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 [19] 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

([19])

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 [20], 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

([20])

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

([20])

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.

Figure 1

Graphs of $B_{n,F} ( x )$, $B_{n}^{F} ( x )$ and $B_{n} ( x )$ for $n=2,3,4,5$

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

([5])

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

$$U_6[x]=\left[\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ x& 1& 0& 0& 0& 0\\ x^2& x& 1& 0& 0& 0\\ x^3& 2x^2& 2x& 1& 0& 0\\ x^4& 3x^3& 6x^2& 3x& 1& 0\\ x^5& 5x^4& 15x^3& 15x^2& 5x& 1\end{array}\right].$$

Definition 5

([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:

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

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:

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

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

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

Proposition 5

([4])

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:

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

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

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

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:

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

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

([5])

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

([16])

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:

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

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. □