Some identities of the generalized bi-periodic Fibonacci and Lucas polynomials

Tingting Du; Zhengang Wu; Tingting Du; Zhengang Wu

doi:10.3934/math.2024363

AIMS Mathematics

2024, Volume 9, Issue 3: 7492-7510. doi: 10.3934/math.2024363

Previous Article Next Article

Research article Special Issues

Some identities of the generalized bi-periodic Fibonacci and Lucas polynomials

Tingting Du ,
Zhengang Wu ^,

Research Center for Number Theory and Its Application, Northwest University, Xi'an 710127, China

Received: 10 January 2024 Revised: 02 February 2024 Accepted: 05 February 2024 Published: 21 February 2024
MSC : 11B37, 11B39

In this paper, we considered the generalized bi-periodic Fibonacci polynomials, and obtained some identities related to generalized bi-periodic Fibonacci polynomials using the matrix theory. In addition, the generalized bi-periodic Lucas polynomial was defined by $L_{n}\left (x \right) = bp\left (x \right) L_{n-1}\left (x \right)+q\left (x \right)L_{n-2}\left (x \right)$ (if $n$ is even) or $L_{n}\left (x \right) = ap\left (x \right) L_{n-1}\left (x \right)+q\left (x \right)L_{n-2}\left (x \right)$ (if $n$ is odd), with initial conditions $L_{0}\left (x \right) = 2$ , $L_{1}\left (x \right) = ap\left (x \right)$ , where $p\left (x \right)$ and $q\left (x \right)$ were nonzero polynomials in $Q \left [ x \right ]$ . We obtained a series of identities related to the generalized bi-periodic Fibonacci and Lucas polynomials.

Keywords:

generalized bi-periodic Fibonacci polynomial,
generalized bi-periodic Lucas polynomial,
Binet formula,
matrix theory

Citation: Tingting Du, Zhengang Wu. Some identities of the generalized bi-periodic Fibonacci and Lucas polynomials[J]. AIMS Mathematics, 2024, 9(3): 7492-7510. doi: 10.3934/math.2024363

Related Papers:

[1]	Tingting Du, Li Wang . On the power sums problem of bi-periodic Fibonacci and Lucas polynomials. AIMS Mathematics, 2024, 9(4): 7810-7818. doi: 10.3934/math.2024379
[2]	Tingting Du, Zhengang Wu . Some identities involving the bi-periodic Fibonacci and Lucas polynomials. AIMS Mathematics, 2023, 8(3): 5838-5846. doi: 10.3934/math.2023294
[3]	Waleed Mohamed Abd-Elhameed, Omar Mazen Alqubori . New expressions for certain polynomials combining Fibonacci and Lucas polynomials. AIMS Mathematics, 2025, 10(2): 2930-2957. doi: 10.3934/math.2025136
[4]	Utkal Keshari Dutta, Prasanta Kumar Ray . On the finite reciprocal sums of Fibonacci and Lucas polynomials. AIMS Mathematics, 2019, 4(6): 1569-1581. doi: 10.3934/math.2019.6.1569
[5]	Ümit Tokeşer, Tuğba Mert, Yakup Dündar . Some properties and Vajda theorems of split dual Fibonacci and split dual Lucas octonions. AIMS Mathematics, 2022, 7(5): 8645-8653. doi: 10.3934/math.2022483
[6]	Waleed Mohamed Abd-Elhameed, Amr Kamel Amin, Nasr Anwer Zeyada . Some new identities of a type of generalized numbers involving four parameters. AIMS Mathematics, 2022, 7(7): 12962-12980. doi: 10.3934/math.2022718
[7]	Can Kızılateş, Halit Öztürk . On parametric types of Apostol Bernoulli-Fibonacci, Apostol Euler-Fibonacci, and Apostol Genocchi-Fibonacci polynomials via Golden calculus. AIMS Mathematics, 2023, 8(4): 8386-8402. doi: 10.3934/math.2023423
[8]	Adikanda Behera, Prasanta Kumar Ray . Determinantal and permanental representations of convolved (u, v)-Lucas first kind p-polynomials. AIMS Mathematics, 2020, 5(3): 1843-1855. doi: 10.3934/math.2020123
[9]	Yulei Chen, Yingming Zhu, Dongwei Guo . Combinatorial identities concerning trigonometric functions and Fibonacci/Lucas numbers. AIMS Mathematics, 2024, 9(4): 9348-9363. doi: 10.3934/math.2024455
[10]	Faik Babadağ . A new approach to Jacobsthal, Jacobsthal-Lucas numbers and dual vectors. AIMS Mathematics, 2023, 8(8): 18596-18606. doi: 10.3934/math.2023946

Abstract

1. Introduction

The Fibonacci and Lucas polynomials are important in various fields such as number theory, probability, numerical analysis, and physics. In addition, many famous polynomials, such as the Pell polynomials, Pell Lucas polynomials, Tribonacci polynomials, etc., are generalizations of the Fibonacci and Lucas polynomials. Many scholars discussed the Fibonacci polynomials and its generalization; see ^[1,2,3,4,5]. This paper mainly extends linear recursive polynomials to nonlinearity and discusses some basic properties of the generalized bi-periodic Fibonacci and Lucas polynomials.

The Fibonacci $\left \{ u_{n} \left (x \right) \right \}$ and Lucas $\left \{v_{n} \left (x \right) \right \}$ polynomials are defined by

$\begin{equation*} u_{0} \left ( x \right ) = 0, \quad u_{1} \left ( x \right ) = 1, \quad u_{n} \left ( x \right ) = xu_{n-1} \left ( x \right )+u_{n-2}\left ( x \right ), \quad n\ge 2 \end{equation*}$

and

$\begin{equation*} v_{0} \left ( x \right ) = 2, \quad v_{1} \left ( x \right ) = x, \quad v_{n} \left ( x \right ) = xv_{n-1} \left ( x \right )+v_{n-2}\left ( x \right ), \quad n\ge 2 . \end{equation*}$

When $x = 1$ , we obtain Fibonacci $\left \{u_{n} \right \}$ and Lucas $\left \{ v_{n} \right \}$ sequences defined by

$\begin{equation*} u_{0} = 0, \quad u_{1} = 1, \quad u_{n} = u_{n-1} +u_{n-2}, \quad n\ge 2 \end{equation*}$

and

$\begin{equation*} v_{0} = 2, \quad v_{1} = 1, \quad v_{n} = v_{n-1}+v_{n-2}, \quad n\ge 2 . \end{equation*}$

The Fibonacci $\left \{ u_{n} \right \}$ ( $\left \{u_{n} \left (x \right) \right \}$ ) or Lucas $\left \{ v_{n} \right \}$ ( $\left \{ v_{n} \left (x \right) \right \}$ ) sequences (polynomials) have more interesting properties and applications; see ^[6,7,8,9,10].

In ^[11], the generalized Fibonacci $\left \{ U_{n} \left (x \right) \right \}$ and Lucas $\left \{ V _{n} \left (x \right) \right \}$ polynomials are defined by

$\begin{equation*} U_{0} \left ( x \right ) = 0, \quad U_{1} \left ( x \right ) = 1, \quad U_{n} \left ( x \right ) = p\left ( x \right ) U_{n-1} \left ( x \right )+q\left ( x \right ) U_{n-2}\left ( x \right ), \quad n\ge 2 \end{equation*}$

and

$\begin{equation*} V _{0} \left ( x \right ) = 2, \quad V _{1} \left ( x \right ) = p\left ( x \right ), \quad V _{n} \left ( x \right ) = p\left ( x \right ) V _{n-1} \left ( x \right )+q\left ( x \right ) V_{n-2}\left ( x \right ), \quad n\ge 2 , \end{equation*}$

where $p\left (x \right)$ and $q\left (x \right)$ are nonzero polynomials in $Q \left [ x \right ]$ . For more consideration of generalized polynomials $\left \{ U_{n} \left (x \right) \right \}$ or $\left \{ V _{n} \left (x \right) \right \}$ , see ^[12,13,14].

In ^[15], the bi-periodic Fibonacci $\left \{ f_{n} \left (x \right) \right \}$ and Lucas $\left \{ l_{n} \left (x \right) \right \}$ polynomials are defined by

$\begin{equation*} f_{0} \left ( x \right ) = 0, \quad f_{1} \left ( x \right ) = 1, \quad {f_{n}\left ( x \right ) = } \begin{cases} axf_{n-1}\left ( x \right )+f_{n-2}\left ( x \right ), \quad & \text{if } n \text{ is even and } n \geq 2 ,\\ bx f_{n-1}\left ( x \right )+f_{n-2}\left ( x \right ), \quad & \text{if } n \text{ is odd and } n \geq 3 , \end{cases} \end{equation*}$

and

$\begin{equation*} l_{0} \left ( x \right ) = 2, \quad l_{1} \left ( x \right ) = ax, \quad {l_{n}\left ( x \right ) = } \begin{cases} bxl_{n-1}\left ( x \right )+l_{n-2}\left ( x \right ), \quad & \text{if } n \text{ is even and } n \geq 2 ,\\ ax l_{n-1}\left ( x \right )+l_{n-2}\left ( x \right ), \quad & \text{if } n \text{ is odd and } n \geq 3 , \end{cases} \end{equation*}$

where $a$ and $b$ are any nonzero real numbers. For more discussions of bi-periodic polynomials $\left \{ f_{n} \left (x \right) \right \}$ and $\left \{ l_{n} \left (x \right) \right \}$ ; see ^[16,17].

In ^[18], the author defined a new kind of Fibonacci polynomials called the generalized bi-periodic Fibonacci polynomial $\left \{ F_{n} \left (x \right) \right \}$ , which is defined by

$\begin{equation} F_{0} \left ( x \right ) = 0, \ \ \ F_{1} \left ( x \right ) = 1, \ \ \ {F_{n}\left ( x \right ) = } \begin{cases} ap\left ( x \right )F_{n-1}\left ( x \right )+q\left ( x \right )F_{n-2}\left ( x \right ), \ \ \ & \text{if } n \text{ is even and } n \geq 2 ,\\ bp\left ( x \right ) F_{n-1}\left ( x \right )+q\left ( x \right )F_{n-2}\left ( x \right ), \ \ \ & \text{if } n \text{ is odd and } n \geq 3 , \end{cases} \end{equation}$

(1.1)

where $a$ , $b$ are nonzero real numbers and $p\left (x \right)$ and $q\left (x \right)$ are nonzero polynomials in $Q \left [ x \right ]$ . They obtained the following Binet formula:

$\begin{equation} F_{m}\left ( x \right ) = \left ( \frac{a^{1-\zeta \left ( m \right ) } }{\left ( ab \right )^{\left \lfloor \frac{m}{2} \right \rfloor } } \right ) \frac{ \sigma ^{m}\left ( x \right )-\tau ^{m} \left ( x \right ) }{ \sigma \left ( x \right )-\tau \left ( x \right ) } , \quad m\ge 0, \end{equation}$

(1.2)

where

$\sigma \left ( x \right ) = \frac{abp\left ( x \right )+\sqrt{a^{2} b^{2}p^{2}\left ( x \right )+4abq\left ( x \right ) } }{2} ,$

$\tau \left ( x \right ) = \frac{abp\left ( x \right )-\sqrt{a^{2} b^{2}p^{2}\left ( x \right )+4abq\left ( x \right ) } }{2},$

and

$\zeta \left ( m \right ) = m-2\left \lfloor \frac{m}{2} \right \rfloor$

is the parity function, with $\left \lfloor \cdot \right \rfloor$ denoting the floor function.

In addition, they obtained a series of classical identities of the generalized bi-periodic Fibonacci polynomial as follows:

(a) Generated functions

$G_{n}\left ( x,t \right ) = \frac{t+ap\left ( x \right )t^{2}-q\left ( x \right )t^{3} }{1-\left ( abp^{2}\left ( x \right )+2q\left ( x \right ) \right )t^{2} +q^{2}\left ( x \right )t^{4} };$

(b) Generalized Catalan's identity

$a^{\zeta \left ( m-r \right ) }b^{1-\zeta \left ( m-r \right ) } F_{m-r}\left ( x \right ) F_{m+r}\left ( x \right ) -a^{\zeta \left ( m \right ) } b^{1-\zeta \left ( m \right ) }F_{m}^{2}\left ( x \right ) = -\left ( -q\left ( x \right ) \right ) ^{m-r} a^{\zeta \left ( r \right ) }b^{1-\zeta \left ( r \right ) }F_{r}^{2} \left ( x \right );$

$a^{\zeta \left ( m-1 \right ) }b^{\zeta \left ( m \right ) } F_{m-1}\left ( x \right ) F_{m+1}\left ( x \right ) -a^{\zeta \left ( m \right ) } b^{1-\zeta \left ( m\right ) }F_{m}^{2}\left ( x \right ) = -a\left ( -q\left ( x \right ) \right ) ^{m-1} ;$

(d) Generalized d'Ocagne's identity

$a^{\zeta \left ( mr+m \right ) }b^{\zeta \left ( mr+r \right ) }F_{m}\left ( x \right ) F_{r+1}\left ( x \right ) - a^{\zeta \left ( mr+r \right ) }b^{\zeta \left ( mr+m \right ) }F_{m+1}\left ( x \right ) F_{r}\left ( x \right ) = -\left ( -q\left ( x \right ) \right )^{r}a^{\zeta \left (m-r \right ) } F_{m-r} \left ( x \right );$

(e) Negative subscript terms

$F_{-m} \left ( x \right ) = \left ( -1 \right )^{m+1} \left ( q\left ( x \right ) \right )^{-m} F_{m} \left ( x \right ).$

2. The generalized bi-periodic Fibonacci polynomial by matrix methods

Recently, some scholars considered the identities of recursive sequences (polynomials) by the matrix theory. For example, in ^[19], the author defined the Fibonacci $Q$ -matrix as follows:

$\begin{equation*} Q = \begin{pmatrix} 1 & 1 \\ 1 &0 \end{pmatrix}, \end{equation*}$

so that

$\begin{equation*} Q^{m} = \begin{pmatrix} u_{m+1} & u_{m} \\ u_{m} & u_{m-1} \end{pmatrix} , \quad n\ge 1, \end{equation*}$

where $\left \{u_{n} \right \}$ is a Fibonacci sequence. For more on considering recursive sequences (polynomials) by the matrix theory; see ^[20,21,22].

In this section, we consider the generalized bi-periodic Fibonacci polynomial defined by a $2\times 2$ matrix $S$ and we give the $m$ th power $S^{m}$ for any integer $m$ .

Theorem 2.1. Let

$\begin{equation} S = \begin{pmatrix} abp\left ( x \right ) & bq\left ( x \right ) \\ a &0 \end{pmatrix}, \end{equation}$

(2.1)

then, we have

$\begin{equation} S^{m} = \left ( ab \right )^{\frac{m-\zeta \left ( m \right ) }{2} } \begin{pmatrix} b^{\zeta \left ( m \right ) } F_{m+1} \left ( x \right ) & a^{-\zeta \left ( m+1 \right ) }bq\left ( x \right )F_{m}\left ( x \right ) \\ a^{\zeta \left ( m \right ) }F_{m} \left ( x \right ) & b^{\zeta \left ( m \right ) }q\left ( x \right ) F_{m-1}\left ( x \right ) \end{pmatrix} , \quad n\ge 1, \end{equation}$

(2.2)

where $a$ and $b$ are nonzero real numbers, and $p\left (x \right)$ and $q\left (x \right)$ are nonzero polynomials in $Q \left [ x \right ]$ ,

$\zeta \left (m \right ) = m-2\left \lfloor \frac{m}{2} \right \rfloor$

is the parity function, with $\left \lfloor \cdot \right \rfloor$ denoting the floor function. $\left \{ F_{n} \left (x \right) \right \}$ is the generalized bi-periodic Fibonacci polynomial.

Proof. We prove (2.2) by mathematical induction. Obviously, the identity is true when $m = 1$ ,

$\begin{equation*} S^{1} = \begin{pmatrix} b F_{2} \left ( x \right ) & bq\left ( x \right )F_{1}\left ( x \right ) \\ aF_{1} \left ( x \right ) & bq\left ( x \right ) F_{0}\left ( x \right ) \end{pmatrix} = \begin{pmatrix} abp\left ( x \right ) & bq\left ( x \right ) \\ a &0 \end{pmatrix} = S. \end{equation*}$

We assume that the identity is true with $m$ . Next, we prove that the identity is true when $m+1$ .

$\begin{equation*} \begin{split} S^{m+1}& = S^{m}\cdot S = \left ( ab \right )^{\frac{m-\zeta \left ( m \right ) }{2} } \begin{pmatrix} b^{\zeta \left ( m \right ) } F_{m+1} \left ( x \right ) & a^{-\zeta \left ( m+1 \right ) }bq\left ( x \right )F_{m}\left ( x \right ) \\ a^{\zeta \left ( m \right ) }F_{m} \left ( x \right ) & b^{\zeta \left ( m \right ) }q\left ( x \right ) F_{m-1}\left ( x \right ) \end{pmatrix} \cdot \begin{pmatrix} abp\left ( x \right ) & bq\left ( x \right ) \\ a &0 \end{pmatrix}\\ & = \begin{pmatrix} a^{\frac{2+m-\zeta \left (m \right ) }{2} } b^{\frac{2+m+\zeta \left ( m \right ) }{2} } p\left ( x \right ) F_{m+1} \left ( x \right )+a^{\frac{m+1-\zeta \left ( m+1 \right ) }{2} } b^{\frac{2+m-\zeta \left (m \right ) }{2} } q\left ( x \right ) F_{m} \left ( x \right ) & a^{\frac{m-\zeta \left ( m \right ) }{2} } b^{\frac{2+m+\zeta \left ( m \right ) }{2} } q\left ( x \right )F_{m+1}\left ( x \right ) \\ a^{\frac{2+m+\zeta \left ( m \right ) }{2} } b^{\frac{2+m-\zeta \left ( m \right ) }{2} } p\left ( x \right ) F_{m} \left ( x \right )+a^{\frac{m+2-\zeta \left ( m \right ) }{2} } b^{\frac{m+\zeta \left ( m \right ) }{2} } q\left ( x \right ) F_{m-1} \left ( x \right ) & a^{\frac{m+\zeta \left ( m \right ) }{2} } b^{\frac{2+m-\zeta \left (m \right ) }{2} } q\left ( x \right )F_{m}\left ( x \right ) \end{pmatrix}\\ & = \left ( ab \right )^{\frac{m+1-\zeta \left ( m+1 \right ) }{2} }\begin{pmatrix} a^{\zeta \left ( m+1 \right ) }bp\left ( x \right )F_{m+1}\left ( x \right )+b^{\zeta \left ( m+1 \right ) }q\left ( x \right ) F_{m} \left ( x \right ) & a^{-\zeta \left ( m \right ) }bq\left ( x \right )F_{m+1}\left ( x \right ) \\ ab^{\zeta \left ( m+1 \right ) }p\left ( x \right )F_{m}\left ( x \right )+a^{\zeta \left ( m+1 \right ) }q\left ( x \right ) F_{m-1} \left ( x \right ) & b^{\zeta \left ( m+1 \right ) }q\left ( x \right ) F_{m}\left ( x \right ) \end{pmatrix} \\ & = \left ( ab \right )^{\frac{m+1-\zeta \left (m+1 \right ) }{2} }\begin{pmatrix} b^{\zeta \left ( m +1\right ) } F_{m+2} \left ( x \right ) & a^{-\zeta \left ( m \right ) }bq\left ( x \right )F_{m+1}\left ( x \right ) \\ a^{\zeta \left ( m+1 \right ) }F_{m+1} \left ( x \right ) & b^{\zeta \left ( m+1 \right ) }q\left ( x \right ) F_{m}\left ( x \right ) \end{pmatrix}, \end{split} \end{equation*}$

where the generalized bi-periodic Fibonacci polynomial is given by

$\begin{equation} F_{0}\left ( x \right ) = 0,\quad F_{1}\left ( x \right ) = 1,\quad F_{m}\left ( x \right ) = a^{\zeta \left ( m+1 \right ) }b^{\zeta \left ( m \right ) } p\left ( x \right ) F_{m-1} \left ( x \right ) +q\left ( x \right ) F_{m-2} \left ( x \right ), \quad m\ge 2, \end{equation}$

(2.3)

where $a$ , $b$ are nonzero real numbers and $p\left (x \right)$ , $q\left (x \right)$ are nonzero polynomials in $Q \left [ x \right ]$ ,

$\zeta \left ( m \right ) = m-2\left \lfloor \frac{m}{2} \right \rfloor$

is the parity function, with $\left \lfloor \cdot \right \rfloor$ denoting the floor function. □

Remark 2.1. The characteristic polynomial of $S$ is

$\lambda^{2}-abp\left ( x \right ) \lambda-abq\left ( x \right ) = 0 .$

Thus

$\sigma \left ( x \right ) = \frac{abp\left ( x \right )+\sqrt{a^{2} b^{2}p^{2}\left ( x \right )+4abq\left ( x \right ) } }{2}$

and

$\tau \left ( x \right ) = \frac{abp\left ( x \right )-\sqrt{a^{2} b^{2}p^{2}\left ( x \right )+4abq\left ( x \right ) } }{2}$

are the eigenvalues of $S$ . We can diagonalize $S$ to get $S = P^{-1}DP$ , so $S^{m} = P^{-1}D^{m}P$ , where $P$ is invertible and

$D = \begin{pmatrix} \sigma \left ( x \right ) & 0 \\ 0 & \tau \left ( x \right ) \end{pmatrix},$

The generalized bi-periodic Fibonacci polynomial may be expressed in the form

$\begin{equation*} F_{m}\left ( x \right ) = A\sigma^{m} \left ( x \right ) +B\tau ^{m} \left ( x \right ), \end{equation*}$

where $A$ and $B$ are constants. When $m = 0$ and $m = 1$ , we use the special value method to get the explicit identity of the generalized bi-periodic Fibonacci polynomial.

Next, we get a series of identities of the generalized bi-periodic Fibonacci polynomial by the matrix $S$ .

Theorem 2.2. (Generalized Cassinis's identity) Let $\left \{ F_{m} \left (x \right) \right \}$ be the generalized bi-periodic Fibonacci polynomial. We have

$\begin{equation} a^{\zeta \left ( m+1 \right ) }b^{\zeta \left ( m \right ) } F_{m+1} \left ( x \right )F_{m-1}\left ( x \right ) -a^{\zeta \left ( m \right ) }b^{\zeta \left ( m+1 \right ) } F_{m}^{2} \left ( x \right ) = -a\left ( -q\left ( x \right ) \right ) ^{m-1}. \end{equation}$

(2.4)

Proof. According to the identity (2.2),

$\begin{equation*} \begin{split} det\left ( S^{m} \right )& = \left ( ab \right )^{m-\zeta \left ( m \right ) } \left ( b^{2\zeta \left ( m \right ) }q\left ( x \right ) F_{m+1}\left ( x \right ) F_{m-1} \left ( x \right ) -a^{\zeta \left ( m \right )-\zeta \left ( m+1 \right ) } bq\left ( x \right )F_{m}^{2} \left ( x \right ) \right ) \\ & = a^{m-1}b^{m} q\left ( x \right )\left ( a^{\zeta \left ( m+1 \right ) } b^{\zeta \left (m \right ) }F_{m+1}\left ( x \right )F_{m-1} \left ( x \right )-a^{\zeta \left ( m \right ) } b^{\zeta \left ( m+1 \right ) }F_{m}^{2}\left ( x \right ) \right ) \\ & = det\left ( S \right )^{m} = \left ( -abq\left ( x \right ) \right )^{m} . \end{split} \end{equation*}$

Thus,

$\begin{equation*} a^{\zeta \left ( m+1 \right ) }b^{\zeta \left ( m \right ) } F_{m+1} \left ( x \right )F_{m-1}\left ( x \right ) -a^{\zeta \left ( m \right ) }b^{\zeta \left ( m+1 \right ) } F_{m}^{2} \left ( x \right ) = -a\left ( -q\left ( x \right ) \right ) ^{m-1}. \end{equation*}$

□

Since $S$ is invertible, and $S^{m}$ is also invertible, we have

$\begin{equation} \left ( S^{m} \right )^{-1} = S^{-m} = \left ( -q\left ( x \right ) \right )^{-m} \left ( ab \right ) ^{\frac{-n-\zeta \left (m \right ) }{2} } \begin{pmatrix} b^{\zeta \left ( m \right ) }q\left ( x \right ) F_{m-1} \left ( x \right ) & -a^{-\zeta \left ( m+1 \right ) }bq\left ( x \right )F_{m}\left ( x \right ) \\ -a^{\zeta \left ( m \right ) }F_{m} \left ( x \right ) & b^{\zeta \left ( m \right ) } F_{m+1}\left ( x \right ) \end{pmatrix} . \end{equation}$

(2.5)

Theorem 2.3. (Generalized d'Ocagne's identity) Let $\left \{ F_{n} \left (x \right) \right \}$ be the generalized bi-periodic Fibonacci polynomial. We have

$\begin{align} F_{m+n+1}\left ( x \right )& = a^{-\zeta \left ( mn \right ) }b^{\zeta \left ( mn \right ) } F_{m+1}\left ( x \right )F_{n+1}\left ( x \right ) +a^{\zeta \left ( mn+m+n \right ) -1}b^{1-\zeta \left ( mn+m+n \right ) }q\left ( x \right ) F_{m} \left ( x \right ) F_{n} \left ( x \right ), \end{align}$

(2.6)

$\begin{align} F_{m+n}\left ( x \right )& = a^{-\zeta \left ( mn +n\right ) }b^{\zeta \left ( mn+n \right ) } F_{m}\left ( x \right )F_{n+1}\left ( x \right ) +a^{-\zeta \left ( mn+m \right ) }b^{\zeta\left ( mn+m \right ) }q\left ( x \right ) F_{m-1} \left ( x \right ) F_{n} \left ( x \right ) , \end{align}$

(2.7)

$\begin{align} F_{m+n-1}\left ( x \right )& = a^{\zeta \left ( mn+m+n \right )-1 }b^{1-\zeta\left ( mn+m+n \right ) } F_{m}\left ( x \right )F_{n}\left ( x \right )+a^{-\zeta \left ( mn \right ) }b^{\zeta \left ( mn \right ) }q\left ( x \right ) F_{m-1} \left ( x \right ) F_{n-1} \left ( x \right ), \end{align}$

(2.8)

$\begin{align} \begin{aligned} F_{m-n+1}\left ( x \right )& = -\left ( -q\left ( x \right ) \right ) ^{-n+1} \left [ a^{-\zeta \left ( mn \right ) }b^{\zeta \left ( mn \right ) } F_{m+1}\left ( x \right )F_{n-1}\left ( x \right )\right. \left. -a^{\zeta\left ( mn+m+n\right )-1 }b^{1-\zeta \left ( mn+m+n \right ) } F_{m} \left ( x \right ) F_{n} \left ( x \right ) \right ], \end{aligned} \end{align}$

(2.9)

$\begin{align} F_{m-n}\left ( x \right )& = \left ( -q\left ( x \right ) \right ) ^{-n} \left [ a^{-\zeta \left ( mn +n\right ) }b^{\zeta\left ( mn+n \right ) } F_{m}\left ( x \right )F_{n+1}\left ( x \right ) \right.\left. -a^{-\zeta\left ( mn+m \right ) }b^{\zeta \left ( mn+m \right ) } F_{m+1} \left ( x \right ) F_{n} \left ( x \right ) \right ] , \end{align}$

(2.10)

$\begin{align} F_{m-n}\left ( x \right )& = -\left ( -q\left ( x \right ) \right ) ^{-n+1} \left [ a^{-\zeta\left ( mn +n\right ) }b^{\zeta \left ( mn+n \right ) } F_{m}\left ( x \right )F_{n-1}\left ( x \right ) \right. \left.-a^{-\zeta\left ( mn+m \right ) }b^{\zeta \left ( mn+m \right ) } F_{m-1} \left ( x \right ) F_{n} \left ( x \right ) \right ] , \end{align}$

(2.11)

$\begin{align} F_{m-n-1}\left ( x \right )& = \left ( -q\left ( x \right ) \right ) ^{-n} \left [ a^{-\zeta \left ( mn \right ) }b^{\zeta \left ( mn\right ) } F_{m-1}\left ( x \right )F_{n+1}\left ( x \right ) \right. \left. -a^{\zeta \left ( mn+m+n \right )-1 }b^{1-\zeta\left ( mn+m+n \right ) } F_{m} \left ( x \right ) F_{n} \left ( x \right ) \right ] . \end{align}$

(2.12)

Proof. According to the identity (2.2),

$\begin{equation} S^{m+n} = \left ( ab \right )^{\frac{m+n-\zeta \left ( m+n \right ) }{2} } \begin{pmatrix} b^{\zeta \left ( m+n \right ) } F_{m+n+1} \left ( x \right ) & a^{-\zeta\left ( m+n+1 \right ) }bq\left ( x \right )F_{m+n}\left ( x \right ) \\ a^{\zeta\left ( m+n \right ) }F_{m+n} \left ( x \right ) & b^{\zeta\left ( m+n \right ) }q\left ( x \right ) F_{m+n-1}\left ( x \right ) \end{pmatrix} \end{equation}$

(2.13)

and

$\begin{equation} \begin{split} S^{m}\cdot S^{n} = & \left ( ab \right )^{\frac{m+n-\zeta \left ( m \right ) -\zeta\left ( n \right )}{2} }\\ & \times\begin{pmatrix} b^{\zeta\left ( m \right )+\zeta \left ( n \right ) }F_{m+1}\left ( x \right ) F_{n+1}\left ( x \right ) &a^{-\zeta\left ( m+1 \right ) }b^{1+\zeta \left ( n \right ) }q\left ( x \right ) F_{m}\left ( x \right ) F_{n+1}\left ( x \right ) \\ +a^{\zeta \left ( m \right )-\zeta\left ( n+1 \right ) }bq\left ( x \right )F_{m } \left ( x \right ) F_{n}\left ( x \right ) &+a^{-\zeta \left ( n+1 \right ) }b^{1+\zeta \left ( m \right ) } q^{2} \left ( x \right )F_{m -1} \left ( x \right ) F_{n}\left ( x \right ) \\ & \\ a^{\zeta \left ( n \right ) } b^{\zeta \left ( m \right ) }F_{m+1}\left ( x \right ) F_{n}\left ( x \right ) &a^{\zeta \left ( n \right )-\zeta\left ( m+1 \right ) }bq\left ( x \right ) F_{m}\left ( x \right ) F_{n}\left ( x \right ) \\ +a^{\zeta\left ( m \right ) }b^{\zeta \left ( n \right ) }q\left ( x \right )F_{m } \left ( x \right ) F_{n-1}\left ( x \right ) &+b^{\zeta\left ( m \right )+\zeta\left ( n \right ) } q^{2} \left ( x \right )F_{m -1} \left ( x \right ) F_{n-1}\left ( x \right ) \end{pmatrix}. \end{split} \end{equation}$

(2.14)

Since

$S^{m+n} = S^{m} S^{n},$

the corresponding entries in identities (2.13) and (2.14) are equal, so we obtain (2.6)–(2.8), where

(f) $\zeta\left (m+n \right) -\zeta \left (m \right)-\zeta \left (n \right) = -2\zeta\left (mn \right)$ ,

(g) $\zeta \left (m+n \right) +\zeta \left (m \right)+\zeta \left (n \right) = 2\zeta \left (mn+m+n \right)$ ,

(h) $\zeta \left (m+n \right) -\zeta \left (m \right)+\zeta \left (n \right) = 2\zeta \left (mn+n \right)$ ,

(i) $\zeta \left (m+n \right) +\zeta \left (m \right)-\zeta \left (n \right) = 2\zeta \left (mn+m \right)$ ,

(j) $\zeta\left (m+n+1 \right) -\zeta \left (m+1 \right)-\zeta \left (n \right) = -2\zeta \left (mn+n \right)$ ,

(k) $\zeta \left (m+n+1 \right) -\zeta \left (m \right)-\zeta \left (n+1 \right) = -2\zeta\left (mn+m \right)$ .

In addition, we have

$\begin{equation} S^{m-n} = \left ( ab \right )^{\frac{m-n-\zeta \left ( m-n \right ) }{2} } \begin{pmatrix} b^{\zeta \left ( m-n \right ) } F_{m-n+1} \left ( x \right ) & a^{-\zeta\left ( m-n+1 \right ) }bq\left ( x \right )F_{m-n}\left ( x \right ) \\ a^{\zeta\left ( m-n \right ) }F_{m-n} \left ( x \right ) & b^{\zeta \left ( m-n \right ) }q\left ( x \right ) F_{m-n-1}\left ( x \right ) \end{pmatrix} \end{equation}$

(2.15)

and

$\begin{equation} \begin{split} S^{m}\cdot S^{-n} = &\left ( -q\left ( x \right ) \right )^{-n+1} \left ( ab \right )^{\frac{m-n-\zeta\left ( m \right ) -\zeta\left ( n \right )}{2} }\\ & \times \begin{pmatrix} b^{\zeta \left ( m \right )+\zeta \left ( n \right ) }F_{m+1}\left ( x \right ) F_{n-1}\left ( x \right ) & a^{-\zeta \left ( m+1 \right ) }b^{1+\zeta \left ( n \right ) } F_{m}\left ( x \right ) F_{n+1}\left ( x \right )\\ -a^{\zeta \left ( n \right )-\zeta \left ( m+1 \right ) }bF_{m } \left ( x \right ) F_{n}\left ( x \right ) & -a^{-\zeta \left ( n+1 \right ) }b^{1+\zeta \left ( m \right ) } F_{m +1} \left ( x \right ) F_{n}\left ( x \right ) \\ &\\ a^{\zeta \left (m \right ) } b^{\zeta \left (n \right ) }F_{m}\left ( x \right ) F_{n-1}\left ( x \right )& b^{\zeta \left ( n \right )+\zeta \left ( m \right ) } F_{m-1}\left ( x \right ) F_{n+1}\left ( x \right )\\ -a^{\zeta\left ( n \right ) }b^{\zeta \left ( m \right ) }F_{m -1} \left ( x \right ) F_{n}\left ( x \right ) & -a^{\zeta \left ( m \right )-\zeta\left ( n+1 \right ) }b F_{m}\left ( x \right )F_{n}\left ( x \right ) \end{pmatrix} . \end{split} \end{equation}$

(2.16)

Since

$S^{m-n} = S^{m} S^{-n} ,$

the corresponding entries in identities (2.15) and (2.16) are equal, so we obtain (2.9)–(2.12), where

(l) $\zeta\left (m-n \right) -\zeta \left (m \right)-\zeta \left (n \right) = -2\zeta\left (mn \right)$ ,

(m) $\zeta \left (m-n \right) +\zeta \left (m \right)+\zeta \left (n \right) = 2\zeta \left (mn+m+n \right)$ ,

(n) $\zeta \left (m-n \right) +\zeta\left (m \right)-\zeta \left (n \right) = 2\zeta\left (mn+m \right)$ ,

(o) $\zeta\left (m-n \right) -\zeta \left (m \right)+\zeta\left (n \right) = 2\zeta \left (mn+n \right)$ . □

Theorem 2.4. (Sum involving binomial coefficients) Let $\left \{ F_{m} \left (x \right) \right \}$ be the generalized bi-periodic Fibonacci polynomial. We have

$\begin{align} F_{2m}\left ( x \right )& = \sum\limits_{k = 0}^{m}\binom{m}{k}a^{\frac{k+\zeta \left ( k \right ) }{2} } b^{\frac{k-\zeta \left ( k \right ) }{2} }p^{k}\left ( x \right )q^{m-k} \left ( x \right )F_{k} \left ( x \right ) , \end{align}$

(2.17)

$\begin{align} F_{2m+1}\left ( x \right )& = \sum\limits_{k = 0}^{m}\binom{m}{k}a^{\frac{k-\zeta\left ( k \right ) }{2} } b^{\frac{k+\zeta \left ( k \right ) }{2} }p^{k}\left ( x \right )q^{m-k} \left ( x \right )F_{k+1} \left ( x \right ) , \end{align}$

(2.18)

$\begin{align} F_{2m-1}\left ( x \right )& = \sum\limits_{k = 0}^{m}\binom{m}{k}a^{\frac{k-\zeta \left ( k \right ) }{2} } b^{\frac{k+\zeta\left ( k \right ) }{2} }p^{k}\left ( x \right )q^{m-k} \left ( x \right )F_{k-1} \left ( x \right ) , \end{align}$

(2.19)

$\begin{align} F_{-2m}\left ( x \right ) & = q ^{-m}\left ( x \right ) \sum\limits_{k = 0}^{m}\binom{m}{k}\left ( -1 \right ) ^{k}a^{\frac{k+\zeta \left ( k \right ) }{2} } b^{\frac{k-\zeta \left ( k \right ) }{2} } p^{k}\left ( x \right ) F_{-k} \left ( x \right ) , \end{align}$

(2.20)

$\begin{align} F_{-2m+1}\left ( x \right ) & = q ^{-m}\left ( x \right ) \sum\limits_{k = 0}^{m}\binom{m}{k}\left ( -1 \right ) ^{k}a^{\frac{k-\zeta \left ( k \right ) }{2} } b^{\frac{k+\zeta \left ( k \right ) }{2} } p^{k}\left ( x \right ) F_{-k+1} \left ( x \right ) , \end{align}$

(2.21)

$\begin{align} F_{-2m-1}\left ( x \right )& = q ^{-m}\left ( x \right ) \sum\limits_{k = 0}^{n}\binom{m}{k}\left ( -1 \right ) ^{k}p^{k}\left ( x \right ) a^{\frac{k-\zeta \left ( k \right ) }{2} } b^{\frac{k+\zeta \left ( k \right ) }{2} } F_{-k-1} \left ( x \right ) . \end{align}$

(2.22)

Proof. According to Cayley Hamilton's theorem, the following matrix $S$ identity is obtained:

$\begin{equation*} S^{2} -abp\left ( x \right )S-abq\left ( x \right )I = 0 , \end{equation*}$

then

$\begin{equation*} \left ( S^{2} \right ) ^{m} = \left ( abp\left ( x \right )S +abq\left ( x \right )I \right )^{m} = \left ( ab \right )^{m} \sum\limits_{k = 0}^{m}\binom{m}{k}\left ( p\left ( x \right ) S \right ) ^{k}\left ( q\left ( x \right ) \right )^{m-k} . \end{equation*}$

We obtain

$\begin{equation} \begin{aligned} \left ( ab \right ) ^{m}& \begin{pmatrix} F_{2m+1}\left ( x \right ) & a^{-1}b q\left ( x \right )F_{2m}\left ( x \right ) \\ F_{2m}\left ( x \right ) & q\left ( x \right )F_{2m-1}\left ( x \right ) \end{pmatrix} = \left ( ab \right ) ^{m}\sum\limits_{k = 0}^{m}\binom{m}{k}p^{k}\left ( x \right )q^{m-k}\left ( x \right ) \left ( ab \right )^{\frac{k-\zeta\left ( k \right ) }{2} } \begin{pmatrix} b^{\zeta \left ( k \right ) }F_{k+1} \left ( x \right ) & a^{-\zeta \left ( k+1 \right ) }bq\left ( x \right )F_{k} \left ( x \right ) \\ a^{\zeta \left ( k \right ) }F_{k} \left ( x \right ) & b^{\zeta \left ( k \right ) }q\left ( x \right )F_{k-1} \left ( x \right ) \end{pmatrix} . \end{aligned} \end{equation}$

(2.23)

The corresponding entries in identity (2.23) are equal, so we obtain (2.17)–(2.19).

Thus,

$S^{-1} = \begin{pmatrix} 0 & a^{-1} \\ b^{-1}q^{-1} \left ( x \right ) &-p\left ( x \right ) q^{-1}\left ( x \right ) \end{pmatrix}.$

According to Cayley Hamilton's theorem, the following matrix $S^{-1}$ identity is obtained:

$\begin{equation*} S^{-2} +p\left ( x \right )q^{-1} \left ( x \right )S ^{-1}-a^{-1} b^{-1} q^{-1} \left ( x \right ) I = 0, \end{equation*}$

then,

$\begin{equation*} \left ( S^{-2} \right ) ^{m} = \left ( a^{-1} b^{-1} q^{-1} \left ( x \right ) I- p\left ( x \right )q^{-1} \left ( x \right )S ^{-1} \right )^{m} = q^{-m} \left ( x \right ) \sum\limits_{k = 0}^{m}\binom{m}{k}\left ( -1 \right ) ^{k} p^{k} \left ( x \right ) S^{-k}\left ( ab \right ) ^{k-m}. \end{equation*}$

We obtain

$\begin{equation} \begin{aligned} \left ( ab \right ) ^{-m}& \begin{pmatrix} F_{-2m+1}\left ( x \right ) & a^{-1}b q\left ( x \right )F_{-2m}\left ( x \right ) \\ F_{-2m}\left ( x \right ) & q\left ( x \right )F_{-2m-1}\left ( x \right ) \end{pmatrix} = \left ( abq\left ( x \right ) \right ) ^{-m} \sum\limits_{k = 0}^{m}\binom{m}{k}\left ( -1 \right ) ^{k} p^{k}\left ( x \right )\left ( ab \right )^{\frac{k-\zeta \left ( k \right ) }{2} } \begin{pmatrix} b^{\zeta\left ( k \right ) }F_{-k+1} \left ( x \right ) & a^{-\zeta\left ( k+1 \right ) }bq\left ( x \right )F_{-k} \left ( x \right ) \\ a^{\zeta \left ( k \right ) }F_{-k} \left ( x \right ) & b^{\zeta \left ( k \right ) }q\left ( x \right )F_{-k-1} \left ( x \right ) \end{pmatrix} .\end{aligned} \end{equation}$

(2.24)

The corresponding entries in identity (2.24) are equal, so we obtain (2.20)–(2.22). □

3. The generalized bi-periodic Lucas polynomial

Inspired by ^[18], in this section, we define the generalized bi-periodic Lucas polynomial $\left \{ L_{n} \left (x \right) \right \}$ as follows:

Definition 3.1. The generalized bi-periodic Lucas polynomial is defined by

$L_{0} \left ( x \right ) = 2,\quad L_{1} \left ( x \right ) = ap\left ( x \right )$

and

$\begin{equation*} {L_{n}\left ( x \right ) = } \begin{cases} bp\left ( x \right ) L_{n-1}\left ( x \right )+q\left ( x \right )L_{n-2}\left ( x \right ), \quad & \text{if } n \text{ is even and } n \geq 2 ,\\ ap\left ( x \right ) L_{n-1}\left ( x \right )+q\left ( x \right )L_{n-2}\left ( x \right ) , \quad & \text{if } n \text{ is odd and } n \geq 3 , \end{cases} \end{equation*}$

where $a$ and $b$ are nonzero real numbers, and $p\left (x \right)$ , $q\left (x \right)$ are nonzero polynomials in $Q \left [ x \right ]$ .

According to the definition, we obtain another expression of the generalized bi-periodic Lucas polynomial as follows:

$\begin{equation} L_{m}\left ( x \right ) = a^{\zeta \left ( m \right ) } b^{1-\zeta \left ( m \right ) }p\left ( x \right )L_{m-1} \left ( x \right ) +q\left ( x \right )L_{m-2}\left ( x \right ), \quad m\ge 2, \end{equation}$

(3.1)

where

$\zeta \left ( m \right ) = m-2\left \lfloor \frac{m}{2} \right \rfloor$

is the parity function, with $\left \lfloor \cdot \right \rfloor$ denoting the floor function. The characteristic polynomial of the generalized bi-periodic Lucas polynomial is

$\begin{equation*} t^{2} -p\left ( x \right ) abt-q\left ( x \right )ab = 0 , \end{equation*}$

and the roots are

$\sigma \left ( x \right ) = \frac{abp\left ( x \right )+\sqrt{a^{2}b^{2}p^{2}\left ( x \right )+4abq\left ( x \right ) } }{2}$

and

$\tau \left ( x \right ) = \frac{abp\left ( x \right )-\sqrt{a^{2}b^{2}p^{2}\left ( x \right )+4abq\left ( x \right ) } }{2} .$

We have:

(p) $\sigma \left (x \right) +\tau \left (x \right) = abp\left (x \right)$ ,

(q) $\sigma \left (x \right) -\tau \left (x \right) = \sqrt{p^{2}\left (x \right)a^{2} b^{2}+4q\left (x \right)ab }$ ,

(r) $\sigma \left (x \right) \tau \left (x \right) = -abq\left (x \right)$ .

Theorem 3.1. The generating functions of the generalized bi-periodic Lucas polynomial $\left \{ L_{m} \left (x \right) \right \}$ are

$\begin{align*} T_{m}\left ( x,t \right )& = \sum\limits_{m = 0}^{\infty } L_{m}\left ( x \right ) t^{m}\\& = \frac{2+ap\left ( x \right )t-\left ( abp^{2}\left ( x \right )+2q\left ( x \right ) \right )t^{2}+ap\left ( x \right )q\left ( x \right )t^{3}}{1-\left ( abp^{2}\left ( x \right ) +2q\left ( x \right ) \right )t^{2}+q^{2}\left ( x \right ) t^{4} } . \end{align*}$

Lemma 3.1. The generalized bi-periodic Lucas $\left \{ L_{m} \left (x \right) \right \}$ polynomial satisfy the following identities

$\begin{equation} L_{2m} \left ( x \right ) = \left (abp^{2} \left ( x \right ) +2q\left ( x \right )\right ) L_{2m-2}\left ( x \right )-q^{2} \left ( x \right )L_{2m-4} \left ( x \right ) \end{equation}$

(3.2)

and

$\begin{equation} L_{2m+1} \left ( x \right ) = \left (abp^{2} \left ( x \right ) +2q \left ( x \right )\right )L_{2m-1}\left ( x \right )-q^{2} \left ( x \right )L_{2m-3} \left ( x \right ) . \end{equation}$

(3.3)

Proof. By identity (3.1),

$\begin{equation*} \begin{split} L_{2m} \left ( x \right )& = bp\left ( x \right )L_{2m-1} \left ( x \right )+q\left ( x \right )L_{2m-2} \left ( x \right ) \\ & = bp\left ( x \right )\left [ ap\left ( x \right )L_{2m-2} \left ( x \right )+q\left ( x \right )L_{2m-3} \left ( x \right ) \right ] +q\left ( x \right )L_{2m-2} \left ( x \right )\\ & = \left [ abp^{2} \left ( x \right ) +q\left ( x \right ) \right ]L_{2m-2} \left ( x \right ) +bp\left ( x \right ) q\left ( x \right )L_{2m-3} \left ( x \right ) \\ & = \left [ abp^{2} \left ( x \right ) +q\left ( x \right ) \right ]L_{2m-2} \left ( x \right ) + q\left ( x \right )L_{2m-2} \left ( x \right )- q^{2} \left ( x \right )L_{2m-4} \left ( x \right ) \\ & = \left [ abp^{2} \left ( x \right ) +2q\left ( x \right ) \right ]L_{2m-2} \left ( x \right ) - q^{2} \left ( x \right )L_{2m-4} \left ( x \right ) \end{split} \end{equation*}$

and

$\begin{equation*} \begin{split} L_{2m+1} \left ( x \right )& = ap\left ( x \right )L_{2m} \left ( x \right )+q\left ( x \right )L_{2m-1} \left ( x \right ) \\ & = ap\left ( x \right )\left [ bp\left ( x \right )L_{2m-1} \left ( x \right )+q\left ( x \right )L_{2m-2} \left ( x \right ) \right ] +q\left ( x \right )L_{2m-1} \left ( x \right )\\ & = \left [ abp^{2} \left ( x \right ) +q\left ( x \right ) \right ]L_{2m-1} \left ( x \right ) +ap\left ( x \right ) q\left ( x \right )L_{2m-2} \left ( x \right ) \\ & = \left [ abp^{2} \left ( x \right ) +q\left ( x \right ) \right ]L_{2m-1} \left ( x \right ) + q\left ( x \right )L_{2m-1} \left ( x \right )- q^{2} \left ( x \right )L_{2m-3} \left ( x \right ) \\ & = \left [ abp^{2} \left ( x \right ) +2q\left ( x \right ) \right ]L_{2m-1} \left ( x \right ) -q^{2} \left ( x \right )L_{2m-3} \left ( x \right ) . \end{split} \end{equation*}$

□

Proof of Theorem 3.1. According to the definition of the generating functions of the generalized bi-periodic Lucas polynomial, we have

$\begin{align*} T_{m}\left ( x,t \right )& = T_{m}^{e} \left ( x,t \right )+T_{m}^{o} \left ( x,t \right )\\& = \sum\limits_{k = 0}^{\infty } L_{2k}\left ( x \right ) t^{2k}+\sum\limits_{k = 0}^{\infty } L_{2k+1}\left ( x \right ) t^{2k+1}. \end{align*}$

To begin, we consider $T_{m}^{e} \left (x, t \right)$ ,

$\begin{equation} T_{m}^{e} \left ( x,t \right ) = \sum\limits_{k = 0}^{\infty } L_{2k}\left ( x \right ) t^{2k} = L_{0} \left ( x \right ) +L_{2} \left ( x \right ) t^{2}+L_{4} \left ( x \right ) t^{4} + \cdots , \end{equation}$

(3.4)

$\begin{equation} -\left (abp^{2} \left ( x \right )+2q\left ( x \right ) \right ) t^{2} T_{m}^{e} \left ( x,t \right ) = -\left (abp^{2} \left ( x \right )+2q\left ( x \right ) \right ) \sum\limits_{k = 0}^{\infty } L_{2k}\left ( x \right ) t^{2k+2}, \end{equation}$

(3.5)

$\begin{equation} q^{2} \left ( x \right ) t^{4} T_{m}^{e} \left ( x,t \right ) = q^{2} \left ( x \right ) \sum\limits_{k = 0}^{\infty } L_{2k}\left ( x \right ) t^{2k+4} . \end{equation}$

(3.6)

Contact (3.4)–(3.6) and Lemma 3.1. We obtain

$\begin{equation*} \begin{split} &\left \{ 1-\left ( abp^{2}\left ( x \right ) +2q\left ( x \right ) \right )t^{2}+q^{2}\left ( x \right ) t^{4} \right \} T_{m}^{e}\left ( x,t \right ) \\ & = L_{0} \left ( x \right )+L_{2}\left ( x \right ) t^{2} + \sum\limits_{k = 2}^{\infty } L_{2k}\left ( x \right ) t^{2k} -\left (abp^{2} \left ( x \right )+2q\left ( x \right ) \right ) \sum\limits_{k = 0}^{\infty } L_{2k}\left ( x \right ) t^{2k+2} +q^{2} \left ( x \right ) \sum\limits_{k = 0}^{\infty } L_{2k}\left ( x \right ) t^{2k+4}\\ & = 2+\left (abp^{2} \left ( x \right )+2q\left ( x \right ) \right ) t^{2} + \sum\limits_{k = 2}^{\infty } L_{2k}\left ( x \right ) t^{2k} - \left ( abp^{2} \left ( x \right )+2q\left ( x \right ) \right ) 2 t^{2}\\ &\quad -\left ( abp^{2} \left ( x \right )+2q\left ( x \right ) \right ) \sum\limits_{k = 2}^{\infty } L_{2k-2}\left ( x \right ) t^{2k}+q^{2} \left ( x \right ) \sum\limits_{k = 2}^{\infty } L_{2k-4}\left ( x \right ) t^{2k}\\ & = 2-\left ( abp^{2} \left ( x \right )+2q\left ( x \right ) \right ) t^{2} + \sum\limits_{k = 2}^{\infty }\left \{ L_{2k}\left ( x \right )-\left ( abp^{2} \left ( x \right )+2q\left ( x \right ) \right ) L_{2k-2}\left ( x \right )+q^{2} \left ( x \right ) L_{2k-4}\left ( x \right ) \right \} t^{2k}\\ & = 2-\left ( abp^{2} \left ( x \right )+2q\left ( x \right ) \right ) t^{2}. \end{split} \end{equation*}$

Therefore,

$\begin{equation} T_{m}^{e}\left ( x,t \right ) = \frac{2-\left ( abp^{2} \left ( x \right )+2q\left ( x \right ) \right ) t^{2}}{ 1-\left ( abp^{2}\left ( x \right ) +2q\left ( x \right ) \right ) t^{2}+q^{2}\left ( x \right ) t^{4} } . \end{equation}$

(3.7)

Next, we consider $T_{m}^{o} \left (x, t \right)$ ,

$\begin{equation} T_{m}^{o} \left ( x,t \right ) = \sum\limits_{k = 0}^{\infty } L_{2k+1}\left ( x \right ) t^{2k+1} = L_{1} \left ( x \right )t +L_{3} \left ( x \right ) t^{3}+L_{5} \left ( x \right ) t^{5} + \cdots , \end{equation}$

(3.8)

$\begin{equation} -\left ( abp^{2} \left ( x \right )+2q\left ( x \right ) \right ) t^{2} T_{m}^{o} \left ( x,t \right ) = -\left ( abp^{2} \left ( x \right )+2q\left ( x \right ) \right ) \sum\limits_{k = 0}^{\infty } L_{2k+1}\left ( x \right ) t^{2k+3}, \end{equation}$

(3.9)

$\begin{equation} q^{2} \left ( x \right ) t^{4} T_{m}^{o} \left ( x,t \right ) = q^{2} \left ( x \right ) \sum\limits_{k = 0}^{\infty } L_{2k+1}\left ( x \right ) t^{2k+5} . \end{equation}$

(3.10)

Contact (3.8)–(3.10) and Lemma 3.1. We obtain

$\begin{equation*} \begin{aligned} \begin{split} &\left \{ 1-\left ( abp^{2}\left ( x \right ) +2q\left ( x \right ) \right ) t^{2}+q^{2}\left ( x \right ) t^{4} \right \} T_{m}^{o}\left ( x,t \right ) \\ & = L_{1} \left ( x \right )t+L_{3}\left ( x \right ) t^{3} + \sum\limits_{k = 2}^{\infty } L_{2k+1}\left ( x \right ) t^{2k+1} -\left (abp^{2} \left ( x \right )+2q\left ( x \right ) \right ) \sum\limits_{k = 0}^{\infty } L_{2k+1}\left ( x \right ) t^{2k+3} +q^{2} \left ( x \right ) \sum\limits_{k = 0}^{\infty } L_{2k+1}\left ( x \right ) t^{2k+5}\\ & = ap\left ( x \right ) t+\left ( a^{2}bp^{3} \left ( x \right ) +3ap\left ( x \right )q\left ( x \right ) \right ) t^{3} + \sum\limits_{k = 2}^{\infty } L_{2k+1}\left ( x \right ) t^{2k+1} - \left ( a^{2}bp^{3} \left ( x \right )+2ap\left ( x \right )q\left ( x \right ) \right ) t^{3}\\ &\quad -\left ( abp^{2} \left ( x \right )+2q\left ( x \right ) \right ) \sum\limits_{k = 2}^{\infty } L_{2k-1}\left ( x \right ) t^{2k+1}+q^{2} \left ( x \right ) \sum\limits_{k = 2}^{\infty } L_{2k-3}\left ( x \right ) t^{2k+1}\\ & = ap\left ( x \right ) t+ap\left ( x \right )q\left ( x \right ) t^{3} + \sum\limits_{k = 2}^{\infty }\left \{L_{2k+1}\left ( x \right )-\left ( abp^{2} \left ( x \right )+2q\left ( x \right ) \right ) L_{2k-1}\left ( x \right )+q^{2} \left ( x \right ) L_{2k-3}\left ( x \right ) \right \} t^{2k+1}\\ & = ap\left ( x \right ) t+ap\left ( x \right )q\left ( x \right ) t^{3} . \end{split}\end{aligned} \end{equation*}$

Thus,

$\begin{equation} T_{m}^{o}\left ( x,t \right ) = \frac{ap\left ( x \right ) t+ap\left ( x \right )q\left ( x \right ) t^{3} }{ 1-\left ( abp^{2}\left ( x \right ) +2q\left ( x \right ) \right )t^{2}+q^{2}\left ( x \right ) t^{4} } . \end{equation}$

(3.11)

By (3.7) and (3.11), we have

$\begin{equation*} T_{m}\left ( x,t \right ) = \frac{2+ap\left ( x \right ) t-\left ( abp^{2} \left ( x \right )+2q\left ( x \right ) \right ) t^{2}+ap\left ( x \right )q\left ( x \right ) t^{3} }{ 1-\left ( abp^{2}\left ( x \right ) +2q\left ( x \right ) \right ) t^{2}+q^{2}\left ( x \right ) t^{4} } . \end{equation*}$

□

Theorem 3.2. The Binet identity of the generalized bi-periodic Lucas polynomial is

$\begin{equation} L_{m}\left ( x \right ) = \frac{a^{\zeta \left ( m \right ) } }{\left ( ab \right )^{\left \lfloor \frac{m+1}{2} \right \rfloor } } \left ( \sigma ^{m} \left ( x \right )+\tau ^{m} \left ( x \right ) \right ), \end{equation}$

(3.12)

where

$\sigma \left ( x \right ) = \frac{abp\left ( x \right )+\sqrt{a^{2}b^{2}p^{2}\left ( x \right )+4abq\left ( x \right ) } }{2} ,$

$\tau \left ( x \right ) = \frac{abp\left ( x \right )-\sqrt{a^{2}b^{2}p^{2}\left ( x \right )+4abq\left ( x \right ) } }{2} ,$

and

$\zeta \left ( m \right ) = m-2\left \lfloor \frac{m}{2} \right \rfloor$

is the parity function, with $\left \lfloor \cdot \right \rfloor$ denoting the floor function.

Proof. We prove (3.12) by mathematical induction. Obviously, the identity is true when $m = 0$ and $m = 1$ . We assume that the identity is true with $m$ . Next, we prove that the identity is true when $m+1$ .

According to the identity (3.1) and mathematical induction, we have

$\begin{equation*} \begin{split} L_{m+1} \left ( x \right )& = a^{\zeta \left ( m+1 \right ) }b^{^{1-\zeta \left ( m+1 \right ) } }p\left ( x \right )L_{m}\left ( x \right )+q\left ( x \right ) L_{m-1}\left ( x \right )\\ & = a^{\zeta \left ( m+1 \right ) }b^{^{1-\zeta \left ( m+1 \right ) } }p\left ( x \right )\left \{ \frac{a^{\zeta \left ( m \right ) } }{\left ( ab \right )^{\left \lfloor \frac{m+1}{2} \right \rfloor } } \left ( \sigma ^{m} \left ( x \right )+\tau ^{m} \left ( x \right ) \right ) \right \} +q\left ( x \right )\left \{\frac{a^{\zeta \left ( m -1\right ) } }{\left ( ab \right )^{\left \lfloor \frac{m}{2} \right \rfloor } } \left ( \sigma ^{m-1} \left ( x \right )+\tau ^{m-1} \left ( x \right ) \right ) \right \} \\ & = a^{\zeta \left ( m+1 \right ) } \sigma ^{m-1} \left ( x \right )\left ( \frac{a^{\zeta \left (m \right ) } b^{1-\zeta \left ( m+1 \right ) }p\left ( x \right )\sigma \left ( x \right ) }{\left ( ab \right )^{\left \lfloor \frac{m+1}{2} \right \rfloor } } +\frac{q\left ( x \right ) }{\left ( ab \right )^{\left \lfloor \frac{m}{2} \right \rfloor } } \right ) \\ & \quad +a^{\zeta \left ( m+1 \right ) } \tau ^{m-1} \left ( x \right )\left ( \frac{a^{\zeta \left ( m \right ) } b^{1-\zeta \left ( m+1 \right ) }p\left ( x \right )\tau \left ( x \right ) }{\left ( ab \right )^{\left \lfloor \frac{m+1}{2} \right \rfloor } } +\frac{q\left ( x \right ) }{\left ( ab \right )^{\left \lfloor \frac{m}{2} \right \rfloor } } \right ) \\ & = a^{\zeta \left ( m+1 \right ) }\sigma ^{m-1}\left ( x \right ) \left ( \frac{abp\left ( x \right )\sigma \left ( x \right ) }{a^{1-\zeta \left ( m \right ) }b^{\zeta \left (m+1 \right ) }\left ( ab \right ) ^{\left \lfloor \frac{m+1}{2} \right \rfloor } } +\frac{abq\left ( x \right ) }{\left ( ab \right )^{\left \lfloor \frac{m}{2} \right \rfloor +1 } } \right ) \\ & \quad +a^{\zeta \left ( m+1 \right ) }\tau ^{m-1}\left ( x \right ) \left ( \frac{abp\left ( x \right )\tau \left ( x \right ) }{a^{1-\zeta \left ( m \right ) }b^{\zeta \left ( m+1 \right ) }\left ( ab \right ) ^{\left \lfloor \frac{m+1}{2} \right \rfloor } }+\frac{abq\left ( x \right ) }{\left ( ab \right )^{\left \lfloor \frac{m}{2} \right \rfloor +1 } } \right ) \\ & = a^{\zeta \left ( m+1 \right ) }\sigma ^{m-1}\left ( x \right )\left \{ \frac{ab\left ( p\left ( x \right )\sigma \left ( x \right ) + q\left ( x \right ) \right )}{\left ( ab \right )^{\left \lfloor \frac{m}{2} \right \rfloor+1 } } \right \} +a^{\zeta \left ( m+1 \right ) }\tau ^{m-1}\left ( x \right )\left \{ \frac{ab\left ( p\left ( x \right )\tau \left ( x \right ) + q\left ( x \right ) \right )}{\left ( ab \right )^{\left \lfloor \frac{m}{2} \right \rfloor+1 } } \right \} \\ & = \frac{a^{\zeta \left ( m+1 \right ) } }{\left ( ab \right )^{\left \lfloor \frac{m}{2} \right \rfloor+1 } }\left [ \sigma ^{m+1}\left ( x \right )+\tau ^{m+1} \left ( x \right ) \right ] , \end{split} \end{equation*}$

where

(s) $a^{1-\zeta \left (m \right) }b^{\zeta \left (m+1 \right) } \left (ab \right)^{\left \lfloor \frac{m+1}{2} \right \rfloor } = \left (ab \right)^{\left \lfloor \frac{m}{2} \right \rfloor+1 }$ ,

(t) $p\left (x \right) \sigma \left (x \right)+q \left (x \right) = \frac{\sigma ^{2} \left (x \right) }{ab}$ ,

(u) $p\left (x \right) \tau \left (x \right)+q \left (x \right) = \frac{\tau ^{2} \left (x \right) }{ab}$ . □

Theorem 3.3. Negative subscript terms of the generalized bi-periodic Lucas polynomial $\left \{ L_{n} \left (x \right) \right \}$ are

$\begin{equation*} L_{-m}\left ( x \right ) = \left ( -1 \right )^{m} q^{-m}\left ( x \right ) L_{m}\left ( x \right ) . \end{equation*}$

Proof. According to the identity (3.12),

$\begin{equation*} \begin{split} L_{-m} \left ( x \right )& = \frac{a^{\zeta \left ( -m \right ) } }{\left ( ab \right )^{\left \lfloor \frac{-m+1}{2} \right \rfloor} }\left ( \sigma ^{-m}\left ( x \right )+\tau ^{-m} \left ( x \right ) \right ) = \left ( -1 \right ) ^{m}\cdot \frac{a^{\zeta \left ( -m \right ) } }{\left ( ab \right )^{\left \lfloor \frac{-m+1}{2} \right \rfloor } }\left \{\frac{\sigma^{m}\left ( x \right ) +\tau ^{m}\left ( x \right ) }{\left ( abq\left ( x \right ) \right )^{m} } \right \} \\ & = \left ( -1 \right )^{m} q^{-m}\left ( x \right ) \left ( \frac{a^{\zeta \left ( m \right ) } }{\left ( ab \right )^{\left \lfloor \frac{m+1}{2} \right \rfloor } } \right ) \left ( \sigma ^{m} \left ( x \right )+\tau ^{m} \left ( x \right ) \right ) = \left ( -1 \right )^{m} q^{-m}\left ( x \right ) L_{m}\left ( x \right ) . \end{split} \end{equation*}$

□

Theorem 3.4. The generalized Catalan's identity of the generalized bi-periodic Lucas polynomial $\left \{ L_{n} \left (x \right) \right \}$ is

$\begin{equation*} a^{1-\zeta \left (m-r \right ) } b^{\zeta\left ( m-r \right ) }L_{m-r}\left ( x \right ) L_{m+r}\left ( x \right ) -a^{1-\zeta \left ( m \right ) }b^{\zeta \left ( m \right ) }L_{m}^{2}\left ( x \right ) = a^{\zeta \left (r+1 \right ) } b^{\zeta \left ( r \right ) }\left (-q\left ( x \right ) \right )^{m-r}L_{r}^{2}\left ( x \right )-4a\left (-q\left ( x \right ) \right ) ^{m}. \end{equation*}$

Proof. According to the identity (3.12),

$\begin{equation*} \begin{split} &a^{1-\zeta \left ( m-r \right ) } b^{\zeta \left ( m-r \right ) }L_{m-r}\left ( x \right ) L_{m+r}\left ( x \right ) -a^{1-\zeta \left ( m \right ) }b^{\zeta \left (m \right ) }L_{m}^{2}\left ( x \right ) \\ & = a^{1-\zeta \left ( m-r \right ) } b^{\zeta \left ( m-r \right ) }\cdot \frac{a^{\zeta \left ( m-r \right ) } }{\left ( ab \right )^{\left \lfloor \frac{m-r+1}{2} \right \rfloor } } \frac{a^{\zeta \left ( m+r \right ) } }{\left ( ab \right )^{\left \lfloor \frac{m+r+1}{2} \right \rfloor } }\left ( \sigma ^{m-r}\left ( x \right )+\tau ^{m-r} \left ( x \right ) \right )\left (\sigma ^{m+r}\left ( x \right )+\tau ^{m+r} \left ( x \right ) \right ) \\ & \quad -a^{1-\zeta \left (m \right ) }b^{\zeta \left ( m \right ) }\left (\frac{a^{\zeta \left ( m \right ) } }{\left ( ab \right )^{\left \lfloor \frac{m+1}{2} \right \rfloor } } \right )^{2} \left ( \sigma ^{m}\left ( x \right )+\tau ^{m} \left ( x \right ) \right ) ^{2} \\ & = \frac{a^{1+\zeta \left ( m+r \right ) }b^{\zeta \left ( m-r \right ) } }{\left ( ab \right ) ^{ m+1-\zeta \left ( m+1-r \right ) }} \left \{ \sigma ^{2m} \left ( x \right )+\left ( \sigma\left ( x \right ) \tau \left ( x \right )\right )^{m-r}\left ( \sigma^{2r} \left ( x \right )+ \tau ^{2r} \left ( x \right ) \right ) +\tau ^{2m}\left ( x \right ) \right \} \\ & \quad - \frac{a^{1+\zeta \left ( m \right ) }b^{\zeta \left ( m\right ) } }{\left ( ab \right ) ^{m+1-\zeta \left ( m+1 \right ) } } \left ( \sigma^{2m} \left ( x \right )+2\sigma ^{m}\left ( x \right ) \tau ^{m}\left ( x \right ) +\tau ^{2m}\left ( x \right )\right ) \\ & = \frac{a}{\left ( ab \right )^{m} } \left \{ \sigma ^{2m} \left ( x \right )+\left ( \sigma\left ( x \right ) \tau \left ( x \right )\right )^{m-r}\left ( \sigma^{2r} \left ( x \right )+ \tau ^{2r} \left ( x \right ) \right ) +\tau ^{2m}\left ( x \right ) \right \} \\ & \quad - \frac{a}{\left ( ab \right )^{m} } \left ( \sigma ^{2m} \left ( x \right )+2\sigma ^{m}\left ( x \right ) \tau ^{m}\left ( x \right ) +\tau ^{2m}\left ( x \right )\right ) \\ & = \frac{a}{\left ( ab \right )^{m} } \left \{ \left [ \sigma\left ( x \right ) \tau \left ( x \right )\right ]^{m-r}\left ( \sigma^{2r} \left ( x \right )+ \tau ^{2r} \left ( x \right ) \right ) -2\sigma ^{m}\left ( x \right ) \tau ^{m} \left ( x \right )\right \} \\ & = \frac{a\left ( \sigma \left ( x \right )\tau \left ( x \right ) \right )^{m-r} }{\left ( ab\right )^{m} }\left ( \sigma ^{2r}\left ( x \right )+ \tau ^{2r} \left ( x \right )- 2\sigma ^{r}\left ( x \right ) \tau ^{r} \left ( x \right ) \right ) \\ & = \frac{a\left ( \sigma \left ( x \right )\tau \left ( x \right ) \right )^{m-r} }{\left ( ab\right )^{m} }\left \{\left (\sigma ^{r}\left ( x \right )+ \tau ^{r} \left ( x \right ) \right ) ^{2} - 4\sigma^{r}\left ( x \right ) \tau ^{r} \left ( x \right ) \right \} \\ & = \frac{a\left ( -q\left ( x \right ) \right )^{m-r} }{\left ( ab \right ) ^{r} } \left (\sigma ^{r}\left ( x \right )+ \tau ^{r} \left ( x \right ) \right ) ^{2} - 4a\left ( -q\left ( x \right ) \right ) ^{m} \\ & = a^{\zeta \left ( r+1 \right ) }b^{\zeta \left ( r \right ) }\left ( -q\left ( x \right ) \right )^{m-r} L_{r}^{2} \left ( x \right ) - 4a\left ( -q\left ( x \right ) \right ) ^{m} , \end{split} \end{equation*}$

where

(v) $\left \lfloor \frac{m-r+1}{2} \right \rfloor +\left \lfloor \frac{m+r+1}{2} \right \rfloor = m+1-\zeta \left (m+1-r \right)$ . □

When $r = 1$ , we have:

Corollary 3.1. The generalized Cassini's identity of the generalized bi-periodic Lucas polynomial $\left \{ L_{n} \left (x \right) \right \}$ is

$\begin{equation*} a^{\zeta \left ( m \right ) } b^{1-\zeta \left ( m \right ) }L_{m-1}\left ( x \right ) L_{m+1}\left ( x \right ) -a^{1-\zeta \left ( m \right ) }b^{\zeta \left ( m \right ) }L_{n}^{2}\left ( x \right ) = a^{2}b\left ( -q\left ( x \right ) \right )^{m-1}p^{2} \left ( x \right )-4a\left ( -q\left ( x \right ) \right ) ^{m} . \end{equation*}$

Theorem 3.5. Let $\left \{ F_{n} \left (x \right) \right \}$ be the generalized bi-periodic Fibonacci and $\left \{ L_{n} \left (x \right) \right \}$ be the Lucas polynomials. We get the relations between $\left \{ F_{n} \left (x \right) \right \}$ and $\left \{ L_{n} \left (x \right) \right \}$ as follows:

$\begin{equation} F_{m+1} \left ( x \right ) +q\left ( x \right )F_{m-1}\left ( x \right ) = L_{m} \left ( x \right ) , \end{equation}$

(3.13)

$\begin{equation} L_{m+1} \left ( x \right ) +q\left ( x \right )L_{m-1}\left ( x \right ) = \left ( p^{2}\left ( x \right ) ab+4q\left ( x \right ) \right ) F_{m} \left ( x \right ), \end{equation}$

(3.14)

$\begin{equation} F_{m+2} \left ( x \right ) -q^{2} \left ( x \right )F_{m-2}\left ( x \right ) = a^{\zeta \left ( m+1 \right ) } b^{\zeta \left ( m \right ) } p\left ( x \right ) L_{m} \left ( x \right ) , \end{equation}$

(3.15)

$\begin{equation} L_{m+2} \left ( x \right )-q^{2} \left ( x \right )L_{m-2}\left ( x \right ) = a^{\zeta \left ( m \right ) } b^{\zeta \left ( m+1 \right ) } \left ( p^{2}\left ( x \right ) ab+4q\left ( x \right ) \right ) p\left ( x \right ) F_{m} \left ( x \right ). \end{equation}$

(3.16)

Proof. We prove only (3.13), and other identities are proved similarly. According to the identities (1.2) and (3.12),

$\begin{equation*} \begin{split} \frac{\left ( ab \right )^{\left \lfloor \frac{m+1}{2} \right \rfloor } }{a^{\zeta \left (m \right ) } }F_{m+1}\left ( x \right )+abq\left ( x \right )\cdot \frac{\left ( ab \right )^{\left \lfloor \frac{m-1}{2} \right \rfloor } }{a^{\zeta \left ( m \right ) } }F_{m-1}\left ( x \right ) & = \frac{\sigma ^{m+1} \left ( x \right )-\tau ^{m+1}\left ( x \right ) }{\sigma \left ( x \right )-\tau \left ( x \right ) }+abq\left ( x \right )\cdot \frac{\sigma ^{m-1} \left ( x \right )-\tau ^{m-1}\left ( x \right ) }{\sigma \left ( x \right )-\tau \left ( x \right ) } \\ & = \frac{\sigma ^{m}\left ( x \right )\left ( \sigma \left ( x \right ) +\frac{abq\left ( x \right ) }{\sigma \left ( x \right ) } \right ) - \tau ^{m}\left ( x \right )\left ( \tau \left ( x \right ) +\frac{abq\left ( x \right ) }{\tau \left ( x \right ) } \right ) }{\sigma \left ( x \right )-\tau \left ( x \right )} \\ & = \sigma ^{m} \left ( x \right ) +\tau ^{m} \left ( x \right ) = \frac{\left ( ab \right )^{\left \lfloor \frac{m+1}{2} \right \rfloor } }{a^{\zeta \left ( n \right ) } }L_{m} \left ( x \right ). \end{split} \end{equation*}$

□

Theorem 3.6. Let $\left \{ F_{n} \left (x \right) \right \}$ be the generalized bi-periodic Fibonacci and $\left \{ L_{n} \left (x \right) \right \}$ be the Lucas polynomials. We have the following identity:

$\begin{equation} \left ( \frac{b}{a} \right ) ^{\zeta \left ( mn+n \right ) }F_{m} \left ( x \right )L_{n}\left ( x \right )+\left ( \frac{b}{a} \right ) ^{\zeta \left ( mn+m \right ) }F_{n} \left ( x \right )L_{m}\left ( x \right ) = 2F_{m+n} \left ( x \right ) , \end{equation}$

(3.17)

$\begin{equation} \left ( \frac{b}{a} \right ) ^{\zeta \left ( mn \right ) }L_{m} \left ( x \right )L_{n}\left ( x \right )+\left ( \frac{a}{b} \right ) ^{\zeta \left ( mn+m+n \right ) } \left ( \frac{a^{2}b^{2}p^{2} \left ( x \right ) +4abq\left ( x \right ) }{a^{2}} \right ) F_{m} \left ( x \right )F_{n}\left ( x \right ) = 2L_{m+n} \left ( x \right ) . \end{equation}$

(3.18)

Proof. According to the identities (1.2) and (3.12),

$\begin{equation*} \begin{aligned} \frac{\left ( ab \right )^{\left \lfloor \frac{m}{2} \right \rfloor +\left \lfloor \frac{n+1}{2} \right \rfloor } }{a^{1-\zeta \left ( m \right )+\zeta \left ( n \right ) } }F_{m}\left ( x \right ) L_{n}\left ( x \right ) +\frac{\left ( ab \right )^{\left \lfloor \frac{m+1}{2} \right \rfloor +\left \lfloor \frac{n}{2} \right \rfloor } }{a^{1-\zeta\left ( n \right )+\zeta \left ( m \right ) } }F_{n}\left ( x \right ) L_{m}\left ( x \right ) = \frac{2\left ( \sigma ^{m+n} \left ( x \right )-\tau ^{m+n}\left ( x \right ) \right ) }{ \sigma \left ( x \right )-\tau \left ( x \right ) } = \frac{2\left ( ab \right )^{\left \lfloor \frac{m+n}{2} \right \rfloor } }{a^{1-\zeta\left ( m+n \right ) } } F_{m+n}\left ( x \right ) .\end{aligned} \end{equation*}$

Similary, we get

$\begin{equation*} \begin{split} &\frac{\left ( ab \right )^{\left \lfloor \frac{m+1}{2} \right \rfloor +\left \lfloor \frac{n+1}{2} \right \rfloor } }{a^{\zeta \left ( m \right )+\zeta \left ( n \right ) } }L_{m}\left ( x \right ) L_{n}\left ( x \right ) +\frac{\left ( ab \right )^{\left \lfloor \frac{m}{2} \right \rfloor +\left \lfloor \frac{n}{2} \right \rfloor }\left ( \sigma \left ( x \right ) -\tau \left ( x \right ) \right ) ^{2} }{a^{2-\zeta\left ( n \right )-\zeta \left ( m \right ) } }F_{m}\left ( x \right ) F_{n}\left ( x \right ) \\ & = \left ( \sigma ^{m}\left ( x \right ) +\tau ^{m}\left ( x \right )\right ) \left ( \sigma ^{n}\left ( x \right ) +\tau ^{n}\left ( x \right )\right )+\left ( \sigma ^{m}\left ( x \right ) -\tau ^{m}\left ( x \right )\right ) \left ( \sigma ^{n}\left ( x \right ) -\tau ^{n}\left ( x \right )\right )\\ & = 2\left ( \sigma ^{n+m}\left ( x \right )+\tau ^{n+m} \left ( x \right ) \right ) = \frac{2\left ( ab \right )^{\left \lfloor \frac{m+n+1}{2} \right \rfloor } }{a^{\zeta \left ( m+n \right ) } } L_{m+n}\left ( x \right ), \end{split} \end{equation*}$

where

(w) $\left \lfloor \frac{m}{2} \right \rfloor+\left \lfloor \frac{n+1}{2} \right \rfloor -\left \lfloor \frac{m+n}{2} \right \rfloor = \zeta \left (mn+n \right)$ ,

(x) $\left \lfloor \frac{n}{2} \right \rfloor+\left \lfloor \frac{m+1}{2} \right \rfloor -\left \lfloor \frac{m+n}{2} \right \rfloor = \zeta \left (mn+m \right)$ ,

(y) $\left \lfloor \frac{n+1}{2} \right \rfloor+\left \lfloor \frac{m+1}{2} \right \rfloor -\left \lfloor \frac{m+n+1}{2} \right \rfloor = \zeta \left (mn \right)$ ,

(z) $\left \lfloor \frac{n}{2} \right \rfloor+\left \lfloor \frac{m}{2} \right \rfloor -\left \lfloor \frac{m+n+1}{2} \right \rfloor = -\zeta\left (mn+m+n \right)$ . □

Theorem 3.7. Let $\left \{ F_{n} \left (x \right) \right \}$ be the generalized bi-periodic Fibonacci and $\left \{ L_{n} \left (x \right) \right \}$ be the Lucas polynomials, then we obtain the following identities

$\begin{equation} \sum\limits_{k = 0}^{m} \binom{m}{k} a^{\zeta \left ( k \right ) } \left ( ab \right )^{\left \lfloor \frac{k}{2} \right \rfloor }p^{k}\left ( x \right )q^{m-k}\left ( x \right )F_{k}\left ( x \right ) = F_{2m}\left ( x \right ) \end{equation}$

(3.19)

and

$\begin{equation} \sum\limits_{k = 0}^{m} \binom{m}{k} a^{\zeta\left ( k+1 \right ) } \left ( ab \right )^{\left \lfloor \frac{k+1}{2} \right \rfloor }p^{k}\left ( x \right )q^{m-k}\left ( x \right )L_{k}\left ( x \right ) = aL_{2m}\left ( x \right ) . \end{equation}$

(3.20)

Proof. We prove only (3.19), and (3.20) is proved similarly. According to the identity (1.2),

$\begin{equation*} \begin{split} &\sum\limits_{k = 0}^{m} \binom{m}{k} a^{\zeta \left ( k \right ) } \left ( ab \right )^{\left \lfloor \frac{k}{2} \right \rfloor }p^{k}\left ( x \right )q^{m-k}\left ( x \right )F_{k}\left ( x \right )\\ & = \sum\limits_{k = 0}^{m} \binom{m}{k} a^{\zeta \left ( k \right ) } \left ( ab \right )^{\left \lfloor \frac{k}{2} \right \rfloor }p^{k}\left ( x \right )q^{m-k}\left ( x \right )\cdot \frac{a^{1-\zeta \left ( k \right ) } }{\left ( ab \right ) ^{\left \lfloor \frac{k}{2} \right \rfloor } }\cdot \frac{ \sigma ^{k} \left ( x \right ) -\tau ^{k}\left ( x \right ) }{ \sigma \left ( x \right )-\tau \left ( x \right ) } \\ & = \sum\limits_{k = 0}^{m} \binom{m}{k} a p^{k}\left ( x \right )q^{m-k}\left ( x \right )\cdot \frac{ \sigma^{k} \left ( x \right ) -\tau ^{k}\left ( x \right ) }{ \sigma \left ( x \right )-\tau \left ( x \right ) } \\ & = \frac{a}{ \sigma \left ( x \right ) -\tau \left ( x \right ) } \left ( \sum\limits_{k = 0}^{m} \binom{m}{k} p^{k} \left ( x \right ) \sigma ^{k}\left ( x \right ) q^{m-k}\left ( x \right ) - \sum\limits_{k = 0}^{m} \binom{m}{k} p^{k}\left ( x \right )\tau ^{k} \left ( x \right ) q^{m-k}\left ( x \right ) \right ) \\ & = \frac{a}{ \sigma \left ( x \right ) -\tau \left ( x \right ) } \left \{ \left ( \sigma \left ( x \right )p\left ( x \right )+q\left ( x \right ) \right )^{m} -\left (\tau \left ( x \right ) p\left ( x \right ) +q\left ( x \right ) \right )^{m} \right \} \\ & = \frac{a}{ \sigma\left ( x \right ) -\tau \left ( x \right ) }\left ( \left ( \frac{ \sigma ^{2} \left ( x \right ) }{ab} \right )^{m}-\left ( \frac{\tau ^{2} \left ( x \right ) }{ab} \right ) ^{m} \right ) \\ & = \frac{a}{\left ( ab \right )^{m} } \left ( \frac{ \sigma ^{2m}\left ( x \right )-\tau ^{2m}\left ( x \right ) }{ \sigma \left ( x \right )-\tau \left ( x \right ) } \right ) = F_{2m}\left ( x \right ) . \end{split} \end{equation*}$

□

Theorem 3.8. The sum of binomial coefficients of generalized bi-periodic Fibonacci $\left \{ F_{m} \left (x \right) \right \}$ and Lucas $\left \{ L_{m} \left (x \right) \right \}$ polynomials are

$\begin{equation} F_{m}\left ( x \right ) = \frac{2a^{1-\zeta \left ( m \right ) } }{2^{m} \left ( ab \right )^{\left \lfloor \frac{m}{2} \right \rfloor } }\cdot\sum\limits_{k = 0}^{\left \lfloor \frac{m-1}{2} \right \rfloor }\binom{m}{2k+1}\left (ab p\left ( x \right ) \right )^{m-2k-1} \left (a^{2}b^{2}p^{2}\left ( x \right )+4abq\left ( x \right ) \right )^{k} , \end{equation}$

(3.21)

$\begin{equation} L_{m}\left ( x \right ) = \frac{2a^{\zeta \left ( m \right ) } }{2^{m} \left ( ab \right )^{\left \lfloor \frac{m+1}{2} \right \rfloor } }\cdot\sum\limits_{k = 0}^{\left \lfloor \frac{m}{2} \right \rfloor }\binom{m}{2k}\left ( abp\left ( x \right ) \right )^{m-2k} \left (a^{2}b^{2}p^{2}\left ( x \right )+4abq\left ( x \right ) \right )^{k} . \end{equation}$

(3.22)

Proof. By

$\sigma \left ( x \right ) = \frac{abp\left ( x \right )+\sqrt{a^{2}b^{2}p^{2}\left ( x \right )+4abq\left ( x \right ) } }{2}$

and

$\tau \left ( x \right ) = \frac{abp\left ( x \right )-\sqrt{a^{2}b^{2}p^{2}\left ( x \right )+4abq\left ( x \right ) } }{2},$

we have

$\begin{equation*} \begin{split} & \sigma ^{m} \left ( x \right )-\tau ^{m} \left ( x \right )\\ & = 2^{-m} \left ( \left ( abp\left ( x \right )+\sqrt{a^{2}b^{2}p^{2}\left ( x \right )+4abq\left ( x \right ) } \right )^{m} - \left (ab p\left ( x \right )-\sqrt{a^{2}b^{2}p^{2}\left ( x \right )+4abq\left ( x \right ) } \right )^{m} \right ) \\ & = 2^{-m} \left ( \sum\limits_{k = 0}^{m} \binom{m}{k} p^{m-k} \left ( x \right ) a^{m-k}b^{m-k}\left (\sqrt{a^{2}b^{2}p^{2}\left ( x \right )+4abq\left ( x \right ) } \right )^{k} \right . \\ &\quad \left. -\sum\limits_{k = 0}^{m} \binom{m}{k} p^{m-k} \left ( x \right ) a^{m-k}b^{m-k}\left (-\sqrt{a^{2}b^{2}p^{2}\left ( x \right )+4abq\left ( x \right ) } \right )^{k} \right )\\ & = 2^{-m+1} \sum\limits_{k = 0}^{\left \lfloor \frac{m-1}{2} \right \rfloor }\binom{m}{2k+1}p^{m-2k-1}\left ( x \right ) a^{m-2k-1}b^{m-2k-1}\left (\sqrt{a^{2}b^{2}p^{2}\left ( x \right )+4abq\left ( x \right ) } \right )^{2k+1} . \end{split} \end{equation*}$

According to the identity (1.2),

$\begin{equation*} \begin{split} F_{m} \left ( x \right )& = \frac{a^{1-\zeta \left ( m \right ) } }{\left ( ab \right )^{\left \lfloor \frac{m}{2} \right \rfloor } } \cdot \frac{\sigma ^{m}\left ( x \right )-\tau ^{m} \left ( x \right ) }{\sigma \left ( x \right )-\tau \left ( x \right ) } \\ & = \frac{2a^{1-\zeta \left ( m \right ) } }{2^{m} \left ( ab \right )^{\left \lfloor \frac{m}{2} \right \rfloor } }\cdot\sum\limits_{k = 0}^{\left \lfloor \frac{m-1}{2} \right \rfloor }\binom{m}{2k+1}\left [ab p\left ( x \right ) \right ]^{m-2k-1} \left (a^{2}b^{2}p^{2}\left ( x \right )+4abq\left ( x \right ) \right )^{k} . \end{split} \end{equation*}$

Similarly, we show that

$\begin{equation*} \sigma ^{m} \left ( x \right )+\tau ^{m} \left ( x \right ) = 2^{-m+1} \sum\limits_{k = 0}^{\left \lfloor \frac{m}{2} \right \rfloor }\binom{m}{2k}p^{m-2k}\left ( x \right ) a^{m-2k}b^{m-2k}\left (\sqrt{p^{2}\left ( x \right )a^{2}b^{2}+4q\left ( x \right ) ab } \right )^{2k} . \end{equation*}$

According to the identity (3.12),

$\begin{equation*} L_{m} \left ( x \right ) = \frac{2a^{\zeta \left (m \right ) } }{2^{m} \left ( ab \right )^{\left \lfloor \frac{m+1}{2} \right \rfloor } }\cdot\sum\limits_{k = 0}^{\left \lfloor \frac{m}{2} \right \rfloor }\binom{m}{2k}\left ( p\left ( x \right )ab \right )^{m-2k} \left (p^{2}\left ( x \right )a^{2}b^{2}+4q\left ( x \right ) ab \right )^{k} . \end{equation*}$

□

Theorem 3.9. Let $\left \{ F_{n} \left (x \right) \right \}$ be the generalized bi-periodic Fibonacci and $\left \{ L_{n} \left (x \right) \right \}$ be the Lucas polynomials. We have the following identities

$\begin{equation} F_{2m}\left ( x \right )F_{2n} \left ( x\right ) = \left ( \frac{a}{b} \right )^{\zeta \left ( m+n \right ) } \left ( F_{m+n}^{2}\left ( x \right )-q^{2n}\left ( x \right )F_{m-n}^{2} \left ( x \right ) \right ) , \end{equation}$

(3.23)

$\begin{equation} F_{2m}\left ( x \right )F_{2n} \left ( x\right ) = \left ( \frac{a}{b} \right ) ^{\zeta \left ( m+n \right ) } F_{m+n}^{2}\left ( x \right ) -\frac{a^{2}q^{2n}\left ( x \right ) }{\left ( \sigma \left ( x \right )-\tau \left ( x \right ) \right ) ^{2} }\cdot \left ( \frac{b}{a} \right ) ^{\zeta\left ( m+n \right ) } L_{m-n}^{2} \left ( x \right ) +\frac{4a^{2}\left ( -q\left ( x \right ) \right )^{m+n} }{\left ( \sigma \left ( x \right )-\tau \left ( x \right ) \right )^{2} } , \end{equation}$

(3.24)

$\begin{equation} \begin{aligned} F_{2m}\left ( x \right )F_{2n} \left ( x\right ) = -q^{2n} \left ( x \right ) \left ( \frac{a}{b} \right ) ^{\zeta\left ( m+n \right ) } F_{m-n}^{2}\left ( x \right )+\frac{a^{2} }{\left ( \sigma \left ( x \right )-\tau \left ( x \right ) \right ) ^{2} }\cdot \left ( \frac{b}{a} \right )^{\zeta \left ( m+n \right ) } L_{m+n}^{2}\left ( x \right )-\frac{4a^{2}\left ( -q\left ( x \right ) \right ) ^{m+n} }{\left ( \sigma \left ( x \right ) -\tau \left ( x \right ) \right )^{2} },\end{aligned} \end{equation}$

(3.25)

$\begin{equation} L_{2m}\left ( x \right )L_{2n} \left ( x\right ) = \left ( \frac{b}{a} \right )^{\zeta \left ( m+n \right ) } \left ( L_{m+n}^{2}\left ( x \right )-q^{2n}\left ( x \right )L_{m-n}^{2} \left ( x \right ) \right )-4\left ( -q\left ( x \right ) \right ) ^{m+n} , \end{equation}$

(3.26)

$\begin{equation} L_{2m}\left ( x \right )L_{2n} \left ( x\right ) = \frac{\left ( \sigma \left ( x \right )-\tau \left ( x \right ) \right )^{2} }{a^{2} } \left ( \frac{a}{b} \right ) ^{\zeta \left ( m+n \right ) } F_{m+n}^{2}\left ( x \right )-q^{2n}\left ( \frac{b}{a} \right ) ^{\zeta \left ( m+n \right ) } L_{m-n}^{2} \left ( x \right ), \end{equation}$

(3.27)

$\begin{equation} L_{2m}\left ( x \right )L_{2n} \left ( x\right ) = \frac{\left (\sigma \left ( x \right )-\tau \left ( x \right ) \right )^{2} q^{2n} \left ( x \right ) }{a^{2} } \cdot \left ( \frac{a}{b} \right ) ^{\zeta \left ( m+n \right ) } F_{m-n}^{2}\left ( x \right )+ \left ( \frac{b}{a} \right )^{\zeta \left ( m+n \right ) } L_{m+n}^{2}\left ( x \right ). \end{equation}$

(3.28)

Proof. We prove only (3.23), and other identities are proved similarly. According to the identity (1.2), we have

$\begin{equation*} \begin{split} &\left ( \frac{a}{b} \right )^{\zeta \left ( m+n \right ) } \left ( F_{m+n}^{2}\left ( x \right )-q^{2n}\left ( x \right )F_{m-n}^{2} \left ( x \right ) \right )\\ & = \left ( \frac{a}{b} \right )^{\zeta \left ( m+n \right ) } \left \{\left ( \left ( \frac{a^{1-\zeta \left ( m+n \right ) } }{\left ( ab \right ) ^{\left \lfloor \frac{m+n}{2} \right \rfloor } } \right ) \frac{\sigma ^{m+n}\left ( x \right ) -\tau ^{m+n}\left ( x \right ) }{\sigma\left ( x \right ) -\tau \left ( x \right ) } \right ) ^{2} \right. \left. -q^{2 n}\left ( x \right )\left ( \left ( \frac{ a^{1-\zeta \left ( m+n \right ) } }{\left ( ab \right ) ^{\left \lfloor \frac{m-n}{2} \right \rfloor } } \right ) \frac{\sigma ^{m-n}\left ( x \right ) -\tau ^{m-n}\left ( x \right ) }{\sigma \left ( x \right )-\tau \left ( x \right ) } \right ) \right \} \\ & = \frac{a^{2} }{\left ( ab \right ) ^{m+n} }\left ( \frac{\sigma ^{2\left ( m+n \right ) }\left ( x \right )+\tau ^{2\left ( m+n \right ) }\left ( x \right ) }{\left ( \sigma \left ( x \right ) -\tau \left ( x \right ) \right )^{2} } \right )-\frac{2a^{2}\left ( -q\left ( x \right ) \right )^{m+n} }{\left ( \sigma\left ( x \right )-\tau \left ( x \right ) \right )^{2} } \\ &\quad -\frac{a^{2} }{\left ( ab \right ) ^{m+n} }\left ( \frac{\sigma ^{2m }\left ( x \right )\tau ^{2n}\left ( x \right ) +\sigma ^{2n} \left ( x \right ) \tau ^{2m }\left ( x \right ) }{\left ( \sigma \left ( x \right ) -\tau \left ( x \right ) \right )^{2} } \right )+\frac{2a^{2}\left ( -q\left ( x \right ) \right )^{m+n} }{\left ( \sigma \left ( x \right )-\tau \left ( x \right ) \right )^{2} } \\ & = \frac{a^{2} }{\left ( ab \right )^{m+n} } \cdot \frac{\sigma ^{2m}\left ( x \right )-\tau ^{2m}\left ( x \right ) }{\sigma \left ( x \right )-\tau \left ( x \right ) } \cdot \frac{\sigma^{2n}\left ( x \right )-\tau ^{2n}\left ( x \right ) }{\sigma \left ( x \right )-\tau \left ( x \right ) }\\ & = F_{2m}\left ( x \right )F_{2n}\left ( x \right ) . \end{split} \end{equation*}$

□

Theorem 3.10. Let $\mathcal{F} _{m}\left (x \right)$ and $\mathcal{L} _{m}\left (x \right)$ denote the $m\times m$ tridiagonal matrix defined by

$\begin{equation} \mathcal{F} _{m} \left ( x \right ) = \begin{bmatrix} ap\left ( x\right ) & q\left ( x \right ) & & & \\ -1 & bp\left ( x \right ) & q\left ( x \right ) & & \\ & -1 & ap\left ( x \right ) & \ddots & \\ & & \ddots & \ddots & q\left ( x \right ) \\ & & & -1 &a^{\zeta\left ( m \right ) }b^{1-\zeta \left ( m \right ) }p\left ( x \right ) \end{bmatrix} , \quad m\ge 1 \end{equation}$

(3.29)

and

$\begin{equation} \mathcal{L} _{m} \left ( x \right ) = \begin{bmatrix} ap\left ( x\right ) & q\left ( x \right ) & & & \\ -2 & bp\left ( x \right ) & q\left ( x \right ) & & \\ & -1 & ap\left ( x \right ) & \ddots & \\ & & \ddots & \ddots & q\left ( x \right ) \\ & & & -1 &a^{\zeta \left ( m \right ) }b^{1-\zeta\left (m \right ) }p\left ( x \right ) \end{bmatrix} , \quad m\ge 1, \end{equation}$

(3.30)

with

$\mathcal{F} _{0}\left ( x \right ) = \begin{bmatrix} 0 \end{bmatrix}$

and

$\mathcal{L} _{0}\left ( x \right ) = \begin{bmatrix} 2 \end{bmatrix} .$

Therefore,

$det \mathcal{F} _{m} \left ( x \right ) = F_{m+1} \left ( x \right )$

and

$det \mathcal{L} _{m} \left ( x \right ) = L_{m} \left ( x \right ).$

Proof. We prove (3.29) and (3.30) by mathematical induction. Obviously, the identity is true when $m = 1$ and $m = 2$ :

$det \mathcal{F} _{1} \left ( x \right ) = ap\left ( x \right ) = F_{2} \left ( x \right ),\quad det\mathcal{F} _{2} \left ( x \right ) = abp^{2}\left ( x \right ) +q\left ( x \right ) = F_{3} \left ( x \right )$

and

$det \mathcal{L} _{1} \left ( x \right ) = ap\left ( x \right ) = L_{1} \left ( x \right ),\quad det \mathcal{L} _{2} \left ( x \right ) = abp^{2}\left ( x \right ) +2q\left ( x \right ) = L_{2} \left ( x \right ).$

We assume that the identity is true when $m-1$ :

$det\mathcal{F} _{m-1} \left ( x \right ) = F_{m} \left ( x \right ),\quad det\mathcal{F} _{m-2} \left ( x \right ) = F_{m-1} \left ( x \right )$

and

$det\mathcal{L} _{m-1} \left ( x \right ) = L_{m-1} \left ( x \right ),\quad det \mathcal{L} _{m-2} \left ( x \right ) = L_{m-2} \left ( x \right ).$

Next, we prove that the identity is true with $m$ .

According to the identities (2.3) and (3.1) and mathematical induction, we have

$\begin{equation*} \begin{split} det\mathcal{F} _{m} \left ( x \right )& = a^{\zeta \left ( m \right ) }b^{1-\zeta \left ( m \right ) }p\left ( x \right ) det \mathcal{F} _{m-1} \left ( x \right )+q\left ( x \right ) det \mathcal{F} _{m-2} \left ( x \right ) \\ & = a^{\zeta \left ( m \right ) }b^{1-\zeta\left ( n \right ) }p\left ( x \right ) {F} _{m} \left ( x \right )+q\left ( x \right ) {F} _{m-1} \left ( x \right ) \\ & = {F} _{m+1} \left ( x \right ) \end{split} \end{equation*}$

and

$\begin{equation*} \begin{split} det \mathcal{L} _{m} \left ( x \right )& = a^{\zeta\left ( m \right ) }b^{1-\zeta \left ( m \right ) }p\left ( x \right ) det \mathcal{L} _{m-1} \left ( x \right )+q\left ( x \right ) det\mathcal{L} _{m-2} \left ( x \right ) \\ & = a^{\zeta\left (m \right ) }b^{1-\zeta \left ( m \right ) }p\left ( x \right ) {L} _{m-1} \left ( x \right )+q\left ( x \right ) {L} _{m-2} \left ( x \right ) \\ & = {L} _{m} \left ( x \right ). \end{split} \end{equation*}$

This completes the proof of Theorem 3.10. □

4. Conclusions

In this paper, we extend the generalized bi-periodic Fibonacci polynomial ${F} _{n} \left (x \right)$ defined in ^[18] and we consider ${F} _{n} \left (x \right)$ using of matrix methods. In addition, we define the generalized bi-periodic Lucas polynomial ${L} _{n} \left (x \right)$ and obtain some identities related to ${L} _{n} \left (x \right)$ . Finally, we obtain a series of identities connecting ${F} _{n} \left (x \right)$ and ${L} _{n} \left (x \right)$ . An interesting idea is that perhaps we can obtain a series of identities related to generalized bi-periodic Lucas polynomials using matrix methods.

Use of AI tools declaration

The authors declare they have not use Artificial Intelligence (AI) tools in the creation of this paper.

Acknowledgments

The authors would like to thank the editors and reviewers for their helpful suggestions. All the authors have contributed equally to this work and have read and approved this final manuscript. This work is supported by the National Natural Science Foundation of China (11701448).

Conflict of interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

References

[1]	W. M. Abd-Elhameed, A. N. Philippou, N. A. Zeyada, Novel results for two generalized classes of Fibonacci and Lucas polynomials and their uses in the reduction of some radicals, Mathematics, 10 (2022), 2342. https://doi.org/10.3390/math10132342 doi: 10.3390/math10132342
[2]	W. M. Abd-Elhameed, N. A. Zeyada, New identities involving generalized Fibonacci and generalized Lucas numbers, Indian J. Pure Appl. Math., 49 (2018), 527–537. https://doi.org/10.1007/s13226-018-0282-7 doi: 10.1007/s13226-018-0282-7
[3]	W. M. Abd-Elhameed, A. Napoli, Some novel formulas of Lucas polynomials via different approaches, Symmetry, 15 (2023), 185. https://doi.org/10.3390/sym15010185 doi: 10.3390/sym15010185
[4]	W. M. Abd-Elhameed, A. Napoli, New formulas of convolved Pell polynomials, AIMS Math., 9 (2024), 565–593. https://doi.org/10.3934/math.2024030 doi: 10.3934/math.2024030
[5]	W. M. Abd-Elhameed, Y. H. Youssri, N. El-Sissi, M. Sadek, New hypergeometric connection formulae between Fibonacci and Chebyshev polynomials, Ramanujan J., 42 (2017), 347–361. https://doi.org/10.1007/s11139-015-9712-x doi: 10.1007/s11139-015-9712-x
[6]	Y. Yi, W. Zhang, Some identities involving the Fibonacci polynomials, Fibonacci Quart, 40 (2002), 314–318.
[7]	V. E. Hoggatt, M. Bicknell, Roots of Fibonacci polynomials, Fibonacci Quart, 11 (1973), 271–274.
[8]	Z. Wu, W. Zhang, The sums of the reciprocals of Fibonacci polynomials and Lucas polynomials, J. Inequal. Appl., 2012 (2012), 134. https://doi.org/10.1186/1029-242X-2012-134 doi: 10.1186/1029-242X-2012-134
[9]	U. K. Dutta, P. K. Ray, On the finite reciprocal sums of Fibonacci and Lucas polynomials, AIMS Math., 4 (2019), 1569–1581. https://doi.org/10.3934/math.2019.6.1569 doi: 10.3934/math.2019.6.1569
[10]	T. Koshy, Fibonacci and Lucas numbers with applications, John Wiley & Sons, Inc., 2001. https://doi.org/10.1002/9781118033067
[11]	R. Flórez, N. McAnally, A. Mukherjee, Identities for the generalized Fibonacci polynomial, arXiv, 2017. https://doi.org/10.48550/arXiv.1702.01855
[12]	A. Nalli, P. Haukkanen, On generalized Fibonacci and Lucas polynomials, Chaos Solitons Fract., 42 (2009), 3179–3186. https://doi.org/10.1016/j.chaos.2009.04.048 doi: 10.1016/j.chaos.2009.04.048
[13]	R. Flórez, R. A. Higuita, A. Mukherjee, Characterization of the strong divisibility property for generalized Fibonacci polynomials, arXiv, 2018. https://doi.org/10.48550/arXiv.1701.06722
[14]	R. Flórez, R. A. Higuita, A. Mukherjee, Alternating sums in the Hosoya polynomial triangle, J. Integer Seq., 17 (2014), 14.9.5.
[15]	N. Yilmaz, A. Coskun, N. Taskara, On properties of bi-periodic Fibonacci and Lucas polynomials, AIP Conf. Proc., 1863 (2017), 310002. https://doi.org/10.1063/1.4992478 doi: 10.1063/1.4992478
[16]	T. Du, Z. Wu, Some identities involving the bi-periodic Fibonacci and Lucas polynomials, AIMS Math., 8 (2023), 5838–5846. https://doi.org/10.3934/math.2023294 doi: 10.3934/math.2023294
[17]	B. Guo, E. Polatli, F. Qi, Determinantal formulas and recurrent relations for bi-periodic Fibonacci and Lucas polynomials, In: S. K. Paikray, H. Dutta, J. N. Mordeson, New trends in applied analysis and computational mathematics, Springer, Singapore, 1356 (2021), 263–276. https://doi.org/10.1007/978-981-16-1402-6_18
[18]	Y. Taşyurdu, Bi-periodic generalized Fibonacci polynomials, Turk. J. Sci., 7 (2022), 157–167.
[19]	H. W. Gould, A history of the Fibonacci Q-matrix and a higher dimensional problem, Fibonacci Quart, 19 (1981), 250–257.
[20]	J. R. Silvester, Fibonacci properties by matrix methods, Math. Gaz., 63 (1979), 188–191. https://doi.org/10.2307/3617892 doi: 10.2307/3617892
[21]	S. P. Jun, K. H. Choi, Some properties of the generalized Fibonacci sequence $q_{n}$ by matrix methods, Korean J. Math., 24 (2016), 681–691. https://doi.org/10.11568/kjm.2016.24.4.681 doi: 10.11568/kjm.2016.24.4.681
[22]	E. Tan, A. B. Ekin, Some identities on conditional sequences by using matrix method, Miskolc Math. Notes, 18 (2017), 469–477. https://doi.org/10.18514/MMN.2017.1321 doi: 10.18514/MMN.2017.1321

This article has been cited by:

1.	Waleed Mohamed Abd-Elhameed, Omar Mazen Alqubori, Anna Napoli, On Convolved Fibonacci Polynomials, 2024, 13, 2227-7390, 22, 10.3390/math13010022
2.	Waleed Mohamed Abd-Elhameed, Omar Mazen Alqubori, Abdulrahim A. Alluhaybi, Amr Kamel Amin, Novel Expressions for Certain Generalized Leonardo Polynomials and Their Associated Numbers, 2025, 14, 2075-1680, 286, 10.3390/axioms14040286

Reader Comments

Your name:*

Email:*
© 2024 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0)

通讯作者: 陈斌, bchen63@163.com

1.
沈阳化工大学材料科学与工程学院沈阳 110142

AIMS Mathematics

1.8 3.1

Metrics

Article views(1305) PDF downloads(70) Cited by(2)

Preview PDF

Download XML

Export Citation

Article outline

Show full outline

AIMS Mathematics

Some identities of the generalized bi-periodic Fibonacci and Lucas polynomials

Related Papers:

Abstract

1. Introduction

2. The generalized bi-periodic Fibonacci polynomial by matrix methods

3. The generalized bi-periodic Lucas polynomial

4. Conclusions

Use of AI tools declaration

Acknowledgments

Conflict of interest

References

This article has been cited by:

Reader Comments

通讯作者: 陈斌, bchen63@163.com

Metrics

Other Articles By Authors

Catalog

AIMS Mathematics

Some identities of the generalized bi-periodic Fibonacci and Lucas polynomials

Related Papers:

Abstract

1. Introduction

2. The generalized bi-periodic Fibonacci polynomial by matrix methods

3. The generalized bi-periodic Lucas polynomial

4. Conclusions

Use of AI tools declaration

Acknowledgments

Conflict of interest

References

This article has been cited by:

Reader Comments

通讯作者: 陈斌, bchen63@163.com

Metrics

Other Articles By Authors

Related pages

Tools

Export File

Citation

Format

Content

Catalog