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\begin{center}
\vskip 1cm{\LARGE\bf On Polycosecant Numbers
}
\vskip 1cm
\large
Masanobu Kaneko \\ 
Faculty of Mathematics\\ 
Kyushu University\\ 
Motooka 744, Nishi-ku \\ 
Fukuoka 819-0395\\ 
Japan\\ 
\href{mailto:mkaneko@math.kyushu-u.ac.jp}{\tt mkaneko@math.kyushu-u.ac.jp} \\
\ \\
Maneka Pallewatta \\ 
Graduate School of Mathematics\\ 
Kyushu University\\ 
Motooka 744, Nishi-ku\\ 
Fukuoka 819-0395\\ 
Japan\\ 
\href{mailto:maneka.osh@gmail.com}{\tt maneka.osh@gmail.com} \\ 
\ \\
Hirofumi Tsumura \\ 
Department of Mathematical Sciences\\ 
Tokyo Metropolitan University\\ 
1-1, Minami-Ohsawa\\ 
Hachioji, Tokyo 192-0397\\ 
Japan\\ 
\href{mailto:tsumura@tmu.ac.jp}{\tt tsumura@tmu.ac.jp}
\end{center}

\vskip .2 in

\begin{abstract}
We introduce and study a ``level two'' generalization of the
poly-Bernoulli numbers, which may also be regarded as a generalization of
the cosecant numbers.  We prove a recurrence relation, two exact formulas,
and a duality relation for negative upper-index numbers.
\end{abstract}



\section{Introduction} \label{sec-1}
The first named author \cite{Kaneko1997} defined the {\it poly-Bernoulli numbers} \seqnum{A099594} and later 
Arakawa and the first named author \cite{AK1999} studied a slightly 
modified version. They are, denoted by $B_n^{(k)}$ and $C_n^{(k)}$ respectively, 
defined by using generating series, as follows. 
For an integer $k\in \mathbb{Z}$, 
let $(B_n^{(k)})_{n \geq 0}$ and $(C_n^{(k)})_{n\geq 0}$ be the sequences of rational numbers given respectively by
\begin{align}
&amp;\frac{{\rm Li}_{k}(1-e^{-t})}{1-e^{-t}}=\sum_{n=0}^\infty B_n^{(k)}\frac{t^n}{n!} \label{eq-1-1}\\
\intertext{and}
&amp;\frac{{\rm Li}_{k}(1-e^{-t})}{e^t-1}=\sum_{n=0}^\infty C_n^{(k)}\frac{t^n}{n!},  \label{eq-1-2}
\end{align}
where ${\rm Li}_{k}(z)$ is the polylogarithm function (or rational function when $k\le0$) defined by
\begin{equation}
{\rm Li}_{k}(z)=\sum_{m=1}^\infty \frac{z^m}{m^k}\quad (|z|&lt;1). \label{eq-1-3}
\end{equation}

Since ${\rm Li}_1(z)=-\log(1-z)$, the generating functions on the left-hand sides of \eqref{eq-1-1}
and \eqref{eq-1-2}   when $k=1$ become 
\[ \frac{te^t}{e^t-1} \quad \text{and} \quad \frac{t}{e^t-1} \] respectively. Hence 
$B_n^{(1)}$ and $C_n^{(1)}$ represent the standard Bernoulli numbers \seqnum{A027641}, 
\seqnum{A027642}, the only difference being $B_1^{(1)}=1/2$ 
and $C_1^{(1)}=-1/2$ and otherwise $B_n^{(1)}=C_n^{(1)}$. 

Several properties of poly-Bernoulli numbers have been found including the following results:
\[ B_n^{(k)} =(-1)^n\sum_{i=0}^n\frac{(-1)^ii!\left\{\begin{matrix} {n}\\{i} \end{matrix}\right\}}{(i+1)^k},\quad
C_n^{(k)}=(-1)^n\sum_{i=0}^n\frac{(-1)^ii!{\left\{\begin{matrix} {n+1}\\{i+1} \end{matrix}\right\}}}{(i+1)^k},\] 
where $k$ is an integer, $n$ a non-negative integer, and let $\left\{\begin{matrix} {n}\\{i} \end{matrix}\right\}$ 
denote the Stirling numbers of the second kind \seqnum{A008277}.
Moreover, their dualities
\begin{align}
&amp; B_n^{(-k)}=B_{k}^{(-n)}, \label{eq-1-4}\\
&amp; C_n^{(-k-1)}=C_{k}^{(-n-1)} \label{eq-1-5}
\end{align}
($k,n\in \mathbb{Z}_{\geq 0}$) are derived in \cite[Theorems 1 and 2]{Kaneko1997} or 
in \cite[Section 2]{Kaneko-Mem}. For combinatorial
applications, see~\cite{Ben2017}.

In this paper, we study the following `level $2$' analog of poly-Bernoulli numbers, denoted by $D_n^{(k)}$,
which we call the {\it polycosecant numbers}. For each $k\in\mathbb{Z}$, define $D_n^{(k)}$ by
\begin{equation}
\frac{\mathrm{A}_k(\tanh(t/2))}{\sinh t}=\sum_{n=0}^\infty D_n^{(k)}\frac{t^n}{n!}, \label{eq-1-6}
\end{equation}
where $\mathrm{A}_k(z)$ is the series 
\begin{equation}
\mathrm{A}_k(z)=2\sum_{n=0}^\infty \frac{z^{2n+1}}{(2n+1)^k} \label{eq-1-7}
\end{equation}
and $\tanh(z)$ and $\sinh(z)$ are the usual hyperbolic tangent and sine functions respectively.
Since $\mathrm{A}_k(z)$, $\tanh(z)$ and $\sinh(z)$ are all odd functions, we immediately see that 
$D_{2n+1}^{(k)}=0$ for all $n\in \mathbb{Z}_{\geq 0}$.
Note that $\mathrm{A}_1(z)=2\tanh^{-1}(z)$, and thus 
\[ \sum_{n=0}^\infty D_n^{(1)}\frac{t^n}{n!} = \frac{t}{\sinh t} = \frac{i t}{\sin(i t)}\quad(i=\sqrt{-1} ).  \]
Hence $D_n^{(1)}$ is the cosecant number $D_n$ which N\"orlund~\cite[p.~27~(39) and p.~32~(52)]{Noe} 
first introduced \seqnum{A001896}, \seqnum{A001897}. 
It should be noted that the terminology `cosecant number' was not used by N\"orlund. 
Apparently Kowalenko \cite{Kowalenko1} was the first to use the terminology, but he adopted the name 
for $D_n/n!$ instead of $D_n$. He gave many applications, and studied a generalization together with 
interesting number-theoretical applications \cite{Kowalenko3, Kowalenko1, Kowalenko2, Kowalenko4}.  

Here, we give a table of $D_n^{(k)}$ for small $k$ and $n$.  A table for $k&lt;0$ will be given in \S3.
\renewcommand{\arraystretch}{1.3}
\begin{table}[h]
\begin{center}
\begin{tabular} {|c|c|c|c|c|c|c|c|c|} \hline
\backslashbox{$\ \ k$}{$n\ \ $} &amp; $ 0 $ &amp; $ 2 $ &amp; $ 4 $ &amp; $ 6 $ &amp; $ 8 $ &amp; $ 10 $ 
 \\ \hline  
$0$ &amp; $1$ &amp; $0$ &amp; $0$ &amp; $0$ &amp; $0$ &amp; $0$ 
\\ \hline

$1$ &amp; $1$ &amp; $-\frac{1}{3}$ &amp; $\frac{7}{15}$ &amp; $-\frac{31}{21}$ &amp; $\frac{127}{15}$ 
&amp; $-\frac{2555}{33}$  
\\ \hline

$2$ &amp; $1$ &amp; $-\frac{4}{9}$ &amp; $\frac{176}{225}$ &amp; $-\frac{6464}{2205}$ 
&amp; $\frac{3328}{175}$ &amp; $-\frac{1037312}{5445}$  
  \\ \hline

$3$ &amp; $1$ &amp; $-\frac{13}{27}$ &amp; $\frac{3103}{3375}$ &amp; $-\frac{859939}{231525}$ 
&amp; $\frac{12761501}{496125}$ &amp; $-\frac{63453851}{232925}$ 
\\ \hline

$4$ &amp; $1$ &amp; $-\frac{40}{81}$ &amp; $\frac{49184}{50625}$ 
&amp; $-\frac{98447744}{24310125}$ &amp; $\frac{4519218688}{156279375}$ &amp; $-\frac{6868861044736}{21791298375}$ 
\\ \hline

$5$ &amp; $1$ &amp; $-\frac{121}{243}$ &amp; $\frac{751927}{759375}$ 
&amp; $-\frac{10665916999}{2552563125}$ &amp; $\frac{1488186370469}{49228003125}$ 
&amp; $-\frac{25213417199300173}{75506848869375}$ 
\\ \hline
\end{tabular}
\end{center}
\label{tab:Bnk}
\caption{$D_n^{(k)}  \: (0\le k \leq 5,\; 0\le n\le 10, {\rm even}) $}
\end{table}

We should also mention that our $D_n^{(k)}$ is (if slightly modified) a special case of a generalization
of the poly-Bernoulli number which Sasaki \cite[Definition 5]{Sasaki2012} introduced. The numbers
$D_n^{(k)}$ are closely connected to the `multiple Hurwitz zeta functions' and the `level 2' multiple zeta functions.
We shall explore these connections in an ongoing project, which will be 
stated in the forthcoming paper.

\section{Recurrence and explicit formulas for polycosecant numbers}

In this section, we  obtain a recurrence and explicit formulas for polycosecant numbers. 

We begin by deriving a recurrence relation. 

\begin{proposition}\label{Pr-3-1} For every integer $ k$  and $n\ge 0 $, the polycosecant numbers obey 
the recurrence relation of 
\begin{equation}\label{rec}
{D}_{n}^{(k-1)}=\sum_{m=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n+1}{2m+1} {D}_{n-2m}^{(k)}.
\end{equation}
\end{proposition}

\begin{proof}
First, differentiate \eqref{eq-1-6}, which yields
\begin{align*}
\frac{ \mathrm{A}_{k-1}(\tanh(t/2))}{\sinh t}&amp;=\cosh t\sum_{n=0}^\infty {D}_n^{(k)}\frac{t^n}{n!}
+\sinh t\sum_{n=1}^\infty {D}_n^{(k)}\frac{t^{n-1}}{(n-1)!}. 
\end{align*}
From the above result we have
\begin{align*}
\sum_{n=0}^\infty {D}_n^{(k-1)}\frac{t^n}{n!}&amp;=\sum_{m=0}^\infty\frac{t^{2m}}{(2m)!} 
\sum_{n=0}^\infty {D}_n^{(k)}\frac{t^n}{n!}+\sum_{m=0}^\infty\frac{t^{2m+1}}{(2m+1)!} 
\sum_{n=1}^\infty {D}_n^{(k)}\frac{t^{n-1}}{(n-1)!} \\
&amp;=\sum_{n=0}^\infty\sum_{m=0} ^{\lfloor\frac{n}{2}\rfloor} {D}_{n-2m}^{(k)}\frac{t^{n}}{(2m)!(n-2m)!}
+\sum_{n=1}^\infty\sum_{m=0} ^{\lfloor\frac{n}{2}\rfloor} {D}_{n-2m}^{(k)}
\frac{t^{n}}{(2m+1)!(n-2m-1)!}\\
&amp;=\sum_{n=0}^\infty\sum_{m=0} ^{\lfloor{\frac{n}{2}}\rfloor} \binom{n}{2m} {D}_{n-2m}^{(k)}
\frac{t^{n}}{n!}+\sum_{n=1}^\infty\sum_{m=0} ^{\lfloor{\frac{n}{2}}\rfloor} 
\binom{n}{2m+1} {D}_{n-2m}^{(k)}\frac{t^{n}}{n!}\\
&amp;=\sum_{n=0}^\infty\sum_{m=0} ^{\lfloor{\frac{n}{2}}\rfloor} 
\binom{n+1}{2m+1} {D}_{n-2m}^{(k)}\frac{t^{n}}{n!}.
\end{align*}
Since $t$ is arbitrary, we can equate like powers of $t$ on both sides, thereby obtaining the desired result.
\end{proof}

Since $\mathrm{A}_0(\tanh(t/2))=\sinh(t)$, we observe $D_0^{(0)}=1$ and $D_n^{(0)}=0$ for  $n\ge1$.
Hence equation \eqref{rec} can be used to compute $D_n^{(k)}$ for $k&lt;0$ recursively starting from 
$D_n^{(0)}$.
For $k&gt;0$, we rewrite \eqref{rec} as 
\[ (n+1) D_n^{(k)}=D_n^{(k-1)}-\sum_{m=1}^{\lfloor\frac{n}{2}\rfloor}\binom{n+1}{2m+1} {D}_{n-2m}^{(k)}\]
in order to compute $D_n^{(k)}$ recursively.
We observe that $D_0^{(k)}=1$ for all $k\in\mathbb{Z}$.

We continue by presenting two formulas for polycosecant numbers. Before doing so, we require the following lemma. 

\begin{lemma}\label{Lem-3-1} For $n\ge1$ we have,
\begin{equation*}
x^n \left(\frac{d}{dx}\right)^n=\sum_{m=1}^{n}(-1)^{n-m}\begin{bmatrix}
n\\m
\end{bmatrix}\left(x\frac{d}{dx}\right)^m. 
\end{equation*}
Here, we let $\begin{bmatrix}n\\ m \end{bmatrix} $  denote the Stirling numbers of the first kind.
\end{lemma}

\begin{proof}
This result can be proved in the same manner as \cite[Proposition~2.6~(4)]{AIK2014}.
Hence we omit here.
\end{proof}

\begin{theorem}\label{Th-3-1} 
For $k\in \mathbb{Z}$ and $ n\ge 0$, the following results hold.
\begin{enumerate}
\item[{\rm (1)}]  \begin{align*}
       {D}_{n}^{(k)}&amp;=4\sum_{m=0}^{\lfloor\frac{n}{2}\rfloor}\frac{1}{(2m+1)^{k+1}}\sum_{p=1}^{2m+1}
       \sum_{q=0}^{n-2m}(2^{p+q+1}-1)\binom{n}{q}\begin{bmatrix}
2m+1\\ p
\end{bmatrix} \left\{\begin{matrix} {n-q}\\{2m} \end{matrix}\right\} \frac{B_{p+q+1}}{p+q+1},
\end{align*}
where $B_n$ ($=C_n^{(1)}$) are the Bernoulli numbers, and
\item[{\rm (2)}]
 \begin{align*}
  {D}_{n}^{(k)}=\sum_{m=0}^{\lfloor\frac{n}{2}\rfloor}\frac{1}{(2m+1)^{k+1}}
  \sum_{p=2m}^{n}\frac{(-1)^{p} (p+1)!}{2^{p}}\binom{p}{2m}\left\{\begin{matrix} {n+1}\\{p+1} \end{matrix}\right\}.
\end{align*}
\end{enumerate}
\end{theorem}
\begin{proof}
We may express \eqref{eq-1-6} as
\begin{align}\label{eq-3-4}
\sum_{n=0}^{\infty}D_n^{(k)}\frac{t^n}{n!}&amp;=\frac{\mathrm{A}_{k}(\tanh (t/2))}{\sinh t}\nonumber \\
&amp;=2 \sum_{m=0}^{\infty} \frac{(\tanh (t/2))^{2m+1}}{(2m+1)^k}\frac{1}{\sinh t}\nonumber\\
&amp;=4 \sum_{m=0}^{\infty}\frac{1}{(2m+1)^k}\frac{e^t(e^t-1)^{2m}}{(e^t+1)^{2m+2}}.&amp;&amp;
\end{align}

Since
\begin{align} \label{eq-3-5}
\frac{1}{(x+1)^{n+1}}&amp;=\frac{(-1)^n}{n!}\left(\frac{d}{dx}\right)^n\frac{1}{x+1},  &amp;&amp;
\end{align}
we see that by setting $x=e^t$ and introducing Lemma~\ref{Lem-3-1},
\begin{align} \label{eq-3-8}
\frac{e^{nt}}{(e^t+1)^{n+1}}&amp;=\frac{1}{n!}\sum_{p=1}^{n}(-1)^{p} \begin{bmatrix}
n\\ p
\end{bmatrix}\left(\frac{d}{dt}\right)^p\frac{1}{e^t+1}.  &amp;&amp;
\end{align}
Moreover, from the generating functions
\begin{align*} 
\frac{t}{e^t-1}&amp;=\sum_{q=0}^\infty  B_q \frac{t^q}{q!},
\end{align*}
and 
\begin{align*} 
\frac{1}{e^t+1}&amp;=\frac{1}{e^t-1}-\frac{2}{e^{2t}-1},
\end{align*}
we find that
\begin{align*} 
\frac{1}{e^t+1}&amp;=\sum_{q=0}^\infty (1-2^q)B_q \frac{t^{q-1}}{q!}.
\end{align*}
Taking the $p$-th derivative on both sides yields
\begin{align*} 
\left(\frac{d}{dt}\right)^p\left(\frac{1}{e^t+1}\right)&amp;=\sum_{q=p+1}^\infty  (1-2^q)
\frac{B_q}q \frac{t^{q-p-1}}{(q-p-1)!}
=\sum_{q=0}^\infty  (1-2^{p+q+1})\frac{B_{p+q+1}}{p+q+1} \frac{t^{q}}{q!}.
\end{align*}
Now we substitute the above result into \eqref{eq-3-8} to obtain
\begin{align*} 
\frac{e^{nt}}{(e^t+1)^{n+1}}&amp;=\frac1{n!}\sum_{p=1}^n (-1)^{p} \begin{bmatrix}
n\\ p
\end{bmatrix}\sum_{q=0}^\infty  (1-2^{p+q+1})\frac{B_{p+q+1}}{p+q+1} \frac{t^{q}}{q!}\\
&amp;=\frac1{n!}\sum_{q=0}^\infty \sum_{p=1}^n (-1)^{p} \begin{bmatrix}
n\\p
\end{bmatrix} (1-2^{p+q+1})\frac{B_{p+q+1}}{p+q+1} \frac{t^{q}}{q!}.
\end{align*}
Hence we obtain
\begin{align*} 
\frac{e^{t}}{(e^{t}+1)^{2m+2}}&amp;=\frac{e^{-(2m+1)t}}{(e^{-t}+1)^{2m+2}}\\
&amp;=\frac1{(2m+1)!}\sum_{q=0}^\infty \sum_{p=1}^{2m+1} (-1)^{p+q} \begin{bmatrix}
2m+1\\
p
\end{bmatrix} (1-2^{p+q+1})\frac{B_{p+q+1}}{p+q+1} \frac{t^{q}}{q!}.
\end{align*}
With the aid of the generating function given by \cite[Proposition~2.6~(7)]{AIK2014}
and noting that $\left\{\begin{matrix} {s}\\{2m} \end{matrix}\right\}=0$ if
$s&lt;2m$ and
\begin{align*} 
(e^{t}-1)^{2m}&amp;=(2m)!\sum_{s=0}^\infty \left\{\begin{matrix} {s}\\{2m} \end{matrix}\right\} \frac{t^{s}}{s!},
\end{align*}
we arrive at
\begin{align*}
&amp;\frac{e^{t}(e^t-1)^{2m}}{(e^{t}+1)^{2m+2}}\\
&amp;=\frac1{2m+1}\sum_{q=0}^\infty \sum_{s=0}^\infty\sum_{p=1}^{2m+1} (-1)^{p+q} (1-2^{p+q+1}) 
\begin{bmatrix}
2m+1\\ p
\end{bmatrix}\left\{\begin{matrix} {s}\\{2m} \end{matrix}\right\}\frac{B_{p+q+1}}{p+q+1} \frac{t^{q+s}}{q!s!} \\
&amp;=\frac{1}{2m+1}\sum_{n=0}^\infty\sum_{q=0}^n \sum_{p=1}^{2m+1} (-1)^{p+q} (1-2^{p+q+1})
\binom{n}{q} \begin{bmatrix}
2m+1 \\p
\end{bmatrix}\left\{\begin{matrix} {n-q}\\{2m} \end{matrix}\right\}\frac{ B_{p+q+1}}{p+q+1}\frac{t^n}{n!}. \nonumber \\
\end{align*}
Substituting the above result into \eqref{eq-3-4}, we have 
\begin{align*}
&amp;\sum_{n=0}^{\infty}D_n^{(k)}\frac{t^n}{n!}\\
&amp;=4 \sum_{m=0}^{\infty}\frac{1}{(2m+1)^{k+1}}\sum_{n=0}^\infty\sum_{q=0}^n 
\sum_{p=1}^{2m+1} (-1)^{p+q} (1-2^{p+q+1})\\
&amp; \qquad \times \binom{n}{q} \begin{bmatrix}
2m+1 \\p
\end{bmatrix}\left\{\begin{matrix} {n-q}\\{2m} \end{matrix}\right\}\frac{ B_{p+q+1}}{p+q+1}\frac{t^n}{n!} \nonumber \\
&amp;=4 \sum_{n=0}^\infty\sum_{m=0}^{\lfloor\frac{n}{2}\rfloor}\frac{1}{(2m+1)^{k+1}}
\sum_{p=1}^{2m+1}\sum_{q=0}^{n-2m} (2^{p+q+1}-1)
\binom{n}{q} \begin{bmatrix}
2m+1 \\p
\end{bmatrix}\left\{\begin{matrix} {n-q}\\{2m} \end{matrix}\right\}\frac{ B_{p+q+1}}{p+q+1}\frac{t^n}{n!}.
\end{align*}
In obtaining the above result, we have used $B_{p+q+1}=0$ for $p+q\ge1$ and even,
while $\left\{\begin{matrix} {n-q}\\{2m} \end{matrix}\right\}=0$ for $n-q&lt;2m$.
By equating like powers, we arrive at the first result in the theorem.

To prove the second result, we require a formula from \cite{Cvi2011} for the higher order 
tangent numbers, $T_{n,m}$, whose generating function is  
\begin{align}\label{eq-3-12}
\frac{\tan^m t}{m!}= \sum_{n=m}^{\infty} T_{n,m}\frac{t^n}{n!}.
\end{align}
The formula is 
\begin{align}\label{eq-3-13}
T_{n,m}=\frac{i^{n-m}}{m!}\sum_{p=m}^{n}(-2)^{n-p}p!\binom{p-1}{m-1} \left\{\begin{matrix} {n}\\{p} \end{matrix}\right\}.
\end{align}
From \eqref{eq-1-6}, 
\begin{align}\label{eq-3-14}
\sum_{n=0}^{\infty}D_n^{(k)}\frac{t^n}{n!}&amp;=\frac{\mathrm{A}_{k}(\tanh (t/2))}{\sinh t}
=\frac{d}{dt}\mathrm{A}_{k+1}(\tanh (t/2))\nonumber \\
&amp;=2\frac{d}{dt} \sum_{m=0}^{\infty}\frac{(\tanh (t/2))^{2m+1}}{(2m+1)^{k+1}}.&amp;&amp;
\end{align}
By using $\tanh t=-i\tan (it)$ and equations \eqref{eq-3-12} and \eqref{eq-3-13}, we can write
\begin{align*}
(\tanh (t/2))^m&amp;=(-i)^m m!\sum_{n=m}^\infty T_{n,m}\frac{i^n}{2^n}\frac{t^n}{n!}\\
&amp;=(-i)^m (-1)^{\frac{n-m}{2}} \sum_{n=m}^\infty \sum_{p=m}^{n}(-2)^{n-p} p!\binom{p-1}{m-1}
\left\{\begin{matrix} {n}\\{p} \end{matrix}\right\}\frac{i^n}{2^n}\frac{t^n}{n!}\\
&amp;=(-1)^m \sum_{n=m}^\infty \sum_{p=m}^{n}(-1)^{p}\frac{p!}{2^p} \binom{p-1}{m-1}\left\{\begin{matrix} {n}\\{p} \end{matrix}\right\}\frac{t^n}{n!}.
\end{align*}
Therefore, we find that
\begin{align*}
\sum_{n=0}^{\infty}D_n^{(k)}\frac{t^n}{n!}&amp;=\sum_{m=0}^{\infty}\frac{1}{(2m+1)^{k+1}}
\sum_{n=2m+1}^\infty  
\sum_{p=2m+1}^{n}(-1)^{p+1} \frac{p!}{2^{p-1}}\binom{p-1}{2m}\left\{\begin{matrix} {n}\\{p} \end{matrix}\right\}\frac{t^{n-1}}{(n-1)!}\\
&amp;=\sum_{m=0}^{\infty}\frac{1}{(2m+1)^{k+1}}\sum_{n=2m}^\infty \sum_{p=2m}^{n}(-1)^{p} 
\frac{(p+1)!}{2^{p}}\binom{p}{2m}\left\{\begin{matrix} {n+1}\\{p+1} \end{matrix}\right\}\frac{t^{n}}{{n}!}\\
&amp;=\sum_{n=0}^{\infty}\sum_{m=0}^{\lfloor\frac{n}{2}\rfloor} \frac{1}{(2m+1)^{k+1}}  
\sum_{p=2m}^{n} \frac{(-1)^{p}(p+1)!}{2^{p}}\binom{p}{2m}\left\{\begin{matrix} {n+1}\\{p+1} \end{matrix}\right\}\frac{t^{n}}{{n}!}.
\end{align*}
By equating like powers of $t$, we arrive at the second result in the theorem.
\end{proof}


\section{Duality}\label{sec-2}

We now turn our attention to the duality property of the polycosecant numbers.
We shall present two different proofs using the same generating function. The first proof is based
on a closed symmetric formula for the generating function, while the second is more indirect
and complicated. However, we have decided to include the latter proof since it reveals
fascinating results, especially regarding hyperbolic trigonometric functions.

Here is a table of $D_{2n}^{(-2k-1)}$ for small $k$ and $n$.
\renewcommand{\arraystretch}{1.1}
\begin{table}[h]
\begin{center}
\begin{tabular} {|c|c|c|c|c|c|c|c|c|} \hline
\backslashbox{$\ \ k$}{$n\ \ $} &amp; $ 0 $ &amp; $ 1 $ &amp; $ 2 $ &amp; $ 3 $ &amp; $ 4 $ &amp; $ 5 $ 
\\ \hline  

$0$ &amp; $1$ &amp; $1$ &amp; $1$ &amp; $1$ &amp; $1$ &amp; $1$  \\ \hline

$1$ &amp; $1$ &amp; $13$ &amp; $121$ &amp; $1093$ &amp; $9841$ &amp; $88573$  \\ \hline

$2$ &amp; $1$ &amp; $121$ &amp; $4081$ &amp; $111721$ &amp; $2880481$ &amp; $72799321$  \\ \hline

$3$ &amp; $1$ &amp; $1093$ &amp; $111721$ &amp; $7256173$ &amp; $403087441$ &amp; $20966597653$  \\ \hline

$4$ &amp; $1$ &amp; $9841$ &amp; $2880481$ &amp; $403087441$ &amp; $42931692481$ &amp; $4032800405041$  \\ \hline

$5$ &amp; $1$ &amp; $88573$ &amp; $72799321$ &amp; $20966597653$ &amp; $4032800405041$ &amp; $638704166793133$ \\ \hline
\end{tabular}
\end{center}
\caption{$ D_{2n}^{(-2k-1)} \: (0\le k \leq 5,\; 0\le n\le 5) $}
\end{table}

\begin{theorem}\label{Th-2-1} For $n,k\in \mathbb{Z}_{\geq 0}$, the polycosecant numbers possess
the duality property of 
\begin{equation}
D_{2n}^{(-2k-1)}=D_{2k}^{(-2n-1)}. \label{eq-2-1}
\end{equation}
\end{theorem}


\begin{proof}[First proof] We show that the generating function of $D_{2n}^{(-2k-1)}$,
\begin{equation}\label{genF}
F(x,y):=\sum_{n=0}^\infty \sum_{k=0}^\infty D_{2n}^{(-2k-1)}\frac{x^{2n}}{(2n)!}\frac{y^{2k}}{(2k)!},
\end{equation}
is symmetric in $x$ and $y$. This is ensured by the following closed formula for $F(x,y)$.
\end{proof}

\begin{proposition}\label{C-2-7}  Let 
\[ G(x,y)=\frac{e^{x+y}}{(1+e^x+e^y-e^{x+y})^2}.\]  Then one finds
$$F(x,y)=G(x,y)+G(x,-y)+G(-x,y)+G(-x,-y).$$
\end{proposition}

\begin{proof} 

We first compute the generating function of all $D_n^{(-k)}$,
\begin{equation}
f(x,y)=\sum_{n=0}^\infty \sum_{k=0}^\infty D_{n}^{(-k)}\frac{x^{n}}{n!}\frac{y^k}{k!}.  \label{eq-2-13}
\end{equation}
We claim that the formula 
\begin{equation}
f(x,y)=\frac{e^x(e^y-1)}{1+e^x+e^{y}-e^{x+y}}+\frac{e^{-x}(e^y-1)}{1+e^{-x}+e^y-e^{-x+y}}\label{eq-2-14}
\end{equation}
holds.  To prove this, we first observe that, by definition,
\begin{align*}
f(x,y)&amp;=\sum_{k=0}^\infty \frac{\mathrm{A}_{-k}(\tanh(x/2))}{\sinh x}\frac{y^k}{k!}\\
&amp; =\frac{2}{\sinh x}\sum_{k=0}^\infty \sum_{n=0}^\infty (2n+1)^k (\tanh(x/2))^{2n+1}\frac{y^k}{k!}.
\end{align*}
Noting that
\begin{equation}\label{eq3-4} 
2\sum_{n=0}^\infty (2n+1)^k t^{2n+1}=2\left(t\frac{d}{dt}\right)^{k}\frac{t}{1-t^2}
=\left(t\frac{d}{dt}\right)^{k}\left(\frac1{1-t}-\frac1{1+t}\right),
\end{equation}
and by the standard formula ({\it cf., e.g.,} \cite[Proposition~2.6~(4)]{AIK2014})
\[ \left(t\frac{d}{dt}\right)^{k}=\sum_{m=1}^{k}\left\{\begin{matrix} {k}\\{m} \end{matrix}\right\}t^m\left(\frac{d}{dt}\right)^{m}, \]
we find that the right-hand side of \eqref{eq3-4} becomes
\begin{align*}
&amp; \sum_{m=1}^{k}\left\{\begin{matrix} {k}\\{m} \end{matrix}\right\}t^m\left(\frac{d}{dt}\right)^{m}\left(\frac{1}{1-t}-\frac{1}{1+t}\right)\\
&amp; \ \ =\sum_{m=1}^{k}\left\{\begin{matrix} {k}\\{m} \end{matrix}\right\} m!\left(\frac{t^m}{(1-t)^{m+1}}-\frac{(-t)^m}{(1+t)^{m+1}}\right).
\end{align*}
Therefore, by setting $t=\tanh(x/2)$ and noting $t/(1-t)=(e^x-1)/2,\,-t/(1+t)=(e^{-x}-1)/2$, 
$(\sinh x) (1-t)=e^{-x}(e^x-1)$, $(\sinh x) (1+t)=e^x-1$, we arrive at
\begin{align*}
f(x,y)&amp;=\frac{1}{\sinh x}\sum_{k=0}^\infty\sum_{m=1}^{k}\left\{\begin{matrix} {k}\\{m} \end{matrix}\right\} m!
\left(\frac{t^m}{(1-t)^{m+1}}-\frac{(-t)^m}{(1+t)^{m+1}}\right)\frac{y^k}{k!}\\
&amp; =\sum_{k=0}^\infty \sum_{m=1}^k \left\{\begin{matrix} {k}\\{m} \end{matrix}\right\}m!\bigg( \frac{e^x}{e^x-1}\left(\frac{e^x-1}{2}\right)^m
-\frac{1}{e^x-1}\left(\frac{e^{-x}-1}{2}\right)^m\bigg)\frac{y^k}{k!}\\
&amp; =\sum_{m=1}^\infty (e^y-1)^m \bigg( \frac{e^x}{e^x-1}\left(\frac{e^x-1}{2}\right)^m
-\frac{1}{e^x-1}\left(\frac{e^{-x}-1}{2}\right)^m\bigg)\\
&amp; = \frac{e^x}{e^x-1}\cdot\frac{(e^y-1)(e^x-1)}{2-(e^y-1)(e^x-1)}-\frac{1}{e^x-1}
\cdot\frac{(e^y-1)(e^{-x}-1)}{2-(e^y-1)(e^{-x}-1)}\\
&amp;=\frac{e^x(e^y-1)}{1+e^x+e^{y}-e^{x+y}}+\frac{e^{-x}(e^y-1)}{1+e^{-x}+e^y-e^{-x+y}}.
\end{align*}
This proves the identity \eqref{eq-2-14}.
From \eqref{eq-2-14} we see that $f(x,y)$ is even in $x$, and so we have 
$$\frac{f(x,y)-f(x,-y)}{2}=\sum_{n=0}^\infty \sum_{k=0}^\infty 
D_{2n}^{(-2k-1)}\frac{x^{2n}}{(2n)!}\frac{y^{2k+1}}{(2k+1)!}.$$
Our generating function $F(x,y)$ is the derivative of this relation with respect to $y$, and 
Proposition~\ref{C-2-7} follows from a straightforward calculation, and by the symmetry
of $F(x,y)$ in $x$ and $y$, Theorem~\ref{Th-2-1} is proved. 

\end{proof}

Before presenting the second proof of Theorem~\ref{Th-2-1}, we require several lemmas.

\begin{lemma}\label{Lem-2-2} 
The function $F(x,y)$ defined by \eqref{genF} can be expressed as
\begin{equation*}
F(x,y)=2\sum_{n=0}^{\infty}\frac{\partial}{\partial x}\left(\tanh^{2n+1}(x/2)\right)\cosh((2n+1)y). 
\end{equation*}
\end{lemma}

\begin{proof}
By \eqref{eq-1-6} and \eqref{genF}, we have
\begin{align*}
F(x,y)&amp;= 2\sum_{k=0}^\infty \frac{\mathrm{A}_{-2k-1}(\tanh(x/2))}{\sinh(x)}\frac{y^{2k}}{(2k)!}\\
&amp;= \frac{2}{\sinh(x)}\sum_{k=0}^\infty \sum_{n=0}^\infty (2n+1)^{2k+1}\tanh^{2n+1}(x/2)\frac{y^{2k}}{(2k)!}\\
&amp;= \frac{2}{\sinh(x)}\sum_{n=0}^\infty (2n+1)\tanh^{2n+1}(x/2)\cosh((2n+1)y)\\
&amp;= \frac{1}{\sinh(x/2)\cosh(x/2)}\sum_{n=0}^\infty (2n+1)\tanh^{2n}(x/2)\frac{\sinh(x/2)}{\cosh(x/2)}\cosh((2n+1)y)\\
&amp;=2\sum_{n=0}^{\infty}\frac{\partial}{\partial x}\left(\tanh^{2n+1}(x/2)\right)\cosh((2n+1)y).
\end{align*}
This completes the proof.
\end{proof}

We write
$$F(x,y)=\sum_{m=0}^\infty g_m(x)\frac{y^{2m}}{(2m)!}
=\sum_{m=0}^\infty h_m(y)\frac{x^{2m}}{(2m)!}.$$
If one can show that $g_m(x)=h_m(x)$ for all $m\geq 0$, then the second proof will be complete. 

First, we consider $g_m(x)$. Expanding $\cosh((2n+1)y)$ in Lemma~\ref{Lem-2-2} and equating like powers of $y$,
we obtain
\begin{align*}
g_m(x)&amp;=\left(\frac{\partial}{\partial y}\right)^{2m}F(x,y)\,\bigg|_{y=0}
=2\frac{d}{dx}\sum_{n=0}^\infty (2n+1)^{2m}\tanh^{2n+1}(x/2).
\end{align*}
Next we note that
\begin{equation}
\sum_{n=0}^\infty (2n+1)^{2m}t^{2n+1}=\left(t\frac{d}{dt}\right)^{2m}\sum_{n=0}^\infty t^{2n+1}
=\left(t\frac{d}{dt}\right)^{2m}\frac{t}{1-t^2}.\label{eq-2-3}
\end{equation}
Setting $t=\tanh(x/2)$ and noting
$$dt=\frac{1}{2}\frac{1}{\cosh^2(x/2)}dx,\quad \frac{t}{1-t^2}
=\frac{\tanh(x/2)}{1-\tanh^2(x/2)}=\frac{1}{2}\sinh x,$$
we have
$$t\frac{d}{dt}=\tanh(x/2)\cdot 2\cosh^2(x/2)\frac{d}{dx}=\sinh x\, \frac{d}{dx}.$$
Therefore we obtain
\begin{equation}
g_m(x)=\frac{d}{dx}\left(\sinh x\,\frac{d}{dx}\right)^{2m}\sinh x. \label{eq-2-4}
\end{equation}
We can explicitly write down the right-hand side by using the following lemma. 

For $m\in \mathbb{Z}_{\geq 0}$, we define sequences $( a_i^{(m)})_{0\leq i\leq m}\subset \mathbb{Q}$ 
inductively by 
\begin{equation}
\begin{split}
&amp; a_0^{(0)}=1, \\
&amp; a_i^{(m)}=\frac{1}{2}\left( i(2i-1)a_{i-1}^{(m-1)}-(2i+1)^2a_{i}^{(m-1)}+(i+1)(2i+3)a_{i+1}^{(m-1)}\right)
\quad (m\geq 1), 
\end{split}
\label{eq-2-5}
\end{equation}
where we formally interpret $a_{i}^{(m)}=0$ for $i&lt;0$ or $i&gt;m$. 


\begin{lemma}\label{Lem-2-3} For $m\in \mathbb{Z}_{\geq 0}$, 
\begin{equation}
\left(\sinh x\,\frac{d}{dx}\right)^{2m}\sinh x=\sum_{i=0}^{m}a_i^{(m)}\sinh((2i+1)x).  \label{eq-2-6}
\end{equation}
\end{lemma}

\begin{proof}
We give the proof by induction on $m$. For $m=0$, the identity trivially holds. We assume 
\begin{equation*}
\left(\sinh x\,\frac{d}{dx}\right)^{2(m-1)}\sinh x=\sum_{i=0}^{m-1}a_i^{(m-1)}\sinh((2i+1)x).  
\end{equation*}
Using 
$$\cosh(kx)\sinh(x)=\frac{1}{2}\left( \sinh((k+1)x)-\sinh((k-1)x)\right),$$
we have
\begin{align*}
&amp; \left(\sinh x\,\frac{d}{dx}\right)^{2m-1}\sinh x=\frac{1}{2}\sum_{i=0}^{m-1}(2i+1)a_i^{(m-1)}
\left(\sinh((2i+2)x)-\sinh(2ix)\right), 
\end{align*}
and  
\begin{align*}
&amp; \left(\sinh x\,\frac{d}{dx}\right)^{2m}\sinh x\\
&amp; \ =\sum_{i=0}^{m-1}(2i+1)a_i^{(m-1)}\bigg( \frac{i+1}{2}\left(\sinh((2i+3)x)-\sinh((2i+1)x)\right)\\
&amp; \qquad \qquad  -\frac{i}{2}\left(\sinh((2i+1)x)-\sinh((2i-1)x)\right)\bigg)\\
&amp; \ =\frac{1}{2}\sum_{i=1}^{m}i(2i-1)a_{i-1}^{(m-1)}\sinh((2i+1)x)\\
&amp; \qquad -\frac{1}{2}\sum_{i=0}^{m-1}(2i+1)^2a_{i}^{(m-1)}\sinh((2i+1)x)\\
&amp; \qquad +\frac{1}{2}\sum_{i=0}^{m-2}(i+1)(2i+3)a_{i+1}^{(m-1)}\sinh((2i+1)x).
\end{align*}
Thus we observe that the coefficients of $\sinh((2i+1)x)$ are in accordance with \eqref{eq-2-5}, 
thereby completing this proof by induction.
\end{proof}
Using this lemma, we obtain
\begin{equation}
g_m(x)=\sum_{i=0}^m (2i+1)a_i^{(m)}\cosh((2i+1)x).  \label{eq-2-7}
\end{equation}

Secondly, we compute $h_m(y)$. Again by using Lemma~\ref{Lem-2-2}, we have
\begin{align}
h_m(y)&amp;=\left(\frac{\partial}{\partial x}\right)^{2m}F(x,y)\,\bigg|_{x=0}\notag\\
      &amp;=2\sum_{n=0}^\infty \left(\frac{d}{dx}\right)^{2m+1}\left(\tanh^{2n+1}(x/2)\right) 
      \cosh((2n+1)y)\,\bigg|_{x=0} \notag\\
      &amp; =2\sum_{n=0}^{m} \left(\frac{d}{dx}\right)^{2m+1}\tanh^{2n+1}(x/2)\,\bigg|_{x=0}\cdot \cosh((2n+1)y) \label{eq-2-8}
\end{align}
because 
$$\tanh^{2n+1}(x/2)=\frac{x^{2n+1}}{2^{2n+1}}+O(x^{2n+2})\ \ (x\to 0).$$
We write down the right-hand side of \eqref{eq-2-8} by using the following lemma. 

\begin{lemma}\label{Lem-2-4} For $n,l\in \mathbb{Z}_{\geq 0}$, there exist sequences 
$(b_j^{(n,l)})_{0\leq j\leq l}\subset \mathbb{Q}$ such that
\begin{equation}
\left(\frac{d}{dx}\right)^{l}\tanh^{2n+1}(x/2)=\sum_{j=0}^{l}b_j^{(n,l)}
\tanh^{2n+1-l+2j}(x/2), \label{eq-2-9}
\end{equation}
where $b_j^{(n,l)}=0$ if $2n+1-l+2j&lt;0$. 
In particular, 
\begin{equation}
\left(\frac{d}{dx}\right)^{2m+1}\tanh^{2n+1}(x/2)\,\bigg|_{x=0}=b_{m-n}^{(n,2m+1)}. \label{eq-2-10}
\end{equation}
\end{lemma}

\begin{proof}
For each $n$, we can immediately obtain the form \eqref{eq-2-9} by induction on $l$, using the relation
$$\frac{d}{dx}\tanh^{2n+1}(x/2)=\frac{2n+1}{2}\left(\tanh^{2n}(x/2)-\tanh^{2n+2}(x/2)\right).$$
\end{proof}

Combining Lemma \ref{Lem-2-4} and \eqref{eq-2-8}, we obtain
\begin{equation}
h_m(y)=2\sum_{n=0}^{m}b_{m-n}^{(n,2m+1)}\cosh((2n+1)y). \label{eq-2-11}
\end{equation}

Now we are going to show $2b_{m-n}^{(n,2m+1)}=(2i+1)a_i^{(m)}$,
which implies $g_m(x)=h_m(x)$. For $m,n\in \mathbb{Z}_{\geq 0}$ with $n\leq m$, set 
$\widetilde{b}_n^{(m)}=2b_{m-n}^{(n,2m+1)}$. Then, by \eqref{eq-2-10}, we have $\widetilde{b}_0^{(0)}=1$. 
Furthermore the following lemma holds.

\begin{lemma}\label{Lem-2-5}\ For $m\in \mathbb{Z}_{\geq 1}$, the $\widetilde{b}_n^{(m)}$ satisfy the recurrence 
relation given by
\begin{equation}
\widetilde{b}_n^{(m)}=\frac{2n+1}{2}\left(n\widetilde{b}_{n-1}^{(m-1)}-(2n+1)\widetilde{b}_n^{(m-1)}+(n+1)
\widetilde{b}_{n+1}^{(m-1)}\right)\quad (n\leq m),  \label{eq-2-12}
\end{equation}
where we interpret $b_{i}^{(k)}=0$ for $i&lt;0$ or $i&gt;k$. 
\end{lemma}

\begin{proof}
It follows from~\eqref{eq-2-9} that
\begin{equation}\label{eq3-14}
\left(\frac{d}{dx}\right)^{2m+1}\tanh^{2n+1}(x/2)=\sum_{j=0}^{2m+1}b_j^{(n,2m+1)}
\tanh^{2n-2m+2j}(x/2).
\end{equation}
Differentiating twice and using~\eqref{eq-2-9}, we see that the left-hand side is equal to
\begin{align*}
&amp; \left(\frac{d}{dx}\right)^{2m}\left(\frac{2n+1}{2}\tanh^{2n}(x/2)-\tanh^{2n+2}(x/2)\right)\\
&amp; =\frac{2n+1}{2}\left(\frac{d}{dx}\right)^{2m-1}\bigg( n\tanh^{2n-1}(x/2)-(2n+1)\tanh^{2n+1}(x/2)
+(n+1)\tanh^{2n+3}(x/2)\bigg)\\
&amp; =\frac{2n+1}{2}\bigg( n\sum_{j=0}^{2m-1}b_j^{(n-1,2m-1)}\tanh^{2n-2m+2j}(x/2)\\
&amp; \qquad\qquad  -(2n+1)\sum_{j=0}^{2m-1}b_j^{(n,2m-1)}\tanh^{2n-2m+2+2j}(x/2)\\
&amp; \qquad\qquad +(n+1)\sum_{j=0}^{2m-1}b_j^{(n+1,2m-1)}\tanh^{2n-2m+4+2j}(x/2)\bigg).
\end{align*}
If we let $x\to 0$, the above result goes to
\begin{align*}
&amp; \frac{2n+1}{2}\bigg( nb_{m-n}^{(n-1,2m-1)}  -(2n+1)b_{m-n-1}^{(n,2m-1)}+(n+1)b_{m-n-2}^{(n+1,2m-1)}\bigg)\\
&amp; \ =\frac{2n+1}{4}\bigg( n\widetilde{b}_{n-1}^{(m-1)}  -(2n+1)\widetilde{b}_{n}^{(m-1)}+(n+1)\widetilde{b}_{n+1}^{(m-1)}\bigg).
\end{align*} 
On the other-hand, the right-hand side of equation~\eqref{eq3-14} 
tends to $b_{m-n}^{(n,2m+1)}=\widetilde{b}_n^{(m)}/2$ as $x\to 0$. Thus we obtain \eqref{eq-2-12}.
\end{proof}

\begin{proof}[Second proof of Theorem \ref{Th-2-1}]  For $(a_i^{(m)})_{0\leq i\leq m}$ defined by \eqref{eq-2-5}, 
set $\widetilde{a}_{i}^{(m)}=(2i+1)a_i^{(m)}$. Then \eqref{eq-2-5} can be written as $\widetilde{a}_0^{(0)}=1$ and 
\begin{equation*}
\widetilde{a}_i^{(m)}=\frac{2i+1}{2}\left(i\widetilde{a}_{i-1}^{(m-1)}-(2i+1)^2\widetilde{a}_i^{(m-1)}+(i+1)\widetilde{a}_{i+1}^{(m-1)}\right)
\end{equation*}
which has exactly the same form as the recurrence relation \eqref{eq-2-12} for $\widetilde{b}_n^{(m)}$.
Therefore one concludes $\widetilde{a}_n^{(m)}=\widetilde{b}_n^{(m)}$. Comparing \eqref{eq-2-7} and \eqref{eq-2-11}, we obtain $g_m(x)=h_m(x)$. Thus we complete our second
proof of Theorem \ref{Th-2-1}.
\end{proof}


\section{Acknowledgments}
The authors wish to express their sincere gratitude to the referee for valuable suggestions and comments. 
This work was supported by Japan Society for the Promotion of Science, Grant-in-Aid for Scientific 
Research (S) 16H06336 (M. Kaneko), and (C) 18K03218 (H. Tsumura).


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\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}: Primary 11B68, Secondary 11M32, 11M99.

\noindent {\it Keywords}: poly-Bernoulli number, multiple zeta value, multiple zeta function, polylogarithm.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A001896},
\seqnum{A001897},
\seqnum{A008277},
\seqnum{A027641}, 
\seqnum{A027642},  and
\seqnum{A099594}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received August 1 2019;
revised versions received  April 12 2020; May 12 2020; May 18 2020.
Published in {\it Journal of Integer Sequences}, June 9 2020.

\bigskip
\hrule
\bigskip

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


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