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\theoremstyle{plain}
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\begin{center}
\vskip 1cm{\LARGE\bf Relating Fibonacci Numbers to Bernoulli \\
\vskip .1in
Numbers via Balancing Polynomials } \\
\vskip 1cm
{\large  Robert Frontczak \\
Landesbank Baden-W{\"u}rttemberg (LBBW) \\ 
Am Hauptbahnhof 2 \\ 
70173 Stuttgart \\ 
Germany \\
\href{mailto:robert.frontczak@lbbw.de}{\tt robert.frontczak@lbbw.de}
}
\end{center}

\vskip .2 in

\begin{abstract}
We present new identities involving Fibonacci and Bernoulli numbers,
and Lucas and Euler numbers, respectively. To achieve this, we derive general relations between
Bernoulli (Euler) polynomials and balancing (Lucas-balancing) polynomials. The derivations make use 
of elementary methods including generating functions and functional equations.
Evaluating these polynomial relations at specific points, we get several new identities for the 
Fibonacci-Bernoulli and Lucas-Euler pairs. We also state some identities involving Bernoulli and balancing numbers.
\end{abstract}

\section{Introduction and Preliminaries}

Bernoulli numbers $(B_n)_{n\geq 0}$ and Fibonacci numbers $(F_n)_{n\geq 0}$ are regarded as the most important and fascinating sequences in mathematics. Relationships between them and between their companion sequences, Euler numbers $(E_n)_{n\geq 0}$ and Lucas numbers $(L_n)_{n\geq 0}$, have been studied by some authors in the past. Zhang and Ma proved \cite{11} a relation between Fibonacci polynomials and Bernoulli numbers. The following identity is a special case of their result:
\begin{equation} \label{zhama1}
\sum_{k=0}^{n} \binom{n}{k} 5^{\frac{n-k}{2}} F_{k} B_{n-k} = n \beta^{n-1},
\end{equation}
where $\beta=(1-\sqrt{5})/2$. Byrd \cite{3} derived a result of similar structure involving Lucas and Euler numbers:
\begin{equation} \label{byrd1}
\sum_{k=0}^{n} \binom{n}{k} \Big (\frac{5}{4}\Big )^{\frac{n-k}{2}} L_{k} E_{n-k} = 2^{1-n}.
\end{equation}
Still other relations are contained in the articles \cite{2, 4, 7, 8}, congruences are studied in \cite{10}. 
Inspired by these elegant results, the present paper is devoted to develop further relations between these famous number sequences. 
To achieve this goal, we study relations between Bernoulli (Euler) polynomials and balancing (Lucas-balancing) polynomials.
The derivations are based on functional equations for the respective generating functions. \\

We begin by introducing some notation. As usual, $B_n(x)$ will be used for the nth Bernoulli polynomial 
and $E_n(x)$ will denote the nth Euler polynomial, where the argument $x$ is assumed to be a complex number. 
These classical polynomials are defined by \cite{1}

\begin{equation}\label{def_b}
H(x,z) = \sum_{n=0}^\infty B_n(x) \frac{z^n}{n!} = \frac{z e^{xz}}{e^z - 1} \qquad (|z|&lt;2\pi) 
\end{equation} 

and

\begin{equation}\label{def_e}
I(x,z) = \sum_{n=0}^\infty E_n(x) \frac{z^n}{n!} = \frac{2e^{xz}}{e^z + 1} \qquad (|z|&lt;\pi). 
\end{equation}
 
The numbers $B_n(0)=B_n$ and $E_n=2^n E_n(1/2)$ are the famous Bernoulli and Euler numbers, respectively. Bernoulli numbers are rational numbers starting with $B_0=1, B_1=-1/2, B_2=1/6, B_4=-1/30$ and so on. Also, $B_{2n+1}=0$ for $n\geq 1$. In contrast, Euler numbers are integers. We have $E_0=1, E_2=-1, E_4=5$ and $E_{2n+1}=0$ for $n\geq 0$. Bernoulli and Euler numbers (polynomials) play a distinguished role in many mathematical branches, such as, number theory, combinatorics and analysis. There is a countless number of articles in which their properties are studied. The basic properties can be found in the textbook \cite{1}, for instance. For more material we refer to a bibliography of Bernoulli numbers (polynomials) \cite{5}. The following properties of Bernoulli and Euler polynomials will be employed in a sequel: The reciprocal relations are
\begin{displaymath}
B_n(1-x) = (-1)^n B_n(x) \quad \mbox{and} \quad E_n(1-x) = (-1)^n E_n(x),
\end{displaymath}
and the difference equations are given by
\begin{displaymath}
n x^{n-1} = (-1)^n B_n(-x) - B_n(x) \,\,\,\,\mbox{and}\,\,\,\, 2 x^{n} = E_n(x) + (-1)^n E_n(-x).
\end{displaymath}
Euler polynomials can be expressed in terms of Bernoulli polynomials via
\begin{displaymath}
E_{n-1}(x) = \frac{2}{n} ( B_n (x) - 2^n B_n(x/2)).
\end{displaymath}

In this article, we choose balancing polynomials $B_n^*(x)$ and Lucas-balancing polynomials $C_n(x)$ to serve as the linking objects between Fibonacci and Bernoulli numbers and Lucas and Euler numbers, respectively. Balancing polynomials are defined by the recurrence (see \cite{9})
\begin{equation}
B_n^*(x) = 6x B_{n-1}^*(x) - B_{n-2}^*(x), \qquad n\geq 2,
\end{equation}
with the initial terms $B_0^*(x)=0$ and $B_1^*(x)=1$. Similarly, Lucas-balancing polynomials are defined by
\begin{equation}
C_n(x) = 6x C_{n-1}(x) - C_{n-2}(x), \qquad n\geq 2,
\end{equation}
with the initial terms $C_0(x)=1$ and $C_1(x)=3x$. The numbers $B_n^*(1)=B_n^*$ and $C_n(1)=C_n$ are called balancing and Lucas-balancing numbers, respectively. It is a routine task to derive the Binet forms for these polynomials. We have
\begin{equation}
B^*_n(x) = \frac{\lambda^n (x) - \lambda^{-n} (x)}{\lambda (x) - \lambda^{-1} (x)}
\quad \mbox{and} \quad 
C_n(x) = \frac{1}{2} \Big ( \lambda^n (x) + \lambda^{-n} (x) \Big ), 
\end{equation}
where $\lambda (x)=3x + \sqrt{9x^2-1}$ and $\lambda^{-1} (x) = 3x - \sqrt{9x^2-1}$. The polynomials are connected to Chebyshev and Legendre polynomials (see \cite{6}). We will exploit two other results about balancing polynomials to prove many of our discoveries. The first result relates balancing polynomials to Fibonacci and Lucas numbers.

\begin{lemma} \label{fundlem1}
Balancing and Lucas-balancing polynomials admit the following evaluations:
\begin{equation}
B_n^*\Big (\frac{L_{2m}}{6} \Big ) = \frac{F_{2mn}}{F_{2m}} \,\, , \qquad C_n\Big (\frac{L_{2m}}{6} \Big ) = \frac{L_{2mn}}{2},
\end{equation}
\begin{equation}
B_n^*\Big (\frac{i}{6} L_{2m+1} \Big ) = i^{n-1} \frac{F_{(2m+1)n}}{F_{2m+1}} \,\, , \qquad C_n\Big (\frac{i}{6}L_{2m+1} \Big ) = i^n \frac{L_{(2m+1)n}}{2},
\end{equation}
where $m$ is an integer, $i=\sqrt{-1}$ is the imaginary unit, and $F_n$ and $L_n$ denote Fibonacci and Lucas numbers, respectively.
\end{lemma}

The second lemma deals with exponential generating functions for balancing and Lucas-balancing polynomials.

\begin{lemma}\label{fundlem2}
Let $F(x,z)$ and $G(x,z)$ be the exponential generating functions of $(B_n^*(x))_{n\geq 0}$ and $(C_n(x))_{n\geq 0}$, respectively. Then
\begin{equation} \label{genf_B}
F(x,z) = \sum_{n=0}^\infty B_n^*(x) \frac{z^n}{n!} = \frac{1}{\sqrt{9x^2-1}}e^{3xz}\sinh (\sqrt{9x^2-1}z),
\end{equation}
and
\begin{equation} \label{genf_C}
G(x,z) = \sum_{n=0}^\infty C_n(x) \frac{z^n}{n!} = e^{3xz}\cosh (\sqrt{9x^2-1}z),
\end{equation}
where $\sinh(\cdot)$ and $\cosh(\cdot)$ are the hyperbolic sine and cosine functions.
\end{lemma} 

\section{Results}

Our first result is the following theorem.

\begin{theorem} \label{thm1}
Let $n\geq 1$. Then
\begin{equation} \label{main1}
\sum_{\substack{k=0 \\ n \equiv k \,(\mathrm {mod} \,2)}}^n \binom{n}{k} B^*_{k}(x) (2\sqrt{9x^2-1})^{n-k} B_{n-k} = n C_{n-1}(x),
\end{equation}
and
\begin{equation} \label{main2}
\sum_{\substack{k=0 \\ n \equiv k \,(\mathrm {mod} \,2)}}^n \binom{n}{k} C_{k}(x) (2\sqrt{9x^2-1})^{n-k} (2^{n-k}-1) B_{n-k} 
= n (9x^2-1) B^*_{n-1}(x).
\end{equation}
\end{theorem}

\begin{proof}
By (\ref{genf_B}) and (\ref{genf_C}) we have
\begin{displaymath}
z G(x,z) = \sqrt{9x^2-1} z \coth (\sqrt{9x^2-1}z) F(x,z).
\end{displaymath}
The left hand side is
\begin{displaymath}
z G(x,z) = \sum_{n=1}^\infty C_{n-1}(x) n \frac{z^n}{n!}. 
\end{displaymath}
Next, form the power series expansion
\begin{displaymath}
\coth(z) = \frac{1}{z} + \sum_{n=1}^\infty \frac{2^{2n}}{(2n)!} B_{2n} z^{2n-1},
\end{displaymath}
we immediately get
\begin{displaymath}
\sqrt{9x^2-1} z \coth (\sqrt{9x^2-1}z) = \sum_{n=0}^\infty (2\sqrt{9x^2-1})^{2n} B_{2n} \frac{z^{2n}}{(2n)!}.
\end{displaymath}
Now, the relation follows from multiplying the series on the right using Cauchy's rule and comparing coefficients. 
The second statement is proved analogously utilizing
\begin{displaymath}
(9x^2-1) z F(x,z) = \sqrt{9x^2-1} z \tanh (\sqrt{9x^2-1}z) G(x,z),
\end{displaymath}
in combination with
\begin{displaymath}
\tanh(z) = \sum_{n=0}^\infty \frac{2^{2n}(2^{2n}-1)}{(2n)!}B_{2n} z^{2n-1}.
\end{displaymath}
\end{proof}

\begin{corollary}
Let $n\geq 1$ be an integer. Then we have the following relations between balancing numbers and Bernoulli numbers:  
\begin{equation}
\sum_{\substack{k=0 \\ n \equiv k  \,(\mathrm {mod} \,2)}}^n \binom{n}{k} 32^{\frac{n-k}{2}} B^*_{k} B_{n-k} = n C_{n-1},
\end{equation}
and
\begin{equation}
\sum_{\substack{k=0 \\ n \equiv k  \,(\mathrm {mod} \,2)}}^n \binom{n}{k} 32^{\frac{n-k}{2}} (2^{n-k}-1) C_{k}  B_{n-k} = 8n B^*_{n-1}.
\end{equation}
Also, for each $m\geq 0$ (or $\geq 1$), the following identities are valid:
\begin{equation}
\sum_{\substack{k=0 \\ n \equiv k  \,(\mathrm {mod} \,2)}}^n \binom{n}{k} 5^{\frac{n-k}{2}} F_{(2m+1)k} F_{2m+1}^{n-1-k} B_{n-k} 
= \frac{n}{2} L_{(2m+1)(n-1)},
\end{equation}
\begin{equation}
\sum_{\substack{k=0 \\ n \equiv k  \,(\mathrm {mod} \,2)}}^n \binom{n}{k} 5^{\frac{n-k}{2}} (2^{n-k}-1) L_{(2m+1)k} F_{2m+1}^{n-1-k} B_{n-k} 
= \frac{5n}{2} F_{(2m+1)(n-1)},
\end{equation}
\begin{equation}
\sum_{\substack{k=0 \\ n \equiv k  \,(\mathrm {mod} \,2)}}^n \binom{n}{k} 5^{\frac{n-k}{2}} F_{2mk} F_{2m}^{n-1-k} B_{n-k} 
= \frac{n}{2} L_{2m(n-1)},
\end{equation}
and
\begin{equation}
\sum_{\substack{k=0 \\ n \equiv k  \,(\mathrm {mod} \,2)}}^n \binom{n}{k} 5^{\frac{n-k}{2}} (2^{n-k}-1) L_{2mk} F_{2m}^{n-1-k} B_{n-k} 
= \frac{5n}{2} F_{2m(n-1)}.
\end{equation}
\end{corollary}
\begin{proof}
Evaluate (\ref{main1}) and (\ref{main2}) at the points $x=1$, $x=i/6 L_{2m+1}$ and $x=1/6 L_{2m}$, respectively and use 
Lemma \ref{fundlem1}. To simplify the square root recall that $L_n^2=5F_n^2+(-1)^{n}4$. 
\end{proof}

Special cases of the corollary are the following sums which should be compared with (\ref{zhama1}):
\begin{equation}
\sum_{\substack{k=0 \\ n \equiv k  \,(\mathrm {mod} \,2)}}^n \binom{n}{k} 5^{\frac{n-k}{2}} F_{k} B_{n-k} = \frac{n}{2} L_{n-1},
\end{equation}
\begin{equation}
\sum_{\substack{k=0 \\ n \equiv k  \,(\mathrm {mod} \,2)}}^n \binom{n}{k} 5^{\frac{n-k}{2}} (2^{n-k}-1) L_{k} B_{n-k} = \frac{5n}{2} F_{n-1},
\end{equation}
\begin{equation}
\sum_{\substack{k=0 \\ n \equiv k  \,(\mathrm {mod} \,2)}}^n \binom{n}{k} 5^{\frac{n-k}{2}} F_{2k} B_{n-k} = \frac{n}{2} L_{2n-2},
\end{equation}
and
\begin{equation}
\sum_{\substack{k=0 \\ n \equiv k  \,(\mathrm {mod} \,2)}}^n \binom{n}{k} 5^{\frac{n-k}{2}} (2^{n-k}-1) L_{2k} B_{n-k} = \frac{5n}{2} F_{2n-2}.
\end{equation}

The next theorem deals with sums of weighted products of balancing and Bernoulli polynomials and Lucas-balancing and Euler polynomials, respectively. 

\begin{theorem} \label{thm2}
Let $n\geq 1$. Then
\begin{equation} \label{main3}
\sum_{k=0}^n \binom{n}{k} (2\sqrt{9x^2-1})^{n-k} B^*_{k}(x) B_{n-k}(x) = n \Big (3x + (2x-1)\sqrt{9x^2-1}\Big )^{n-1}. 
\end{equation}
Also, for $n\geq 0$, we have the identity
\begin{equation} \label{main4}
\sum_{k=0}^n \binom{n}{k} (2\sqrt{9x^2-1})^{n-k} C_{k}(x) E_{n-k}(x) = \Big (3x + (2x-1)\sqrt{9x^2-1}\Big )^{n}. 
\end{equation}
\end{theorem}
\begin{proof}
Since
\begin{displaymath}
H(x,z) = \frac{z e^{(x-1/2)z}}{e^{z/2} - e^{-z/2}} = \frac{z}{2} \frac{e^{(x-1/2)z}}{\sinh(z/2)},
\end{displaymath}
we have the functional relation
\begin{displaymath}
F(x,z)H(x,2\sqrt{9x^2-1}z) = z e^{(3x + (2x-1)\sqrt{9x^2-1})z}.
\end{displaymath}
Expanding both sides in form of power series and comparing the coefficients of $z^n$ yields the first formula. 
The second formula corresponds to the similar functional relation
\begin{displaymath}
G(x,z)I(x,2\sqrt{9x^2-1}z) = e^{(3x + (2x-1)\sqrt{9x^2-1})z}.
\end{displaymath}
\end{proof}

Theorem \ref{thm2} also contains some interesting sums as special cases. 
In order to give a more detailed exposition, we state each example as a separate corollary.
\begin{corollary}
For $n\geq 1$
\begin{equation}
\sum_{k=0}^{\left \lfloor{\frac{n-1}{2}}\right \rfloor} \binom{n}{2k+1} 2^{n-2k-1}B_{n-2k-1} = n (-1)^{n-1},
\end{equation}
and for $n\geq 0$
\begin{equation}
\sum_{k=0}^{\left \lfloor{\frac{n}{2}}\right \rfloor} \binom{n}{2k} \frac{B_{n-2k+1}}{n-2k+1} 2^{n-2k+1} (1-2^{n-2k+1})= (-1)^{n}.
\end{equation}
\end{corollary} 
\begin{proof}
Both expressions follow from (\ref{main3}) and (\ref{main4}) evaluated at $x=0$ in combination with 
\begin{displaymath}
B^*_n(0) = \left.
  \begin{cases}
    0, &amp; \mbox{$n$ even}; \\
    (-1)^{\frac{n-1}{2}} &amp; \mbox{$n$ odd};
  \end{cases}
  \right. \quad \mbox{and} \quad
C_n(0) =  \left.
	\begin{cases}
		(-1)^{\frac{n}{2}}, &amp; \mbox{$n$ even}; \\
		 0 &amp; \mbox{$n$ odd.}
	\end{cases} 
	\right.		
\end{displaymath}
To complete the proof of the second expression we have used 
\begin{displaymath}
E_n(0) = \frac{2}{n+1} B_{n+1} (1-2^{n+1}).
\end{displaymath}
\end{proof}

\begin{corollary}
Let $m\geq 1$. Then, for each $n\geq 1$
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} 5^{\frac{n-k}{2}} F_{2mk} F_{2m}^{n-k-1} B_{n-k}\Big (\frac{L_{2m}}{6} \Big ) = n \Big ( \frac{3L_{2m} + \sqrt{5}F_{2m}(L_{2m}-3)}{6}\Big )^{n-1},
\end{equation}
and for $n\geq 0$
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} 5^{\frac{n-k}{2}} L_{2mk} F_{2m}^{n-k} E_{n-k}\Big (\frac{L_{2m}}{6} \Big ) = 2 \Big ( \frac{3L_{2m} + \sqrt{5}F_{2m}(L_{2m}-3)}{6}\Big )^{n}.
\end{equation}
\end{corollary} 
\begin{proof}
Evaluate (\ref{main3}) and (\ref{main4}) at $x=L_{2m}/6.$
\end{proof}

It is worth to remark that the following two results are contained in the corollary as special cases for $m=1$:
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} \Big (\frac{5}{4}\Big )^{\frac{n-k}{2}} \Big (1 - 2^{n-k-1}\Big ) F_{2k} B_{n-k} = 
\frac{n}{2} \Big (\frac{3}{2}\Big )^{n-1},
\end{equation}
and 
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} \Big (\frac{5}{4}\Big )^{\frac{n-k}{2}} L_{2k} E_{n-k} = 2 \Big (\frac{3}{2}\Big )^{n}.
\end{equation}
Compare with (\ref{zhama1}) and (\ref{byrd1}).
Also, using the relations $B^*_n(-x)=(-1)^{n+1}B^*_n(x)$ and $C_n(-x)=(-1)^{n}C_n(x)$ in combination with the 
difference relations for $B_n(x)$ and $E_n(x)$, explicit evaluations at negative points can be derived easily. 
For instance, setting $x=-1/2$ in (\ref{main3}) and (\ref{main4}) leads to
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} \Big (\frac{5}{4}\Big )^{\frac{n-k}{2}} F_{2k} \Big ( \Big (1 - 2^{n-k-1}\Big ) B_{n-k} + n-k \Big ) 
= \frac{n}{2} \Big (\alpha^2 + \frac{\sqrt{5}}{2}\Big )^{n-1},
\end{equation}
and 
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} \Big (\frac{5}{4}\Big )^{\frac{n-k}{2}} L_{2k} \Big ( 1 - \frac{E_{n-k}}{2}\Big ) 
= \Big (\alpha^2 + \frac{\sqrt{5}}{2}\Big )^{n},
\end{equation}
where $\alpha = (1+\sqrt{5})/2$ is the golden ratio.

\begin{corollary}
Let $m\geq 0$ and $i=\sqrt{-1}$. Then, for each $n\geq 1$
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} 5^{\frac{n-k}{2}} F_{(2m+1)k} F_{2m+1}^{n-k-1} B_{n-k}\Big (i \frac{L_{2m+1}}{6} \Big ) = n \Big ( \frac{3L_{2m+1} + \sqrt{5}F_{2m+1}(i L_{2m+1}-3)}{6}\Big )^{n-1},
\end{equation}
and for $n\geq 0$
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} 5^{\frac{n-k}{2}} L_{(2m+1)k} F_{2m+1}^{n-k} E_{n-k}\Big (i \frac{L_{2m+1}}{6} \Big ) = 2 \Big ( \frac{3L_{2m+1} + \sqrt{5}F_{2m+1}(i L_{2m+1}-3)}{6}\Big )^{n}.
\end{equation}
\end{corollary} 
\begin{proof}
Evaluate (\ref{main3}) and (\ref{main4}) at $x=i L_{2m+1}/6.$
\end{proof}

When $m=0$, then
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} 5^{\frac{n-k}{2}} F_{k} B_{n-k}\Big (\frac{i}{6}\Big ) = n \Big ( \beta + \frac{\sqrt{5}i}{6} \Big )^{n-1},
\end{equation}
and 
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} 5^{\frac{n-k}{2}} L_{k} E_{n-k}\Big (\frac{i}{6}\Big ) = 2 \Big ( \beta + \frac{\sqrt{5}i}{6} \Big )^{n},
\end{equation}
where $\beta=(1-\sqrt{5})/2=-1/\alpha$. \\

Indeed, we can prove more for these special sums.

\begin{theorem}\label{thm3}
For $n\geq 1$ we have
\begin{equation}\label{main5}
\sum_{k=0}^{n} \binom{n}{k} 5^{\frac{n-k}{2}} F_{k} B_{n-k}(x) = n \Big (\beta + \sqrt{5}x \Big )^{n-1},
\end{equation}
and for $n\geq 0$ the analogue relation for the Lucas-Euler pair is
\begin{equation}\label{main6}
\sum_{k=0}^{n} \binom{n}{k} 5^{\frac{n-k}{2}} L_{k} E_{n-k}(x) = 2 \Big (\beta + \sqrt{5}x \Big )^{n}.
\end{equation}
\end{theorem}
\begin{proof}
We start with
\begin{displaymath}
H(x,\sqrt{5} z) = \frac{\sqrt{5} z}{2} \frac{e^{(x-1/2)\sqrt{5} z}}{\sinh(\sqrt{5}z/2)}
\end{displaymath}
and
\begin{displaymath}
\sum_{n=0}^\infty F_n \frac{z^n}{n!} = \frac{2}{\sqrt{5}} e^{z/2} \sinh(\sqrt{5}z/2).
\end{displaymath}
Hence, 
\begin{displaymath}
\Big (\sum_{n=0}^\infty F_n \frac{z^n}{n!}\Big ) \Big (\sum_{n=0}^\infty B_n(x) 5^{n/2} \frac{z^n}{n!}\Big ) 
= z e^{z/2} e^{(x-1/2)\sqrt{5} z} = z e^{(\beta + \sqrt{5}x)z},
\end{displaymath}
and the proof of the first identity is completed. To prove the second identity we use
\begin{displaymath}
I(x,\sqrt{5} z) = \frac{e^{(x-1/2)\sqrt{5} z}}{\cosh(\sqrt{5}z/2)}
\end{displaymath}
and
\begin{displaymath}
\sum_{n=0}^\infty L_n \frac{z^n}{n!} = 2 e^{z/2} \cosh(\sqrt{5}z/2).
\end{displaymath}
\end{proof}
If we set $x=0$ in (\ref{main5}), then we recover (\ref{zhama1}). Setting $x=0$ in (\ref{main6}) gives
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} 5^{\frac{n-k}{2}} \frac{1-2^{n-k+1}}{n-k+1} L_{k} B_{n-k+1} = \beta^{n}.
\end{equation}
Similarly, setting $x=1/2$ in (\ref{main5}) results in 
\begin{equation} \label{fb1x}
\sum_{k=0}^{n} \binom{n}{k} 5^{\frac{n-k}{2}} \big ( 2^{1-(n-k)} - 1 \big ) F_{k} B_{n-k} = n 2^{1-n},
\end{equation}
whereas setting $x=1/2$ in (\ref{main6}) gives (\ref{byrd1}). Also, using the reciprocal relations for $B_n(x)$ and $E_n(x)$ 
and $\alpha-\beta = \sqrt{5}$, we immediately get the alternating version of Theorem \ref{thm3}:

\begin{corollary}
The alternating variants for the Fibonacci-Bernoulli and Lucas-Euler pairs equal
\begin{equation}\label{altmain5}
\sum_{k=0}^{n} \binom{n}{k} 5^{\frac{n-k}{2}} (-1)^{n-k} F_{k} B_{n-k}(x) = n \Big (\alpha - \sqrt{5}x \Big )^{n-1},
\end{equation}
and
\begin{equation}\label{altmain6}
\sum_{k=0}^{n} \binom{n}{k} 5^{\frac{n-k}{2}} (-1)^{n-k} L_{k} E_{n-k}(x) = 2 \Big (\alpha - \sqrt{5}x \Big )^{n}.
\end{equation}
\end{corollary}

Now, the value $x=0$ produces
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} 5^{\frac{n-k}{2}} (-1)^{n-k} F_{k} B_{n-k} = n \alpha ^{n-1},
\end{equation}
which is the alternating version of (\ref{zhama1}) and
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} 5^{\frac{n-k}{2}} (-1)^{n-k} \frac{1-2^{n-k+1}}{n-k+1} L_{k} B_{n-k+1} = \alpha^{n}.
\end{equation}

\begin{corollary}
Let $n$ and $q$ be two integers with $n\geq 1$ and $q\geq 2$. Then it holds that
\begin{eqnarray} \label{fbcor1}
&amp; &amp; \sum_{k=0}^{n} \binom{n}{k} 5^{\frac{n-k}{2}} \big ( q^{1-(n-k)} - 1 \big ) F_{k} B_{n-k} \nonumber \\
&amp; = &amp; \sum_{k=0}^{n} \binom{n}{k} 5^{\frac{n-k}{2}} (-1)^{n-k} \big ( q^{1-(n-k)} - 1 \big ) F_{k} B_{n-k} \nonumber \\
&amp; = &amp; n q^{1-n} \sum_{r=1}^{q-1} \big ( \beta q + \sqrt{5} r \big )^{n-1} 
= n q^{1-n} \sum_{r=1}^{q-1} \big ( \alpha q - \sqrt{5} r \big )^{n-1}.
\end{eqnarray}
\end{corollary}
\begin{proof}
The statement follows from (\ref{main5}) and (\ref{altmain5}) combined with Raabe's formula
\begin{displaymath}
\frac{1}{q} \sum_{r=0}^{q-1} B_n \Big (x + \frac{r}{q}\Big ) = q^{-n} B_n(qx).
\end{displaymath}
\end{proof}
When $q=2$, then (\ref{fbcor1}) gives (\ref{fb1x}). When $q=3$, then we obtain
\begin{eqnarray}
&amp; &amp; \sum_{k=0}^{n} \binom{n}{k} 5^{\frac{n-k}{2}} \big ( 3^{1-(n-k)} - 1 \big ) F_{k} B_{n-k} \nonumber \\
&amp; = &amp; \sum_{k=0}^{n} \binom{n}{k} 5^{\frac{n-k}{2}} (-1)^{n-k} \big ( 3^{1-(n-k)} - 1 \big ) F_{k} B_{n-k} = n 3^{1-n} L_{2n-2}.
\end{eqnarray}

\begin{theorem}\label{thm4}
The following relation between even-indexed Fibonacci numbers and Bernoulli polynomials holds for $n\geq 1$:
\begin{equation}\label{main7}
\sum_{k=0}^{n} \binom{2n}{2k} 5^{n-k} F_{2k} B_{2(n-k)}(x) = 2^{2(1-n)} \sum_{k=0}^n \binom{2n}{2k} k 5^{n-k} (2x-1)^{2(n-k)}.
\end{equation}
For $n\geq 0$, the analogue relation involving even-indexed Lucas numbers and Euler polynomials is
\begin{equation}\label{main8}
\sum_{k=0}^{n} \binom{2n}{2k} 5^{n-k} L_{2k} E_{2(n-k)}(x) = 2^{1-2n} \sum_{k=0}^n \binom{2n}{2k} 5^{n-k} (2x-1)^{2(n-k)}.
\end{equation}
\end{theorem}
\begin{proof}
We start with (\ref{main7}). By (\ref{def_b}) we have
\begin{displaymath}
\sum_{n=0}^{\infty} B_{2n}(x) \frac{z^{2n}}{(2n)!} = \frac{1}{2} ( H(x,z) + H(x,-z) ) = \frac{z}{2} \frac{\cosh((x-1/2)z)}{\sinh(z/2)}.
\end{displaymath}
Also, it is easy to show that
\begin{displaymath}
\sum_{n=0}^\infty F_{2n} \frac{z^{2n}}{(2n)!} = \frac{2}{\sqrt{5}} \sinh(\sqrt{5}z/2) \sinh(z/2).
\end{displaymath}
Thus, applying Cauchy's rule, we arrive at
\begin{displaymath}
\sum_{n=0}^\infty \sum_{k=0}^n \binom{2n}{2k} 5^{n-k} F_{2k} B_{2(n-k)}(x) \frac{z^{2n}}{(2n)!} = z \sinh(z/2) \cosh((2x-1)\sqrt{5} z/2).
\end{displaymath}
To finish the proof of (\ref{main7}) we must apply Cauchy's rule a second time and keep in mind that
\begin{displaymath}
z \sinh(z/2) = \sum_{n=0}^\infty (2n) 2^{1-2n} \frac{z^{2n}}{(2n)!},
\end{displaymath} 
and
\begin{displaymath}
\cosh((2x-1)\sqrt{5} z/2) = \sum_{n=0}^\infty 5^{n} 2^{-2n} (2x-1)^{2n} \frac{z^{2n}}{(2n)!}.
\end{displaymath} 
The proof of (\ref{main8}) uses similar arguments. The key components are
\begin{displaymath}
\sum_{n=0}^{\infty} E_{2n}(x) \frac{z^{2n}}{(2n)!} = \frac{\cosh((x-1/2)z)}{\cosh(z/2)},
\end{displaymath}
which follows from (\ref{def_e}) and
\begin{displaymath}
\sum_{n=0}^\infty L_{2n} \frac{z^{2n}}{(2n)!} = 2 \cosh(\sqrt{5}z/2) \cosh(z/2).
\end{displaymath}
\end{proof}

A slight modification in the above proof leads to the next theorem.

\begin{theorem} \label{thm5}

\begin{equation} \label{main9}
\sum_{k=0}^{n} \binom{2n}{2k} F_{2k} B_{2(n-k)}(x) = 2^{2(1-n)} \sum_{k=0}^n \binom{2n}{2k} k 5^{k-1} (2x-1)^{2(n-k)},
\end{equation}
and
\begin{equation} \label{main10}
\sum_{k=0}^{n} \binom{2n}{2k} L_{2k} E_{2(n-k)}(x) = 2^{1-2n} \sum_{k=0}^n \binom{2n}{2k} 5^{k} (2x-1)^{2(n-k)}.
\end{equation}
\end{theorem}

\begin{corollary}
For each $n\geq 1$ we have the following relations:
\begin{equation}\label{thm4c1}
\sum_{k=0}^{n} \binom{2n}{2k} 5^{n-k} F_{2k} B_{2(n-k)} = 2^{2(1-n)} \sum_{k=0}^n \binom{2n}{2k} k 5^{n-k},
\end{equation}
\begin{equation} \label{thm4c2}
\sum_{k=0}^{n} \binom{2n}{2k} 5^{n-k} \big ( 2^{1-2(n-k)} - 1 \big ) F_{2k} B_{2(n-k)} = n 2^{2(1-n)},
\end{equation}
\begin{equation} \label{thm4c3}
\sum_{k=0}^{n} \binom{2n}{2k} 5^{n-k} \big ( 3^{1-2(n-k)} - 1 \big ) F_{2k} B_{2(n-k)} = 
2^{2(1-n)+1} \sum_{k=0}^n \binom{2n}{2k} k \Big (\frac{5}{9}\Big )^{n-k},
\end{equation}
\begin{equation}\label{thm5c1}
\sum_{k=0}^{n} \binom{2n}{2k} F_{2k} B_{2(n-k)} = 2^{2(1-n)} \sum_{k=0}^n \binom{2n}{2k} k 5^{k-1},
\end{equation}
\begin{equation} \label{thm5c2}
\sum_{k=0}^{n} \binom{2n}{2k} \big ( 2^{1-2(n-k)} - 1 \big ) F_{2k} B_{2(n-k)} = n 5^{n-1} 2^{2(1-n)},
\end{equation}
and
\begin{equation} \label{thm5c3}
\sum_{k=0}^{n} \binom{2n}{2k} \big ( 3^{1-2(n-k)} - 1 \big ) F_{2k} B_{2(n-k)} = 
\frac{8}{5} 6^{-2n} \sum_{k=0}^n \binom{2n}{2k} k 45^{k}.
\end{equation}

In addition, we have the identities
\begin{equation} \label{thm4c4}
L_{2n} = 2^{1-2n} \sum_{k=0}^n \binom{2n}{2k} 5^{n-k},
\end{equation}
\begin{equation} \label{thm4c5}
\sum_{k=0}^{n} \binom{2n}{2k} \Big ( \frac{5}{4}\Big )^{n-k} L_{2k} E_{2(n-k)} = 2^{1-2n},
\end{equation}
and
\begin{equation} \label{thm4c6}
\sum_{k=0}^{n} \binom{2n}{2k} 2^{-2(n-k)} L_{2k} E_{2(n-k)} = 5 ^{n} 2^{1-2n}.
\end{equation}
\end{corollary}
\begin{proof}
Equations (\ref{thm4c1}), (\ref{thm4c2}) and (\ref{thm4c3}) follow from (\ref{main7}) for $x=0$ (or $x=1$), $x=1/2$ and for $x=1/3$, respectively, where we have also used $B_{2n}(1/3)=2^{-1}(3^{1-2n}-1)B_{2n}$. To get equations (\ref{thm5c1}), (\ref{thm5c2}) 
and (\ref{thm5c3}) set $x=0$ (or $x=1$), $x=1/2$ and for $x=1/3$ in (\ref{main9}).
Identity (\ref{thm4c4}) is (\ref{main8}) (or (\ref{main10})) evaluated at $x=0$ (or $x=1$).
Identities (\ref{thm4c5}) and (\ref{thm4c6}) are (\ref{main8}) and (\ref{main10}) evaluated at $x=1/2$, respectively. 
\end{proof}

Interestingly, it is known that
\begin{displaymath}
\sum_{k=0}^{n} \binom{2n}{2k} = 2^{2n-1}.
\end{displaymath}
This leads to somewhat curious versions of the previous results. For instance, $L_{2n}$ can be expressed as
\begin{equation} 
L_{2n} = \frac{ \sum_{k=0}^n \binom{2n}{2k} 5^{k} }{\sum_{k=0}^{n} \binom{2n}{2k}}.
\end{equation}

We conclude this study with the following theorem.

\begin{theorem} \label{thm6}
Let $n\geq 1$. Then
\begin{equation} \label{main11}
\sum_{k=1}^n \binom{n}{k} \frac{B_{k}^*(x) B_{n-k}(x)}{(2\sqrt{9x^2-1})^{k}}  = \sum_{k=1}^{n} \binom{n}{k} k \frac{C_{k-1}(x) E_{n-k}(x)}{(2\sqrt{9x^2-1})^{k}}. 
\end{equation}
\end{theorem}
\begin{proof}
From (\ref{def_b}) and (\ref{def_e}) we have
\begin{displaymath}
\frac{H(x,z)}{I(x,z)} = \frac{z}{2} \frac{e^{z/2} + e^{-z/2}}{e^{z/2} - e^{-z/2}} = \frac{z}{2} \coth (z/2).
\end{displaymath}
This gives the functional relation
\begin{displaymath}
F(x,z) H(x,2\sqrt{9x^2-1}z) = z G(x,z) I(x,2\sqrt{9x^2-1}z).
\end{displaymath}
Expanding both sides in form of power series and comparing the coefficients of $z^n$ yields
\begin{displaymath}
\sum_{k=0}^n \binom{n}{k} \frac{B^*_{k}(x) B_{n-k}(x)}{(2\sqrt{9x^2-1})^{k}}  = n \sum_{k=0}^{n-1} \binom{n-1}{k} \frac{C_{k}(x) E_{n-1-k}(x)}{(2\sqrt{9x^2-1})^{k+1}}. 
\end{displaymath}
The proof is completed noting that $B^*_0(x)=0$ and
\begin{displaymath}
n \binom{n-1}{k} = (k+1) \binom{n}{k+1}.
\end{displaymath}
\end{proof}

\begin{corollary}
The following identities are immediate consequences of Theorem \ref{thm6}:
\begin{equation}
\sum_{k=1}^n \binom{n}{k} (-1)^{n-k} 32^{-\frac{k}{2}} B_k^* B_{n-k}  = 
\sum_{k=1}^{n} \binom{n}{k} (-1)^{n-k} 32^{-\frac{k}{2}} \frac{2k}{n-k+1} \big (1- 2^{n-k+1} \big ) C_{k-1} B_{n-k+1},
\end{equation}
\begin{equation}
\sum_{k=1}^n \binom{n}{k} 5^{-\frac{k}{2}} \big ( 2^{1-(n-k)} - 1 \big ) F_{2k} B_{n-k}  = 
\sum_{k=1}^{n} \binom{n}{k} k 5^{-\frac{k}{2}} 2^{-(n-k+1)} L_{2k-2} E_{n-k},
\end{equation}
and more generally
\begin{equation}
\sum_{k=1}^n \binom{n}{k} 5^{-\frac{k}{2}} F_{2mk} F^{-(k+1)}_{2m} B_{n-k}\Big ( \frac{L_{2m}}{6}\Big )  = 
\sum_{k=1}^{n} \binom{n}{k} \frac{k}{2} 5^{-\frac{k}{2}} F^{-k}_{2m} L_{2m(k-1)} E_{n-k} \Big ( \frac{L_{2m}}{6}\Big ),
\end{equation}
and
\begin{equation}
\sum_{k=1}^n \binom{n}{k} 5^{-\frac{k}{2}} F_{(2m+1)k} F^{-(k+1)}_{2m+1} B_{n-k}\Big ( i \frac{L_{2m+1}}{6}\Big )  = 
\sum_{k=1}^{n} \binom{n}{k} \frac{k}{2} 5^{-\frac{k}{2}} F^{-k}_{2m+1} L_{(2m+1)(k-1)} E_{n-k} \Big ( i \frac{L_{2m+1}}{6}\Big ).
\end{equation}
\end{corollary} 


\section{Acknowledgments}

The author wishes to thank Professor Harris Kwong from State University of New York at Fredonia for his detailed comments
on the first draft of this manuscript. Moreover, the author is grateful to the anonymous referee and the editor-in-chief 
for constructive criticism and helpful comments and suggestions on the manuscript.


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D. Castellanos, A generalization of Binet's formula and some of its
consequences, {\it Fibonacci Quart.} {\bf 27} (1989), 424--438.

\bibitem{5} 
K. Dilcher and I. Sh. Slavutskii, {\it A Bibliography of Bernoulli
Numbers}, \url{https://www.mscs.dal.ca/~dilcher/bernoulli.html}.

\bibitem{6}
R. Frontczak, On balancing polynomials, {\it Appl. Math. Sci.} {\bf 13}
(2019), 57--66.

\bibitem{7}
G. Ozdemir and Y. Simsek, Identities and relations associated
with Lucas and some special sequences, {\it AIP Conference
Proceedings} {\bf 1863}, 300003 (2017).  Available at
\url{https://aip.scitation.org/doi/abs/10.1063/1.4992452}.

\bibitem{8}
G. Ozdemir, Y. Sim\c{s}ek, and G. V. Milovanovi\'{c}, Generating functions for
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\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B37, 11B65; Secondary 05A15.

\noindent \emph{Keywords: }
Bernoulli polynomial, Bernoulli
number, Euler polynomial, Euler number, balancing polynomial,
balancing number,
Fibonacci number, Lucas number, generating function.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000045},
\seqnum{A000032},
\seqnum{A100615},
\seqnum{A122045},
\seqnum{A001109}, and 
\seqnum{A001541}.) 

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received January 15 2019; revised version received July 10 2019.
Published in {\it Journal of Integer Sequences}, August 22 2019.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in

\end{document}

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