal{B}}_{1,k}{\mathcal{A}}_{2,m}{V}_1-{\mathcal{B}}_{1,k}{\mathcal{B}}_{2,m}\left(V_1{V}_2\right).$$

Rearranging these terms into standard order and dividing by the scalar denominator ${\left(C\left(Z_2\right)\right)}^m$ perfectly yields identity (b).

(c)

For the mixed product $Z_1^{-k}{Z}_2^m$, we multiply the negative expansion of $Z_1^{-k}$ by the positive expansion of $Z_2^m$:

$$Z_1^{-k}{Z}_2^m=\left({\displaystyle \frac{{\mathcal{A}}_{1,k}-{\mathcal{B}}_{1,k}{V}_1}{{\left(C\left(Z_1\right)\right)}^k}}\right)({\mathcal{A}}_{2,m}+{\mathcal{B}}_{2,m}{V}_2).$$

Factoring out the scalar denominator ${\left(C\left(Z_1\right)\right)}^k$ and expanding the numerator yields:

$$({\mathcal{A}}_{1,k}-{\mathcal{B}}_{1,k}{V}_1)({\mathcal{A}}_{2,m}+{\mathcal{B}}_{2,m}{V}_2)={\mathcal{A}}_{1,k}{\mathcal{A}}_{2,m}+{\mathcal{A}}_{1,k}{\mathcal{B}}_{2,m}{V}_2-{\mathcal{B}}_{1,k}{\mathcal{A}}_{2,m}{V}_1-{\mathcal{B}}_{1,k}{\mathcal{B}}_{2,m}\left(V_1{V}_2\right).$$

Rearranging into standard order and applying the denominator directly produces identity (c), completing the proof.

□

Remark 8.

To practically apply the mixed product identities established in Theorem 21 and Corollary 4, one must explicitly evaluate the non-commutative pure vector product $\left(V_1{V}_2\right)$. Let the isolated vector parts be given by $V_1=b_1\mathbf{i}+c_1\mathbf{\epsilon }+d_1\mathbf{h}$ and $V_2=b_2\mathbf{i}+c_2\mathbf{\epsilon }+d_2\mathbf{h}$. By applying Definition 2 and setting the scalar terms to zero ($a_1=0$ and $a_2=0$), we obtain the exact algebraic expansion:

$$\begin{array}{cc}\hfill V_1{V}_2& =(-b_1{b}_2+b_1{c}_2+c_1{b}_2+d_1{d}_2)\hfill \\ \hfill & +(b_1{d}_2-d_1{b}_2)\mathbf{i}\hfill \\ \hfill & +(b_1{d}_2-d_1{b}_2+d_1{c}_2-c_1{d}_2)\mathbf{\epsilon }\hfill \\ \hfill & +(-b_1{c}_2+c_1{b}_2)\mathbf{h}.\hfill \end{array}$$

(5.10)

Notably, if $V_1=V_2=V$ (i.e., we are squaring a single vector), all the cross-terms in the $\mathbf{i}$, $\mathbf{\epsilon }$, and $\mathbf{h}$ components perfectly cancel to zero, leaving only the real scalar $(-{\mathbf{b}}^{\mathbf{2}}+\mathbf{2}\mathbf{bc}+{\mathbf{d}}^{\mathbf{2}})$, which is strictly consistent with our definition of the vector discriminant Δ.

6. Generalization to Complex Hybrid Numbers and Higher-Dimensional Tensor Algebras

Throughout the preceding sections, we restricted the components of the hybrid number $\mathbf{Z}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$ to the field of real numbers ($a,b,c,d\in \mathbb{R}$). However, the structural methodologies we developed—specifically the left-regular matrix representation, the quadratic characteristic equation, and the combinatorial vector decomposition—do not strictly require a real base field.

In this section, we extend our investigation to the ring of complex hybrid numbers (often referred to as bi-hybrid numbers), where the coefficients are drawn from the complex field $\mathbb{C}$. To avoid notational ambiguity with the hybrid imaginary unit $\mathbf{i}$ (where ${\mathbf{i}}^2=-1$), we explicitly denote the standard complex imaginary unit as I, such that $I^2=-1$. Thus, a complex hybrid number ${\mathbf{Z}}_{\mathbb{C}}$ is formally defined as:

$${\mathbf{Z}}_{\mathbb{C}}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$$

(6.1)

where $a,b,c,d\in \mathbb{C}$ take the form $x+yI$ for $x,y\in \mathbb{R}$.

Because the complex field $\mathbb{C}$ is a commutative field, the complex scalars $a,b,c,d$ commute freely with the non-commutative primary hybrid units $\{\mathbf{i},\mathbf{\epsilon },\mathbf{h}\}$. Consequently, the algebra of complex hybrid numbers is strictly isomorphic to the ring of $2\times 2$ complex matrices, ${\mathcal{M}}_2\left(\mathbb{C}\right)$.

This isomorphism guarantees a profound algebraic continuity: the fundamental theorems established for real hybrid numbers analytically extend to the complex domain. The hybrid conjugate remains a perfect anti-involution, the characteristic equation ${\mathbf{Z}}_{\mathbb{C}}^2-2a{\mathbf{Z}}_{\mathbb{C}}+C\left({\mathbf{Z}}_{\mathbb{C}}\right)=0$ holds identically with complex traces and determinants, and the fundamental sequence $G_k$ transitions seamlessly into a linear recurrence sequence over $\mathbb{C}$. As we will establish throughout the latter half of this section, this commutative framework serves as the foundational blueprint for projecting hybrid numbers into a vast spectrum of generalized spaces. Whether a researcher intends to customize coefficients with Gaussian numbers, expand dimensionality via dual units, or construct hypercomplex tensor spaces such as Hybrid-Quaternions, the mathematical principles of strict unit separation and global commutativity introduced here remain absolutely non-negotiable for generating higher dimensions. Conversely, we also demonstrate that purposefully equating these units serves as a highly effective analytical tool in its own regard. By intentionally inducing dimensional collapse—such as in the case of the Gaussian Fibonacci Hybrid sequence—researchers can seamlessly leverage the robust 4-dimensional real framework established in our preceding sections to resolve complex-coefficient sequences with remarkable efficiency.

6.1. Complex Recursive Framework and Fundamental Sequence (Method 1)

Just as the combinatorial expansion extends to the complex plane, the recursive sequence framework detailed in Theorem 11 also generalizes flawlessly. When evaluating a complex hybrid number, the real trace $u=2a$ and the real norm $v=-C\left({\mathbf{Z}}_{\mathbb{C}}\right)$ transition into complex scalars. Consequently, the fundamental sequence $G_k$ transitions from a real integer sequence into a complex linear recurrence sequence.

Theorem 22.

Let ${\mathbf{Z}}_{\mathbb{C}}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$ be a complex hybrid number satisfying the complex characteristic equation:

$${\mathbf{Z}}_{\mathbb{C}}^2-u{\mathbf{Z}}_{\mathbb{C}}-v=0$$

(6.2)

where $u=2a\in \mathbb{C}$ and $v=-C\left({\mathbf{Z}}_{\mathbb{C}}\right)\in \mathbb{C}$.

Let $G_k$ be the complex fundamental sequence defined by the linear recurrence relation:

$$G_k=u{G}_{k-1}+v{G}_{k-2}$$

(6.3)

with standard initial conditions $G_0=0$ and $G_1=1$.

For any integer $k\ge 1$, the k-th power of ${\mathbf{Z}}_{\mathbb{C}}$ is given by the identity:

$${\mathbf{Z}}_{\mathbb{C}}^k=G_k{\mathbf{Z}}_{\mathbb{C}}+v{G}_{k-1}.$$

(6.4)

Furthermore, if ${\mathbf{Z}}_{\mathbb{C}}$ is invertible over $\mathbb{C}$ (i.e., its complex character $C\left({\mathbf{Z}}_{\mathbb{C}}\right)\ne 0$, implying $v\ne 0$), this identity holds for all integers $k\in \mathbb{Z}$, where the complex sequence $G_k$ is extended to negative indices via the backward relation $G_{k-2}={\displaystyle \frac{1}{v}}(G_k-u{G}_{k-1})$.

Proof. 

Proof. The proof is structurally identical to that of Theorem 11. The validity of the recursive power identity relies exclusively on two foundational algebraic properties: first, that the element satisfies a quadratic characteristic equation, and second, that the coefficients u and v commute with the element itself. Because $u,v\in \mathbb{C}$ are scalars belonging to a commutative field, they globally commute with the non-commutative hybrid units. Thus, both the forward mathematical induction (for $k\ge 1$) and the backward induction requiring the inverse element (for $k\le 0$) execute perfectly without requiring non-commutative adjustments. □

6.2. Complex Lucas-Type Sequences and Structural Equivalence (Method 2)

Building upon the successful extension of both the recursive fundamental sequence (Theorem 22) and the combinatorial expansion (Theorem 25) to the complex domain, we establish that the structural equivalence unifying these frameworks also remains perfectly intact over $\mathbb{C}$.

Just as in the real domain, we define the complex Lucas-type sequence associated with the complex hybrid number ${\mathbf{Z}}_{\mathbb{C}}$.

Definition 23.

Let ${\mathbf{Z}}_{\mathbb{C}}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$ be a complex hybrid number. The complex Lucas-type sequence $H_{n,\mathbb{C}}$ associated with ${\mathbf{Z}}_{\mathbb{C}}$ is defined by the recurrence relation:

$$H_{n,\mathbb{C}}=u{H}_{n-1,\mathbb{C}}+v{H}_{n-2,\mathbb{C}}=2a{H}_{n-1,\mathbb{C}}-C\left({\mathbf{Z}}_{\mathbb{C}}\right)H_{n-2,\mathbb{C}}$$

(6.5)

with the complex initial conditions $H_{0,\mathbb{C}}=2$ and $H_{1,\mathbb{C}}=u=2a$, where $a\in \mathbb{C}$.

With this sequence formally extended, the equivalence mapping the combinatorial expansions to the recursive sequences holds identically for bi-hybrid numbers.

Theorem 24.

Let ${\mathbf{Z}}_{\mathbb{C}}=a+V$ be a complex hybrid number. Let $G_{n,\mathbb{C}}$ and $H_{n,\mathbb{C}}$ be its associated complex fundamental and Lucas-type sequences, and let ${\mathcal{A}}_n$ and ${\mathcal{B}}_n$ be its complex combinatorial scaling sequences. For any non-negative integer n, the exact equivalences hold:

$${\mathcal{A}}_n={\displaystyle \frac{1}{2}}{H}_{n,\mathbb{C}}\mathrm{and}{\mathcal{B}}_n=G_{n,\mathbb{C}}$$

(6.6)

Consequently, the n-th power of ${\mathbf{Z}}_{\mathbb{C}}$ can be explicitly evaluated as:

$${\mathbf{Z}}_{\mathbb{C}}^n={\displaystyle \frac{1}{2}}{H}_{n,\mathbb{C}}+G_{n,\mathbb{C}}V$$

(6.7)

Proof. 

Proof. The proof is structurally identical to the real derivation presented in Theorem 13. Because the complex field $\mathbb{C}$ is globally commutative with the primary hybrid basis, the complex Binet formulas for $G_{n,\mathbb{C}}$ and $H_{n,\mathbb{C}}$ behave identically to the real case. The algebraic equivalence ${\mathcal{A}}_n={\displaystyle \frac{1}{2}}{H}_{n,\mathbb{C}}$ and ${\mathcal{B}}_n=G_{n,\mathbb{C}}$ relies entirely on evaluating the complex characteristic roots ${\lambda }_{1,2}=a\pm \sqrt{\Delta }$, which maintains rigorous structural validity in the complex domain. □

6.3. Structural and Complex Combinatorial Expansions (Methods 3 & 4)

When extending to $\mathbb{C}$, the hybrid character $C\left({\mathbf{Z}}_{\mathbb{C}}\right)=a^2+b^2-2bc-d^2$ and the vector square $\Delta =-b^2+2bc+d^2$ are no longer restricted to the real line; they evaluate to complex scalars. Despite this, the combinatorial expansion remains structurally identical.

Remark 9.

To explicitly verify that the vector square evaluates cleanly to a complex scalar, let $V=b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$ be the vector part of a complex hybrid number, where the coefficients $b,c,d\in \mathbb{C}$. When evaluating the vector square $V^2$, we distribute the terms. Crucially, because the complex coefficients $b,c,d$ are drawn from a commutative field and we explicitly assume the complex imaginary unit I is distinct from the hybrid unit $\mathbf{i}$, the complex coefficients $b,c,d$ act as central elements. They globally commute with the non-commutative hybrid basis $\{\mathbf{i},\mathbf{\epsilon },\mathbf{h}\}$, which allows us to cleanly factor the scalars out from the unit products:

$$\begin{array}{cc}\hfill V^2& =(b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h})(b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h})\hfill \\ \hfill & =b^2{\mathbf{i}}^2+c^2{\mathbf{\epsilon }}^2+d^2{\mathbf{h}}^2+bc(\mathbf{i}\mathbf{\epsilon }+\mathbf{\epsilon }\mathbf{i})+bd(\mathbf{i}\mathbf{h}+\mathbf{h}\mathbf{i})+cd(\mathbf{\epsilon }\mathbf{h}+\mathbf{h}\mathbf{\epsilon }).\hfill \end{array}$$

Applying Özdemir’s fundamental hybrid multiplication rules, we evaluate the pure squares and the anti-commutator interaction brackets:

• Pure Squares:${\mathbf{i}}^2=-1$, ${\mathbf{\epsilon }}^2=0$, and ${\mathbf{h}}^2=1$.
• Cross-Term Interactions:

  −

  $\mathbf{i}\mathbf{\epsilon }+\mathbf{\epsilon }\mathbf{i}=(1-\mathbf{h})+(1+\mathbf{h})=2$

  −

  $\mathbf{i}\mathbf{h}+\mathbf{h}\mathbf{i}=(\mathbf{\epsilon }+\mathbf{i})+(-\mathbf{\epsilon }-\mathbf{i})=0$

  −

  $\mathbf{\epsilon }\mathbf{h}+\mathbf{h}\mathbf{\epsilon }=(-\mathbf{\epsilon })+\mathbf{\epsilon }=0$

Substituting these evaluations back into the expansion elegantly eliminates all non-commutative basis vectors:

$$\begin{array}{cc}\hfill V^2& =b^2(-1)+c^2\left(0\right)+d^2\left(1\right)+bc\left(2\right)+bd\left(0\right)+cd\left(0\right)\hfill \\ \hfill & =-b^2+2bc+d^2.\hfill \end{array}$$

Because the complex field $\mathbb{C}$ is closed under standard algebraic addition and multiplication, and given that $b,c,d\in \mathbb{C}$, it strictly follows that the resulting scalar combination must also belong to the complex field. Therefore, the complex vector discriminant evaluates exactly to

$$\Delta =V^2=-b^2+2bc+d^2\in \mathbb{C}.$$

With the rigorous guarantee that this non-commutative vector square cleanly collapses into a pure complex scalar—a result strictly dependent on our assumption that the complex imaginary unit I is distinct from the hybrid unit $\mathbf{i}$—we can now confidently apply the combinatorial expansion framework directly to the complex domain.

Theorem 25

(Complex Combinatorial Expansion). Let ${\mathbf{Z}}_{\mathbb{C}}=a+V$ be a complex hybrid number with scalar part $a\in \mathbb{C}$ and vector part $V=b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$, where $b,c,d\in \mathbb{C}$. Let the complex vector discriminant be $\Delta =V^2=-b^2+2bc+d^2\in \mathbb{C}$.

Define the complex combinatorial scalar sequences ${\mathcal{A}}_n$ and ${\mathcal{B}}_n$ for any non-negative integer n:

$${\mathcal{A}}_n={\displaystyle \sum _{k=0}^{\lfloor n/2\rfloor }}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)a^{n-2k}{\Delta }^k\mathit{and}{\mathcal{B}}_n={\displaystyle \sum _{k=0}^{\lfloor (n-1)/2\rfloor }}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k+1}}\right)a^{n-1-2k}{\Delta }^k.$$

(6.8)

The positive powers of ${\mathbf{Z}}_{\mathbb{C}}$ are explicitly given by:

$${\mathbf{Z}}_{\mathbb{C}}^n={\mathcal{A}}_n+{\mathcal{B}}_{n}V.$$

(6.9)

Furthermore, if ${\mathbf{Z}}_{\mathbb{C}}$ is invertible over the complex field (i.e., its complex character $C\left({\mathbf{Z}}_{\mathbb{C}}\right)=a^2-\Delta \ne 0$), the negative powers exactly satisfy:

$${\mathbf{Z}}_{\mathbb{C}}^{-n}={\displaystyle \frac{{\mathcal{A}}_n-{\mathcal{B}}_{n}V}{{(a^2-\Delta )}^n}}.$$

(6.10)

Proof. 

Proof. The proof follows identically to the derivations in Theorem 17. The fundamental mechanism governing the binomial expansion of ${(a+V)}^n$ is the mutual commutativity of a and V. Because $a\in \mathbb{C}$ commutes with V, the algebraic distribution over the binomial coefficients holds universally. Similarly, taking the algebraic conjugate to evaluate the negative powers relies exclusively on the matrix adjugate identity, which is universally valid for ${\mathcal{M}}_2\left(\mathbb{C}\right)$. Thus, the real identities map directly to $\mathbb{C}$ without modification. □

Remark 10.

While the combinatorial formulas (Theorem 25) bypass the need to explicitly calculate the characteristic roots ${\lambda }_{1,2}=a\pm \sqrt{\Delta }$, the structural root formulas from Theorem 15 also remain valid for complex hybrid numbers. In the complex domain, $\sqrt{\Delta }$ must be evaluated as the principal complex square root.

6.4. Complex Mixed Products

Because the combinatorial scalar sequences ${\mathcal{A}}_n$ and ${\mathcal{B}}_n$ are perfectly well-defined over $\mathbb{C}$, the mixed product identities for evaluating expressions like $Z_{1,\mathbb{C}}^k{Z}_{2,\mathbb{C}}^m$ transition directly to the complex domain.

Theorem 26.

Let $Z_1=a_1+V_1$ and $Z_2=a_2+V_2$ be two complex hybrid numbers, with $a_1,a_2\in \mathbb{C}$. Let ${\mathcal{A}}_{1,k},{\mathcal{B}}_{1,k}$ and ${\mathcal{A}}_{2,m},{\mathcal{B}}_{2,m}$ be their respective complex scaling sequences as defined in Theorem 25. For any positive integers k and m, the mixed products are explicitly given by the following four cases:

(a)

Both Positive Powers:

$$Z_1^k{Z}_2^m={\mathcal{A}}_{1,k}{\mathcal{A}}_{2,m}+{\mathcal{B}}_{1,k}{\mathcal{A}}_{2,m}{V}_1+{\mathcal{A}}_{1,k}{\mathcal{B}}_{2,m}{V}_2+{\mathcal{B}}_{1,k}{\mathcal{B}}_{2,m}\left(V_1{V}_2\right).$$

(6.11)

(b)

Both Negative Powers(requires $C\left(Z_1\right)\ne 0$ and $C\left(Z_2\right)\ne 0$):

$$Z_1^{-k}{Z}_2^{-m}={\displaystyle \frac{{\mathcal{A}}_{1,k}{\mathcal{A}}_{2,m}-{\mathcal{B}}_{1,k}{\mathcal{A}}_{2,m}{V}_1-{\mathcal{A}}_{1,k}{\mathcal{B}}_{2,m}{V}_2+{\mathcal{B}}_{1,k}{\mathcal{B}}_{2,m}\left(V_1{V}_2\right)}{{\left(C\left(Z_1\right)\right)}^k{\left(C\left(Z_2\right)\right)}^m}}.$$

(6.12)

(c)

First Positive, Second Negative(requires $C\left(Z_2\right)\ne 0$):

$$Z_1^k{Z}_2^{-m}={\displaystyle \frac{{\mathcal{A}}_{1,k}{\mathcal{A}}_{2,m}+{\mathcal{B}}_{1,k}{\mathcal{A}}_{2,m}{V}_1-{\mathcal{A}}_{1,k}{\mathcal{B}}_{2,m}{V}_2-{\mathcal{B}}_{1,k}{\mathcal{B}}_{2,m}\left(V_1{V}_2\right)}{{\left(C\left(Z_2\right)\right)}^m}}.$$

(6.13)

(d)

First Negative, Second Positive(requires $C\left(Z_1\right)\ne 0$):

$$Z_1^{-k}{Z}_2^m={\displaystyle \frac{{\mathcal{A}}_{1,k}{\mathcal{A}}_{2,m}-{\mathcal{B}}_{1,k}{\mathcal{A}}_{2,m}{V}_1+{\mathcal{A}}_{1,k}{\mathcal{B}}_{2,m}{V}_2-{\mathcal{B}}_{1,k}{\mathcal{B}}_{2,m}\left(V_1{V}_2\right)}{{\left(C\left(Z_1\right)\right)}^k}}.$$

(6.14)

In all cases, the non-commutative pure vector product $\left(V_1{V}_2\right)$ evaluates exactly as in the real case:

$$V_1{V}_2=(-b_1{b}_2+b_1{c}_2+c_1{b}_2+d_1{d}_2)+(b_1{d}_2-d_1{b}_2)\mathbf{i}+(b_1{d}_2-d_1{b}_2+d_1{c}_2-c_1{d}_2)\mathbf{\epsilon }+(-b_1{c}_2+c_1{b}_2)\mathbf{h}$$

(6.15)

where $b_j,c_j,d_j\in \mathbb{C}$.

Proof. 

As established in Theorem 21 and Corollary 4, the separation of the mixed product into scalar and vector interaction terms depends solely on the property that the scalar sequences ($\mathcal{A},\mathcal{B}$) globally commute across the equation. Because $\mathbb{C}$ is a commutative field, ${\mathcal{A}}_{1,k}{\mathcal{B}}_{2,m}={\mathcal{B}}_{2,m}{\mathcal{A}}_{1,k}$, ensuring that the scalars can be cleanly factored out from the non-commutative interaction of the complex vectors $\left(V_1{V}_2\right)$. Since this algebraic distribution holds universally regardless of sign (provided the required inverses exist), the structural decompositions for all four mixed sign configurations map identically from the real to the complex domain. □

6.5. Dimensional Collapse and Higher-Dimensional Tensor Constructions

The successful extension of the exponentiation and mixed-product theorems to the complex domain—as achieved in the preceding subsections—relies entirely on a single, non-negotiable algebraic property: the explicit distinction of the introduced complex unit I from the primary hybrid basis. This distinction alone guarantees the global commutativity required for the combinatorial expansion.

However, a pervasive structural pitfall arises in contemporary literature when researchers attempt to define complex, dual, or hypercomplex coefficients without maintaining this strict separation. If the introduced scalar units are conflated with the existing non-commutative hybrid imaginary units $\{\mathbf{i},\mathbf{\epsilon },\mathbf{h}\}$, the fundamental algebraic distribution rules fail, and the intended higher-dimensional space suffers a complete structural collapse.

The following remarks systematically detail the profound algebraic consequences of failing to maintain this strict unit separation. By examining complex, Gaussian, Dual, and Cayley-Dickson algebras, we illustrate why constructing these generalized number systems must be treated formally as tensor products of algebras over $\mathbb{R}$. Furthermore, we provide a critical review of recent literature to highlight both the mathematical utility of intentional dimensional collapse and the dangers of notational ambiguity when operating in these generalized non-commutative spaces.

Remark 11

(The Bi-Hybrid Distinction and Algebraic Collapse). In defining the complex hybrid number ${\mathbf{Z}}_{\mathbb{C}}$, it is structurally imperative to introduce the complex imaginary unit I (where $I^2=-1$) as a central element that strictly commutes with all hybrid basis units ($I\mathbf{i}=\mathbf{i}I$, $I\mathbf{\epsilon }=\mathbf{\epsilon }I$, etc.), and crucially, $I\ne \mathbf{i}$. This creates an 8-dimensional algebra over the real numbers, formally known as the bi-hybrid numbers.

If this distinction is not strictly maintained—that is, if one attempts to identify the complex scalar unit I directly with the non-commutative hybrid unit $\mathbf{i}$—the 8-dimensional complex space immediately collapses back into the standard 4-dimensional real hybrid space.

To observe this structural collapse explicitly, let the components of ${\mathbf{Z}}_{\mathbb{C}}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$ be defined using the hybrid unit $\mathbf{i}$ instead of an independent complex I. Let $a=a_1+a_2\mathbf{i}$, $b=b_1+b_2\mathbf{i}$, $c=c_1+c_2\mathbf{i}$, and $d=d_1+d_2\mathbf{i}$, where $a_k,b_k,c_k,d_k\in \mathbb{R}$. Substituting these into the standard form yields:

$${\mathbf{Z}}_{\mathbb{C}}=(a_1+a_2\mathbf{i})+(b_1+b_2\mathbf{i})\mathbf{i}+(c_1+c_2\mathbf{i})\mathbf{\epsilon }+(d_1+d_2\mathbf{i})\mathbf{h}$$

Expanding this expression from the left requires the application of Özdemir’s non-commutative hybrid multiplication rules (specifically ${\mathbf{i}}^2=-1$, $\mathbf{i}\mathbf{\epsilon }=1-\mathbf{h}$, and $\mathbf{i}\mathbf{h}=\mathbf{i}+\mathbf{\epsilon }$):

$$\begin{array}{cc}\hfill {\mathbf{Z}}_{\mathbb{C}}& =a_1+a_2\mathbf{i}+b_1\mathbf{i}+b_2{\mathbf{i}}^2+c_1\mathbf{\epsilon }+c_2\left(\mathbf{i}\mathbf{\epsilon }\right)+d_1\mathbf{h}+d_2\left(\mathbf{i}\mathbf{h}\right)\hfill \\ \hfill & =a_1+a_2\mathbf{i}+b_1\mathbf{i}+b_2(-1)+c_1\mathbf{\epsilon }+c_2(1-\mathbf{h})+d_1\mathbf{h}+d_2(\mathbf{i}+\mathbf{\epsilon }).\hfill \end{array}$$

By grouping the real scalar terms and factoring the standard hybrid basis units $\mathbf{i},\mathbf{\epsilon },$ and $\mathbf{h}$, the expression simplifies elegantly:

$${\mathbf{Z}}_{\mathbb{C}}=(a_1-b_2+c_2)+(a_2+b_1+d_2)\mathbf{i}+(c_1+d_2)\mathbf{\epsilon }+(d_1-c_2)\mathbf{h}.$$

We observe that the resulting algebraic entity is no longer a complex hybrid number, but simply a standard real hybrid number ${\mathbf{Z}}_{new}=A+B\mathbf{i}+C\mathbf{\epsilon }+D\mathbf{h}$, with translated real components:

$$\begin{array}{cc}\hfill A& =a_1-b_2+c_2,\hfill \\ \hfill B& =a_2+b_1+d_2,\hfill \\ \hfill C& =c_1+d_2,\hfill \\ \hfill D& =d_1-c_2.\hfill \end{array}$$

This explicit algebraic collapse demonstrates exactly why extending the mathematical framework to an 8-dimensional space strictly necessitates a central, independent complex unit I. However, this dimension reduction is also highly valuable in its own right; if researchers intentionally equate the units, they can directly apply all the real 4-dimensional hybrid theorems developed in the preceding sections to efficiently evaluate sequences with complex components.

This phenomenon of algebraic collapse is not merely a theoretical curiosity; it represents a fundamental structural pitfall in contemporary research when defining new generalized number systems. A highly relevant example is the recent interest in constructing hybrid algebras using Gaussian coefficients, where the failure to explicitly separate the imaginary units leads to the exact dimensional collapse demonstrated above.

Remark 12

(Hybrid Numbers with Gaussian Coefficients). In modern mathematical literature concerning generalized number systems, it is increasingly common to investigate structures whose coefficients are Gaussian numbers. While classical Gaussian integers restrict components to $\mathbb{Z}$, the generalized continuous definition expresses a Gaussian number as $x+yi$, where $x,y\in \mathbb{R}$ and $i^2=-1$.

If a researcher attempts to construct a “Gaussian Hybrid Number” by substituting Gaussian numbers for the coefficients $a,b,c,d$ in $\mathbf{Z}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$, a critical structural bifurcation occurs based strictly on how the imaginary unit i is defined:

• Case 1: The Central Commutative Unit ($i=I\ne \mathbf{i}$).If the Gaussian unit i is defined as an independent, central complex unit (denoted as I in this section), it commutes globally with the hybrid basis ($\mathbf{i},\mathbf{\epsilon },\mathbf{h}$). This construction successfully generates the 8-dimensional Complex Hybrid (or Bi-hybrid) space. Because the coefficients commute, the binomial expansions and mixed-power theorems established in this section remain rigorously valid.
• Case 2: The Non-Commutative Hybrid Unit ($i=\mathbf{i}$).Conversely, if the Gaussian unit i is identified directly with the non-commutative hybrid imaginary unit $\mathbf{i}$, the coefficients take the form $a=a_1+a_2\mathbf{i}$, $b=b_1+b_2\mathbf{i}$, etc. In this case, the coefficients are drawn strictly from the 2-dimensional hybrid subring $\mathbb{R}+\mathbb{R}\mathbf{i}$. Because $\mathbf{i}$ does not commute with $\mathbf{\epsilon }$ and $\mathbf{h}$, the standard algebraic distribution rules necessary for tensor expansion fail. As explicitly derived in Remark 11, this “Gaussian Hybrid Number” collapses entirely back into a standard 4-dimensional real hybrid number:

  $${\mathbf{Z}}_{\mathbf{new}}=\mathbf{A}+\mathbf{B}\mathbf{i}+\mathbf{C}\mathbf{\epsilon }+\mathbf{D}\mathbf{h}$$

  where the real components shift to $\mathbf{A}={\mathbf{a}}_{\mathbf{1}}-{\mathbf{b}}_{\mathbf{2}}+{\mathbf{c}}_{\mathbf{2}}$, $\mathbf{B}={\mathbf{a}}_{\mathbf{2}}+{\mathbf{b}}_{\mathbf{1}}+{\mathbf{d}}_{\mathbf{2}}$, $\mathbf{C}={\mathbf{c}}_{\mathbf{1}}+{\mathbf{d}}_{\mathbf{2}}$, and $\mathbf{D}={\mathbf{d}}_{\mathbf{1}}-{\mathbf{c}}_{\mathbf{2}}$.

This distinction highlights a profound algebraic caveat: one cannot trivially substitute Gaussian numbers into a non-commutative algebra with the expectation of generating a new, higher-dimensional space without explicitly establishing the commutativity of the interacting imaginary units. Consequently, the choice between these two assumptions depends entirely on the structural objective of the research. Both approaches open rich areas of research: Case 1 constructs new topological spaces, while Case 2 leverages the robust toolkit of standard 4-dimensional hybrid algebra to establish parametrizations of existing spaces. For the generalized combinatorial and mixed-product theorems presented in this section to maintain their extended validity, we strictly operate under the assumption of Case 1.

The critical necessity of establishing a central, commuting unit is not restricted solely to complex or Gaussian coefficients. In fact, this structural bifurcation serves as a universal algebraic law that governs the creation of any higher-dimensional hybrid space, encompassing the vast array of generalized algebras currently proliferating in the literature.

Remark 13

(Generalization to Dual, Hyperbolic, and Hybrid-Hybrid Algebras).

It is not a coincidence that the most customary generalizations in contemporary literature—such as Complex-Hybrid, Dual-Hybrid, and Hyperbolic-Hybrid numbers—draw their extended coefficients from the exact same fundamental number systems (complex, dual, and hyperbolic) that originally compose the hybrid number itself.

The structural bifurcation identified for Gaussian coefficients in Remark 12, however, represents a universal algebraic law for successfully generating any of these extended spaces. When researchers replace the real coefficients of $\mathbf{Z}=a+b\mathbf{i}+c\mathbf{\epsilon }+d\mathbf{h}$ with elements from the dual plane ($\mathbb{R}\left[x\right]/\langle x^2\rangle$), the split-complex/hyperbolic plane ($\mathbb{R}\left[x\right]/\langle x^2-1\rangle$), or even secondary hybrid numbers, they face the exact same structural choice.

To successfully construct these higher-dimensional spaces (e.g., an 8-dimensional Dual-Hybrid space or a 16-dimensional Hybrid-Hybrid space) without suffering algebraic collapse, the construction must formally be a tensor product of algebras over $\mathbb{R}$. Practically, this means the basis units of the coefficient algebra (e.g., the dual unit ${\epsilon }^{\prime }$, the hyperbolic unit $h^{\prime }$, or the secondary hybrid units ${\mathbf{i}}^{\prime },{\mathbf{\epsilon }}^{\prime },{\mathbf{h}}^{\prime }$) must be defined as central elements that strictly commute with the primary hybrid basis $\{\mathbf{i},\mathbf{\epsilon },\mathbf{h}\}$.

If this global commutativity is guaranteed, the coefficient algebra acts identically to a commutative scalar field with respect to the primary hybrid units. Consequently, the quadratic characteristic equation, the sequence $G_k$, the structural root formulas, and the closed-form combinatorial expansions developed in this work remain rigorously valid and applicable universally across all such extended Dual, Hyperbolic, and Hybrid-Hybrid number systems.

This delicate choice of unit interaction becomes even more critical—and strictly limited—when extending hybrid numbers into hypercomplex algebras.

Remark 14

(Extensions to Cayley-Dickson Algebras: Quaternions, Octonions, and Sedenions). Beyond the foundational systems of complex, dual, and hyperbolic numbers, there is significant active research into combining hybrid numbers with higher-dimensional hypercomplex algebras generated via the Cayley-Dickson construction—specifically Quaternions ($\mathbb{H}$), Octonions ($\mathbb{O}$), and Sedenions ($\mathbb{S}$).

When attempting to define Hybrid-Quaternions, Hybrid-Octonions, or Hybrid-Sedenions, the choice of how to handle the interacting units becomes even more mathematically restricted. This strict restriction explains the hidden algebraic necessity behind the notational illusions that will be explored in Remark 15.

• The Impossibility of Equating Units:Unlike complex or dual combinations where equating units merely results in a manageable dimensional collapse, attempting to algebraically equate the standard quaternion units $\{i,j,k\}$ with the primary hybrid units $\{\mathbf{i},\mathbf{\epsilon },\mathbf{h}\}$ yields fatal structural contradictions. If one attempts a full bijective mapping ($i=\mathbf{i}$, $j=\mathbf{\epsilon }$, $k=\mathbf{h}$), evaluating the square of the second unit under quaternion rules yields $j^2=-1$, whereas hybrid rules yield ${\mathbf{\epsilon }}^2=0$, forcing the mathematical absurdity $-1=0$.
  Furthermore, quaternion units strictly anti-commute ($ij=-ji$), whereas hybrid units do not ($\mathbf{i}\mathbf{\epsilon }=1-\mathbf{h}$ while $\mathbf{\epsilon }\mathbf{i}=1+\mathbf{h}$).
  More insidiously, researchers might be tempted by a partial conflation. Because the quaternion i and the hybrid $\mathbf{i}$ share the identical property $i^2={\mathbf{i}}^2=-1$, it is a common pitfall to erroneously assume $i=\mathbf{i}$ while leaving the remaining units distinct. However, this utterly destroys the associative property of the algebra. To prove this, consider the associative law $\left(ji\right)\mathbf{\epsilon }=j\left(i\mathbf{\epsilon }\right)$, keeping in mind that to form a valid tensor space, elements from distinct bases must commute (e.g., $j\mathbf{\epsilon }=\mathbf{\epsilon }j$ and $j\mathbf{h}=\mathbf{h}j$).
  Evaluating the left side using quaternion non-commutative rules yields:

  $$\left(ji\right)\mathbf{\epsilon }=-k\mathbf{\epsilon }.$$
  Evaluating the right side, we substitute our assumption $i=\mathbf{i}$ and apply Özdemir’s hybrid multiplication rules (specifically $\mathbf{i}\mathbf{\epsilon }=1-\mathbf{h}$):

  $$j\left(i\mathbf{\epsilon }\right)=j\left(\mathbf{i}\mathbf{\epsilon }\right)=j(1-\mathbf{h})=j-j\mathbf{h}.$$
  Because associativity must hold, we equate the two sides:

  $$-k\mathbf{\epsilon }=j-j\mathbf{h}.$$
  Squaring both sides exposes the structural collapse. Since components from distinct bases commute, we can square the terms directly:

  $$\begin{array}{cc}\hfill {(-k\mathbf{\epsilon })}^2& ={(j-j\mathbf{h})}^2,\hfill \\ \hfill k^2{\mathbf{\epsilon }}^2& =j^2{(1-\mathbf{h})}^2,\hfill \\ \hfill (-1)\left(0\right)& =(-1)(1-2\mathbf{h}+{\mathbf{h}}^2).\hfill \end{array}$$
  Because ${\mathbf{h}}^2=1$ in hybrid algebra, the equation simplifies to:

  $$\begin{array}{cc}\hfill 0& =(-1)(1-2\mathbf{h}+1)\hfill \\ \hfill 0& =-2+2\mathbf{h}\hfill \\ \hfill \mathbf{h}& =1.\hfill \end{array}$$
  This forces the non-commutative hybrid basis unit $\mathbf{h}$ to improperly collapse into the real scalar 1, destroying the entire hybrid number system. Consequently, these coefficient algebras simply cannot be merged into or overlapped with the primary hybrid basis; the units must remain strictly and entirely distinct.
• Constructing the Tensor Spaces:To successfully form these algebras without contradiction, they must be constructed as formal tensor products over $\mathbb{R}$ (e.g., $\mathbb{H}\otimes \mathbb{K}$). The units of the Cayley-Dickson algebra must be defined as central elements relative to the hybrid basis, meaning every quaternion or octonion unit must strictly commute with every hybrid unit ($\mathbf{i},\mathbf{\epsilon },\mathbf{h}$). This requirement safely generates 16-dimensional Hybrid-Quaternions, 32-dimensional Hybrid-Octonions, and 64-dimensional Hybrid-Sedenions.
• The Associativity Caveat:While ensuring commutativity between the bases allows the tensor product to form, researchers must navigate a severe algebraic limitation inherent to the Cayley-Dickson construction itself. Quaternions ($\mathbb{H}$) are non-commutative but remain associative; thus, the 16-dimensional Hybrid-Quaternion space preserves associativity. This guarantees that characteristic equations, left-regular matrix isomorphisms, and Binet-style exponentiation formulas remain valid.
  However, Octonions ($\mathbb{O}$) and Sedenions ($\mathbb{S}$) are strictly non-associative (e.g., $\left(xy\right)z\ne x\left(yz\right)$). Because associativity is lost, the standard matrix representations and exponential generating functions fundamentally break down. Researchers working in Hybrid-Octonion or Hybrid-Sedenion spaces cannot reliably utilize standard recurrence methodologies without introducing complex, non-associative algebraic corrections.

In this work, we denote the primary hybrid basis using bold characters as $\{\mathbf{i},\mathbf{\epsilon },\mathbf{h}\}$. When there is no risk of confusion, the fundamental units of other generalized systems—such as the complex unit I, the dual unit ${\epsilon }^{\prime }$, and the hyperbolic unit $h^{\prime }$—are often denoted in the literature simply as $i,\epsilon ,h$, and their combinations such as $\epsilon h$.

In the literature concerning combined number systems, authors frequently face a pivotal choice: whether to assume these secondary units are identically equal to the primary hybrid basis, or to treat them as distinct. When authors assume the units are identically equal (i.e., $\mathbf{i}=i$, $\mathbf{\epsilon }=\epsilon$, $\mathbf{h}=h$, and $\mathbf{\epsilon }\mathbf{h}=\epsilon h$), the resulting combined numbers remain completely within the standard 4-dimensional space.

This dimensional restriction is highly significant and mathematically advantageous. By evaluating the new sequences within the same 4-dimensional framework, the newly defined combined sequences—whether based on second-order relations (Fibonacci, Lucas, Pell, Jacobsthal) or higher-order relations (Tribonacci, Tetranacci)-satisfy recurrence relations identical to the original real sequences. Consequently, they share the exact same characteristic equations, generating functions, and fundamental identities (such as Cassini’s, Catalan’s, and Honsberger’s). The ability to seamlessly inherit these structural properties is a primary reason why many researchers intentionally equate the units. Conversely, choosing to distinguish the units generates entirely new, higher-dimensional tensor spaces, opening up vastly different avenues of research.

Remark 15.

The n-th standard Fibonacci number is defined by $F_n=F_{n-1}+F_{n-2}$ for $n\ge 2$ with initial values $F_0=0$ and $F_1=1$ (with n naturally extending to all integers). The associated characteristic equation is $z^2-z-1=0$, yielding the roots $\alpha ={\displaystyle \frac{1+\sqrt{5}}{2}}$ and $\beta ={\displaystyle \frac{1-\sqrt{5}}{2}}$. Binet’s formula is elegantly given by $F_n={\displaystyle \frac{{\alpha }^n-{\beta }^n}{\alpha -\beta }}$.

Similarly, the n-th Leonardo number is defined by the recurrence relation $L{e}_n=L{e}_{n-1}+L{e}_{n-2}+1$ for $n\ge 2$, with initial conditions $L{e}_0=L{e}_1=1$ [1]. The associated characteristic equation is ${\lambda }^3-2{\lambda }^2+1=0$, which yields the roots $\alpha ={\displaystyle \frac{1+\sqrt{5}}{2}}$, $\beta ={\displaystyle \frac{1-\sqrt{5}}{2}}$, and 1. Its Binet formula is explicitly given by $L{e}_n={\displaystyle \frac{2{\alpha }^{n+1}-2{\beta }^{n+1}}{\alpha -\beta }}-1$.

For algebraic convenience in the following examples, we define the auxiliary hybrid elements:

$$\widehat{\alpha }=1+\alpha \mathbf{i}+{\alpha }^2\mathbf{\epsilon }+{\alpha }^3\mathbf{h}\mathrm{and}\widehat{\beta }=1+\beta \mathbf{i}+{\beta }^2\mathbf{\epsilon }+{\beta }^3\mathbf{h}.$$

(a)

Base Generalizations:

• Gaussian Fibonacci Numbersextend the Fibonacci sequence into the complex plane, formally defined as $G{F}_n=F_n+i{F}_{n-1}$ for $n\ge 1$ with initial values $G{F}_0=i$ and $G{F}_1=1$, where i is the standard imaginary unit. They satisfy the second-order recurrence relation $G{F}_n=G{F}_{n-1}+G{F}_{n-2}$, and their Binet formula is:

  $$G{F}_n={\displaystyle \frac{{\alpha }^n-{\beta }^n}{\alpha -\beta }}+i{\displaystyle \frac{{\alpha }^{n-1}-{\beta }^{n-1}}{\alpha -\beta }}={\displaystyle \frac{(1-\beta i){\alpha }^n-(1-\alpha i){\beta }^n}{\alpha -\beta }}.$$
• Hybrid Fibonacci Numberscombine standard Fibonacci numbers with the non-commutative hybrid basis (see Szynal-Liana and Włoch [18]). The n-th Hybrid Fibonacci number is defined as $F{H}_n=F_n+F_{n+1}\mathbf{i}+F_{n+2}\mathbf{\epsilon }+F_{n+3}\mathbf{h}$. They satisfy the recurrence relation $F{H}_n=F{H}_{n-1}+F{H}_{n-2}$, and their Binet formula is concisely written as $F{H}_n={\displaystyle \frac{\widehat{\alpha }{\alpha }^n-\widehat{\beta }{\beta }^n}{\alpha -\beta }}$.
• Hybrid Leonardo Numbersare formally defined as $HL{e}_n=L{e}_n+L{e}_{n+1}\mathbf{i}+L{e}_{n+2}\mathbf{\epsilon }+L{e}_{n+3}\mathbf{h}$ [1]. They satisfy the second-order recurrence relation $HL{e}_{n+1}=2HL{e}_n-HL{e}_{n-2}$. The Binet formula for hybrid Leonardo numbers is defined as $HL{e}_n={\displaystyle \frac{2\widehat{\alpha }{\alpha }^{n+1}-2\widehat{\beta }{\beta }^{n+1}}{\alpha -\beta }}-A$, where $A=1+\mathbf{i}+\mathbf{\epsilon }+\mathbf{h}$.

(b)

Gaussian Fibonacci Hybrid Numbers:

• This structure combines Gaussian Fibonacci coefficients with the hybrid basis. The n-th Gaussian Fibonacci hybrid number is defined as $GF{H}_n=G{F}_n+G{F}_{n+1}\mathbf{i}+G{F}_{n+2}\mathbf{\epsilon }+G{F}_{n+3}\mathbf{h}$, where $G{F}_n$ represents the n-th Gaussian Fibonacci number.
• In their work, Uysal and Özkan [20] explicitly assume that the imaginary unit i and the hybrid unit $\mathbf{i}$ are equal ($\mathbf{i}=i$). Because this assumption collapses the structure into a 4-dimensional space, the sequence flawlessly inherits the standard recurrence relation $GF{H}_n=GF{H}_{n-1}+GF{H}_{n-2}$. Consequently, the authors were able to effortlessly establish Binet’s formula, $GF{H}_n={\displaystyle \frac{(1-\beta i)\widehat{\alpha }{\alpha }^n-(1-\alpha i)\widehat{\beta }{\beta }^n}{\alpha -\beta }}$, alongside classical identities such as Cassini’s and Catalan’s.
• The Distinction:If a researcher were to assume the units are distinct ($i\ne \mathbf{i}$), the resulting combination would generate an 8-dimensional tensor space (e.g., 8D Gaussian-Hybrid Fibonacci numbers). While this distinct choice prevents the trivial inheritance of classical 4D identities and requires a newly derived 8D multiplication table, it opens the door to uncovering entirely new structural symmetries.

(c)

Hybrid Numbers with Fibonacci & Leonardo Hybrid Coefficients:

• Investigated by Polatlı [13], Hybrid numbers with Fibonacci hybrid coefficients are defined by placing hybrid Fibonacci sequences onto a secondary hybrid basis: ${\mathbb{F}}_n=F{H}_n+F{H}_{n+1}{\mathbf{i}}^{\prime }+F{H}_{n+2}{\mathbf{\epsilon }}^{\prime }+F{H}_{n+3}{\mathbf{h}}^{\prime }$. In this study, the author assumes that the primary hybrid units ($\mathbf{i},\mathbf{\epsilon },\mathbf{h}$) inside the coefficients $F{H}_n$ are identically equal to the secondary basis units (${\mathbf{i}}^{\prime },{\mathbf{\epsilon }}^{\prime },{\mathbf{h}}^{\prime }$). By enforcing this equality, the structure remains perfectly 4-dimensional. Thus, the sequence obeys the standard second-order linear recurrence relation ${\mathbb{F}}_n={\mathbb{F}}_{n-1}+{\mathbb{F}}_{n-2}$, and yields the compact Binet formula ${\mathbb{F}}_n={\displaystyle \frac{{\widehat{\alpha }}^2{\alpha }^n-{\widehat{\beta }}^2{\beta }^n}{\alpha -\beta }}$.
• Similarly, in their study of Hybrid Numbers with Hybrid Leonardo Number Coefficients, Spreafico, Costa, and Catarino [17] define the sequence $L{e}_n=HL{e}_n+HL{e}_{n+1}{\mathbf{i}}^{\prime }+HL{e}_{n+2}{\mathbf{\epsilon }}^{\prime }+HL{e}_{n+3}{\mathbf{h}}^{\prime }$. The authors explicitly equate the primary hybrid units within the coefficients to the secondary hybrid units ($\mathbf{i}={\mathbf{i}}^{\prime }$, $\mathbf{\epsilon }={\mathbf{\epsilon }}^{\prime }$, $\mathbf{h}={\mathbf{h}}^{\prime }$). By collapsing the space to 4 dimensions, they are able to utilize standard 4D hybrid identities, such as $A^2=2A+1$ (where $A=1+\mathbf{i}+\mathbf{\epsilon }+\mathbf{h}$), to elegantly derive their Binet formulas and Cassini identities without needing higher-dimensional non-commutative tensor algebra.
• The Distinction:If we choose to keep the primary and secondary units distinct, the resulting algebra forms a 16-dimensional tensor space known as Hybrid-Hybrid numbers. While mathematically much more complex, distinguishing the units provides a rigorous pathway to exploring higher-order non-commutative algebras without dimensional collapse.

(d)

Hybrid Quaternions and the Notational Trap:

• The extension of hybrid numbers into 16-dimensional spaces by combining them with quaternions is a highly active area of contemporary research. However, it introduces a severe notational pitfall in the literature.
• Explicitly Distinct Notation:Several studies construct true 16-dimensional hybrid quaternion spaces by explicitly distinguishing the quaternion basis from the hybrid basis in both algebra and notation. For instance, Gül [5] defines generalized k-order Fibonacci hybrid quaternions by using hybrid numbers as coefficients for distinct quaternion units. Similarly, Uysal and Özkan [19] define Padovan hybrid quaternions utilizing distinctly labeled quaternion units $e_1,e_2,e_3$ alongside Padovan hybrid numbers.
• The Illusion of Equality:Conversely, some literature implicitly conflates the notation while mathematically operating in a 16-dimensional tensor space. For example, Eser, Kuloğlu, and Özkan [4] define Mersenne hybrinomial quaternions, Mangueira, Alves, and Catarino [8] study Hybrid Quaternions of Leonardo, and Özkan and Uysal [9] define Mersenne and Mersenne-Lucas Hybrid Quaternions. In these works, the authors overload the standard imaginary symbol i, using it to represent both the hybrid imaginary unit and the quaternion imaginary unit within the exact same expressions. For instance, Özkan and Uysal [9] define the quaternions as $MH{Q}_n=M{H}_n+M{H}_{n+1}i+M{H}_{n+2}j+M{H}_{n+3}k$, where the coefficient $M{H}_n$ already contains a hybrid i. This creates the visual illusion that the units are equated. However, their Binet formula proofs factor the hybrid and quaternion components symmetrically (e.g., separating them into distinct commuting brackets), revealing that they algebraically treat the two i’s as completely independent, commuting tensor components. As established in Remark 14, if the units were truly equated, their algebraic operators would fatally contradict one another.

(e)

Explicitly Distinct Tensor Constructions:

While the aforementioned studies often equate units to preserve 4-dimensional properties (or create illusions of doing so), recent literature also actively explores the alternative pathway: strictly distinguishing the units to explore higher-dimensional tensor spaces.

A prime example of distinguishing units to preserve dimensionality is the recent work by Seçgin, Tan, and Gök [14] entitled “Dual Hybrid Numbers with Horadam Number Coefficients”. The authors define the set of dual hybrid numbers by drawing coefficients directly from the dual numbers, forming the set $DK=\{a+bi+cϵ+dh:a,b,c,d\in D\}$, where D represents the dual numbers. Crucially, they explicitly assume that the dual unit ε (where ${\epsilon }^2=0$) is completely distinct from the hybrid units $i,ϵ,h$. By keeping these units separate, they preserve the expanded dimensionality, establishing that the algebra of dual hybrid numbers has an 8-element basis: $\{1,i,ϵ,h,\epsilon ,\epsilon i,\epsilon ϵ,\epsilon h\}$.

Building on this 8-dimensional framework, they define Horadam hybrid numbers as $Q_n=H_n+H_{n+1}i+H_{n+2}ϵ+H_{n+3}h$. From this, they construct the n-th dual Horadam hybrid number as ${\mathbb{Q}}_n={\mathbb{H}}_n+{\mathbb{H}}_{n+1}i+{\mathbb{H}}_{n+2}ϵ+{\mathbb{H}}_{n+3}h$, where the coefficients ${\mathbb{H}}_n=H_n+H_{n+1}\epsilon$ are dual Horadam numbers. This relationship is beautifully condensed as ${\mathbb{Q}}_n=Q_n+Q_{n+1}\epsilon$.

Compared to the methodologies discussed previously—where equating units collapses the system to allow native inheritance of classical 4D identities—Seçgin et al. demonstrate the vast richness of the alternative choice. By maintaining distinct units, they successfully derive generalized Binet formulas, generating functions, and structural identities (such as Vajda’s, Catalan’s, and d’Ocagne’s) specifically adapted for this higher-dimensional, commutative dual-hybrid tensor space.