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\begin{center}
\vskip 1cm{\LARGE\bf Some Recurrence Relations for Cauchy\\
\vskip .1in
Numbers of the First Kind} \vskip 1cm \large
Hong-Mei Liu, Shu-Hua Qi and Shu-Yan Ding\\
School of Science\\
Dalian Nationalities University\\
Dalian 116600 \\
China\\
\href{mailto:hmliu99@yahoo.com.cn}{\tt hmliu99@yahoo.com.cn}
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\begin{abstract}
In this paper, we present some recurrence relations for the Cauchy
numbers of the first kind, by making use of the Stirling numbers of the
first kind.
\end{abstract}

\newtheorem{theorem}{Theorem}
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\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
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\section{Introduction}

The Cauchy numbers of the first kind $b_n$, $n=0,1,2,\cdots,$ are
defined by the generating function \cite{Com74}
\begin{align}
\sum_{n\geq 0}b_n\frac{t^n}{n!}=\frac{t}{\log{(1+t)}}\,.\label{bn}
\end{align}
These numbers are also called the Bernoulli numbers of the second
kind, and are represented by sequences \seqnum{A006232} and
\seqnum{A006233} in Sloane's {\it Encyclopedia}.
The Cauchy numbers play important roles in some fields, and one of
the most important applications is the so-called Laplace summation
formula \cite{MeSpVe06}. These numbers are related to various
special combinatorial numbers, such as the Stirling numbers of both
kinds, the Bernoulli numbers (of the first kind) and the harmonic numbers
\cite{Howard96,MeSpVe06,zhao}. Furthermore, many generalizations of
these numbers are also introduced \cite{ChHwLee07,wang2,zheng}. The
most basic recurrence relation of these numbers is
\begin{align*}
\sum_{j=0}^n\frac{(-1)^{n-j}b_j}{j!(n+1-j)}=0 \quad\quad(n\geq 1)\,,
\end{align*}
with $b_0=1$. It can be derived from (\ref{bn}) by multiplying both
sides by $\log{(1+t)}$, and then equating the coefficients of the
powers of $x$.

The classical Bernoulli numbers (or the Bernoulli numbers of the first kind) $B_n$, $n=0,1,2,\cdots,$
are defined by the generating function \cite{Com74}
\begin{align*}
\sum_{n\geq 0}B_n\frac{t^n}{n!}=\frac{t}{e^t-1};
\end{align*}
see sequences \seqnum{A000367} and \seqnum{A002445}.
There is much work devoted to the Bernoulli numbers. For example, Agoh and Dilcher
\cite{agoh} derived recently the following recurrence relations:
\begin{align}
&amp;\sum_{j=0}^n\binom{n+k}{j}S(n-j+k,k)B_j=\frac{n+k}{k}S(n+k-1,k-1)\,,\label{Bernoulli1}\\
&amp;\sum_{j=m}^n\binom{n+k}{j-m}S(n-j+k+m, k+m)B_j
    =\frac{(-1)^m}{k+m}\sum_{j=1}^{m+1}\binom{k+m-1}{j-1}^{-1}N_1(n,m,k,j)\,
    ,\label{Bernoulli2}
\end{align}
where $S(n,k)$ are the Stirling numbers of the second kind and $N_1(n,m,k,j)$ is defined by
\[
(n+k)S(m+1,j)S(n+k-1,k+m-j)-mS(m,j)S(n+k, k+m-j).
\]

From these relations, we can see that the classical Bernoulli numbers
are closely related to the Stirling numbers of the second kind, and we can also find that the classical Bernoulli numbers and the Sirling numbers of the second kind are both associated with the delta series $e^t-1$.

On the other hand, it is well known that the Stirling numbers of the first kind $s(n,k)$ are defined by the generating function
\begin{align}
\sum_{n\geq
k}s(n,k)\frac{t^n}{n!}=\frac{\log^k(1+t)}{k!}\,.\label{snk}
\end{align}
They have the basic recurrence relation
\begin{align}
s(n,k)=s(n-1,k-1)-(n-1)s(n-1,k)\quad\quad (n,k\geq 1)\label{sr}
\end{align}
as well as the following special values:
\begin{align}
s(0,0)=1,\ s(n,0)=s(0,n)=0,\ s(n,n)=1,\quad (n\geq 1)\,.
\label{value}
\end{align}
From the generating functions, we can find that the Cauchy numbers of the first kind and the Stirling numbers of the first kind are both associated with the delta series $\log(1+t)$, which is the compositional inverse of $e^t-1$. Motivated by this fact, in this
paper, we will give some recurrence relations on the Cauchy numbers of
the first kind, in which the Stirling numbers of the first kind are
involved. The results of this paper can be viewed as the analogs of
those on the classical Bernoulli numbers and the Stirling numbers of
the second kind.

\section{The main results}

Firstly, we give a relation similar to the identity (\ref{Bernoulli1}).

\begin{theorem}\label{the1}
For any integers $k\geq 1$ and $n\geq 0$ we have
\begin{align*}
\sum_{j=0}^n\binom{n+k}{j}s(n-j+k,k)b_j=\frac{n+k}{k}s(n+k-1,k-1)\,.
\end{align*}
\end{theorem}

\begin{proof}
We can write
\begin{align}
\left(\frac{\log(1+t)}{t}\right)^k=\frac{k!}{t^k}\frac{\log^k(1+t)}{k!}=\sum_{i\geq
0}\frac{s(i+k,k)}{\binom{i+k}{k}}\frac{t^i}{i!}\,.\label{s1}
\end{align}
Putting (\ref{s1}) in the following identity
\begin{align*}
\left(\frac{\log(1+t)}{t}\right)^k\frac{t}{\log(1+t)}=\left(\frac{\log(1+t)}{t}\right)^{k-1}\,,
\end{align*}
we have
\begin{align}
\sum_{i\geq 0}
\frac{s(i+k,k)}{\binom{i+k}{k}}\frac{t^i}{i!}\sum_{j\geq
0}b_j\frac{t^j}{j!} &amp;=\sum_{n\geq
0}\left(\sum_{j=0}^n\binom{n}{j}\frac{s(n-j+k,k)}{\binom{n-j+k}{k}}b_j\right)\frac{t^n}{n!}\nonumber\\
&amp;=\sum_{n\geq
0}\frac{s(n+k-1,k-1)}{\binom{n+k-1}{k-1}}\frac{t^n}{n!}\,.\label{s2}
\end{align}
Thus, comparing the coefficients of $t^n/n!$ in (\ref{s2}) gives the
final result.
\end{proof}

In order to give a relation similar to (\ref{Bernoulli2}), we need
the following lemma.

\begin{lemma}\label{lem}For any $m\geq0$ we have
\begin{align}
&amp;\frac{d^m}{dt^m}\frac{1}{\log(1+t)}=\smbb{-\frac{1}{1+t}}^m\sum_{j=0}^mj!\frac{(-1)^{m-j}s(m,j)}{\log^{j+1}(1+t)}\,,\label{lem1}\\
&amp;\frac{d^m}{dt^m}\frac{t}{\log(1+t)}=\smbb{-\frac{1}{1+t}}^m\sum_{j=0}^m(-1)^{m-j}j!\frac{ts(m,j)+m(1+t)s(m-1,j)}{\log^{j+1}(1+t)}\,.\label{lem2}
\end{align}
\end{lemma}

\begin{proof}
We prove (\ref{lem1}) by induction. According to (\ref{value}), it is true
for $m=0$. Suppose it is true for $m$, then
\begin{align*}
&amp;(-1)^{m+1}\frac{d^{m+1}}{dt^{m+1}}\frac{1}{\log(1+t)}\\
&amp;\quad=(-1)^{m+1}\frac{d}{dt}\smbb{\frac{-1}{1+t}}^m
\sum_{j=0}^m\frac{(-1)^{m-j}s(m,j)j!}{\log^{j+1}(1+t)}\\
&amp;\quad=\frac{m}{(1+t)^{m+1}}\sum_{j=0}^m\frac{(-1)^{m-j}s(m,j)j!}{\log^{j+1}(1+t)}
+\frac{1}{(1+t)^{m+1}}\sum_{j=0}^m\frac{(-1)^{m-j}s(m,j)(j+1)!}{\log^{j+2}(1+t)}\\
&amp;\quad=\frac{1}{(1+t)^{m+1}}\sum_{j=0}^{m+1}\frac{(-1)^{m+1-j}s(m+1,j)j!}{\log^{j+1}(1+t)}\,,
\end{align*}
where the last equation is obtained by relation (\ref{sr}). This gives (\ref{lem1}) finally. Now, using the Leibniz's rule in the form
\begin{align*}
\frac{d^m}{dt^m}\frac{t}{\log(1+t)}=t\frac{d^m}{dt^m}\frac{1}{\log(1+t)}+m\frac{d^{m-1}}{dt^{m-1}}\frac{1}{\log(1+t)}\,,
\end{align*}
and using identity (\ref{lem1}), we can immediately obtain
(\ref{lem2}).
\end{proof}

From (\ref{bn}) we have
\begin{align*}
&amp;\frac{d^{m-1}}{dt^{m-1}}\frac{1}{\log(1+t)}=\frac{(-1)^{m-1}(m-1)!}{t^m}+\sum_{n\geq
0}\frac{b_{n+m}}{n+m}\frac{t^n}{n!}\,.
\end{align*}
Then, based on Lemma \ref{lem} and the above identity, we obtain
the following recurrence relation on the Cauchy numbers of the first kind.

\begin{theorem}\label{the2}
For any integers $1\leq m\leq n$ and $k\geq 0$ we have
\begin{align*}
&amp;\sum_{j=m}^n\binom{n+k}{j-m}s(n-j+k+m,
k+m)\frac{b_j}{j}=\frac{(-1)^m}{m}\frac{s(n+k+m,
k+m)}{\binom{n+k+m}{m}}\\
&amp;\quad+\sum_{i=0}^n\sum_{j=0}^{m-1}(-1)^j\binom{n+k}{i}\binom{k+m}{j}^{-1}\frac{(-m+1)_i}{k+m-j}s(m-1,
j)s(n+k-i, k+m-j-1)\,,
\end{align*}
where $(n)_i=n(n-1)\cdots(n-i+1)$\,.
\end{theorem}

\begin{proof}
Replacing $k$ by $k+m$ in (\ref{s1}) and multiplying the resulting
identity with (\ref{lem1}), we have
\begin{align}
&amp;\left(\frac{\log(1+t)}{t}\right)^{k+m}\frac{d^{m-1}}{dt^{m-1}}\frac{1}{\log(1+t)}\nonumber\\
&amp;\quad\quad=\sum_{i\geq
0}\frac{s(i+k+m,k+m)}{\binom{i+k+m}{k+m}}\frac{(-1)^{m-1}(m-1)!}{i!}t^{i-m}\nonumber\\
&amp;\quad\quad+\sum_{n\geq
0}\left(\sum_{j=0}^n\binom{n}{j}\frac{s(n-j+k+m,k+m)}{\binom{n-j+k+m}{k+m}}\frac{b_{j+m}}{j+m}\right)\frac{t^n}{n!}\,.\label{s3}
\end{align}
Denote the left-hand side of (\ref{s3}) by $A(t)$. By (\ref{lem1}) we
obtain
\begin{align*}
A(t)=\frac{(-1)^{m-1}}{t^{k+m}}(t+1)^{-m+1}\sum_{j=0}^{m-1}j!(-1)^{m-1-j}s(m-1,j)(\log(1+t))^{k+m-j-1}\,,
\end{align*}
and by (\ref{snk}) we obtain
\begin{align}
A(t)&amp;=\frac{(-1)^{m-1}}{t^{k+m}}\sum_{i\geq
0}\binom{-m+1}{i}t^i\sum_{j=0}^{m-1}j!(-1)^{m-1-j}s(m-1,j)\nonumber\\
&amp;\quad(k+m-j-1)!\sum_{l\geq 0}s(l,k+m-j-1)\frac{t^l}{l!}\nonumber\\
&amp;=\sum_{n\geq
0}\sum_{i=0}^{n+m}\sum_{j=0}^{m-1}\binom{-m+1}{i}(-1)^{j}j!\frac{(k+m-j-1)!}{(n+k+m-i)!}s(m-1,j)\nonumber\\
&amp;\quad s(n+k+m-i,k+m-j-1)t^n\,.\label{s4}
\end{align}
Comparing the coefficients of $t^n$ in (\ref{s3}) and (\ref{s4}),
and replacing $n$ by $n-m$, we have the desired result.
\end{proof}

Let us consider the extreme cases. When $m=1$, we return to
Theorem \ref{the1}. When $m=n$, we have the following corollary.

\begin{corollary}
For any integers $n\geq 1$ and $k\geq 0$ we have
\begin{align*}
b_n&amp;=(-1)^n\frac{s(2n+k,
n+k)}{\binom{2n+k}{n}}\\
&amp;+\sum_{i=0}^n\sum_{j=0}^{n-1}(-1)^j\binom{n+k}{i}\binom{n+k}{j}^{-1}\frac{n(-n+1)_i}{n+k-j}s(n-1,
j)s(n+k-i, n+k-j-1)\,.
\end{align*}
In particular when $k=0$, we have
\begin{align*}
b_n=(-1)^n\frac{s(2n, n)}{\binom{2n}{n}}
+\sum_{i=0}^n\sum_{j=0}^{n-1}(-1)^j\binom{n}{i}\binom{n}{j}^{-1}\frac{n(-n+1)_i}{n-j}s(n-1,
j)s(n-i, n-j-1)\,.
\end{align*}
\end{corollary}

The last identity indicates that many of the results in this section
can also be seen as convolution identities for the Stirling numbers of
the first kind.

Finally, from (\ref{bn}) we also have
\begin{equation}\label{diff.identity}
\frac{d^{m}}{dt^{m}}\frac{t}{\log(1+t)}=\sum_{n\geq
0}b_{n+m}\frac{t^n}{n!}\,.
\end{equation}
Similarly to the proof of Theorem \ref{the2}, by identities
(\ref{lem2}) and (\ref{diff.identity}), we can easily obtain the
following theorem.

\begin{theorem}\label{the3}
For any integers $0\leq m\leq n$ and $k\geq 1$ we have
\begin{align*}
&amp;\sum_{j=m}^n\binom{n+k}{j-m}s(n-j+k+m, k+m)b_j\\
&amp;\quad=\sum_{j=0}^m\sum_{l=m-j-1}^{n}(-1)^j
    \binom{n+k}{n-l}\binom{k+m}{j}^{-1}\frac{(-m)_{n-l}}{k+m-j}N_2(l,m,k,j)\,,
\end{align*}
where the number $N_2(l,m,k,j)$ is
\begin{gather*}
(l+k)(s(m,j)+ms(m-1,j))s(l-1+k,k+m-j-1)\\
+ms(m-1,j)s(l+k,k+m-j-1)\,.
\end{gather*}
\end{theorem}
Setting $m=0$ in this theorem, we can obtain Theorem \ref{the1} once again.

\section{Acknowledgments}

The authors would like to thank the anonymous referees for their
valuable comments and Dr. Weiping Wang for his kind help. The present
investigation is supported by the Dalian Nationalities University
foundation (20076210).

\begin{thebibliography}{99}
\bibitem{agoh} T. Agoh, K. Dilcher, Shortened recurrence relations for Bernoulli numbers,
        \textit{Discrete Math.}, \textbf{309} (2009) 887--898.

\bibitem{ChHwLee07} G.-S. Cheon, S.-G. Hwang, S.-G. Lee, Several polynomials associated with the harmonic numbers,
        \textit{Discrete Appl. Math.}, \textbf{155} (2007) 2573--2584.

\bibitem{Com74} L. Comtet, \textit{Advanced Combinatorics}, D. Reidel Publishing Co., Dordrecht, 1974.

\bibitem{Howard96} F. T. Howard, Explicit formulas for degenerate Bernoulli numbers,
         \textit{Discrete Math.}, \textbf{162} (1996) 175--185.

\bibitem{MeSpVe06} D. Merlini, R. Sprugnoli, M. C. Verri, The Cauchy numbers, \textit{Discrete Math.}, \textbf{306} (2006) 1906--1920.

\bibitem{wang2}W. Wang, Generalized higher order Bernoulli number pairs and generalized Stirling number pairs,
       \textit{J. Math. Anal. Appl.}, \textbf{364} (2010) 255--274.

\bibitem{zhao} F.-Z. Zhao, Sums of products of Cauchy numbers, \textit{Discrete Math.},
       \textbf{209} (2009) 3830--3842.

\bibitem{zheng} D.-Y. Zheng, Generalization of two kinds of Cauchy numbers, \textit{J. Dalian Univ. Technol.},
      \textbf{44} (2004) 606--609.

\end{thebibliography}

\bigskip
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\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11B83; Secondary 42A16.\\
\noindent \emph{Keywords: } Cauchy numbers;
Stirling numbers; combinatorial identities.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000367},
\seqnum{A002445},
\seqnum{A006232}, and
\seqnum{A006233}.)

\bigskip
\hrule
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\vspace*{+.1in}
\noindent
Received December 1 2009;
revised version received  March  16 2010.
Published in {\it Journal of Integer Sequences}, March 18 2010.

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