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Bernoulli F-polynomials and Fibo–Bernoulli matrices

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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.

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].

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. □

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

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

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
figure1

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

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]= [ 1 0 0 0 0 0 x 1 0 0 0 0 x 2 x 1 0 0 0 x 3 2 x 2 2 x 1 0 0 x 4 3 x 3 6 x 2 3 x 1 0 x 5 5 x 4 15 x 3 15 x 2 5 x 1 ] .

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)= [ 1 0 0 0 0 0 x 1 0 0 0 0 0 x 1 0 0 0 x 3 0 2 x 1 0 0 x 4 3 x 3 0 3 x 1 0 6 x 5 5 x 4 15 x 3 0 5 x 1 ] .

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:

B(x,F)= [ 1 0 0 0 x + 1 1 0 0 x 2 + x + 1 2 x + 1 1 0 x 3 + 2 x 2 + x + 1 3 2 x 2 + 2 x + 1 2 x + 2 1 ] .

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)= [ 1 0 0 0 0 0 1 1 0 0 0 0 1 2 1 1 0 0 0 1 3 1 2 1 0 0 1 5 1 3 3 1 0 1 8 1 5 15 2 5 1 ] .

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)= [ 1 0 0 0 0 0 1 1 0 0 0 0 1 2 1 1 0 0 0 1 3 1 2 1 0 0 3 10 1 3 3 1 0 5 8 3 2 5 15 2 5 1 ] .

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

U n + 1 [ x ] W ( F ) = [ 1 0 0 0 x 1 0 0 x 2 x 1 0 x 3 2 x 2 2 x 1 ] × [ 1 0 0 0 1 1 0 0 1 2 1 1 0 1 3 1 2 1 ] = [ 1 0 0 0 x + 1 1 0 0 x 2 + x + 1 2 x + 1 1 0 x 3 + 2 x 2 + x + 1 3 2 x 2 + 2 x + 1 2 x + 2 1 ] = B ( x , F ) .

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:

D ( x , F ) = [ 1 0 0 0 0 1 1 0 0 0 1 2 1 1 0 0 1 3 1 2 1 0 3 10 1 3 3 1 ] × [ 1 0 0 0 0 x 1 0 0 0 0 x 1 0 0 x 3 0 2 x 1 0 x 4 3 x 3 0 3 x 1 ] = [ 1 0 0 0 0 x 1 1 0 0 0 x + 1 2 x 1 1 0 0 x 3 x 1 3 2 x + 1 2 x 2 1 0 x 4 3 x 3 + x + 3 10 3 x 3 3 x 1 6 x + 3 3 x 3 1 ] .

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 = [ 1 0 0 0 1 2 1 0 0 1 4 1 2 1 0 1 4 1 2 1 1 ] .

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)= [ 1 0 0 0 0 x 2 1 0 0 0 x 2 4 x 2 1 0 0 x 3 4 x 2 2 x 1 0 5 x 4 8 3 x 3 4 3 x 2 2 3 x 2 1 ] .

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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Correspondence to Naim Tuglu.

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MSC

  • 11B68
  • 11B39
  • 15A60

Keywords

  • Bernoulli polynomials
  • Bernoulli F-polynomials
  • Euler–Fibonacci numbers
  • Bernoulli matrices
  • Generating function