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\begin{center}
\vskip 1cm{\LARGE\bf 
Some Formulas for a Family of Numbers \\
\vskip .02in
Analogous to the Higher-Order Bernoulli \\
\vskip .12in
Numbers
}
\vskip 1cm
{\large}
Guo-Dong Liu \\ 
Department of Mathematics \\
Huizhou University \\
Huizhou 516007, Guangdong \\
People's Republic of China\\
\href{mailto:gdliu@pub.huizhou.gd.cn}{\tt gdliu@pub.huizhou.gd.cn} \\
\ \\
H. M. Srivastava\footnote{Corresponding author.}\\
Department of Mathematics and Statistics \\
University of Victoria \\
Victoria, BC V8W 3R4 \\
Canada\\
\href{mailto:harimsri@math.uvic.ca}{\tt harimsri@math.uvic.ca} \\
\ \\
Hai-Qing Wang \\
Department of Mathematics \\
Huizhou University\\
Huizhou 516007, Guangdong \\
People's Republic of China\\ 
\href{mailto:whq@hzu.edu.cn}{\tt whq@hzu.edu.cn} \\
\end{center}

\vskip .2 in

\begin{abstract}
In this paper the authors establish
several formulas and results for the $D$ numbers $D_{2n}^{(k)}$ and
$d_{2n}^{(k)}$, which are analogous to the higher-order Bernoulli numbers. Some applications of these families of $D$ numbers are also presented.
\end{abstract}


\section{Introduction}

The main purpose of this paper is to prove several formulas and results
for the $D$ numbers $D_{2n}^{(k)}$ and $d_{2n}^{(k)}$, which are (in a
sense) analogous to the higher-order Bernoulli numbers. We also discuss
some applications of the various results which are presented here for
these $D$ numbers. With a view to making our presentation as much
self-contained as possible, and for the convenience of (and ready
reference by) the interested reader, we have closely followed and
chosen to freely reproduce here some basic definitions and preliminary
results from such recent publications as (for example) \cite{8} (and
indeed also from the book \cite{9}). We thus begin this paper by
introducing the Bernoulli polynomials $B_n^{(k)}(x)$ of order $k$ and
degree $n$, which may be defined (for any integer $k$) by means of the
following generating function (see, for example,
\cite{2,4,5,9,12,13,15}):
\begin{equation}\label{1.1}
\left(\frac{t}{e^t-1}\right)^k
e^{xt}=\sum_{n=0}^{\infty}B_n^{(k)}(x)\frac{t^n}{n!}\qquad
(|t|&lt;2\pi;\; k\in \mathbb{Z}),
\end{equation}
where $\mathbb{Z}$ denotes the set of integers.
In partucular, the numbers $B_n^{(k)}=B_n^{(k)}(0)$ are the Bernoulli numbers of
order $k$ and the numbers $B_n^{(1)}=B_n$ are referred to as the ordinary Bernoulli numbers. By using the generating function \eqref{1.1}, we can get 
\begin{equation*}
\frac{{\rm d}}{{\rm d}x}B_n^{(k)}(x)=nB_{n-1}^{(k)}(x),
\end{equation*}
\begin{equation*}
B_n^{(k+1)}(x)=\frac{k-n}{k}B_n^{(k)}(x)+(x-k)\frac{n}{k}B_{n-1}^{(k)}(x)
\end{equation*}
and
\begin{equation*}
B_n^{(k+1)}(x+1)=\frac{nx}{k}B_{n-1}^{(k)}(x)-\frac{n-k}{k}B_n^{(k)}(x),
\end{equation*}
where 
$$n\in \mathbb{N}\qquad (\mathbb{N}:=\{1,2,3,\cdots\}
=\mathbb{N}_0\setminus\{0\}).$$ 
Specifically, the numbers $B_n^{(n)}$ are called the N\"{o}rlund numbers (see
\cite{1,3,4}). A generating function for the numbers $B_n^{(n-k)}$ is given by (see \cite{9})
\begin{equation*}
\frac{1}{1+t}\left(\frac{t}{\log(1+t)}\right)^{k+1}=\sum_{n=0}^\infty
B_n^{(n-k)}\frac{t^n}{n!}.
\end{equation*}
For many interesting applications of the numbers $B_{n}^{(n)}$ and $B_{n}^{(n-1)}$, one may refer to \cite{9} (see also \cite{8}).

The so-called $D$ numbers $D_{2n}^{(k)}$ of the first kind are defined by means of the following generating function (see \cite{6,7,8,5,9,10}):
\begin{equation}\label{1.9}
(t\csc t)^k=\sum_{n=0}^\infty(-1)^nD_{2n}^{(k)}\frac{t^{2n}}{(2n)!}\qquad 
(|t|&lt;\pi).
\end{equation}
Indeed, by using \eqref{1.1} and \eqref{1.9}, and by observing that 
$$\csc t=\frac{2i}{e^{it}-e^{-it}},$$ 
we easily find that
\begin{equation*}
\sum_{n=0}^\infty(-1)^nD_{2n}^{(k)}\frac{t^{2n}}{(2n)!}=\left(\frac{2it}{e^{it}-e^{-it}}\right)^k=\left(\frac{2it}{e^{2it}-1}\right)^ke^{kit}
=\sum_{n=0}^{\infty}B_n^{(k)}\left(\frac{k}{2}\right)\frac{(2it)^n}{n!}.\nonumber
\end{equation*}
Therefore, we have
\begin{equation}\label{1.10}
 D_{2n}^{(k)}= 4^nB_{2n}^{(k)}\left(\frac{k}{2}\right).
\end{equation}
Upon setting $k=1$ and $k=2$ in this last equation \eqref{1.10}, if we note that 
$$B_{2n}^{(1)}\left(\frac{1}{2}\right)=(2^{1-2n}-1)B_{2n}\qquad \text{and} \qquad
B_{2n}^{(2)}(1)=(1-2n)B_{2n},$$ 
we have
\begin{equation*}
D_{2n}^{(1)}=(2-2^{2n})B_{2n}\qquad \text{and}\qquad
D_{2n}^{(2)}=4^n(1-2n)B_{2n}.
\end{equation*}
On the other hand, if we put $k=-1$ in \eqref{1.9} and note that 
$$\sin t=\sum_{n=1}^\infty
(-1)^{n-1}\frac{t^{2n-1}}{(2n-1)!}\qquad \text{and} \qquad \cos t=\sum_{n=0}^\infty
(-1)^n\frac{t^{2n}}{(2n)!},$$ 
then it is easily seen that
\begin{equation*}
D_{2n}^{(-1)}=\frac{1}{2n+1}\qquad \text{and} \qquad
D_{2n}^{(-2)}=\frac{4^n}{(n+1)(2n+1)}.
\end{equation*}

The $D$ numbers $D_{2n}^{(k)}$ satisfy the following recurrence relation
(see \cite{6}):
\begin{equation}\label{1.12}
D_{2n}^{(k)}=\frac{(2n-k+2)(2n-k+1)}{(k-2)(k-1)}D_{2n}^{(k-2)}-\frac{2n(2n-1)(k-2)}{k-1}D_{2n-2}^{(k-2)}.
\end{equation}
In light of \eqref{1.12}, we can immediately deduce the following known results (see \cite{9}):
\begin{equation*}
D_{2n}^{(2n+1)}=\frac{(-1)^n(2n)!}{4^n}\binom{2n}{n}\qquad \text{and} \qquad
D_{2n}^{(2n+2)}=\frac{(-1)^n4^n}{2n+1}(n!)^2
\end{equation*}
and
\begin{equation*}
D_{2n}^{(2n+3)}=\frac{(-1)^n(2n)!}{2\cdot4^{2n}}\binom{2n+2}{n+1}\left(1+\frac{1}{3^2}+\frac{1}{5^2}+\cdots+\frac{1}{(2n+1)^2}\right).
\end{equation*}

Recently, Liu \cite{8} derived the following exponential generating function for
$D_{2n}^{(2n-k)}$:
\begin{equation}\label{1.15}
\sum_{n=0}^\infty
D_{2n}^{(2n-k)}\frac{t^{2n}}{(2n)!}=\frac{1}{\sqrt{1+t^2}}\left(\frac{t}{\log(t+\sqrt{1+t^2})}\right)^{k+1}.
\end{equation}
Some of the important applications of the numbers $D_{2n}^{(2n)}$ and $D_{2n}^{(2n-1)}$ include (for example) each of the following known results:
\begin{equation}\label{1.16}
\int_0^{\frac{\pi}{2}}\frac{\sin t}{t}\;{\rm d}t=\sum_{n=0}^\infty
\frac{(-1)^nD_{2n}^{(2n)}}{(2n+1)!},
\end{equation}
\begin{equation}\label{1.17}
\int_0^{\frac{\pi}{2}}\frac{\sin
t}{t}\;{\rm d}t=\frac{\pi}{2}\sum_{n=0}^\infty
\frac{(-1)^{n+1}D_{2n}^{(2n-1)}}{2^{2n}(2n-1)(n!)^2}
\end{equation}
and
\begin{equation}\label{1.18}
\frac{2}{\pi}=\sum_{n=0}^{\infty}
\frac{(-1)^{n+1}\;D_{2n}^{(2n-1)}}{(2n-1)(2n)!}.
\end{equation}

The $D$ numbers $d_{2n}^{(k)}$ of the second kind may be defined
by means of the following generating function:
\begin{equation}\label{1.19}
\left(\frac{t}{\log(t+\sqrt{1+t^2})}\right)^{k}=\sum_{n=0}^\infty d_{2n}^{(k)}t^{2n}.
\end{equation}
The numbers $d_n=d_n^{(1)}$ are referred to as the ordinary $D$ numbers of the
second kind. Some consequences of the generating function \eqref{1.19} are given below: 
$$d_0=1, \;\;d_2=\frac{1}{6},\;\;
d_4=-\frac{17}{360},\;\; d_6=\frac{367}{15120},\;\;
d_8=-\frac{195013}{27216000}\quad \text{and} \quad d_{10}=\frac{1295803}{252806400}.$$
Indeed, by using the generating function \eqref{1.19}, we also find that
\begin{equation*}
\sum_{n=1}^\infty 2nd_{2n}^{(k)}t^{2n-1}=k\left(\frac{t}{\log(t+\sqrt{1+t^2})}\right)^{k-1}
\frac{\log(t+\sqrt{1+t^2})-\frac{t}{\sqrt{1+t^2}}}{\left[\log(t+\sqrt{1+t^2})\right]^2},
\end{equation*}
that is, that
\begin{equation}
\sum_{n=1}^\infty 2nd_{2n}^{(k)}t^{2n}=k\sum_{n=0}^\infty
d_{2n}^{(k)}t^{2n}-k\sum_{n=0}^\infty
D_{2n}^{(2n-k)}\frac{t^{2n}}{(2n)!}.
\end{equation}
Consequently, we have (see \cite[Theorem 4]{8})
\begin{equation}\label{1.22}
(2n)!d_{2n}^{(k)}=\frac{k}{k-2n}D_{2n}^{(2n-k)}.
\end{equation}
By applying \eqref{1.22}, we obtain
\begin{equation*}
(2n)!d_{2n}^{(2n+1)}=1\qquad \text{and} \qquad
(2n)!d_{2n}^{(2n+2)}=\frac{4^n}{2n+1},
\end{equation*}
and
\begin{equation*}
(2n)!d_{2n}^{(2n-1)}=(2n-1)(2^{2n}-2)B_{2n}\qquad \text{and} \qquad
(2n)!d_{2n}^{(2n-2)}=4^n(n-1)(2n-1)B_{2n}.
\end{equation*}

We turn now to the {\it central factorial numbers} $\mathfrak{t}(n, k)$ of the
first kind, which are usually defined by (see \cite{11})
\begin{equation}\label{1.23}
x\left(x+\frac{n}{2}-1\right)\left(x+\frac{n}{2}-2\right)\cdots\left(x+\frac{n}{2}-n+1\right)=\sum_{k=0}^n\mathfrak{t}(n,k)x^k
\end{equation}
or, equivalently, by means of the following generating function:
\begin{equation}\label{1.24}
\left[2\log\left(\frac{x}{2}+\sqrt{1+\frac{x^2}{4}}\right)\right]^k=k!\sum_{n=k}^\infty
\mathfrak{t}(n,k)\frac{x^n}{n!}.
\end{equation}
By appealing to \eqref{1.23} or \eqref{1.24}, we can show that
\begin{equation}\label{1.25}
\mathfrak{t}(n,k)=\mathfrak{t}(n-2,k-2)-\frac{1}{4}(n-2)^2\;\mathfrak{t}(n-2,k)
\end{equation}
and that
\begin{equation*}
\mathfrak{t}(n,0)=\delta_{n,0}\quad (n\in
\mathbb{N}_0:=\mathbb{N}\cup\{0\})\qquad \text{and} \qquad  \mathfrak{t}(n,n)=1 \quad (n\in
\mathbb{N})
\end{equation*}
and
\begin{equation*}
\mathfrak{t}(n,k)=0\quad (n+k\quad \text{odd})\quad \text{and}\quad \mathfrak{t}(n,k)=0
\quad (k&gt;n \quad\text{or}\quad k&lt;0),
\end{equation*}
where $\delta_{m,n}$ denotes the Kronecker symbol.

Next, by making use of \eqref{1.23}, we obtain
\begin{equation}\label{1.26}
(x^2-1^2)(x^2-3^2)\cdots[x^2-(2n-1)^2]=\sum_{k=0}^n4^{n-k}\mathfrak{t}(2n+1,2k+1)x^{2k}
\end{equation}
and
\begin{equation}\label{1.27}
x^2(x^2-1^2)(x^2-2^2)\cdots[x^2-(n-1)^2]=\sum_{k=0}^n\mathfrak{t}(2n,2k)x^{2k}.
\end{equation}

By applying \eqref{1.26} and \eqref{1.27}, we find for $n\in \mathbb{N}_0$ that
\begin{equation*}
\mathfrak{t}(2n+1,1)=\left(-\frac{1}{4}\right)^n\cdot 1^2\cdot3^2\cdots(2n-1)^2 \quad \text{and} \quad \mathfrak{t}(2n+2,2)=(-1)^n(n!)^2
\end{equation*}
and
\begin{equation*}
\mathfrak{t}(2n+2,4)=(-1)^{n+1}(n!)^2\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\right)\qquad
(n\in \mathbb{N}).
\end{equation*}

By using \eqref{1.12} and \eqref{1.25}, we have
\begin{equation}\label{1.30}
D_{2n-2k}^{(2n+1)}=\frac{4^{n-k}}{\dbinom{2n}{2k}}\;\mathfrak{t}(2n+1,2k+1)\qquad
(n\geqq k\geqq 0)
\end{equation}
and
\begin{equation}\label{1.31}
D_{2n-2k}^{(2n)}=\frac{4^{n-k}}{\dbinom{2n-1}{2k-1}}\;\mathfrak{t}(2n,2k)\qquad(n\geqq k\geqq 1).
\end{equation}

In Sections 2 and 3 of this paper, we shall state and prove several
formulas and results for the $D$ numbers $D_{2n}^{(k)}$ and
$d_{2n}^{(k)}$, respectively. Then, in Section 4, we discuss some
applications of these families of  $D$ numbers.

\section{Formulas and Results Involving the Numbers $D_{2n}^{(k)}$}

\begin{theorem}\label{thm1}
Let $n\in\mathbb{N}$. Then
\begin{equation}\label{2.1}
D_{2n}^{(2n)}=\int_0^1(x^2-1^2)(x^2-3^2)\cdots[x^2-(2n-1)^2]{\rm d}x.
\end{equation}
\end{theorem}

\begin{proof} 
By applying \eqref{1.26} and \eqref{1.30} and noting that (see \cite{6})
$$D_{2k}^{(-1)}=\frac{1}{2k+1}\qquad \text{and} \qquad
D_{2n}^{(k)}=\sum_{j=0}^n\binom{2n}{2j}D_{2n-2j}^{(k-l)}D_{2j}^{(l)},$$
we get
\begin{align*}
D_{2n}^{(2n)}&amp;=\sum_{k=0}^n\binom{2n}{2k}D_{2n-2k}^{(2n+1)}D_{2k}^{(-1)}
=\sum_{k=0}^n\binom{2n}{2k}D_{2n-2k}^{(2n+1)}D_{2k}^{(-1)}\int_0^1(2k+1)x^{2k}\;{\rm d}x\nonumber\\
&amp;=\int_0^1\sum_{k=0}^n\binom{2n}{2k}D_{2n-2k}^{(2n+1)}x^{2k}\;{\rm d}x
=\int_0^1\sum_{k=0}^n4^{n-k}\mathfrak{t}(2n+1,2k+1)x^{2k}\;{\rm d}x\nonumber\\
&amp;=\int_0^1(x^2-1^2)(x^2-3^2)\cdots[x^2-(2n-1)^2]{\rm d}x,\nonumber
\end{align*}
which completes the proof of the assertion \eqref{2.1} of Theorem \ref{thm1}.
\end{proof}

\begin{theorem}\label{thm2}
Let $n\in\mathbb{N}$. Then
\begin{equation}\label{2.2}
D_{2n}^{(2n-1)}=(1-2n)\sum_{k=0}^n\binom{2n}{2k}D_{2n-2k}^{(2n)}.
\end{equation}
\end{theorem}

\begin{proof}
By using \eqref{1.15}, \eqref{1.31} and \eqref{1.24}, we have
\begin{align*}
&amp;\sum_{n=0}^\infty\sum_{k=0}^n\binom{2n}{2k}D_{2n-2k}^{(2n)}\frac{t^{2n}}{(2n)!}=
\sum_{n=0}^\infty D_{2n}^{(2n)}\frac{t^{2n}}{(2n)!}+
\sum_{n=1}^\infty\sum_{k=1}^n\frac{n4^{n-k}}{k}\;\mathfrak{t}(2n,2k)\frac{t^{2n}}{(2n)!}\nonumber\\
&amp;\qquad =\frac{t}{\sqrt{1+t^2}\log\left(t+\sqrt{1+t^2}\right)}
+\sum_{k=1}^\infty\frac{t}{2k}\sum_{n=k}^\infty\frac{{\rm d}}{{\rm d}t}\left\{4^{n-k}\;\mathfrak{t}(2n,2k)\;\frac{t^{2n}}{(2n)!}\right\}\nonumber\\
&amp;\qquad =\frac{t}{\sqrt{1+t^2}\;\log\left(t+\sqrt{1+t^2}\right)}
+\sum_{k=1}^\infty\frac{t}{(2k)\cdot(2k)!}\frac{{\rm d}}{{\rm d}t}\left\{\left[\log\left(t+\sqrt{1+t^2}\right)\right]^{2k}\right\}\nonumber\\
&amp;\qquad=\frac{t}{\sqrt{1+t^2}\;\log\left(t+\sqrt{1+t^2}\right)}
\sum_{k=0}^\infty\frac{\left[\log\left(t+\sqrt{1+t^2}\right)\right]^{2k}}{(2k)!}\nonumber \\
&amp; \qquad =\frac{t}{\log\left(t+\sqrt{1+t^2}\right)}
=\sum_{n=0}^\infty d_{2n}t^{2n},
\end{align*}
which, in view of \eqref{1.22}, yields
\begin{equation*}
\sum_{k=0}^n\binom{2n}{2k}D_{2n-2k}^{(2n)}=(2n)!d_{2n}=\frac{1}{1-2n}D_{2n}^{(2n-1)}.
\end{equation*}
This completes the proof of the assertion \eqref{2.2} of Theorem \ref{thm2}.
\end{proof}

\begin{theorem}\label{thm3}
Let $n\in\mathbb{N}$. Then
\begin{equation}\label{2.3}
D_{2n}^{(2n-1)}=(1-2n)\int_0^1x^2(x^2-2^2)\cdots[x^2-(2n-2)^2]{\rm d}x.
\end{equation}
\end{theorem}

\begin{proof}
By using \eqref{1.27}, \eqref{1.31} and \eqref{2.2}, we have
\begin{align*}
&amp;\int_0^1x^2(x^2-2^2)\cdots[x^2-(2n-2)^2]{\rm d}x\nonumber\\
&amp;\qquad =2\int_0^{\frac{1}{2}}4x^2(4x^2-2^2)\cdots[4x^2-(2n-2)^2]{\rm d}x\nonumber\\
&amp;\qquad =2^{2n+1}\int_0^{\frac{1}{2}}\sum_{k=0}^n\mathfrak{t}(2n,2k)x^{2k}\;{\rm d}x=\sum_{k=0}^n\frac{4^{n-k}}{2k+1}\mathfrak{t}(2n,2k)
\nonumber \\
&amp;\qquad =\sum_{k=0}^n\frac{\binom{2n-1}{2k-1}}{2k+1}D_{2n-2k}^{(2n)}
=\frac{1}{n}\sum_{k=0}^n\frac{k}{2k+1}\binom{2n}{2k}D_{2n-2k}^{(2n)}\nonumber \\
&amp;\qquad =\frac{1}{2n}\sum_{k=0}^n\binom{2n}{2k}D_{2n-2k}^{(2n)}-\frac{1}{2n}\sum_{k=0}^n\frac{1}{2k+1}\binom{2n}{2k}D_{2n-2k}^{(2n)}\nonumber\\
&amp;\qquad =\frac{1}{2n(1-2n)}D_{2n}^{(2n-1)}-\frac{1}{2n}\sum_{k=0}^n\binom{2n}{2k}D_{2n-2k}^{(2n)}D_{2k}^{(-1)}\nonumber\\
&amp;\qquad=\frac{1}{2n(1-2n)}D_{2n}^{(2n-1)}-\frac{1}{2n}D_{2n}^{(2n-1)}=\frac{1}{1-2n}D_{2n}^{(2n-1)},
\end{align*}
which completes the proof of the assertion \eqref{2.3} of Theorem \ref{thm3}.
\end{proof}

\section{Formulas and Results Involving the Numbers $d_{2n}^{(k)}$}

\begin{theorem}\label{thm4}
Let $n,k\in\mathbb{N}$. Then
\begin{equation}\label{3.1}
k(k+1)d_{2n}^{(k+2)}=(2n-k)(2n-k-1)d_{2n}^{(k)}+(2n-k-2)^2d_{2n-2}^{(k)}.
\end{equation}
\end{theorem}

\begin{proof} By making use of \eqref{1.19}, we find that
\begin{align}\label{3.2}
&amp;\sum_{n=0}^\infty
(2n-k)(2n-k-1)d_{2n}^{(k)}t^{2n-k-2}
=\frac{{\rm d}^2}{{\rm d}t^2}\left(\frac{1}{\log\left(t
+\sqrt{1+t^2}\right)}\right)^k\nonumber\\
&amp;\qquad =\frac{k(k+1)}{1+t^2}\left(\frac{1}{\log\left(t
+\sqrt{1+t^2}\right)}\right)^{k+2}
+\left(\frac{1}{\log\left(t
+\sqrt{1+t^2}\right)}\right)^{k+1}\frac{kt}{(1+t^2)^{3/2}}.
\end{align}
On the other hand, we also have
\begin{align*}
\sum_{n=1}^\infty
(2n-k-2)d_{2n-2}^{(k)}t^{2n-k-3}&amp;=\frac{{\rm d}}{{\rm d}t}\left(\frac{1}{\log\left(t
+\sqrt{1+t^2}\right)}\right)^k\nonumber\\
&amp;=-\frac{k}{\sqrt{1+t^2}}\left(\frac{1}{\log\left(t
+\sqrt{1+t^2}\right)}\right)^{k+1}.
\end{align*}
Therefore, we get
\begin{align*}
&amp;\sum_{n=1}^\infty
(2n-k-2)^2d_{2n-2}^{(k)}t^{2n-k-3}
=-\frac{{\rm d}}{{\rm d}t}\frac{kt}{\sqrt{1+t^2}}\left(\frac{1}{\log\left(t
+\sqrt{1+t^2}\right)}\right)^{k+1}\nonumber\\
&amp;\qquad =\frac{k(k+1)t}{1+t^2}\left(\frac{1}{\log\left(t
+\sqrt{1+t^2}\right)}\right)^{k+2}
-\left(\frac{1}{\log\left(t
+\sqrt{1+t^2}\right)}\right)^{k+1}\frac{k}{(1+t^2)^{3/2}},
\end{align*}
that is,
\begin{align}\label{3.4}
&amp;\sum_{n=1}^\infty (2n-k-2)^2d_{2n-2}^{(k)}t^{2n-k-2}
=\frac{k(k+1)t^2}{1+t^2}\left(\frac{1}{\log\left(t
+\sqrt{1+t^2}\right)}\right)^{k+2}\nonumber \\
&amp; \qquad \qquad \qquad -\left(\frac{1}{\log\left(t
+\sqrt{1+t^2}\right)}\right)^{k+1}\frac{kt}{(1+t^2)^{3/2}}.
\end{align}
Now, by using \eqref{3.2} and \eqref{3.4}, we obtain
\begin{align}\label{3.5}
&amp;\sum_{n=0}^\infty
(2n-k)(2n-k-1)d_{2n}^{(k)}t^{2n-k-2}+\sum_{n=1}^\infty
(2n-k-2)^2d_{2n-2}^{(k)}t^{2n-k-2}\nonumber\\
&amp;\qquad =k(k+1)\left(\frac{1}{\log\left(t
+\sqrt{1+t^2}\right)}\right)^{k+2}
=k(k+1)\sum_{n=0}^\infty
d_{2n}^{(k+2)}t^{2n-k-2}.
\end{align}
Finally, by comparing the coefficients of $t^{2n-k-2}$ on both sides of \eqref{3.5}, we are led easily to \eqref{3.1}.
This completes the proof of Theorem \ref{thm4}.
\end{proof}

\begin{remark}\label{rem0}
Upon setting $k=2n-2$ in Theorem \ref{thm4}, if we make use of
\eqref{1.22}, we immediately obtain the following result:
\begin{equation}\label{3.14}
(2n)!d_{2n}^{(2n)}=4^nB_{2n} \qquad (n\in\mathbb{N}_0).
\end{equation}
A generalization of the above result \eqref{3.14} and other analogous results can be found in the recent work by Liu \cite[Corollary 1 and Theorem 4]{8}.
\end{remark}

\begin{theorem}\label{thm5}
Let $n,k\in\mathbb{N}$ and $n \geqq k+1$.
Then
\begin{equation}\label{3.6}
(2k)!(2n-2k-1)!d_{2n}^{(2k+1)}=\sum_{j=0}^k(2n-1-2j)!\sigma(n,k,j)d_{2n-2j},
\end{equation}
where
\begin{equation*}
\sigma(n,k,j)=\mathop{\sum_
{v_1,\cdots,v_{k-j+1}\in\mathbb{N}_0}}_{(v_1+\cdots
+v_{k-j+1}=j)}(2n-2j-1)^{2v_1}(2n-2j-3)^{2v_2}\cdots
(2n-2k-1)^{2v_{k-j+1}}.
\end{equation*}
\end{theorem}

\begin{proof}
We prove the assertion \eqref{3.6} of Theorem \ref{thm5} by using the 
principle of mathematical induction. Indeed, when $k=1$, \eqref{3.6} is true by virtue of \eqref{3.1}. Suppose now that \eqref{3.6} is true for some natural number $k\in \mathbb{N}\setminus\{1\}$. Then, by
the superposition of \eqref{3.1}, we have
\begin{align}\label{3.8}
&amp;(2k+1)(2k+2)d_{2n}^{(2k+3)}\nonumber\\
&amp;\qquad =(2n-2k-1)(2n-2k-2)d_{2n}^{(2k+1)}+(2n-2k-3)^2d_{2n-2}^{(2k+1)}\nonumber\\
&amp;\qquad =\frac{(2n-2k-1)(2n-2k-2)}{(2k)!(2n-2k-1)!}\sum_{j=0}^k(2n-1-2j)!\sigma(n,k,j)d_{2n-2j}\nonumber\\
&amp;\qquad \qquad +
\frac{(2n-2k-3)^2}{(2k)!(2n-2k-3)!}\sum_{j=0}^k(2n-3-2j)!\sigma(n-1,k,j)d_{2n-2-2j}\nonumber\\
&amp;\qquad =\frac{1}{(2k)!(2n-2k-3)!}\sum_{j=0}^k(2n-1-2j)!\sigma(n,k,j)d_{2n-2j}\nonumber\\
&amp;\qquad \qquad +
\frac{(2n-2k-3)^2}{(2k)!(2n-2k-3)!}\sum_{j=1}^{k+1}(2n-2-2j)!\sigma(n-1,k,j-1)d_{2n-2j}.
\end{align}
In light of this last result \eqref{3.8}, and by noting that
$$\sigma(n,k+1,0)=\sigma(n,k,0),$$
$$\sigma(n,k+1,k+1)=(2n-2k-3)^2\sigma(n-1,k,k)
$$
and
$$\sigma(n,k+1,j)=\sigma(n,k,j)+(2n-2k-3)^2\sigma(n-1,k,j-1),$$
we find that
\begin{align*}
&amp;(2k+2)!(2n-2k-3)!d_{2n}^{(2k+3)}\nonumber\\
&amp;\qquad =\sum_{j=0}^k(2n-1-2j)!\sigma(n,k,j)d_{2n-2j}\nonumber\\
&amp;\qquad \qquad +(2n-2k-3)^2\sum_{j=1}^{k+1}(2n-1-2j)!\sigma(n-1,k,j-1)d_{2n-2j}\nonumber\\
&amp;\qquad =(2n-1)!\sigma(n,k,0)d_{2n}+(2n-2k-3)^2(2n-3-2k)!\sigma(n-1,k,k)d_{2n-2k-2}\nonumber\\
&amp;\qquad \qquad +\sum_{j=1}^k(2n-1-2j)!\left[\sigma(n,k,j)+(2n-2k-3)^2\sigma(n-1,k,j-1)\right]d_{2n-2j}\nonumber\\
&amp;\qquad =(2n-1)!\sigma(n,k+1,0)d_{2n}+(2n-3-2k)!\sigma(n,k+1,k+1)d_{2n-2k-2}\nonumber\\
&amp;\qquad \qquad +\sum_{j=1}^k(2n-1-2j)!\sigma(n,k+1,j)d_{2n-2j}\nonumber\\
&amp;\qquad =\sum_{j=0}^{k+1}(2n-1-2j)!\sigma(n,k+1,j)d_{2n-2j},
\end{align*}
which shows that \eqref{3.6} is also true for the natural number
$k+1$. Thus, by the principle of mathematical induction, \eqref{3.6} holds true
for all $k\in \mathbb{N}$. This completes
the proof of Theorem \ref{thm5}.
\end{proof}

\begin{remark} \label{rem1}
{\rm Upon setting $k=1,2,3$ in \eqref{3.6}, we can immediately
deduce 
$$2!d_{2n}^{(3)}=(2n-1)(2n-2)d_{2n}+(2n-3)^2d_{2n-2},$$

\begin{align*}
4!d_{2n}^{(5)}&amp;=(2n-1)(2n-2)(2n-3)(2n-4)d_{2n}\nonumber\\
&amp;\qquad +(2n-3)(2n-4)\big[(2n-3)^2+(2n-5)^2\big]d_{2n-2}+(2n-5)^4d_{2n-4}
\end{align*}
and 

\begin{align*}
6!d_{2n}^{(7)}&amp;=(2n-1)(2n-2)(2n-3)(2n-4)(2n-5)(2n-6)d_{2n}\nonumber\\
&amp;\qquad +(2n-3)(2n-4)(2n-5)(2n-6)\big[(2n-3)^2+(2n-5)^2\nonumber \\
&amp;\qquad  +(2n-7)^2\big]d_{2n-2}
+(2n-5)(2n-6)\big[(2n-5)^4\nonumber \\
&amp; \qquad +(2n-5)^2(2n-7)^2+(2n-7)^4\big]d_{2n-4}
+(2n-7)^6d_{2n-6}.
\end{align*}}
\end{remark}

\begin{theorem}\label{thm6}
Let $n,k\in\mathbb{N}$ and $n\geqq k$. Then
\begin{equation}\label{3.10}
(2k-1)!(2n-2k)!d_{2n}^{(2k)}=\sum_{j=0}^{k-1}(2n-2-2j)!\tau(n,k,j)d_{2n-2j}^{(2)},
\end{equation}
where
\begin{equation*}
\tau(n,k,j)=\mathop{\sum_
{v_1,\cdots,v_{k-j}\in\mathbb{N}_0}}_{(v_1+\cdots
+v_{k-j}=j)}(2n-2j-2)^{2v_1}(2n-2j-4)^{2v_2}\cdots
(2n-2k)^{2v_{k-j}}.
\end{equation*}
\end{theorem}

\begin{proof}
We prove the assertion \eqref{3.10} of Theorem \ref{thm6} by using the principle of mathematical induction. In fact, when $k=1$ and $k=2$, \eqref{3.10} is true by \eqref{3.1}.
Suppose now that \eqref{3.10} is true for some natural number $k\in \mathbb{N}\setminus\{1\}$. Then, by the
superposition of \eqref{3.1}, we have
\begin{align}\label{3.12}
&amp;2k(2k+1)d_{2n}^{(2k+2)}\nonumber\\
&amp;\qquad =(2n-2k)(2n-2k-1)d_{2n}^{(2k)}+(2n-2k-2)^2d_{2n-2}^{(2k)}\nonumber\\
&amp;\qquad =\frac{(2n-2k)(2n-2k-1)}{(2k-1)!(2n-2k)!}\sum_{j=0}^{k-1}(2n-2-2j)!\tau(n,k,j)d_{2n-2j}^{(2)}\nonumber\\
&amp;\qquad \qquad +
\frac{(2n-2k-2)^2}{(2k-1)!(2n-2k-2)!}\sum_{j=0}^{k-1}(2n-4-2j)!\tau(n-1,k,j)d_{2n-2-2j}^{(2)}\nonumber\\
&amp;\qquad =\frac{1}{(2k-1)!(2n-2k-2)!}\sum_{j=0}^{k-1}(2n-2-2j)!\tau(n,k,j)d_{2n-2j}^{(2)}\nonumber\\
&amp;\qquad\qquad +
\frac{(2n-2k-2)^2}{(2k-1)!(2n-2k-2)!}\sum_{j=1}^k(2n-2-2j)!\tau(n-1,k,j-1)d_{2n-2j}^{(2)}.
\end{align}

By using \eqref{3.12}, and noting that
$$\tau(n,k+1,0)=\tau(n,k,0),$$
$$\tau(n,k+1,k)=(2n-2k-2)^2\tau(n-1,k,k-1)$$
and
$$\tau(n,k+1,j)=\tau(n,k,j)+(2n-2k-2)^2\tau(n-1,k,j-1),$$
we find that
\begin{align*}
&amp;(2k+1)!(2n-2k-2)!d_{2n}^{(2k+2)}\nonumber \\
&amp;\qquad =\sum_{j=0}^{k-1}(2n-2-2j)!\tau(n,k,j)d_{2n-2j}^{(2)}\nonumber\\
&amp;\qquad \qquad+(2n-2k-2)^2\sum_{j=1}^k(2n-2-2j)!\tau(n-1,k,j-1)d_{2n-2j}^{(2)}\nonumber\\
&amp;\qquad=(2n-2)!\tau(n,k,0)d_{2n}^{(2)}+(2n-2k-2)^2(2n-2-2k)!\tau(n-1,k,k-1)d_{2n-2k}^{(2)}\nonumber\\
&amp;\qquad \qquad+\sum_{j=1}^{k-1}(2n-2-2j)!\left[\tau(n,k,j)+(2n-2k-2)^2\tau(n-1,k,j-1)\right]d_{2n-2j}^{(2)}\nonumber\\
&amp;\qquad=(2n-2)!\tau(n,k+1,0)d_{2n}^{(2)}+(2n-2-2k)!\tau(n,k+1,k)d_{2n-2k}^{(2)}\nonumber\\
&amp;\qquad \qquad+\sum_{j=1}^{k-1}(2n-2-2j)!\left[\tau(n,k,j)+(2n-2k-2)^2\tau(n-1,k,j-1)\right]d_{2n-2j}^{(2)}\nonumber\\
&amp;\qquad=\sum_{j=0}^k(2n-2-2j)!\tau(n,k+1,j)d_{2n-2j}^{(2)},
\end{align*}
which shows that \eqref{3.10} holds true also for the natural number
$k+1$. This evidently completes
the proof of Theorem \ref{thm6} by the principle of mathematical induction on $k\in\mathbb{N}$.
\end{proof}

\begin{remark}\label{rem2}
By setting $k=2,3,4$ in \eqref{3.10}, we can immediately
deduce 
$$3!d_{2n}^{(4)}=(2n-2)(2n-3)d_{2n}^{(2)}+(2n-4)^2d_{2n-2}^{(2)},$$
\begin{align*}
&amp;5!d_{2n}^{(6)}=(2n-2)(2n-3)(2n-4)(2n-5)d_{2n}^{(2)}\nonumber\\
&amp;\qquad \qquad +(2n-4)(2n-5)[(2n-4)^2+(2n-6)^2]d_{2n-2}^{(2)}+(2n-6)^4d_{2n-4}^{(2)}
\end{align*}
and
\begin{align*}
&amp;7!d_{2n}^{(8)}=(2n-2)(2n-3)(2n-4)(2n-5)(2n-6)(2n-7)d_{2n}^{(2)}\nonumber\\
&amp;\qquad\qquad+(2n-4)(2n-5)(2n-6)(2n-7)[(2n-4)^2+(2n-6)^2+(2n-8)^2]d_{2n-2}^{(2)}\nonumber\\
&amp;\qquad\qquad +(2n-6)(2n-7)[(2n-6)^4+(2n-6)^2(2n-8)^2+(2n-8)^4]d_{2n-4}^{(2)}\nonumber\\
&amp;\qquad\qquad +(2n-8)^6d_{2n-6}^{(2)}.
\end{align*}
\end{remark}

\section{A Set of Applications}

We first give the following application of the results presented in the preceding sections.

\begin{theorem}\label{thm8} 
Let $k\in\mathbb{Z}$. Then
\begin{equation}\label{4.1}
\int_0^{\frac{\pi}{2}}\left(\frac{\sin
t}{t}\right)^{k+1}\;{\rm d}t=\sum_{n=0}^\infty
\frac{(-1)^nD_{2n}^{(2n-k)}}{(2n+1)!}.
\end{equation}
\end{theorem}

\begin{proof}
By using \eqref{1.15}, we find that
\begin{equation*}
\sum_{n=0}^\infty
(-1)^nD_{2n}^{(2n-k)}\frac{x^{2n}}{(2n)!}=\frac{1}{\sqrt{1-x^2}}\left(\frac{ix}{\log(ix+\sqrt{1-x^2})}\right)^{k+1}.
\end{equation*}
Therefore, we have
\begin{align*}
\sum_{n=0}^\infty
\frac{(-1)^nD_{2n}^{(2n-k)}}{(2n+1)!}&amp;=\int_0^1\frac{1}{\sqrt{1-x^2}}\left(\frac{ix}{\log(ix+\sqrt{1-x^2})}\right)^{k+1}\;{\rm d}x\notag \\
&amp;=\int_0^{\frac{\pi}{2}}\frac{1}{\cos\theta}\left(\frac{i\sin\theta}{\log(i\sin\theta+\cos\theta)}\right)^{k+1}d(\sin\theta)
=\int_0^{\frac{\pi}{2}}\left(\frac{\sin\theta}{\theta}\right)^{k+1}d\theta,
\end{align*}
which obviously completes the proof of Theorem \ref{thm8}.
\end{proof}

\begin{remark}\label{rem4} 
Setting $k=0$ in \eqref{4.1}, we immediately
obtain \eqref{1.16}. Moreover, if we set $k=-1$ and $k= -2$ in \eqref{4.1} and note that
$$D_{2n}^{(2n+1)}=\frac{(-1)^n(2n)!}{4^n}\binom{2n}{n}\qquad \text{and} \qquad
D_{2n}^{(2n+2)}=\frac{(-1)^n4^n}{2n+1}(n!)^2,$$ 
we get
\begin{equation*}
\sum_{n=0}^\infty \frac{1}{(2n+1)4^n}\binom{2n}{n}=\frac{\pi}{2}
\end{equation*}
and
\begin{equation*}
\int_0^{\frac{\pi}{2}}\frac{t}{\sin t}\;{\rm d}t=\sum_{n=0}^\infty
\frac{4^n}{(2n+1)^2\binom{2n}{n}},
\end{equation*}
The sum on the right-hand side of this last result can be expressed as a Clausenian hypergeometric series as follows:
$$\;_3F_2\left(\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2};1\right).$$ Moreover, by using a computer algebra system, the integral on the left-hand side can be found to be twice the Catalan constant $\mathbf{G}$ which is defined by (see, for details, \cite[p.~43 et seq.]{13})
\begin{equation*}
\mathbf{G}:=\frac{1}{2} \int_0^1 \textbf{\textrm{K}}(\kappa) {\rm d}\kappa=\sum_{n=0}^\infty
\frac{(-1)^n}{(2n+1)^2} \doteq 0.91596\, 55941 \, 77219\,015 \cdots,
\end{equation*}
where $\textbf{\textrm{K}}(\kappa)$ is the complete elliptic integral of the first
kind, given by 
\begin{equation*} 
\textbf{\textrm{K}}(\kappa) :=
\int_0^{\frac{\pi}{2}} \frac{{\rm d}t}{\sqrt{1-\kappa^2 \sin^2 t}} \qquad (|\kappa|&lt;1).
\end{equation*}
In fact, the above observation can be verified {\it analytically} by setting $\tau=\tan\left(\frac{t}{2}\right)$ and ${\rm d}\tau=\frac{1}{2}\sec^2\left(\frac{t}{2}\right){\rm d}t$ in the following known result \cite[p. 110, Equation 2.3(37)]{12}:
\begin{equation*}
\int_{0}^{1}\frac{\arctan \tau}{\tau}\; {\rm d}\tau=\mathbf{G}.
\end{equation*}
\end{remark}

\begin{theorem}\label{thm9} 
Let $k\in\mathbb{Z}$. Then
\begin{equation}\label{4.4}
\int_0^{\frac{\pi}{2}}\left(\frac{\sin t}{t}\right)^k\cos
t\;{\rm d}t=\sum_{n=0}^\infty
\frac{(-1)^{n+1}kD_{2n}^{(2n-k)}}{(2n-k)(2n+1)!}.
\end{equation}
\end{theorem}

\begin{proof}
By using \eqref{1.19} and \eqref{1.22}, we find that
\begin{align*}
\sum_{n=0}^\infty
\frac{(-1)^{n+1}kD_{2n}^{(2n-k)}}{(2n-k)(2n)!}x^{2n}&amp;=\sum_{n=0}^\infty
(-1)^nd_{2n}^{(k)}x^{2n}
=\left(\frac{ix}{\log(ix+\sqrt{1-x^2})}\right)^{k}.
\end{align*}
Therefore, we have
\begin{align*}
\sum_{n=0}^\infty
\frac{(-1)^{n+1}kD_{2n}^{(2n-k)}}{(2n-k)(2n+1)!}&amp;=\int_0^1\left(\frac{ix}{\log(ix+\sqrt{1-x^2})}\right)^k
{\rm d}x\nonumber\\
&amp;=\int_0^{\frac{\pi}{2}}\left(\frac{i\sin t}{\log(i\sin t+\cos
t)}\right)^k\;{\rm d}(\sin t) =\int_0^{\frac{\pi}{2}}\left(\frac{\sin
t}{t}\right)^k\;\cos t\; {\rm d}t,
\end{align*}
which evidently completes the proof of Theorem \ref{thm9}.
\end{proof}

\begin{remark}\label{rem5} 
By setting $k=1$ and $k= -1$ in \eqref{4.4}, we have
\begin{equation*}
\int_0^\pi\frac{\sin t}{t}\;{\rm d}t=2\sum_{n=0}^\infty
\frac{(-1)^{n+1}D_{2n}^{(2n-1)}}{(2n-1)(2n+1)!}
\end{equation*}
and
\begin{equation*}
\int_0^{\frac{\pi}{2}}t\cot t \;{\rm d}t=\sum_{n=0}^\infty
\frac{1}{4^n\;(2n+1)^2}\binom{2n}{n}
\end{equation*}
or, equivalently,
\begin{equation*}
\int_0^{1}\frac{\arcsin \tau}{\tau} \;{\rm d}\tau=\sum_{n=0}^\infty
\frac{1}{4^n\;(2n+1)^2}\binom{2n}{n}
\end{equation*}
by setting $\sin t=\tau$ and $\cos t\; {\rm d}t={\rm d}\tau$. The sum on the right-hand side of this last result can be expressed in a closed form via a Clausenian hypergeometric series as follows:
$$\int_0^{1}\frac{\arcsin \tau}{\tau} \;{\rm d}\tau=\sum_{n=0}^\infty
\frac{1}{4^n\;(2n+1)^2}\binom{2n}{n}=\;_3F_2\left(\frac{1}{2}, \frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};1\right)=\frac{\pi}{2}\; \ln 2.$$
\end{remark}

\begin{theorem}\label{thm10} 
Let $k\in\mathbb{Z}$. Then
\begin{equation}\label{4.7}
\sum_{n=0}^\infty\frac{(-1)^{n+1}k}{(2n-k)(2n)!}D_{2n}^{(2n-k)}=\left(\frac{2}{\pi}\right)^k,
\end{equation}
provided that each member of $\eqref{4.7}$ exists.
\end{theorem}

\begin{proof}
By applying \eqref{1.19} and \eqref{1.22}, we obtain
\begin{align*}
&amp;\sum_{n=0}^\infty\frac{(-1)^{n+1}k}{(2n-k)(2n)!}D_{2n}^{(2n-k)}x^{2n}
=\sum_{n=0}^\infty
(-1)^nd_{2n}^{(k)}x^{2n}\notag \\
&amp; \qquad =\left(\frac{ix}{\log(ix+\sqrt{1-x^2})}\right)^{k}
=\left(\frac{1}{\sum\limits_{n=0}^\infty
\frac{1}{(2n+1)4^n}\binom{2n}{n}x^{2n}}\right)^{k}.
\end{align*}
Therefore, we get
\begin{equation*}
\sum_{n=0}^\infty\frac{(-1)^{n+1}k}{(2n-k)(2n)!}D_{2n}^{(2n-k)}=\left(\frac{1}{\sum\limits_{n=0}^\infty
\frac{1}{(2n+1)4^n}\binom{2n}{n}}\right)^{k}=\left(\frac{2}{\pi}\right)^k.
\end{equation*}
This completes the proof of Theorem \ref{thm10}.
\end{proof}

\begin{remark}\label{rem6}
Upon setting $k=1$ in \eqref{4.7}, we immediately
obtain \eqref{1.18}. If, on the other hand, we put $k=-2$ in \eqref{4.7} and note that
$$D_{2n}^{(2n+2)}=\frac{(-1)^n4^n}{2n+1}(n!)^2,$$ 
then we find that
\begin{equation*}
\sum_{n=0}^\infty \frac{4^n(n!)^2}{(2n+2)!}=\frac{\pi^2}{8}.
\end{equation*}
This last identity follows also from a known power-series expansion for the function $(\arcsin x)^2$.
\end{remark}

\bigskip

\section{Acknowledgements}
 
This work was supported by the Guangdong
Provincial Natural Science Foundation of the People's Republic of China 
(Grant No.\ 8151601501000002).


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Publ. Math. Debrecen} {\bf 79} (2011), 41--53.

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\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B68, 11B83; Secondary 05A19.

\noindent \emph{Keywords: } 
Bernoulli polynomial, Bernoulli
number, $D$ number, $D$ number of the second kind,
N\"{o}rlund number; central factorial number of the first kind,
Catalan's constant, Clausenian hypergeometric series.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000367},
\seqnum{A002445},
\seqnum{A002657},
\seqnum{A002790},
\seqnum{A027641}, and
\seqnum{A027642}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received October 23 2013; revised versions received
February 15 2014; February 21 2014.
Published in {\it Journal of Integer Sequences}, February 22 2014.

\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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