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\vskip 1cm{\LARGE\bf 
Two New Explicit Formulas for the \\
\vskip .1in
Even-Indexed Bernoulli Numbers
}
\vskip 1cm
\large
Sumit Kumar Jha\\
International Institute of Information Technology\\
Hyderabad-500 032 \\
India\\
\href{mailto:kumarjha.sumit@research.iiit.ac.in}{\tt kumarjha.sumit@research.iiit.ac.in}\\
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\begin{abstract}
We give two new explicit formulas for the even-indexed Bernoulli numbers in terms of the Stirling numbers of the second kind.
\end{abstract}
\section{Introduction}
\begin{definition}
The \emph{Bernoulli numbers} $B_{n}$ can be defined by the 
following generating function:
\begin{equation*}
    \frac{t}{e^{t}-1}=\sum_{n\geq 0}\frac{B_{n}t^{n}}{n!},
\end{equation*}
where $|t|<2\pi$.
\end{definition}
\begin{definition}
The \emph{Stirling number of the second kind}, denoted by $S(n,m)$, is the number of ways of partitioning a set of $n$ elements into $m$ nonempty sets.
\end{definition}
There are many explicit formulas known for the Bernoulli numbers \cite{Gould, Gould2, Jha1, Jha3, Jha2}.  For example, all of the formulas below express the Bernoulli numbers explicitly in terms of the Stirling numbers of the second 
kind:
\begin{align}
   B_{r}&=\sum_{k=1}^{r}(-1)^{k}\cdot k! \cdot \frac{S(r,k)}{k+1}, \nonumber \\
   B_{r}&=\sum_{k=1}^{r}(-1)^{k-1}\cdot (k-1)! \cdot \frac{S(r,k)}{k+1}, \nonumber \\
 B_{r+1}&=\sum_{k=1}^{r}\frac{(-1)^{k-1}\, k!\, S(r,k)}{(k+1)(k+2)},\nonumber  \\
B_{r}&=\frac{r}{1-2^{r}}\sum_{k=1}^{r-1}\frac{(-1)^{k}\,k!\, S(r-1,k)}{2^{k+1}}\nonumber , \\
B_{r+1}&=\frac{(-1)^r\cdot (r+1)\cdot 2^{r-1} }{2^{r+1}-1}\sum_{k=1}^{r}\frac{(-1)^{k}\, k!\, S(r,k)}{k+1}\cdot 2^{-2k}\binom{2k}{k}, \nonumber \\
B_{r}&=\sum_{i=0}^{r}(-1)^{i}\frac{\binom{r+1}{i+1}}{\binom{r+i}{i}}S(r+i,i),  \nonumber\\
B_{r+1}&=-\frac{r+1}{4(1+2^{-(r+1)}(1-2^{-r}))}\left(\sum_{k=1}^{r}(-1)^{k}\cdot \frac{S(r,k)}{k+1}\cdot \left(\frac{3}{4}\right)^{(k)}+4^{-r}E_{r}\right),
    \label{main3}
\end{align}
where $(E_{r})$ denotes the \emph{Euler numbers} defined by the 
following generating function:
$${\displaystyle {\frac {1}{\cosh t}}={\frac {2}{e^{t}+e^{-t}}}=\sum _{n=0}^{\infty }{\frac {E_{n}}{n!}}\cdot t^{n}}.$$
In the following section,
we derive two new explicit formulas for the even-indexed Bernoulli numbers in terms of the Stirling numbers of the second kind.
\section{Main results}
Our main results are the following.
\begin{theorem}
\label{mainTheorem}
For all positive integers $r$ we have
\begin{equation}
\label{main1}
    B_{2r}=\frac{-4r}{3(3-3^{1-2r})}\sum_{k=1}^{2r-1}(-1)^{k}\frac{S(2r-1,k)}{k+1}\left(\frac{2}{3}\right)^{(k)},
\end{equation}
and
\begin{equation}
\label{main2}
    B_{2r}=\frac{2r}{3((1-2^{1-2r})(1-3^{1-2r})-2)}\sum_{k=1}^{2r-1}(-1)^{k}\frac{S(2r-1,k)}{k+1}\left(\frac{5}{6}\right)^{(k)},
\end{equation}
where $S(2r-1,k)$ denotes the Stirling numbers of the second kind,
and $$x^{(n)}=x(x+1)(x+2)\cdots (x+n-1)$$ denotes the rising factorial.
\end{theorem}
To prove the above result, we first recall the following fact
\begin{theorem}\cite{Sun1}
The $n$th Bernoulli polynomial, $B_{n}(t)$, defined by 
\begin{equation*}
B_{n}(t)=\sum_{j=0}^{n}\binom{n}{j}B_{n-j} t^{j}
\end{equation*}
takes the following `special' values at certain rational numbers with small denominators
\begin{equation*}
B_{n}(1)=B_{n}(0)=B_{n}\qquad \text{for $n\geq 1$},
\end{equation*}
\begin{equation*}
B_{n}\left(\frac{1}{2}\right)=(2^{1-n}-1)B_{n},
\end{equation*}
\begin{equation}
\label{3/4}
B_{2n}\left(\frac{1}{4}\right)=B_{2n}\left(\frac{3}{4}\right)=\frac{2-2^{2n}}{4^{2n}}B_{2n},
\end{equation}
\begin{equation}
\label{2/3}
B_{2n}\left(\frac{1}{3}\right)=B_{2n}\left(\frac{2}{3}\right)=\frac{3-3^{2n}}{2\cdot 3^{2n}}B_{2n},
\end{equation}
\begin{equation}
\label{5/6}
B_{2n}\left(\frac{1}{6}\right)=B_{2n}\left(\frac{5}{6}\right)=\frac{(2-2^{2n})(3-3^{2n})}{2\cdot 6^{2n}}B_{2n}.
\end{equation}
\end{theorem}
\begin{remark}
Granville and Sun \cite{Sun2} noted that, ``It is not known if $B_{n}(a/q)$ has as simple a `closed form' for any other rational $a/q$ with $1\leq a\leq q-1$ and $(a,q)=1$, though this has long been considered an interesting question."
\end{remark}
Our proof also requires the following result.
\begin{lemma}
For all $0<t<1$ and $l\in \mathbb{N}$ we have
\begin{equation}
\label{integral}
\normalfont
    \int_{0}^{\infty}x^{t-1}\frac{\Li_{-l}(-x)}{1+x}\, dx=\frac{\pi}{\sin{t\pi}}\left(\frac{B_{l+1}(1-t)-B_{l+1}}{l+1}\right),
\end{equation}
where $\normalfont \Li_{-l}(-x)$ is the negative polylogarithm function, and $B_{r}(1-t)$ denotes the Bernoulli polynomial.
\end{lemma}
\begin{proof}
Consider the following generating function \cite{polyLog}
\begin{equation}
\label{help}
    \frac{\Li_{-r}(-x)}{1+x}=\sum_{n=0}^{\infty}S_{r}(n)(-x)^{n},
\end{equation}
where $S_{r}(n)=\sum_{k=1}^{n}k^{r}$ for $n\geq 1$, and $S_{r}(0)=0$.

We use Ramanujan's master theorem (\textsc{RMT}) from \cite{Hardy} 
that states that
\begin{equation*}
    \int_{0}^{\infty} x^{t-1}\{ \phi(0)-x \phi(1)+x^{2} \phi(2)-\cdots\}\, dx=\frac{\pi}{\sin{t \pi}}\phi(-t),
\end{equation*}
where the integral is convergent for $0<\Re(t)<1$, and after certain conditions are satisfied by $\phi$.

Using \textsc{RMT} with Eq.~\eqref{help} gives us Eq.~\eqref{integral}.
\end{proof}
\begin{proof}[Proof of Theorem \ref{mainTheorem}]
Substituting  $t=2/3$ and $l=2r-1$ in Eq.~\eqref{integral} gives us
$$
\int_{0}^{\infty}x^{-1/3}\frac{\Li_{1-2r}(-x)}{1+x}\, dx=\frac{2\pi}{\sqrt{3}}\left(\frac{B_{2r}(1/3)-B_{2r}}{2r}\right).
$$
We use the following representation from the note \cite{Landsburg}
\begin{equation}
\label{LiRep}
    \Li_{1-2r}(-x)=\sum_{k=1}^{2r-1}\,k!S(2r-1,k)\left(\frac{1}{1+x}\right)^{k+1}(-x)^{k}
\end{equation}
to conclude that
\begin{align*}
    \int_{0}^{\infty}x^{-1/3}\frac{\Li_{1-2r}(-x)}{1+x}\, dx=\sum_{k=1}^{2r-1}(-1)^{k}\cdot k!\, S(2r-1,k)\int_{0}^{\infty}\frac{x^{k-1/3}}{(1+x)^{k+2}}\, dx\\
=\sum_{k=1}^{2r-1}(-1)^{k}\cdot k!\, S(2r-1,k)\frac{\Gamma(k+2/3)\Gamma(4/3)}{\Gamma(k+2)}\\
=\sum_{k=1}^{2r-1}(-1)^{k}\cdot \frac{S(2r-1,k)}{k+1}\left(\frac{2}{3}\right)^{(k)}\cdot \frac{2\pi}{3\sqrt{3}},
\end{align*}
where $\Gamma(\cdot)$ denotes the Gamma function. Recalling Eq.~\eqref{2/3} we have
\begin{align*}
    B_{2r}(1/3)-B_{2r}=-\left(3-3^{1-2r}\right)\frac{B_{2r}}{2}.
\end{align*}
Now we can readily conclude Eq.~\eqref{main1}.

To prove Eq.~\eqref{main2} we substitute $t=5/6$ and $l=2r-1$ in the 
Eq.~\eqref{integral} to get
$$
\int_{0}^{\infty}x^{-1/6}\frac{\Li_{1-2r}(-x)}{1+x}\, dx=2\pi\left(\frac{B_{2r}(1/6)-B_{2r}}{2r}\right).
$$
The representation \eqref{LiRep} also lets us to conclude that 
\begin{align*}
    \int_{0}^{\infty}x^{-1/6}\frac{\Li_{1-2r}(-x)}{1+x}\, dx=\sum_{k=1}^{2r-1}(-1)^{k}\cdot k!\, S(2r-1,k)\int_{0}^{\infty}\frac{x^{k-1/6}}{(1+x)^{k+2}}\, dx\\
=\sum_{k=1}^{2r-1}(-1)^{k}\cdot k!\, S(2r-1,k)\frac{\Gamma(k+5/6)\Gamma(7/6)}{\Gamma(k+2)}\\
=\sum_{k=1}^{2r-1}(-1)^{k}\cdot \frac{S(2r-1,k)}{k+1}\left(\frac{5}{6}\right)^{(k)}\cdot \frac{\pi}{3}.
\end{align*}
Recalling Eq.~\eqref{5/6} we have
\begin{align*}
    B_{2r}(1/6)-B_{2r}=\frac{B_{2r}}{2}((1-2^{1-2r})(1-3^{1-2r})-2).
\end{align*}
Now we can readily conclude Eq.~\eqref{main2}.
\end{proof}
\begin{remark}
Substituting $t=3/4$ and $l=2r-1$ in Eq.~\eqref{integral}, and using 
Eq.~\eqref{3/4} we can obtain
$$
B_{2r}=-\frac{2r}{4(1+2^{-2r}(1-2^{-(2r-1)}))}\sum_{k=1}^{2r-1}(-1)^{k}\cdot \frac{S(2r-1,k)}{k+1}\cdot \left(\frac{3}{4}\right)^{(k)}.
$$
The above is just a special case of Eq.~\eqref{main3} which was obtained in \cite{Jha2}.
\end{remark}

\begin{thebibliography}{10}

\bibitem{Gould}
H. W. Gould, Explicit formulas for Bernoulli numbers, \emph{Amer. Math. Monthly} \textbf{79} (1972), 44--51. 

\bibitem{Sun2} A. Granville and Z. W. Sun, Values of Bernoulli
polynomials, \emph{Pacific J. Math.} \textbf{172} (1996), 117--137.

\bibitem{Hardy}
G. H. Hardy, \emph{Ramanujan. Twelve Lectures on Subjects Suggested by His Life and Work},  3rd edition, Chelsea, 1978.

\bibitem{Jha1}
S. K. Jha, Two new explicit formulas for the Bernoulli numbers, preprint, 2019. Available at \url{https://arxiv.org/abs/1905.11216}.

\bibitem{Jha3}
S. K. Jha, A new explicit formula for the Bernoulli numbers in terms of the Stirling numbers of the second kind, preprint, 2019. Available at \url{https://osf.io/95gkw}.

\bibitem{Jha2}
S. K. Jha, A new explicit formula for Bernoulli numbers involving the Euler number, \emph{Mosc. J. Comb. Number Theory} \textbf{8} (2019), 385--387.

\bibitem{Landsburg}
S. E. Landsburg, Stirling numbers and polylogarithms, preprint, 2009. Available at \url{http://www.landsburg.com/query.pdf}.

\bibitem{polyLog}
L. Lewin, \emph{Polylogarithms and Associated Functions}, North-Holland, 1981.

\bibitem{Gould2}
J. Quaintance and H. W. Gould, \emph{Combinatorial Identities For Stirling
Numbers: The Unpublished Notes Of H. W. Gould}, World Scientific, 2015.

\bibitem{Sun1}
Z. H. Sun, Congruences involving Bernoulli and Euler numbers, \emph{J. Number Theory} \textbf{128} (2008), 280--312.

\end{thebibliography}

\bigskip
\hrule
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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B68. 

\noindent \emph{Keywords:} Bernoulli number, Stirling number of the
second kind, polylogarithm function, Ramanujan's master theorem.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000367},
\seqnum{A002445},
\seqnum{A008277},
\seqnum{A027641}, and
\seqnum{A027642}.)

\bigskip
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\vspace*{+.1in}
\noindent
Received November 6 2019;
revised versions received November 7 2019;
January 19 2020; January 20 2020; January 23 2020; 
January 27 2020; February 10 2020; February 20 2020.
Published in {\it Journal of Integer Sequences}, 
February 20 2020.

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