Theory and Modern Applications

# Bernoulli F-polynomials and Fibo–Bernoulli matrices

## Abstract

In this article, we define the Euler–Fibonacci numbers, polynomials and their exponential generating function. Several relations are established involving the Bernoulli F-polynomials, the Euler–Fibonacci numbers and the Euler–Fibonacci polynomials. A new exponential generating function is obtained for the Bernoulli F-polynomials. Also, we describe the Fibo–Bernoulli matrix, the Fibo–Euler matrix and the Fibo–Euler polynomial matrix by using the Bernoulli F-polynomials, the Euler–Fibonacci numbers and the Euler–Fibonacci polynomials, respectively. Factorization of the Fibo–Bernoulli matrix is obtained by using the generalized Fibo–Pascal matrix and a special matrix whose entries are the Bernoulli–Fibonacci numbers. The inverse of the Fibo–Bernoulli matrix is also found.

## 1 Introduction

Many mathematicians have recently studied various matrices and analogs of these matrices. Especially, these matrices are the Bernoulli, Pascal and Euler matrices [1,2,3,4,5,6,7,8,9,10,11]. These matrices and their analogs are obtained using numbers and polynomials such as the Bernoulli, Euler, q-Bernoulli, and q-Euler expressions [5, 12,13,14,15,16,17,18].

In this study we are interested in some matrices whose entries are the Bernoulli F-polynomials, Bernoulli–Fibonacci numbers, Euler–Fibonacci numbers and Euler–Fibonacci polynomials.

The Fibonacci sequence $$\{ F_{n} \} _{n\geq 0}$$ is defined by

$$F_{n}= \textstyle\begin{cases} F_{n+2}=F_{n+1}+F_{n}, \\ F_{0}=0 ,\qquad F_{1}=1. \end{cases}$$

For convenience of the reader, we provide a summary of the mathematical notations and some basic definitions of the Fibonomial coefficient.

The F-factorial is defined as follows:

$$F_{n}!=F_{n} F_{n-1} F_{n-2}\cdots F_{1},\qquad F_{0}!=1.$$

The Fibonomial coefficients are defined $$n\geq k\geq 1$$ as

$$\binom{n}{k}_{F}=\frac{F_{n}!}{F_{n-k}!F_{k}!},$$

with $$\binom{n}{0}_{F}=1$$ and $$\binom{n}{k}_{F}=0$$ for $$n< k$$. Fibonomial coefficients have the following properties:

$$\binom{n}{k}_{F}=\binom{n}{n-k}_{F}$$

and

$$\binom{n}{k}_{F}\binom{k}{j}_{F}= \binom{n}{j}_{F}\binom{n-j}{k-j}_{F}.$$

The binomial theorem for the F-analog is given by

$$( x+_{F}y ) ^{n}=\sum_{k=0}^{n} \binom{n}{k}_{F}x ^{k}y^{n-k}.$$
(1)

The F-exponential function $$e_{F}^{t}$$ is defined by

$$e_{F}^{t}=\sum_{n=0}^{\infty } \frac{t^{n}}{F_{n}!}$$
(2)

in [19, 20].

## 2 The Bernoulli F-polynomials and some of its properties

Firstly, we mention the Bernoulli F-polynomials. Krot  defined the Bernoulli F-polynomials. In this section, we obtain an exponential generating function of the Bernoulli F-polynomials. Then we give some properties of the Bernoulli F-polynomials.

### Definition 1

()

Let $$\binom{n}{k}_{F}$$ be Fibonomial coefficients and $$F_{n}$$ be the nth Fibonacci numbers, and we use Bernoulli’s F-polynomials of order 1; we define

$$B_{n,F} ( x ) =\sum_{k\geq 0}\frac{1}{F_{k+1}} \binom{n}{k}_{F} x^{n-k}.$$
(3)

The first few Bernoulli’s F-polynomials are as follows:

\begin{aligned}& B_{0,F} ( x ) = 1 , \\& B_{1,F} ( x ) = x+1 , \\& B_{2,F} ( x ) = x^{2}+x+\frac{1}{2} , \\& B_{3,F} ( x ) = x^{3}+2x^{2}+x+\frac{1}{3} , \\& B_{4,F} ( x ) = x^{4}+3x^{3}+3x^{2}+x+ \frac{1}{5} , \\& B_{5,F} ( x ) = x^{5}+5x^{4}+\frac{15}{2}x^{3}+5x^{2}+x+ \frac{1}{8}. \end{aligned}

### Theorem 1

The exponential generating function of the Bernoulli F-polynomial $$B_{n,F} ( x )$$ is

$$g ( x ) =\frac{e_{F}^{xt} ( e_{F}^{t}-1 ) }{t}.$$
(4)

### Proof

For the proof, we use the F-exponential function $$e_{F}^{t}$$.

\begin{aligned} \frac{e_{F}^{xt} ( e_{F}^{t}-1 ) }{t} &=\frac{1}{t} \Biggl(\sum _{n=0}^{\infty }x^{n}\frac{t^{n}}{F_{n}!} \Biggr) \Biggl(\sum_{n=0}^{\infty }\frac{t^{n}}{F_{n}!}-1 \Biggr) \\ &= \Biggl( \sum_{n=0}^{\infty }\frac{1}{F_{n+1}} \frac{t^{n}}{F _{n}!} \Biggr) \Biggl( \sum_{n=0}^{\infty }x^{n} \frac{t^{n}}{F _{n}!} \Biggr) \\ &=\sum_{n=0}^{\infty } \Biggl( \sum _{k=0}^{n}\frac{1}{F _{k+1}!} \frac{x^{n-k}}{F_{n-k}!} \Biggr) t^{n} \\ &=\sum_{n=0}^{\infty } \Biggl( \sum _{k=0}^{n}\frac{1}{F _{k+1}}\binom{n}{k}_{F}x^{n-k} \Biggr) \frac{t^{n}}{F_{n}!} \\ &=\sum_{n=0}^{\infty }B_{n,F} ( x ) \frac{t^{n}}{F _{n}!} . \end{aligned}

□

### Theorem 2

Let $$B_{n,F} ( x+y )$$ be the Bernoulli F-polynomials, we have

$$B_{n,F} ( x+y ) =\sum_{k=0}^{n} \binom{n}{k}_{F}B _{k,F} (x ) y^{n-k},$$
(5)

where $$B_{n,F} ( x+y ) =\sum_{k\geq 0}\frac{1}{F _{k+1}}\binom{n}{k}_{F} ( x+_{F}y ) ^{n-k}$$ for all nonnegative integers n.

### Proof

By virtue of the definition of the Bernoulli F-polynomials we get

\begin{aligned} \Biggl( \sum_{n=0}^{\infty }B_{n,F} ( x ) \frac{t ^{n}}{F_{n}!} \Biggr) \Biggl( \sum_{n=0}^{\infty }y^{n} \frac{t ^{n}}{F_{n}!} \Biggr) &=\sum_{n=0}^{\infty } \Biggl( \sum_{k=0}^{n}\frac{B_{k,F} ( x ) }{F_{k}!} \frac{y^{n-k}}{F _{n-k}!} \Biggr) t^{n} \\ &=\sum_{n=0}^{\infty } \Biggl( \sum _{k=0}^{n} \binom{n}{k}_{F}B_{k,F} ( x ) y^{n-k} \Biggr) \frac{t^{n}}{F _{n}!}. \end{aligned}
(6)

On the other hand,

\begin{aligned} \Biggl( \sum_{n=0}^{\infty }B_{n,F} ( x ) \frac{t ^{n}}{F_{n}!} \Biggr) \Biggl( \sum_{n=0}^{\infty }y^{n} \frac{t ^{n}}{F_{n}!} \Biggr) &= \Biggl( \sum_{n=0}^{\infty } \sum_{k=0}^{n}\frac{1}{F_{k+1}} \binom{n}{k}_{F}x^{n-k}\frac{t^{n}}{F _{n}!} \Biggr) \Biggl( \sum_{n=0}^{\infty }y^{n} \frac{t^{n}}{F _{n}!} \Biggr) \\ &= \Biggl( \sum_{n=0}^{\infty }\frac{1}{F_{n+1}} \frac{t^{n}}{F _{n}!} \Biggr) \Biggl( \sum_{n=0}^{\infty }x^{n} \frac{t^{n}}{F _{n}!} \Biggr) \Biggl( \sum_{n=0}^{\infty }y^{n} \frac{t^{n}}{F _{n}!} \Biggr) \\ &= \Biggl( \sum_{n=0}^{\infty }\frac{t^{n}}{F_{n+1}!} \Biggr) \Biggl(\sum_{n=0}^{\infty } ( x+_{F}y ) ^{n}\frac{t ^{n}}{F_{n}!} \Biggr) \\ &=\sum_{n=0}^{\infty } \Biggl( \sum _{k=0}^{n}\frac{1}{F _{k+1}}\binom{n}{k}_{F} ( x+_{F}y ) ^{n-k} \Biggr) \frac{t ^{n}}{F_{n}!} \\ &=\sum_{n=0}^{\infty }B_{n,F} ( x+y ) \frac{t^{n}}{F _{n}!}. \end{aligned}
(7)

Comparing the coefficients of $$\frac{t^{n}}{F_{n}!}$$ on both sides of Eqs. (6) and (7), we arrive at the desired result. □

## 3 The Euler–Fibonacci polynomials and their relation with Bernoulli F-polynomials

In this section, we define the Euler–Fibonacci numbers and the Euler–Fibonacci polynomials. Then we obtain their exponential functions and the relationship between the Bernoulli F-polynomials and these polynomials.

### Definition 2

For all nonnegative integer n, the Euler–Fibonacci numbers $$E_{n,F}$$ are defined by

$$E_{n,F}=-\sum_{k=0}^{n} \binom{n}{k}_{F}E_{k,F},$$
(8)

where $$E_{0,F}=1$$.

The first few Euler–Fibonacci numbers are as follows:

$$\textstyle\begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} E_{0,F} & E_{1,F} & E_{2,F} & E_{3,F} & E_{4,F} & E_{5,F} \\ 1 & -\frac{1}{2} & -\frac{1}{4} & -\frac{1}{4} & \frac{11}{8} & \frac{17}{16} \end{array}$$

### Theorem 3

The exponential generating function of Euler–Fibonacci numbers $$E_{n,F}$$ is defined by

$$\sum_{n=0}^{\infty } E_{n,F} \frac{t^{n}}{F_{n}!}=\frac{2}{e _{F}^{t}+1}.$$
(9)

### Proof

For the proof, we show that

$$\Biggl( \sum_{n=0}^{\infty }E_{n,F} \frac{t^{n}}{F_{n}!} \Biggr) \bigl(e_{F}^{t}+1 \bigr) =2.$$

From (2), we have

\begin{aligned} \Biggl( \sum_{n=0}^{\infty }E_{n,F} \frac{t^{n}}{F_{n}!} \Biggr) \Biggl(\sum_{n=0}^{\infty } \frac{t^{n}}{F_{n}!}+1 \Biggr) &= \Biggl(\sum_{n=0}^{\infty }E_{n,F} \frac{t^{n}}{F_{n}!} \Biggr) \Biggl(2+\sum_{n=1}^{\infty } \frac{t^{n}}{F_{n}!} \Biggr) \\ &= 2\sum_{n=0}^{\infty }E_{n,F} \frac{t^{n}}{F_{n}!}+\sum_{n=1}^{\infty } \Biggl( \sum_{k=0}^{n-1}\frac{E_{k,F}}{F _{k}!} \frac{1}{F_{n-k}!} \Biggr) t^{n} \\ &= 2\sum_{n=0}^{\infty }E_{n,F} \frac{t^{n}}{F_{n}!}+\sum_{n=1}^{\infty } \Biggl( \sum_{k=0}^{n}\binom{n}{k}_{F}E _{k,F}-E_{n,F} \Biggr) \frac{t^{n}}{F_{n}!} \\ &=2\sum_{n=0}^{\infty }E_{n,F} \frac{t^{n}}{F_{n}!}+\sum_{n=1}^{\infty } ( -2E_{n,F} ) \frac{t^{n}}{F_{n}!} \\ &= 2, \end{aligned}

which is the desired result. □

### Definition 3

The Euler–Fibonacci polynomials $$E_{n,F} ( x )$$ are defined by

$$E_{n,F} (x )=\sum_{k=0}^{n} \binom{n}{k}_{F}E_{k,F} x^{n-k},$$

where $$E_{0,F} ( x ) =1$$ and $$E_{n,F}$$ are the nth Euler–Fibonacci numbers.

The first few Euler–Fibonacci polynomials are as follows:

\begin{aligned}& E_{0,F} ( x ) = 1 , \\& E_{1,F} ( x ) = x-\frac{1}{2} , \\& E_{2,F} ( x ) = x^{2}-\frac{x}{2}-\frac{1}{4} , \\& E_{3,F} ( x ) = x^{3}-x^{2}-\frac{x}{2}- \frac{1}{4} , \\& E_{4,F} ( x ) = x^{4}-\frac{3}{2}x^{3}- \frac{3}{2}x^{2}- \frac{3}{4}x+\frac{11}{8} , \\& E_{5,F} ( x ) = x^{5}-\frac{5}{2}x^{4}- \frac{15}{4}x^{3}- \frac{15}{4}x^{2}+ \frac{55}{8}x+\frac{17}{16} . \end{aligned}

### Theorem 4

The exponential generating function of Euler–Fibonacci polynomials $$E_{n,F} ( x )$$ is defined by

$$\sum_{n=0}^{\infty }E_{n,F} ( x ) \frac{t^{n}}{F _{n}!}=\frac{2e_{F}^{xt}}{ ( e_{F}^{t}+1 ) } .$$
(10)

### Proof

By virtue of the definition of the Euler–Fibonacci polynomials, we get

\begin{aligned} \frac{2e_{F}^{xt}}{ ( e_{F}^{t}+1 ) } &= \sum_{n=0} ^{\infty }E_{n,F} \frac{t^{n}}{F_{n}!}\sum_{n=0}^{\infty }x^{n} \frac{t ^{n}}{F_{n}!} \\ &= \sum_{n=0}^{\infty } \Biggl( \sum _{k=0}^{n}\frac{E _{k,F}}{F_{k}!} \frac{x^{n-k}}{F_{n-k}!} \Biggr) t^{n} \\ &= \sum_{n=0}^{\infty } \Biggl( \sum _{k=0}^{n} \binom{n}{k}_{F}E_{k,F} x^{n-k} \Biggr) \frac{t^{n}}{F_{n}!} \\ &= \sum_{n=0}^{\infty }E_{n,F} ( x ) \frac{t^{n}}{F _{n}!}. \end{aligned}

□

In the following proposition, we will give a relationship between the Bernoulli F-polynomials $$B_{n,F} ( x )$$ and the Euler–Fibonacci polynomials $$E_{n,F} ( x )$$.

### Proposition 1

Let n be a nonnegative integer,

$$B_{n,F} ( x ) =\frac{x^{n+1}-E_{n+1,F} ( x ) }{F _{n+1}}+\sum_{k=0}^{n} \frac{1}{F_{k+1}}\binom{n}{k}_{F} \bigl(x ^{k+1}-E_{k+1,F} ( x ) \bigr).$$
(11)

### Proof

For the proof, we use the exponential generating functions for the Bernoulli F-polynomial and the Euler–Fibonacci polynomials. We have

\begin{aligned} &\sum_{n=0}^{\infty }B_{n,F} ( x ) \frac{t^{n}}{F _{n}!} \\ &\quad =\frac{e_{F}^{xt} ( e_{F}^{t}-1 ) }{t} \\ &\quad =\frac{ ( e_{F}^{t}+1 ) }{t} \biggl( e_{F}^{xt}- \frac{2e _{F}^{xt}}{e_{F}^{t}+1} \biggr) \\ &\quad = \Biggl( \sum_{n=0}^{\infty } \frac{t^{n}}{F_{n}!}+1 \Biggr) \Biggl(\sum_{n=0}^{\infty } \frac{x^{n}t^{n-1}}{F_{n}!}-\sum_{n=0}^{\infty }E_{n,F} ( x ) \frac{t^{n-1}}{F_{n}!} \Biggr) \\ &\quad = \Biggl( \sum_{n=0}^{\infty } \frac{t^{n}}{F_{n}!}+1 \Biggr) \Biggl(\sum_{n=0}^{\infty } \bigl( x^{n+1}-E_{n+1,F} ( x ) \bigr)\frac{t^{n}}{F_{n+1}!} \Biggr) \\ &\quad =\sum_{n=0}^{\infty } \Biggl( \sum _{k=0}^{n}\frac{x ^{k+1}-E_{k+1,F} ( x ) }{F_{k+1}!} \frac{1}{F_{n-k}!} \Biggr)t ^{n}+\sum_{n=0}^{\infty } \biggl( \frac{x^{n+1}-E_{n+1,F} (x ) }{F_{n+1}} \biggr) \frac{t ^{n}}{F_{n}!} \\ &\quad =\sum_{n=0}^{\infty } \Biggl( \sum _{k=0}^{n}\frac{1}{F _{k+1}} \binom{n}{k}_{F} \bigl( x^{k+1}-E_{k+1,F} ( x ) \bigr) \Biggr)t^{n}+\sum_{n=0}^{\infty } \biggl( \frac{x^{n+1}-E _{n+1,F} (x ) }{F_{n+1}} \biggr) \frac{t^{n}}{F_{n}!} \\ &\quad =\sum_{n=0}^{\infty } \Biggl( \frac{x^{n+1}-E_{n+1,F} ( x ) }{F_{n+1}}+\sum_{k=0}^{n} \frac{1}{F_{k+1}} \binom{n}{k}_{F} \bigl(x^{k+1}-E_{k+1,F} ( x ) \bigr) \Biggr) \frac{t^{n}}{F_{n}!}. \end{aligned}

Comparing the coefficients of $$t^{n}/F_{n}!$$ on both sides of the above equations we arrive at the desired result. □

Also,

$$B_{n,F} ( x ) =2 \biggl( \frac{x^{n+1}-E_{n+1,F} ( x ) }{F_{n+1}} \biggr) +\sum _{k=0}^{n-1} \frac{1}{F_{k+1}}\binom{n}{k}_{F} \bigl( x^{k+1}-E_{k+1,F} ( x ) \bigr).$$
(12)

For example, if we take $$n=2$$ in Proposition 1, we have

\begin{aligned} B_{2,F} ( x ) &=\frac{x^{3}-E_{3,F} ( x ) }{F _{3}}+\sum_{k=0}^{2} \frac{1}{F_{k+1}}\binom{2}{k}_{F} \bigl(x ^{k+1}-E_{k+1,F} ( x ) \bigr) \\ &=\frac{1}{2} \biggl( x^{2}+\frac{x}{2}- \frac{1}{4} \biggr) +x- \biggl( x-\frac{1}{2} \biggr) +x^{2}- \biggl( x^{2}-\frac{x}{2}-\frac{1}{4} \biggr) \\ &\quad {} +\frac{1}{2} \biggl( x^{3}- \biggl( x^{3}-x^{2}+\frac{x}{2}+\frac{1}{4} \biggr) \biggr) \\ &=x^{2}+x+\frac{1}{2}. \end{aligned}

### Proposition 2

Let $$E_{n,F}$$ be the nth Euler–Fibonacci number. Then we have

$$\sum_{k=0}^{n}\binom{n}{k}_{F}B_{k,F} (x )E_{n-k,F}= \sum_{k=0}^{n} \frac{1}{F_{k+1}}\binom{n}{k}_{F}E_{n-k,F} ( x ).$$
(13)

### Proof

We have

\begin{aligned}& \begin{aligned}[b] \Biggl(\sum_{n=0}^{\infty }E_{n,F} ( x ) \frac{t ^{n}}{F_{n}!} \Biggr) \biggl( \frac{e_{F}^{t}-1}{t} \biggr) &= \Biggl( \sum_{n=0}^{\infty }E_{n,F} ( x ) \frac{t^{n}}{F_{n}!} \Biggr) \Biggl(\sum_{n=1}^{\infty } \frac{t^{n-1}}{F_{n}!} \Biggr) \\ &=\sum_{n=0}^{\infty } \Biggl( \sum _{k=0}^{n}\frac{1}{F _{k+1}!}\frac{E_{n-k,F} ( x ) }{F_{n-k}!} \Biggr) t^{n} \\ &=\sum_{n=0}^{\infty } \Biggl( \sum _{k=0}^{n}\frac{1}{F _{k+1}}\binom{n}{k}_{F}E_{n-k,F} ( x ) \Biggr) \frac{t ^{n}}{F_{n}!} , \end{aligned} \end{aligned}
(14)
\begin{aligned}& \begin{aligned}[b] \Biggl(\sum_{n=0}^{\infty }E_{n,F} ( x ) \frac{t ^{n}}{F_{n}!} \Biggr) \biggl( \frac{e_{F}^{t}-1}{t} \biggr) &= \Biggl( \sum_{n=0}^{\infty }E_{n,F} \frac{t^{n}}{F_{n}!} \Biggr) \Biggl(\sum_{n=0}^{\infty }x^{n} \frac{t^{n}}{F_{n}!} \Biggr) \Biggl(\sum_{n=1}^{\infty } \frac{t^{n-1}}{F_{n}!} \Biggr) \\ &= \Biggl( \sum_{n=0}^{\infty }E_{n,F} \frac{t^{n}}{F_{n}!} \Biggr) \Biggl( \sum_{n=0}^{\infty } \Biggl( \sum_{k=0}^{n}\frac{1}{F _{k+1}!} \frac{x^{n-k}}{F_{n-k}!} \Biggr) t^{n} \Biggr) \\ &=\sum_{n=0}^{\infty }E_{n,F} \frac{t^{n}}{F_{n}!}\sum_{n=0}^{\infty }B_{n,F} ( x ) \frac{t^{n}}{F_{n}!} \\ &=\sum_{n=0}^{\infty } \Biggl( \sum _{k=0}^{n} \binom{n}{k}_{F}B_{k,F} ( x ) E_{n-k,F} \Biggr) \frac{t ^{n}}{F_{n}!} . \end{aligned} \end{aligned}
(15)

From (14) and (15), we get

$$\sum_{k=0}^{n}\binom{n}{k}_{F}B_{k,F} ( x )E_{n-k,F}= \sum_{k=0}^{n} \frac{1}{F_{k+1}}\binom{n}{k}_{F}E_{n-k,F} ( x ).$$

□

For example

\begin{aligned} \sum_{k=0}^{2}\binom{2}{k}_{F}B_{k,F} ( x ) E_{2-k,F} &=-\frac{1}{4}+ ( x+1 ) \biggl( -\frac{1}{2} \biggr) + \biggl( x^{2}+x+\frac{1}{2} \biggr) 1 \\ &=x^{2}+\frac{1}{2}x-\frac{1}{4} \end{aligned}

and

\begin{aligned} \sum_{k=0}^{2}\frac{1}{F_{k+1}} \binom{2}{k}_{F}E_{2-k,F} (x ) &=x^{2}- \frac{x}{2}-\frac{1}{4}+x-\frac{1}{2}+\frac{1}{2} \\ &=x^{2}+\frac{1}{2}x-\frac{1}{4}. \end{aligned}

## 4 The Bernoulli–Fibonacci numbers and the Bernoulli–Fibonacci polynomials

In , the author defined the nth Bernoulli–Fibonacci numbers and the Bernoulli–Fibonacci polynomials. For all nonnegative integers n, the nth Bernoulli–Fibonacci polynomials $$B_{n}^{F} (x )$$ are given with the exponential generating function as follows:

$$\sum_{n=0}^{\infty }B_{n}^{F} ( x ) \frac{t^{n}}{F _{n}!}=\frac{te_{F}^{tx}}{e_{F}^{t}+1},$$
(16)

where $$B_{n}^{F} ( 0 )=B_{n}^{F}$$.

Let the nth Bernoulli–Fibonacci number be $$B_{n}^{F} (0 )=B _{n}^{F}$$, its exponential generating function is

$$\sum_{n=0}^{\infty }B_{n}^{F} \frac{t^{n}}{F_{n}!}=\frac{t}{e _{F}^{t}+1}.$$
(17)

### Proposition 3

()

Let the nth Bernoulli–Fibonacci numbers be $$B_{n}^{F}$$ having defined $$B_{0}^{F}=1$$ and

$$B_{n}^{F}=-\sum_{k=0}^{n} \frac{1}{F_{n-k+1}}\binom{n}{k}_{F}B _{k}^{F}.$$
(18)

The first few Bernoulli–Fibonacci numbers are as follows:

$$\textstyle\begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} B_{0}^{F} & B_{1}^{F} & B_{2}^{F} & B_{3}^{F} & B_{4}^{F} & B_{5}^{F} & B_{6}^{F} & B_{7}^{F} \\ 1 & -1 & \frac{1}{2} & -\frac{1}{3} & \frac{3}{10} & -\frac{5}{8} & \frac{101}{39} & -\frac{323}{21} \end{array}$$

### Proposition 4

()

The recurrence formula of the nth Bernoulli–Fibonacci polynomials is

$$B_{n}^{F} ( x ) =\sum_{k=0}^{n} \binom{n}{k}_{F}B _{k}^{F}x^{n-k}.$$
(19)

The first few Bernoulli–Fibonacci polynomials are as follows:

\begin{aligned}& B_{0}^{F} ( x ) =1, \\& B_{1}^{F} ( x ) =x+1, \\& B_{2}^{F} ( x ) =x^{2}-x+\frac{1}{2}, \\& B_{3}^{F} ( x ) =x^{3}-2x^{2}+x- \frac{1}{3}, \\& B_{4}^{F} ( x ) = x^{4}-3x^{3}+3x^{2}-x+ \frac{3}{10} , \\& B_{5}^{F} ( x ) =x^{5}-5x^{4}+ \frac{15}{2}x^{3}-5x^{2}+ \frac{3}{2}x- \frac{5}{8}. \end{aligned}

Now, we give the relationship of the first few Bernoulli F-polynomials $$B_{n,F} ( x )$$ and Bernoulli–Fibonacci polynomials $$B_{n}^{F} (x )$$ and the classical Bernoulli polynomials $$B_{n} (x )$$ with graphics in Fig. 1.

## 5 Fibo–Bernoulli matrices

In this section, we define an interesting Fibo–Bernoulli matrix by using the Bernoulli F-polynomials. Then we obtain a factorization of the Fibo–Bernoulli matrix by using a generalized Fibo–Pascal matrix. Moreover, we obtain the inverse of the Fibo–Bernoulli matrix. We define the Fibo–Euler matrix, the Fibo–Euler polynomial matrix and their inverses. Also, we show a relationship of the Fibo–Bernoulli matrix, Fibo–Euler matrix and Fibo–Euler polynomial matrix.

### Definition 4

()

The generalized Fibo–Pascal matrix $$U_{n+1} [ x ] = ( U_{n+1} ( x;i,j ) )$$ is defined by

$$U_{n+1} ( x;i,j ) = \textstyle\begin{cases} \binom{i}{j}_{F} x^{i-j} & \text{if }i\geqslant j, \\ 0 & \text{otherwise}. \end{cases}$$
(20)

### Example 1

We have

${U}_{6}\left[x\right]=\left[\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ x& 1& 0& 0& 0& 0\\ {x}^{2}& x& 1& 0& 0& 0\\ {x}^{3}& 2{x}^{2}& 2x& 1& 0& 0\\ {x}^{4}& 3{x}^{3}& 6{x}^{2}& 3x& 1& 0\\ {x}^{5}& 5{x}^{4}& 15{x}^{3}& 15{x}^{2}& 5x& 1\end{array}\right].$

### Definition 5

()

For $$n\geq 2$$, the inverse of the generalized Fibo–Pascal matrix $$V ( F ) = ( v_{ij} )$$ is defined by

$$v_{ij}= \textstyle\begin{cases} b_{i-j+1}\binom{i}{j}_{F} x^{i-j} & \text{if } i\geq j, \\ 0 & \text{otherwise}, \end{cases}$$
(21)

where $$b_{1}=1$$ and $$b_{n}=-\sum_{k=1}^{n-1}b_{k}\binom{n}{k} _{F}$$.

### Example 2

For $$n=5$$, the inverse of the generalized Fibo–Pascal matrix $$V ( F )$$ is as follows:

$V\left(F\right)=\left[\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ -x& 1& 0& 0& 0& 0\\ 0& -x& 1& 0& 0& 0\\ {x}^{3}& 0& -2x& 1& 0& 0\\ -{x}^{4}& 3{x}^{3}& 0& -3x& 1& 0\\ -6{x}^{5}& -5{x}^{4}& 15{x}^{3}& 0& -5x& 1\end{array}\right].$

### Definition 6

Let $$B_{n,F} ( x )$$ be the nth Bernoulli’s F-polynomial. $$(n+1 ) \times ( n+1 )$$; the Fibo–Bernoulli matrix $$\mathcal{B} ( x,F ) = [ b _{ij} ( x,F ) ]$$ is defined by

$$b_{ij} ( x,F ) = \textstyle\begin{cases} \binom{i}{j}_{F} B_{i-j,F} ( x ) & \text{if } i\geq j, \\ 0 & \text{otherwise}, \end{cases}$$
(22)

where $$0\leq i,j\leq n$$.

For $$n=3$$, the Fibo–Bernoulli matrix is as follows:

$\mathcal{B}\left(x,F\right)=\left[\begin{array}{cccc}1& 0& 0& 0\\ x+1& 1& 0& 0\\ {x}^{2}+x+\frac{1}{2}& x+1& 1& 0\\ {x}^{3}+2{x}^{2}+x+\frac{1}{3}& 2{x}^{2}+2x+1& 2x+2& 1\end{array}\right].$

Now, we define a special matrix by using the Fibonomial coefficient. Then we obtain the factorization Fibo–Bernoulli matrix by using the generalized Fibo–Pascal matrix.

### Definition 7

Let the nth Fibonacci numbers be $$F_{n}$$. For $$1 \leq i,j \leq n+1$$, the $$W(F)= [ w_{ij} ]$$ matrix is defined as follows:

$$w_{ij}= \textstyle\begin{cases} \frac{1}{F_{i-j+1}} \binom{i}{j}_{F} & \text{if } i\geqslant j, \\ 0 & \text{otherwise}. \end{cases}$$
(23)

For $$n=5$$, the $$W(F)$$ matrix is

$W\left(F\right)=\left[\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 1& 1& 0& 0& 0& 0\\ \frac{1}{2}& 1& 1& 0& 0& 0\\ \frac{1}{3}& 1& 2& 1& 0& 0\\ \frac{1}{5}& 1& 3& 3& 1& 0\\ \frac{1}{8}& 1& 5& \frac{15}{2}& 5& 1\end{array}\right].$

### Proposition 5

()

We have

$$\sum_{k=0}^{n}\binom{n}{k}_{F}B_{n-k}^{F} \frac{1}{F_{k+1}}=F_{n}! \delta _{n,0} .$$
(24)

### Theorem 5

Let $$B_{n}^{F}$$ be the nth Bernoulli–Fibonacci numbers. $$T ( F ) = [ t_{ij} ] _{ ( n+1 ) \times ( n+1 ) }$$, the inverse of the $$W(F)$$ matrix, is

$$t_{ij}= \textstyle\begin{cases} \binom{i}{j}_{F} B_{i-j}^{F} & \textit{if } i\geqslant j, \\ 0 & \textit{otherwise}. \end{cases}$$
(25)

### Proof

We have

\begin{aligned} \bigl( T ( F ) W(F) \bigr) _{ij} &=\sum_{k=j}^{i}t_{ik} w_{kj} \\ &=\sum_{k=j}^{i}\binom{i}{k}_{F}B_{i-k}^{F} \frac{1}{F_{k-j+1}} \binom{k}{j}_{F} \\ &=\sum_{k=j}^{i}\binom{i}{j}_{F} \binom{i-j}{k-j}_{F}B_{i-k}^{F} \frac{1}{F _{k-j+1}} \\ &=\binom{i}{j}_{F}\sum_{k=0}^{i-j} \binom{i-j}{k}_{F}B_{i-j-k}^{F} \frac{1}{F_{k+1}} \\ &=\binom{i}{j}_{F}F_{i-j}! \delta _{i-j,0} . \end{aligned}

Hence, $$( T ( F ) W(F) ) _{ij}=1$$ for $$i=j$$ and $$(T ( F ) W(F) ) _{ij}=0$$ for $$i\neq j$$. □

For $$n=5$$, $$T ( F )$$ is as follows:

$T\left(F\right)=\left[\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ -1& 1& 0& 0& 0& 0\\ \frac{1}{2}& -1& 1& 0& 0& 0\\ -\frac{1}{3}& 1& -2& 1& 0& 0\\ \frac{3}{10}& -1& 3& -3& 1& 0\\ -\frac{5}{8}& \frac{3}{2}& -5& \frac{15}{2}& -5& 1\end{array}\right].$

### Theorem 6

Let $$\mathcal{B} ( x,F )$$ be the Fibo–Bernoulli matrix and $$U_{n+1} [ x ]$$ be a generalized Fibo–Pascal matrix, then

$$\mathcal{B} ( x,F ) =U_{n+1} [ x ] W(F).$$

### Proof

We have

\begin{aligned} \bigl( U [ x ] \cdot W(F) \bigr) _{ij} &=\sum _{k=j} ^{i}u_{ik}w_{kj} \\ &=\sum_{k=j}^{i}\binom{i}{k}_{F}x^{i-k} \frac{1}{F_{k-j+1}} \binom{k}{j}_{F} \\ &=\binom{i}{j}_{F}\sum_{k=j}^{i} \frac{1}{F_{k-j+1}} \binom{i-j}{k-j}_{F}x^{i-k} \\ &=\binom{i}{j}_{F}\sum_{k=0}^{i-j} \frac{1}{F_{k+1}} \binom{i-j}{k}_{F}x^{i-j-k} \\ &=\binom{i}{j}_{F}B_{i-j,F} ( x ) \\ &= \bigl[ \mathcal{B} ( x,F ) \bigr] _{ij}. \end{aligned}

□

### Example 3

For $$n=3$$, we have

$\begin{array}{rl}{U}_{n+1}\left[x\right]W\left(F\right)& =\left[\begin{array}{cccc}1& 0& 0& 0\\ x& 1& 0& 0\\ {x}^{2}& x& 1& 0\\ {x}^{3}& 2{x}^{2}& 2x& 1\end{array}\right]×\left[\begin{array}{cccc}1& 0& 0& 0\\ 1& 1& 0& 0\\ \frac{1}{2}& 1& 1& 0\\ \frac{1}{3}& 1& 2& 1\end{array}\right]\\ & =\left[\begin{array}{cccc}1& 0& 0& 0\\ x+1& 1& 0& 0\\ {x}^{2}+x+\frac{1}{2}& x+1& 1& 0\\ {x}^{3}+2{x}^{2}+x+\frac{1}{3}& 2{x}^{2}+2x+1& 2x+2& 1\end{array}\right]\\ & =\mathcal{B}\left(x,F\right).\end{array}$

### Theorem 7

Let $$\mathcal{D} ( x,F ) = [ d_{ij} ]$$ be the $$(n+1 ) \times ( n+1 )$$ matrix defined by

$$d_{ij}= \textstyle\begin{cases} \binom{i}{j}_{F} \sum_{k=0}^{i-j}\binom{i-j}{k}_{F} B_{i-j-k} ^{F} b_{k+1} x^{k} & \textit{if } i\geqslant j, \\ 0 & \textit{otherwise}. \end{cases}$$
(26)

Then $$\mathcal{D} ( x,F )$$ is the inverse of the Fibo–Bernoulli matrix. Thus,

$$\mathcal{B}^{-1} ( x,F ) =\mathcal{D} ( x,F ).$$

### Proof

Let $$U_{n+1} [ x ]$$ be a generalized Fibo–Pascal matrix. Using the factorization of $$\mathcal{B} ( x,F )$$ in Theorem 6

$$\mathcal{B}^{-1} ( x,F ) =W^{-1}(F) U_{n+1}^{-1} [ x ]=T ( F ) V ( F )$$

and the inverse of the generalized Fibo–Pascal matrix in (21), we obtain

\begin{aligned} \bigl[ T ( F ) V ( F ) \bigr] _{ij} &=\sum _{k=j}^{i} \binom{i}{k}_{F} B_{i-k}^{F}\binom{k}{j}_{F}b_{k-j+1} x ^{k-j} \\ &=\binom{i}{j}_{F} \sum_{k=j}^{i} \binom{i-j}{k-j}_{F} B_{i-k} ^{F} b_{k-j+1} x^{k-j} \\ &=\binom{i}{j}_{F} \sum_{k=0}^{i-j} \binom{i-j}{k}_{F}B_{i-j-k}^{F} b_{k+1} x^{k-j} \\ &= \bigl[ \mathcal{D} ( x,F ) \bigr] _{ij} . \end{aligned}

□

### Example 4

For $$n=4$$, $$\mathcal{D} ( x,F )$$ is as follows:

$\begin{array}{rl}\mathcal{D}\left(x,F\right)& =\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ -1& 1& 0& 0& 0\\ \frac{1}{2}& -1& 1& 0& 0\\ -\frac{1}{3}& 1& -2& 1& 0\\ \frac{3}{10}& -1& 3& -3& 1\end{array}\right]×\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ -x& 1& 0& 0& 0\\ 0& -x& 1& 0& 0\\ {x}^{3}& 0& -2x& 1& 0\\ -{x}^{4}& 3{x}^{3}& 0& -3x& 1\end{array}\right]\\ & =\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ -x-1& 1& 0& 0& 0\\ x+\frac{1}{2}& -x-1& 1& 0& 0\\ {x}^{3}-x-\frac{1}{3}& 2x+1& -2x-2& 1& 0\\ -{x}^{4}-3{x}^{3}+x+\frac{3}{10}& 3{x}^{3}-3x-1& 6x+3& -3x-3& 1\end{array}\right].\end{array}$

### Definition 8

Let $$E_{n,F}$$ be the Euler–Fibonacci number. For $$1 \leq i,j \leq n+1$$, then the Fibo–Euler matrix $$E_{F} = ( e_{F} ) _{ij}$$ is defined as follows:

$$( e_{F} ) _{ij}= \textstyle\begin{cases} \binom{i}{j}_{F} E_{i-j,F} & \text{if } i\geqslant j, \\ 0 & \text{otherwise}. \end{cases}$$
(27)

### Example 5

For $$n=3$$, the Fibo–Euler matrix is

${E}_{F}=\left[\begin{array}{cccc}1& 0& 0& 0\\ -\frac{1}{2}& 1& 0& 0\\ -\frac{1}{4}& -\frac{1}{2}& 1& 0\\ \frac{1}{4}& -\frac{1}{2}& -1& 1\end{array}\right].$

### Definition 9

()

The Fibo–Pascal matrix $$U_{n+1,F}= [ u_{i,j} ] _{ (n+1 ) \times ( n+1 ) }$$ is defined by

$$u_{i,j}= \textstyle\begin{cases} \binom{i}{j}_{F} & \text{if } i\geqslant j, \\ 0 & \text{otherwise}. \end{cases}$$

### Proposition 6

()

Let $$E_{n,F}$$ be the Euler–Fibonacci number

$$\sum_{k=0}^{n}\binom{n}{k}_{F}E_{n-k,F}+E_{n,F}=2 \delta _{0,n}.$$
(28)

### Theorem 8

Let $$U_{n+1,F}= [ u_{i,j} ]$$ be the $$( n+1 ) \times (n+1 )$$ the Fibo–Pascal matrix, $$I_{n+1}$$ be the identity matrix, and $$E_{F}$$ be the Fibo–Euler matrix, then we get

$$\frac{1}{2} ( U_{n+1,F}+I_{n+1} ) =E_{F}^{-1}.$$

### Proof

We have

\begin{aligned} \biggl( E_{F}\frac{1}{2} ( U_{n+1,F}+I_{n+1} ) \biggr) _{ij} &=\frac{1}{2} ( E_{F} U_{n+1,F}+E_{F} ) _{ij} \\ &=\sum_{k=j}^{i}\binom{i}{k}_{F}E_{i-k,F} \frac{1}{2} \binom{k}{j}_{F}+\binom{i}{j}_{F}E_{i-j,F} \\ &=\frac{1}{2}\binom{i}{j}_{F}\sum _{k=j}^{i}\binom{i-j}{k-j} _{F}E_{i-k,F}+ \binom{i}{j}_{F}E_{i-j,F} \\ &=\frac{1}{2}\binom{i}{j}_{F} \Biggl[ \sum _{k=0}^{i-j} \binom{i-j}{k}_{F}E_{i-j-k,F}+E_{i-j,F} \Biggr] \\ &=\frac{1}{2}\binom{i}{j}_{F}2 \delta _{0,i-j} \\ &=\binom{i}{j}_{F} \delta _{0,i-j}. \end{aligned}

Thus, for $$i=j$$, $$\binom{i}{j}_{F} \delta _{0,i-j}=1$$ and for $$i\neq j$$ $$\binom{i}{j}_{F} \delta _{0,i-j}=0$$. Hence,

$$\frac{1}{2} ( U_{n+1,F}+I_{n+1} ) =E_{F}^{-1}.$$

□

### Definition 10

Let $$E_{n,F}$$ be the Euler–Fibonacci number. For $$1 \leq i,j \leq n+1$$, then the Fibo–Euler polynomial matrix $$E_{F} ( x ) = [ ( \varepsilon _{F} ) _{ij} ]$$ is defined as follows:

$$( \varepsilon _{F} ) _{ij}= \textstyle\begin{cases} \binom{i}{j}_{F} E_{i-j,F} x^{i-j} & \text{if }i\geqslant j, \\ 0 & \text{otherwise}. \end{cases}$$
(29)

### Example 6

$$5\times 5$$ For $$n=4$$, the Fibo–Euler polynomial matrix is as follows:

${E}_{F}\left(x\right)=\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ -\frac{x}{2}& 1& 0& 0& 0\\ -\frac{{x}^{2}}{4}& -\frac{x}{2}& 1& 0& 0\\ \frac{{x}^{3}}{4}& -\frac{{x}^{2}}{2}& -x& 1& 0\\ \frac{5{x}^{4}}{8}& \frac{3{x}^{3}}{4}& -\frac{3{x}^{2}}{2}& -\frac{3x}{2}& 1\end{array}\right].$

### Theorem 9

Let $$H_{F} ( x ) = [ ( h_{F} ) _{ij} ]$$ be the inverse of the Fibo–Euler polynomial matrix, then we have

$$H_{F} (x )=\frac{1}{2} \bigl( U_{n+1} [ x ] +I _{n+1} \bigr),$$
(30)

where $$U_{n+1,F}$$ is $${ ( n+1 ) \times ( n+1 ) }$$ Fibo–Pascal matrix and $$I_{n+1}$$ is the identity matrix.

### Proof

\begin{aligned} \bigl( E_{F} ( x ) \bigl( U_{n+1} [ x ]+I_{n+1} \bigr) \bigr) _{ij} &= \sum_{k=j}^{i} \binom{i}{k}_{F}E_{i-k,F} x^{i-k} \binom{k}{j}_{F}x^{k-j}+ \binom{i}{j}_{F}E_{i-j,F} x^{i-j} \\ &= \binom{i}{j}_{F}\sum_{k=j}^{i} \binom{i-j}{k-j}_{F}E_{i-k,F} x^{i-j}+ \binom{i}{j}_{F}E_{i-j,F} x^{i-j} \\ &=\binom{i}{j}_{F} x^{i-j} \Biggl[ \sum _{k=0}^{i-j} \binom{i-j}{k}_{F}E_{i-j-k,F}+E_{i-j,F} \Biggr] \\ &=2 \binom{i}{j}_{F} x^{i-j} \delta _{0,i-j} \end{aligned}

for $$i=j$$ $$\binom{i}{j}_{F}x^{i-j} \delta _{0,i-j}=1$$ and for $$i\neq j$$ $$\binom{i}{j}_{F}x^{i-j} \delta _{0,i-j}=0$$. Thus the proof is completed. □

Now, we obtain the Fibo–Bernoulli matrix factorization by using the inverse of the Fibo–Euler polynomial matrix.

### Theorem 10

Let $$\mathcal{B} ( x,F )$$ be $${ ( n+1 ) \times ( n+1 ) }$$ the Fibo–Bernoulli matrix, then we have

$$\mathcal{B} ( x,F ) = \bigl[ 2H_{F} ( x ) -I _{n+1} \bigr] W(F).$$
(31)

### Proof

We have

$$\bigl( \bigl[ 2H_{F} ( x ) -I_{n+1} \bigr] W(F) \bigr) _{ij}=\sum_{k=j}^{i} \biggl( 2 \frac{1}{2}\binom{i}{k}_{F}x ^{i-k}- \delta _{ik} \biggr) \binom{k}{j}_{F}\frac{1}{F_{k-j+1}}$$

for $$j< k< i$$ $$\delta _{ik}=0$$, then we get

\begin{aligned} \bigl( \bigl[ 2H_{F} ( x ) -I_{n+1} \bigr] W(F) \bigr) _{ij} &=\sum_{k=j}^{i} \binom{i}{k}_{F}x^{i-k}\binom{k}{j}_{F} \frac{1}{F_{k-j+1}} \\ &=\binom{i}{j}\sum_{k=j}^{i} \binom{i-j}{k-j}_{F} \frac{1}{F _{k-j+1}}x^{i-k} \\ &=\binom{i}{j}\sum_{k=0}^{i-j} \binom{i-j}{k}_{F}\frac{1}{F _{k+1}}x^{i-j-k} \\ &=\binom{i}{j}_{F}B_{i-j,F} ( x ) \\ &= \bigl[ \mathcal{B} ( x,F ) \bigr] _{ij} \end{aligned}

and

$$\bigl( \bigl[ 2H_{F} ( x ) - \delta \bigr] W(F) \bigr) _{ij}=0$$

for $$i=j=k$$ and $$i< k< j$$. Thus the proof is completed. □

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### Acknowledgements

The authors are grateful to two anonymous referees and the associate editor for their careful reading, helpful comments and constructive suggestions, which improved the presentation of results.

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Each of the authors contributed to each part of this work equally and read and approved the final version of the manuscript.

### Corresponding author

Correspondence to Naim Tuglu.

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