Selection of artificial intelligence provider via multi-attribute decision-making technique under the model of complex intuitionistic fuzzy rough sets

Tahir Mahmood; Ahmad Idrees; Majed Albaity; Ubaid ur Rehman; Tahir Mahmood; Ahmad Idrees; Majed Albaity; Ubaid ur Rehman

doi:10.3934/math.20241581

AIMS Mathematics

2024, Volume 9, Issue 11: 33087-33138. doi: 10.3934/math.20241581

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Research article Topical Sections

Selection of artificial intelligence provider via multi-attribute decision-making technique under the model of complex intuitionistic fuzzy rough sets

1.
Department of Mathematics and Statistics, International Islamic University Islamabad, Pakistan
2.
SK-Research-Oxford Business College, Oxford, OX1 2EP, UK
3.
Department of Mathematics, Faculty of Science, King Abdulaziz University, P. O. Box 80348, Jeddah 22254, Saudi Arabia
4.
Department of Mathematics, University of Management and Technology, C-II, Johar Town, Lahore, 54700, Punjab, Pakistan

Received: 24 July 2024 Revised: 29 October 2024 Accepted: 05 November 2024 Published: 21 November 2024
MSC : 03E52, 03E72, 94D05

Choosing an optimal artificial intelligence (AI) provider involves multiple factors, including scalability, cost, performance, and dependability. To ensure that decisions align with organizational objectives, multi-attribute decision-making (MADM) approaches aid in the systematic evaluation and comparison of AI vendors. Therefore, in this article, we propose a MADM technique based on the framework of the complex intuitionistic fuzzy rough model. This approach effectively manages the complex truth grade and complex false grade along with lower and upper approximation. Furthermore, we introduced aggregation operators based on Dombi t-norm and t-conorm, including complex intuitionistic fuzzy rough (CIFR) Dombi weighted averaging (CIFRDWA), CIFR Dombi ordered weighted averaging (CIFRDOWA), CIFR Dombi weighted geometric (CIFRDWG), and CIFR Dombi ordered weighted geometric (CIFRDOWG) operators, which were integrated into our MADM technique. We then demonstrated the application of this technique in a case study on AI provider selection. To highlight its advantages, we compared our proposed method with other approaches, showing its superiority in handling complex decision-making scenarios.

Keywords:

Citation: Tahir Mahmood, Ahmad Idrees, Majed Albaity, Ubaid ur Rehman. Selection of artificial intelligence provider via multi-attribute decision-making technique under the model of complex intuitionistic fuzzy rough sets[J]. AIMS Mathematics, 2024, 9(11): 33087-33138. doi: 10.3934/math.20241581

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Abstract

1. Introduction

Fibonacci polynomials and Lucas polynomials are important in various fields such as number theory, probability theory, numerical analysis, and physics. In addition, many well-known polynomials, such as Pell polynomials, Pell Lucas polynomials, Tribonacci polynomials, etc., are generalizations of Fibonacci polynomials and Lucas polynomials. In this paper, we extend the linear recursive polynomials to nonlinearity, that is, we discuss some basic properties of the bi-periodic Fibonacci and Lucas polynomials.

The bi-periodic Fibonacci $\left \{ f_{n} \left (t\right) \right \}$ and Lucas $\left \{ l_{n} \left (t \right) \right \}$ polynomials are defined recursively by

$\begin{equation*} f_{0}\left (t \right ) = 0, \quad f_{1}\left (t\right ) = 1, \quad {f_{n}\left ( t\right ) = } \begin{cases} ayf_{n-1}\left ( t \right ) +f_{n-2}\left ( t \right ) \quad & n \equiv 0 \pmod{2}, \\ byf_{n-1}\left (t \right ) +f_{n-2}\left (t \right ) \quad & n \equiv 1 \pmod{2}, \end{cases} \quad n\ge 2, \end{equation*}$

and

$\begin{equation*} l_{0}\left ( t \right ) = 2, \quad l_{1}\left (t \right ) = at, \quad {l_{n}\left ( t \right ) = } \begin{cases} byl_{n-1}\left ( t \right ) +l_{n-2}\left ( t \right ) \quad & n \equiv 0 \pmod{2}, \\ ayl_{n-1}\left ( t \right ) +l_{n-2}\left ( t \right ) \quad & n \equiv 1 \pmod{2}, \end{cases} \quad n\ge 2, \end{equation*}$

where $a$ and $b$ are nonzero real numbers. For $t = 1$ , the bi-periodic Fibonacci and Lucas polynomials are, respectively, well-known bi-periodic Fibonacci $\left \{f_{n} \right \}$ and Lucas $\left \{l_{n} \right \}$ sequences. We let

$\begin{equation*} {\varsigma \left ( n \right ) = } \begin{cases} 0 \quad & n \equiv 0 \pmod{2}, \\ 1 \quad & n \equiv 1 \pmod{2}, \end{cases} \quad n\ge 2. \end{equation*}$

In ^[1], the scholars give the Binet formulas of the bi-periodic Fibonacci and Lucas polynomials as follows:

$\begin{equation} f_{n}\left (t \right ) = \frac{a^{\varsigma \left (n+1 \right) }}{ \left( ab \right )^{\left \lfloor \frac{n}{2} \right \rfloor } }\left (\frac{\sigma^{n} \left ( t\right ) -\tau ^{n}\left ( t \right ) }{\sigma\left (t \right ) -\tau \left ( t \right ) }\right), \end{equation}$

(1.1)

and

$\begin{equation} l_{n}\left (t \right ) = \frac{a^{\varsigma \left (n \right) }}{ \left( ab \right )^{\left \lfloor \frac{n+1}{2} \right \rfloor } }\left (\sigma ^{n}\left (t\right ) +\tau ^{n}\left ( t\right ) \right), \end{equation}$

(1.2)

where $n\ge 0$ , $\sigma\left (t \right)$ , and $\tau \left (t \right)$ are zeros of $\lambda ^{2}-abt\lambda -ab$ . This is $\sigma \left (t \right) = \frac{abt +\sqrt{a^{2} b^{2} t^{2}+4ab} }{2}$ and $\tau \left (t \right) = \frac{abt -\sqrt{a^{2} b^{2} t^{2}+4ab} }{2}$ . We note the following algebraic properties of $\sigma\left (t \right)$ and $\tau \left (t \right)$ :

$\begin{equation*} \sigma \left ( t \right ) +\tau \left (t \right ) = abt, \quad \sigma \left (t \right ) -\tau \left ( t \right ) = \sqrt{a^{2} b^{2} t^{2} +4ab}, \quad \sigma\left ( t \right ) \tau \left ( t\right ) = -ab. \end{equation*}$

Many scholars studied the properties of bi-periodic Fibonacci and Lucas polynomials; see ^[2,3,4,5,6]. In addition, many scholars studied the power sums problem of second-order linear recurrences and its divisible properties; see ^[7,8,9,10].

Taking $a = b = 1$ and $t = 1$ , we obtain the Fibonacci $\left \{ F_{n} \right \}$ or Lucas $\left \{ L_{n} \right \}$ sequence. Melham ^[11] proposed the following conjectures:

Conjecture 1. Let $m\ge 1$ be an integer, then the sum

$\begin{equation*} L_{1} L_{3} L_{5}\cdots L_{2m+1}\sum\limits_{k = 1}^{n} F_{2k}^{2m+1} \end{equation*}$

can be represented as $\left (F_{2n+1}-1 \right)^{2}R_{2m-1}\left (F_{2n+1} \right)$ , including $R_{2m-1}\left (t \right)$ as a polynomial with integer coefficients of degree $2m-1$ .

Conjecture 2. Let $m\ge 1$ be an integer, then the sum

$\begin{equation*} L_{1} L_{3} L_{5}\cdots L_{2m+1}\sum\limits_{k = 1}^{n} L_{2k}^{2m+1} \end{equation*}$

can be represented as $\left (L_{2n+1}-1 \right)Q_{2m}\left (L_{2n+1} \right)$ , where $Q_{2m}\left (t \right)$ is a polynomial with integer coefficients of degree $2m$ .

In ^[12], the authors completely solved the Conjecture $2$ and discussed the Conjecture $1$ . Using the definition and properties of bi-periodic Fibonacci and Lucas polynomials, the power sums problem and their divisible properties are studied in this paper. The results are as follows:

Theorem 1. We get the identities

$\begin{align} & \sum\limits_{k = 1}^{n}f_{2k}^{2m+1} \left (t\right ) = \frac{a^{2m+1} }{b\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits _{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m+1}{m-j}\left ( \frac{f_{\left ( 2n+1 \right )\left ( 2j+1\right ) }\left ( t \right ) - f_{2j+1}\left ( t \right ) }{l_{2j+1} \left ( t \right ) } \right ), \end{align}$

(1.3)

$\begin{align} & \sum\limits_{k = 1}^{n}f_{2k+1}^{2m+1} \left ( t \right ) = \frac{\left ( ab \right ) ^{m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits_{j = 0}^{m} \binom{2m+1}{m-j} \left ( \frac{f_{\left ( 2n+2\right )\left (2j+1 \right ) }\left (t\right ) -f_{2\left ( 2j+1 \right ) }\left (t\right ) }{l_{2j+1}\left ( t \right ) } \right ), \end{align}$

(1.4)

$\begin{align} & \sum\limits_{k = 1}^{n}l_{2k}^{2m+1} \left (t \right ) = \sum\limits_{j = 0}^{m} \binom{2m+1}{m-j} \left ( \frac{l_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left (t \right ) -l_{2j+1}\left ( t\right ) }{l_{2j+1}\left ( t\right ) } \right ), \end{align}$

(1.5)

$\begin{align} &\sum\limits_{k = 1}^{n}l_{2k+1}^{2m+1} \left (t\right ) = \frac{ a ^{m+1} }{b^{m+1} } \sum\limits_{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m+1}{m-j} \left ( \frac{l_{\left ( 2n+2\right )\left ( 2j+1\right ) }\left ( t \right ) -l_{2\left ( 2j+1 \right ) }\left ( t \right ) }{l_{2j+1}\left ( t \right ) } \right ), \end{align}$

(1.6)

where $n$ and $m$ are positive integers.

Theorem 2. We get the identities

$\begin{align} & \sum\limits_{k = 1}^{n}f_{2k}^{2m} \left ( t \right ) = \frac{a^{2m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits_{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m}{m-j} \frac{f_{2j\left ( 2n+1 \right ) }\left (t \right ) }{f_{2j}\left (t\right ) } -\frac{a^{2m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} }\binom{2m}{m}\left ( -1 \right )^{m}\left ( n+ \frac{1}{2 } \right ) , \end{align}$

(1.7)

$\begin{align} & \sum\limits_{k = 1}^{n}f_{2k+1}^{2m} \left ( t \right ) = \frac{\left ( ab \right ) ^{m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits_{j = 0}^{m} \binom{2m}{m-j} \left ( \frac{f_{2j\left ( 2n+2\right ) }\left ( t \right ) -f_{4j }\left ( t \right ) }{f_{2j}\left ( t \right ) } \right )-\frac{\left ( ab \right ) ^{m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} }\binom{2m}{m}n , \end{align}$

(1.8)

$\begin{align} & \sum\limits_{k = 1}^{n}l_{2k}^{2m} \left ( t \right ) = \sum\limits_{j = 0}^{m} \binom{2m}{m-j} \frac{f_{2j\left ( 2n+1 \right ) }\left ( t \right ) }{l_{2j+1}\left (t \right ) }-2^{2m-1}-\binom{2m}{m}\left ( n+\frac{1}{2} \right ) , \end{align}$

(1.9)

$\begin{align} &\sum\limits_{k = 1}^{n}l_{2k+1}^{2m} \left ( t \right ) = \frac{ a ^{m} }{b^{m} } \sum\limits_{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m}{m-j} \left ( \frac{f_{2j\left ( 2n+2\right ) }\left ( t \right ) -f_{4j }\left (t \right ) }{f_{2j}\left ( t \right ) } \right ) -\frac{a^{m} }{b^{m} } \binom{2m}{m} \left ( -1 \right )^{m}n , \end{align}$

(1.10)

where $n$ and $m$ are positive integers.

As for application of Theorem 1, we get the following:

Corollary 1. We get the congruences:

$\begin{equation} bl_{1} \left ( t \right ) l_{3} \left ( t \right ) \cdots l_{2m+1}\left ( t \right )\sum\limits _{k = 1}^{n}f_{2k}^{2m+1}\left ( t \right ) \equiv 0\pmod{f_{2n+1}\left ( t \right ) -1}, \end{equation}$

(1.11)

and

$\begin{equation} al_{1} \left ( t\right ) l_{3} \left (t \right ) \cdots l_{2m+1}\left ( t \right )\sum\limits _{k = 1}^{n}l_{2k}^{2m+1}\left ( t \right ) \equiv 0\pmod{l_{2n+1}\left (t \right ) -at}, \end{equation}$

(1.12)

where $n$ and $m$ are positive integers.

Taking $t = 1$ in Corollary 1, we have the following conclusions for bi-periodic Fibonacci $\left \{ f_{n} \right \}$ and Lucas $\left \{ l_{n} \right \}$ sequences.

Corollary 2. We get the congruences:

$\begin{equation} bl_{1} l_{3} \cdots l_{2m+1} \sum\limits _{k = 1}^{n}f_{2k}^{2m+1} \equiv 0\pmod{f_{2n+1} -1}, \end{equation}$

(1.13)

and

$\begin{equation} al_{1} l_{3} \cdots l_{2m+1} \sum\limits _{k = 1}^{n}l_{2k}^{2m+1} \equiv 0\pmod{l_{2n+1} -a }, \end{equation}$

(1.14)

where $n$ and $m$ are nonzero real numbers.

Taking $a = b = 1$ and $t = 1$ in Corollary 1, we have the following conclusions for bi-periodic Fibonacci $\left \{ F_{n} \right \}$ and Lucas $\left \{ L_{n} \right \}$ sequences.

Corollary 3. We get the congruences:

$\begin{equation} L_{1} L_{3} \cdots L_{2m+1} \sum\limits _{k = 1}^{n}F_{2k}^{2m+1} \equiv 0\pmod{F_{2n+1} -1}, \end{equation}$

(1.15)

and

$\begin{equation} L_{1} L_{3} \cdots L_{2m+1} \sum\limits _{k = 1}^{n}L_{2k}^{2m+1} \equiv 0\pmod{L_{2n+1} -1 }, \end{equation}$

(1.16)

where $n$ and $m$ are nonzero real numbers.

2. Proofs of theorems

To begin, we will give several lemmas that are necessary in proving theorems.

Lemma 1. We get the congruence

$\begin{equation*} f_{\left ( 2n+1 \right )\left ( 2j+1\right ) } \left ( t \right ) -f_{2j+1 } \left ( t \right ) \equiv 0 \pmod{f_{2n+1 } \left ( t \right )-1}, \end{equation*}$

where $n$ and $m$ are nonzero real numbers.

Proof. We prove it by complete induction for $j\ge 0$ . This clearly holds when $j = 0$ . If $j = 1$ , we note that $abf_{3\left (2n+1 \right) } \left (t \right) = \left (a^{2} b^{2} t^{2} +4ab \right) f_{2n+1}^{3}\left (t \right)-3abf_{2n+1} \left (t\right)$ and we obtain

$\begin{equation*} \begin{split} & f_{3\left ( 2n+1 \right ) }\left ( t \right )-f_{3 }\left ( t \right ) = \left ( abt^{2}+4 \right ) f_{2n+1}^{3}\left ( t\right )-3f_{2n+1} \left (t \right )-\left ( abt^{2}+4 \right )f_{1}^{3} \left (t \right )+3f_{1} \left (t\right ) \\ & = \left ( abt^{2}+4 \right )\left ( f_{2n+1}\left ( t \right )-f_{1}\left ( t \right )\right )\left ( f_{2n+1}^{2}\left (t \right )+ f_{2n+1}\left ( t \right )f_{1}\left ( t \right )+ f_{1}^{2}\left ( t \right ) \right )-3\left ( f_{2n+1} \left ( t\right ) -f_{1}\left ( t \right ) \right ) \\ & = \left ( abt^{2}+4 \right )\left ( f_{2n+1}\left ( t\right )-1\right )\left ( f_{2n+1}^{2}\left (t \right )+ f_{2n+1}\left ( t \right )f_{1}\left ( t \right )+ f_{1}^{2}\left ( t \right ) \right )-3\left ( f_{2n+1} \left ( t \right ) -1 \right ) \\ & \equiv 0 \pmod{f_{2n+1}\left (t \right ) -1 } . \end{split} \end{equation*}$

This is obviously true when $j = 1$ . Assuming that Lemma 1 holds if $j = 1, 2, \dots, k$ , that is,

$\begin{equation*} f_{\left ( 2n+1 \right )\left ( 2j+1\right ) } \left ( t \right ) -f_{2j+1 } \left ( t \right ) \equiv 0 \pmod{f_{2n+1 } \left ( t \right )-1}. \end{equation*}$

If $j = k+1 \ge 2$ , we have

$\begin{equation*} l_{2\left ( 2n+1 \right ) } \left ( t \right ) f_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left ( t \right ) = f_{\left ( 2n+1 \right )\left ( 2j+3 \right ) }\left ( t \right ) +abf_{\left ( 2n+1 \right )\left ( 2j-1 \right ) }\left ( t \right ), \end{equation*}$

and

$\begin{equation*} \begin{split} abl_{2\left ( 2n+1 \right ) } \left ( t \right ) = \left ( a^{2} b^{2} t^{2} +4ab \right )f_{2n+1}^{2}\left ( t \right )-2ab \equiv \left ( a^{2} b^{2} t^{2} +4ab \right )f_{1}^{2}\left ( t \right )-2ab \pmod{f_{2n+1}\left ( t \right ) -1 }. \end{split} \end{equation*}$

We have

$\begin{equation*} \begin{split} &f_{\left ( 2n+1 \right )\left ( 2k+3 \right ) } \left (t\right ) -f_{ 2k+3 } \left ( t\right )\\ & = l_{2\left ( 2n+1 \right ) }\left ( t \right ) f_{\left ( 2n+1 \right )\left ( 2k+1 \right ) }\left ( t \right )-abf_{\left (2n+1\right )\left ( 2k-1 \right ) } \left ( t \right ) -l_{2} \left ( t \right )f_{2k+1}\left ( t \right )+abf_{2k-1} \left ( t \right ) \\ & \equiv \left ( \left ( abt^{2}+4 \right )f_{ 1}^{2} \left (t \right ) -2 \right )f_{\left ( 2n+1 \right )\left ( 2k+1 \right ) }\left (t \right )-abf_{\left ( 2n+1 \right )\left ( 2k-1 \right ) } \left (t \right ) \\ &\quad -\left ( \left ( abt^{2}+4 \right )f_{ 1}^{2} \left ( t \right ) -2 \right ) f_{2k+1} \left ( t \right )+abf_{2k-1}\left ( t \right ) \\ & \equiv \left ( \left ( abt^{2}+4 \right )f_{ 1}^{2} \left ( t \right ) -2 \right )\left ( f_{\left ( 2n+1 \right ) \left ( 2k+1 \right ) }\left ( t \right )-f_{2k+1}\left ( t\right ) \right ) -ab\left ( f_{\left ( 2n+1 \right )\left ( 2k-1 \right ) } \left (t \right ) -f_{2k-1}\left ( t \right ) \right ) \\ & \equiv 0 \pmod{f_{2n+1}\left (t \right ) -1 } . \end{split} \end{equation*}$

This completely proves Lemma 1. □

Lemma 2. We get the congruence

$\begin{equation*} al_{\left ( 2n+1 \right )\left ( 2j+1\right ) } \left ( t \right ) -al_{2j+1 } \left ( t \right ) \equiv 0 \pmod{l_{2n+1 } \left ( t \right )-at}, \end{equation*}$

where $n$ and $m$ are nonzero real numbers.

Proof. We prove it by complete induction for $j\ge 0$ . This clearly holds when $j = 0$ . If $j = 1$ , we note that $al_{3\left (2n+1 \right) } \left (t \right) = b l_{2n+1}^{3}\left (t \right)+3al_{2n+1} \left (t \right)$ and we obtain

$\begin{equation*} \begin{split} & a l_{3\left ( 2n+1 \right ) }\left (t \right )-al_{3 }\left ( t \right ) = b l_{2n+1}^{3}\left ( t \right )+3al_{2n+1} \left ( t \right )-bl_{1}^{3} \left (t \right )-3al_{1} \left ( t \right ) \\ & = \left ( l_{2n+1}\left ( t \right )-l_{1}\left ( t \right )\right )\left ( bl_{2n+1}^{2}\left ( t \right )+ bl_{2n+1}\left ( t \right )l_{1}\left ( t \right )+ bl_{1}^{2}\left (t \right ) \right )-3a\left ( l_{2n+1} \left ( t \right ) -l_{1}\left ( t\right ) \right ) \\ & = \left ( l_{2n+1}\left ( t \right )-at\right )\left ( bl_{2n+1}^{2}\left ( t \right )+ bayl_{2n+1}\left ( t \right )+ ba^{2}t^{2} \right )-3a\left ( l_{2n+1} \left ( t \right ) -at \right )\\ & \equiv 0 \pmod{l_{2n+1}\left ( t \right ) -at } . \end{split} \end{equation*}$

This is obviously true when $j = 1$ . Assuming that Lemma 2 holds if $j = 1, 2, \dots, k$ , that is,

$\begin{equation*} al_{\left ( 2n+1 \right )\left ( 2j+1\right ) } \left (t \right ) -al_{2j+1 } \left ( t \right ) \equiv 0 \pmod{l_{2n+1 } \left ( t \right )-at}. \end{equation*}$

If $j = k+1 \ge 2$ , we have

$\begin{equation*} l_{2\left ( 2n+1 \right ) } \left (t \right ) l_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left ( t\right ) = l_{\left ( 2n+1 \right )\left ( 2j+3 \right ) }\left ( t\right ) +l_{\left ( 2n+1 \right )\left ( 2j-1 \right ) }\left ( t \right ), \end{equation*}$

and

$\begin{equation*} \begin{split} & al_{2\left ( 2n+1 \right ) } \left ( t \right ) = bl_{2n+1}^{2}\left ( t \right )+2a \equiv bl_{1}^{2}\left ( t \right )+2a \pmod{l_{2n+1}\left ( t \right ) -at }. \end{split} \end{equation*}$

We have

$\begin{equation*} \begin{split} &al_{\left ( 2n+1 \right )\left ( 2k+3 \right ) } \left ( t\right ) -al_{\left ( 2k+3 \right ) } \left ( t \right )\\ & = a\left ( l_{2\left ( 2n+1 \right ) }\left ( t \right ) l_{\left ( 2n+1 \right )\left ( 2k+1 \right ) }\left (t \right )-l_{\left (2n+1\right )\left ( 2k-1 \right ) } \left ( t \right ) \right )-a\left ( l_{2} \left (t \right )l_{2k+1}\left ( t \right )-l_{2k-1} \left (t \right ) \right ) \\ & \equiv \left ( bl_{ 1}^{2} \left ( t \right ) +2a \right )l_{\left ( 2n+1 \right )\left ( 2k+1 \right ) }\left ( t \right )-al_{\left ( 2n+1 \right )\left ( 2k-1 \right ) } \left ( t \right ) -\left ( bl_{ 1}^{2} \left ( t \right ) +2a \right ) l_{2k+1} \left ( t \right )+al_{2k-1}\left ( t \right ) \\ & \equiv \left ( abt^{2} +2 \right )\left ( al_{\left ( 2n+1 \right ) \left ( 2k+1 \right ) }\left ( t\right )-al_{2k+1}\left ( t \right ) \right ) - \left ( al_{\left ( 2n+1 \right )\left ( 2k-1 \right ) } \left ( t\right ) -al_{2k-1}\left (t \right ) \right ) \\ & \equiv 0 \pmod{l_{2n+1}\left ( t \right ) -at } . \end{split} \end{equation*}$

This completely proves Lemma 2. □

Proof of Theorem 1. We only prove (1.3), and the proofs for other identities are similar.

$\begin{equation*} \begin{split} \sum\limits _{k = 1}^{n}f_{2k}^{2m+1}\left ( t \right )& = \sum\limits_{k = 1}^{n}\left (\frac{a^{\varsigma \left ( 2k+1 \right ) } }{\left (ab \right )^{ \left \lfloor \frac{2k}{2} \right \rfloor} }\cdot \left ( \frac{\sigma ^{2k}\left ( t \right ) - \tau ^{2k} \left (t \right ) }{\sigma\left ( t \right ) - \tau \left (t \right ) } \right ) \right ) ^{2m+1} \\ & = \frac{a^{2m+1} }{\left ( \sigma\left (t \right ) -\tau \left ( t \right ) \right )^{2m+1} }\sum\limits _{k = 1}^{n}\frac{\left ( \sigma ^{2k}\left (t \right ) - \tau ^{2k} \left ( t \right ) \right )^{2m+1} }{\left ( ab \right )^{\left ( 2m+1 \right )k } } \\ & = \frac{a^{2m+1} }{\left ( \sigma\left (t\right ) -\tau \left ( t \right ) \right )^{2m+1} }\sum\limits _{k = 1}^{n}\sum\limits _{j = 0}^{2m+1}\left ( -1 \right )^{j} \binom{2m+1}{j} \frac{ \sigma^{2k\left ( 2m+1-j \right ) } \left ( t \right ) \tau ^{2kj} \left ( t \right ) }{\left ( ab \right )^{\left ( 2m+1 \right )k } } \\ & = \frac{a^{2m+1} }{\left ( \sigma\left ( t \right ) -\tau \left (t \right ) \right )^{2m+1} }\sum\limits _{j = 0}^{2m+1}\left ( -1 \right )^{j} \binom{2m+1}{j} \left ( \frac{1-\frac{\sigma ^{2n\left ( 2m+1-2j \right ) }\left (t \right ) }{\left ( ab \right )^{\left ( 2m+1-2j \right )n } } }{\frac{\left ( ab \right )^{2m+1-2j} }{\sigma ^{2\left ( 2m+1-2j \right ) }\left (t \right ) } -1} \right )\\ & = \frac{a^{2m+1} }{\left (\sigma\left (t \right ) -\tau \left (t\right ) \right )^{2m+1} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m+1}{j} \left ( \frac{1-\frac{\sigma ^{2n\left ( 2m+1-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2m+1-2j \right )n } } }{\frac{\left ( ab \right )^{2m+1-2j} }{\sigma ^{2\left ( 2m+1-2j \right ) }\left (t \right ) } -1} - \frac{1-\frac{\sigma ^{2n\left ( 2j-1-2m \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2j-1-2m\right )n } } }{\frac{\left ( ab \right )^{2j-1-2m} }{\sigma ^{2\left ( 2j-1-2m \right ) }\left ( t \right ) } -1}\right )\\ & = \frac{a^{2m+1} }{\left ( \sigma\left ( t \right ) -\tau \left ( t \right ) \right )^{2m+1} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m+1}{j} \left ( \frac{\frac{\sigma^{2 \left ( 2m+1-2j \right ) }\left ( t\right ) }{\left ( ab \right )^{ 2m+1-2j } }- \frac{\sigma ^{\left ( 2n+2 \right ) \left ( 2m+1-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( n+1 \right ) \left ( 2m+1-2j \right ) } } +1- \frac{\sigma ^{2n \left ( 2j-1-2m \right ) }\left (t \right ) }{\left ( ab \right )^{ \left ( 2j-1-2m \right ) n }} }{1- \frac{\sigma^{2 \left ( 2m+1-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{ \left ( 2m+1-2j \right ) }}} \right )\\ & = \frac{a^{2m+1} }{\left ( \sigma\left ( t \right ) -\tau \left (t \right ) \right )^{2m+1} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m+1}{j}\\ & \quad \times \left ( \frac{\sigma^{2m+1-2j}\left ( t \right ) - \tau ^{2m+1-2j}\left ( t \right ) -\frac{\sigma^{\left ( 2n+1 \right ) \left ( 2m+1-2j \right ) }\left ( t \right )}{\left ( ab \right )^{\left ( 2m+1-2j \right )n } } +\frac{\tau ^{\left ( 2n+1 \right ) \left ( 2m+1-2j \right ) }\left (t \right )}{\left ( ab \right )^{ \left ( 2m+1-2j \right )n } }}{-\sigma ^{2m+1-2j}\left ( t \right ) -\tau ^{2m+1-2j}\left ( t \right ) } \right ) \\ & = \frac{a^{2m+1} }{b\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits _{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m+1}{m-j}\left ( \frac{f_{\left ( 2n+1 \right )\left ( 2j+1\right ) }\left (t \right ) - f_{2j+1}\left (t \right ) }{l_{2j+1} \left ( t \right ) } \right ) . \end{split} \end{equation*}$

□

Proof of Theorem 2. We only prove (1.7), and the proofs for other identities are similar.

$\begin{equation*} \begin{split} \sum\limits _{k = 1}^{n}f_{2k}^{2m }\left ( t \right )& = \sum\limits_{k = 1}^{n}\left (\frac{a^{\varsigma \left ( 2k +1\right ) } }{\left (ab \right )^{ \left \lfloor \frac{2k}{2} \right \rfloor} }\cdot \left ( \frac{\sigma ^{2k}\left ( t\right ) - \tau ^{2k} \left ( t \right ) }{\sigma\left ( t\right ) - \tau \left (t \right ) } \right ) \right ) ^{2m} \\ & = \frac{a^{2m} }{\left ( \sigma\left (t \right ) -\tau \left (t \right ) \right )^{2m} }\sum\limits _{k = 1}^{n}\frac{\left ( \sigma^{2k}\left ( t\right ) - \tau ^{2k} \left ( t \right ) \right )^{2m} }{\left ( ab \right )^{ 2mk } } \\ & = \frac{a^{2m} }{\left ( \sigma\left ( t \right ) -\tau \left (t \right ) \right )^{2m} }\sum\limits _{k = 1}^{n}\sum\limits _{j = 0}^{2m}\left ( -1 \right )^{j} \binom{2m}{j} \frac{ \sigma^{2k\left ( 2m-j \right ) } \left (t \right ) \tau ^{2kj} \left ( t \right ) }{\left ( ab \right )^{2mk } }\\ & = \frac{a^{2m} }{\left (\sigma\left (t \right ) -\tau \left ( t \right ) \right )^{2m} }\sum\limits _{j = 0}^{2m}\left ( -1 \right )^{j} \binom{2m }{j} \left ( \frac{1-\frac{\sigma ^{2n\left ( 2m-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2m-2j \right )n } } }{\frac{\left ( ab \right )^{2m-2j} }{\sigma ^{2\left ( 2m-2j \right ) }\left ( t \right ) } -1} \right ) \end{split} \end{equation*}$

$\begin{equation*} \begin{split} & = \frac{a^{2m} }{\left (\sigma\left (t \right ) -\tau \left ( t \right ) \right )^{2m} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m}{j} \left ( \frac{1-\frac{\sigma ^{2n\left ( 2m-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2m-2j \right )n } } }{\frac{\left ( ab \right )^{2m-2j} }{\sigma ^{2\left ( 2m-2j \right ) }\left ( t \right ) } -1} +\frac{1-\frac{\sigma ^{2n\left ( 2j-2m \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2j-2m\right )n } } }{\frac{\left ( ab \right )^{2j-2m} }{\sigma ^{2\left ( 2j-2m \right ) }\left (t\right ) } -1}\right )\\ & \quad +\frac{a^{2m} }{\left ( \sigma \left ( t \right ) -\tau \left (t \right ) \right )^{2m} } \left ( -1 \right ) ^{m+1}\binom{2m}{m}n\\ & = \frac{a^{2m} }{\left (\sigma\left ( t \right ) -\tau \left (t \right ) \right )^{2m} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m}{j} \left ( \frac{\frac{\sigma ^{2 \left ( 2m -2j \right ) }\left ( t \right ) }{\left ( ab \right )^{ 2m -2j } }- \frac{\sigma ^{\left ( 2n+2 \right ) \left ( 2m -2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( n+1 \right ) \left ( 2m -2j \right ) } } -1+\frac{\sigma^{2n \left ( 2j -2m \right ) }\left ( t \right ) }{\left ( ab \right )^{ \left ( 2j -2m \right ) n }} }{1- \frac{\sigma^{2 \left ( 2m -2j \right ) }\left ( t \right ) }{\left ( ab \right )^{ 2m-2j }}} \right )\\ & \quad +\frac{a^{2m} }{\left (\sigma \left ( t \right ) -\tau \left ( t \right ) \right )^{2m} } \left ( -1 \right ) ^{m+1}\binom{2m}{m}n \\ & = \frac{a^{2m } }{\left ( \sigma\left ( t \right ) -\tau \left ( t \right ) \right )^{2m } }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m }{j} \left ( \frac{\sigma^{2m -2j}\left ( t \right ) - \tau ^{2m -2j}\left (t\right ) -\frac{\sigma^{\left ( 2n+1 \right ) \left ( 2m -2j \right ) }\left (t \right )}{\left ( ab \right )^{n\left ( 2m -2j \right ) } } +\frac{\tau ^{\left ( 2n+1 \right ) \left ( 2m-2j \right ) }\left (t \right )}{\left ( ab \right )^{n\left ( 2m-2j \right ) } }}{ \tau ^{2m -2j}\left ( t \right )-\sigma ^{2m -2j}\left ( t \right ) } \right ) \\ & \quad +\frac{a^{2m} }{\left ( \sigma \left (t \right ) -\tau \left ( t \right ) \right )^{2m} } \left ( -1 \right ) ^{m+1}\binom{2m}{m}n\\ & = \frac{a^{2m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits _{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m}{m-j}\left ( \frac{f_{2j\left ( 2n+1 \right ) }\left ( t \right ) - f_{ 2j}\left (t \right ) }{f_{ 2j} \left (t \right ) } \right ) +\frac{a^{2m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \left ( -1 \right ) ^{m+1}\binom{2m}{m}n . \end{split} \end{equation*}$

□

Proof of Corollary 1. First, from the definition of $f_{n}\left (t \right)$ and binomial expansion, we easily prove $\left (f_{2n+1}\left (t \right)-1, a^{2} b^{2} t^{2} +4ab \right) = 1$ . Therefore, $\left (f_{2n+1}\left (t \right)-1, \left (a^{2} b^{2} t^{2} +4ab \right)^{m} \right) = 1$ . Now, we prove (1.11) by Lemma 1 and (1.3):

$\begin{equation*} \begin{split} &b l_{1} \left ( t\right ) l_{3} \left ( t \right ) \cdots l_{2m+1}\left ( t \right )\sum\limits _{k = 1}^{n}f_{2k}^{2m+1}\left ( t\right ) \\ & = l_{1} \left ( t\right ) l_{3} \left (t \right ) \cdots l_{2m+1}\left ( t \right ) \left ( \frac{a^{2m+1} }{\left ( \sigma\left (t \right ) -\tau \left (t \right ) \right )^{2m} } \sum\limits_{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m+1}{m-j} \left ( \frac{f_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left ( t \right ) -f_{2j+1}\left ( t \right ) }{l_{2j+1}\left (t\right ) } \right ) \right )\\ & \equiv 0\pmod{f_{2n+1}\left ( t \right ) -1}. \end{split} \end{equation*}$

Now, we use Lemma 2 and (1.5) to prove (1.12):

$\begin{equation*} \begin{split} &al_{1} \left ( t\right ) l_{3} \left ( t \right ) \cdots l_{2m+1}\left ( t\right )\sum\limits _{k = 1}^{n}l_{2k}^{2m+1}\left ( t \right ) \\ & = l_{1} \left ( t \right ) l_{3} \left (t \right ) \cdots l_{2m+1}\left ( t \right ) \left ( \sum\limits_{j = 0}^{m} \binom{2m+1}{m-j} \left ( \frac{al_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left ( t\right ) -al_{2j+1}\left (t \right ) }{l_{2j+1}\left ( t \right ) } \right ) \right )\\ & \equiv 0\pmod{l_{2n+1}\left ( t \right ) -at}. \end{split} \end{equation*}$

□

3. Conclusions

In this paper, we discuss the power sums of bi-periodic Fibonacci and Lucas polynomials by Binet formulas. As corollaries of the theorems, we extend the divisible properties of the sum of power of linear Fibonacci and Lucas sequences to nonlinear Fibonacci and Lucas polynomials. An open problem is whether we extend the Melham conjecture to nonlinear Fibonacci and Lucas polynomials.

Use of AI tools declaration

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

Acknowledgments

The authors would like to thank the editor and referees for their helpful suggestions and comments, which greatly improved the presentation of this work. All authors contributed equally to the work, and they have read and approved this final manuscript. This work is supported by Natural Science Foundation of China (12126357).

Conflict of interest

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

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Selection of artificial intelligence provider via multi-attribute decision-making technique under the model of complex intuitionistic fuzzy rough sets

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Selection of artificial intelligence provider via multi-attribute decision-making technique under the model of complex intuitionistic fuzzy rough sets

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2. Proofs of theorems

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