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
$$ U_{n+1} ( x;i,j ) = \textstyle\begin{cases} \binom{i}{j}_{F} x^{i-j} & \text{if }i\geqslant j, \\ 0 & \text{otherwise}. \end{cases} $$
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Example 1
We have
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:
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
$$ b_{ij} ( x,F ) = \textstyle\begin{cases} \binom{i}{j}_{F} B_{i-j,F} ( x ) & \text{if } i\geq j, \\ 0 & \text{otherwise}, \end{cases} $$
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where \(0\leq i,j\leq n\).
For \(n=3\), the Fibo–Bernoulli matrix is as follows:
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:
$$ w_{ij}= \textstyle\begin{cases} \frac{1}{F_{i-j+1}} \binom{i}{j}_{F} & \text{if } i\geqslant j, \\ 0 & \text{otherwise}. \end{cases} $$
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For \(n=5\), the \(W(F)\) matrix is
Proposition 5
([4])
We have
$$ \sum_{k=0}^{n}\binom{n}{k}_{F}B_{n-k}^{F} \frac{1}{F_{k+1}}=F_{n}! \delta _{n,0} . $$
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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
$$ t_{ij}= \textstyle\begin{cases} \binom{i}{j}_{F} B_{i-j}^{F} & \textit{if } i\geqslant j, \\ 0 & \textit{otherwise}. \end{cases} $$
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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:
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
Theorem 7
Let
\(\mathcal{D} ( x,F ) = [ d_{ij} ] \)
be the
\((n+1 ) \times ( n+1 ) \)
matrix defined by
$$ d_{ij}= \textstyle\begin{cases} \binom{i}{j}_{F} \sum_{k=0}^{i-j}\binom{i-j}{k}_{F} B_{i-j-k} ^{F} b_{k+1} x^{k} & \textit{if } i\geqslant j, \\ 0 & \textit{otherwise}. \end{cases} $$
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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:
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:
$$ ( e_{F} ) _{ij}= \textstyle\begin{cases} \binom{i}{j}_{F} E_{i-j,F} & \text{if } i\geqslant j, \\ 0 & \text{otherwise}. \end{cases} $$
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Example 5
For \(n=3\), the Fibo–Euler matrix is
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
$$ \sum_{k=0}^{n}\binom{n}{k}_{F}E_{n-k,F}+E_{n,F}=2 \delta _{0,n}. $$
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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:
$$ ( \varepsilon _{F} ) _{ij}= \textstyle\begin{cases} \binom{i}{j}_{F} E_{i-j,F} x^{i-j} & \text{if }i\geqslant j, \\ 0 & \text{otherwise}. \end{cases} $$
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Example 6
\(5\times 5\) For \(n=4\), the Fibo–Euler polynomial matrix is as follows:
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
$$ \mathcal{B} ( x,F ) = \bigl[ 2H_{F} ( x ) -I _{n+1} \bigr] W(F). $$
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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. □