\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \usepackage{cite} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://www.research.att.com/cgi-bin/access.cgi/as/~njas/sequences/eisA.cgi?Anum=#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \begin{center} {\LARGE Some Properties of Associated Stirling Numbers} \end{center} \begin{center} Feng-Zhen Zhao\\ Department of Applied Mathematics\\ Dalian University of Technology\\ Dalian 116024 \\ China \\ \href{mailto:fengzhenzhao@yahoo.com.cn}{\tt fengzhenzhao@yahoo.com.cn} \end{center} \begin{abstract} In this paper, we discuss the properties of associated Stirling numbers. By means of the method of coefficients, we establish a series of identities involving associated Stirling numbers, Bernoulli numbers, harmonic numbers, and the Cauchy numbers of the first kind. In addition, we give the asymptotic expansion of certain sums involving $2$-associated Stirling numbers of the second kind by Darboux's method. \end{abstract} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{thm}{Theorem}[section] \newtheorem{lema}{Lemma}[section] \newtheorem{prop}{Proposition}[section] \newtheorem{ex}{Example}[section] \newtheorem{rem}{Remark}[section] \newtheorem{defi}{Definition}[section] \newtheorem{alg}{Algorithm}[section] \newtheorem{assu}{Assumption}[section] \newtheorem{cor}{Corollary}[section] \newtheorem{propt}{Property}[section] \renewcommand{\Box}{\framebox{\rule{0.3em}{0.0em}}} \newtheorem{remark}{Remark} \renewcommand{\Box}{\hfill \rule{2.3mm}{2.3mm}} \newcommand{\afterBox}{\Box\\ \\ \noindent} \bibliographystyle{plain} \setlength{\textwidth}{16cm} \setlength{\textheight}{23cm} \setlength{\oddsidemargin}{0.0cm} \setlength{\evensidemargin}{0.0cm} \setlength{\topmargin}{-1cm} %\setlength{\parskip}{0.25cm} \renewcommand{\textfraction}{0.1} \renewcommand{\bottomfraction}{0.1} \renewcommand{\topfraction}{1} %\renewcommand{\baselinestretch}{1.18} \makeatletter \renewcommand{\theequation}{\thesection.\arabic{equation}} \@addtoreset{equation}{section} \makeatother %\newenvironment{proof}{\noindent{\bf Proof. }}{\hfill $\Box$\medskip} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Introduction} Stirling numbers are generalized by many forms. See for instance \cite{ref1, ref2, ref3, ref4, ref5} and \cite{ref9}. In this paper, we are interested in associated Stirling numbers. The associated Stirling numbers of the first kind $s_2(n, k)$ \cite{ref3} are given by \begin{eqnarray*} \sum_{n=k}^{\infty}s_2(n, k)\frac{t^n}{n!}=\frac{[\ln(1+t)-t]^k}{k!} \end{eqnarray*} and the $r$-associated Stirling numbers of the second kind $S_r(n, k)$ \cite{ref3} are given by \begin{eqnarray*} \sum_{n=k}^{\infty}S_r(n, k)\frac{t^n}{n!}=\frac{1}{k!}\bigg[e^t-\sum_{j=0}^{r-1}\frac{t^j}{j!}\bigg]^k, \end{eqnarray*} where $k$ and $r$ are positive integers. It is clear that \begin{eqnarray*} \sum_{n=k}^{\infty}S_2(n, k)\frac{t^n}{n!}&=&\frac{(e^t-1-t)^k}{k!},\\ \sum_{n=k}^{\infty}S_3(n, k)\frac{t^n}{n!}&=&\frac{(e^t-1-t-t^2/2)^k}{k!}. \end{eqnarray*} Like the ordinary Stirling numbers, the associated Stirling numbers also play important roles in combinatorics. For example, $|s_2(n, k)|$ equals the number of derangements of a set ${\it N}$ ($|{\it N}|=n$), with $k$ orbits, and $S_r(n, k)$ is the number of partitions of the set ${\it N}$ ($|{\it N}|=n$), into $k$ blocks, all of cardinality $\geq r$. It is clear that $S_1(n, k)$ is the Stirling number of the second kind $S(n, k)$. Therefore, associated Stirling numbers deserve to be investigated. The aim of this paper is to investigate the properties of associated Stirling numbers by making use of the method of coefficients \cite{ref7}. We establish a series of identities relating associated Stirling numbers with Bernoulli, harmonic, and Cauchy numbers of the first kind. In addition, we give the asymptotic expansion of certain sums involving $r$-associated Stirling numbers by Darboux's method. The paper is organized as follows. In Section 2, we establish a series of identities involving associated Stirling, Bernoulli, harmonic and Cauchy numbers of the first kind. In Section 3, we give the asymptotic expansion of certain sums involving $r$-associated Stirling numbers by Darboux's method. For convenience, we recall some definitions of combinatorial numbers involved in the paper. Throughout, we denote Stirling numbers of the first kind by $s(n, k)$, and let $B_n$, $B_n^{(k)}$, and $E_n$ stand for Bernoulli, generalized Bernoulli, and Euler numbers respectively. That is, \begin{eqnarray*} &&\sum_{n=k}^{\infty}s(n, k)\frac{t^n}{n!}=\frac{\ln^k(1+t)}{k!},\quad\quad \sum_{n=0}^{\infty}B_n\frac{t^n}{n!}=\frac{t}{e^t-1}, \\ &&\sum_{n=0}^{\infty}B_n^{(k)}\frac{t^n}{n!}=\frac{t^k}{(e^t-1)^k} \quad (k\geq 1), \quad\sum_{n=0}^{\infty}E_n\frac{t^n}{n!}=\frac{2}{e^t+e^{-t}}. \end{eqnarray*} The Cauchy numbers of the first kind $a_n$ are given by \begin{eqnarray*} \sum_{n=0}^{\infty}a_n\frac{t^n}{n!}=\frac{t}{\ln(1+t)}. \end{eqnarray*} The harmonic numbers $H_n$ are given by \begin{eqnarray*} \sum_{n=1}^{\infty}H_nt^n=-\frac{\ln(1-t)}{1-t}. \end{eqnarray*} In this paper, $[t^n]f(t)$ denotes the coefficient of $t^n$ in $f(t)$, where $$ f(t)=\sum_{n=0}^{\infty}f_nt^n. $$ The expression $[t^n]$ is called the {\it ``coefficient of'' functionals} \cite{ref7}. If $f(t)$ and $g(t)$ are formal power series, the following relations hold \cite{ref7}: \begin{eqnarray} &&[t^n](\alpha f(t)+\beta g(t))=\alpha[t^n]f(t)+\beta[t^n]g(t), \label{pf-1.1}\\ &&[t^n]tf(t)=[t^{n-1}]f(t), \label{pf-1.2}\\ &&[t^n]f^{\prime}(t)=(n+1)[t^{n+1}]f(t), \label{pf-1.3}\\ &&[t^n]f(t)g(t)=\sum_{k=0}^n([y^k]f(y))[t^{n-k}]g(t). \label{pf-1.4} \end{eqnarray} In Section 2, we obtain a series of identities related to associated Stirling numbers by using (\ref{pf-1.1})-(\ref{pf-1.4}). %\mbox{}\\[4pt] %\begin{flushleft} \section{Identities involving associated Stirling, Bernoulli, and harmonic numbers} %\end{flushleft} Bernoulli numbers and harmonic numbers are important in combinatorics, and Stirling numbers are related to them. From \cite{ref3}, we know that Stirling numbers and Bernoulli numbers satisfy \begin{eqnarray*} \sum_{j=0}^n\frac{(-1)^jj!S(n, j)}{j+1}=B_n, \ \ \sum_{j=0}^ns(n, j)B_j=\frac{(-1)^nn!}{n+1}. \end{eqnarray*} By the generating functions of $S_2(n, k)$, $S(n, k)$, and $B_n$, we observe that $S_2(n, k)$ is also related to $B_n$, and we have \begin{thm} For $n\geq 1$ and $k\geq 1$, $S_2(n, k)$, $B_n$, and $S(n, k)$ satisfy the equations \begin{eqnarray} \sum_{j=0}^nS_2(n-j+k, k){n+k\choose j}B_j&=&(n+k)\sum_{j=1}^k\frac{(-1)^{k-j}}{j}{n+k-1\choose k-j}S(n+j-1, j-1)\nonumber\\ &&+(-1)^k{n+k\choose k}B_n, \label{pf-2.1} \end{eqnarray} \begin{eqnarray} \sum_{j=0}^n{n+k-1\choose j}S_2(n-j+k, k)B_j&=&(n+k-1)S_2(n+k-2, k-1) \ \ k\geq 2. \label{pf-2.2} \end{eqnarray} \end{thm} \begin{proof} By the definitions of $S_2(n, k)$, $B_n$, and $S(n , k)$, we have \begin{eqnarray*} \sum_{j=0}^nS_2(n-j+k, k){n+k\choose j}B_j&=&(n+k)!\sum_{j=0}^n\frac{S_2(n-j+k, k)}{(n-j+k)!} \cdot\frac{B_j}{j!} \\ &=&(n+k)!\sum_{j=0}^n[t^{n-j+k}]\frac{(e^t-1-t)^k}{k!}[t^j]\frac{t}{e^t-1}\\ &=&(n+k)!\sum_{j=0}^n[t^{n-j}]\frac{(e^t-1-t)^k}{k!t^k}[t^j]\frac{t}{e^t-1}\\ &=&(n+k)![t^n]\frac{(e^t-1-t)^kt}{k!t^k(e^t-1)}\\ &=&(n+k)![t^n]\sum_{j=0}^k(-1)^{k-j}{k\choose j}\frac{(e^t-)^{j-1}t^{-j+1}}{k!}\\ &=&\frac{(-1)^k(n+k)!}{k!}[t^n]\frac{t}{e^t-1}+(n+k)!\sum_{j=1}^k[t^n]\frac{(-1)^{k-j}(e^t-1)^{j-1} }{j(k-j)!(j-1)!t^{j-1}}\\ &=&(-1)^k{n+k\choose k}B_n+(n+k)!\sum_{j=1}^k\frac{(-1)^{k-j}S(n+j-1, j-1)}{j(k-j)!(n+j-1)!}. \end{eqnarray*} Then (\ref{pf-2.1}) holds. Now we give the proof of (\ref{pf-2.2}). \begin{eqnarray*} \sum_{j=0}^n{n+k-1\choose j}S_2(n-j+k, k)B_j&=&(n+k-1)!\sum_{j=0}^n\frac{S_2(n-j+k, k)}{(n-j+k-1)!}\cdot\frac{B_j}{j!}\\ &=&(n+k-1)!\sum_{j=0}^n\frac{(n-j+k)S_2(n-j+k, k)}{(n-j+k)!}\cdot\frac{B_j}{j!}\\ &=&(n+k-1)!\sum_{j=0}^n(n-j+k)[t^{n-j+k}]\frac{(e^t-1-t)^k}{k!}[t^j]\frac{t}{e^t-1}\\ &=&(n+k-1)!\sum_{j=0}^n[t^{n-j+k-1}]\frac{(e^t-1-t)^{k-1}(e^t-1)}{(k-1)!}[t^j]\frac{t}{e^t-1}\\ &=&(n+k-1)!\sum_{j=0}^n[t^{n-j}]\frac{k(e^t-1-t)^{k-1}(e^t-1)}{t^{k-1}k!}[t^j]\frac{t}{e^t-1}\\ &=&(n+k-1)![t^n]\frac{(e^t-1-t)^{k-1}}{t^{k-2}(k-1)!}\\ &=&(n+k-1)![t^{n+k-2}]\frac{(e^t-1-t)^{k-1}}{(k-1)!}\\ &=&(n+k-1)S_2(n+k-2, k-1). \end{eqnarray*} This completes the proof. \end{proof} Formula (\ref{pf-2.1}) relates associated Stirling, Bernoulli, and Stirling numbers of the second kind. The generating functions of generalized Bernoulli numbers $B_n^{(k)}$ implies that they are related to associated Stirling numbers. For $S_2(n, k)$ and $B_n^{(k)}$, we get \begin{cor} For $n\geq 1$ and $k\geq 1$, $2$-associated Stirling numbers $S_2(n, k)$ and generalized Bernoulli numbers $B_n^{(k)}$ satisfy \begin{eqnarray} \sum_{j=0}^nS_2(n-j+k, k){n+k\choose j}B_j^{(k)}={n+k\choose k}\sum_{j=0}^k(-1)^{k-j}{k\choose j}B_n^{(k-j)}. \label{pf-2.3} \end{eqnarray} \end{cor} \begin{proof} \begin{eqnarray*} \sum_{j=0}^nS_2(n-j+k, k){n+k\choose j}B_j^{(k)}&=&(n+k)!\sum_{j=0}^n\frac{S_2(n-j+k, k)}{(n-j+k)!}\cdot\frac{B_j^{(k)}}{j!}\\ &=&(n+k)!\sum_{j=0}^n[t^{n-j+k}]\frac{(e^t-1-t)^k}{k!}[t^j]\frac{t^k}{(e^t-1)^k}\\ &=&\frac{(n+k)!}{k!}\sum_{j=0}^n[t^{n-j}]\frac{(e^t-1-t)^k}{t^k}[t^j]\frac{t^k}{(e^t-1)^k}\\ &=&\frac{(n+k)!}{k!}[t^n]\frac{(e^t-1-t)^k}{(e^t-1)^k}\\ &=&\frac{(n+k)!}{k!}\sum_{j=0}^k(-1)^{k-j}{k\choose j}\frac{B_n^{(k-j)}}{n!}. \end{eqnarray*} Hence (\ref{pf-2.3}) holds. \end{proof} For $S_3(n, k)$ and Bernoulli numbers $B_n$, we have \begin{thm} For $n\geq k$ and $k\geq 1$, $S_3(n, k)$ and $B_n$ satisfy \begin{eqnarray} \sum_{j=0}^n{n+k\choose j+k}S_3(j+k, k)B_{n-j}&=&(n+k)!\sum_{j=1}^k\frac{(-1)^{k-j}}{j(k-j)!}\sum_{j_1=0}^{k-j}{k-j\choose j_1}\frac{S(n-j_1+j-1, j-1)}{2^{j_1}(n-j_1+j-1)!}\nonumber\\ &&+\frac{(-1)^k(n+k)!}{k!}\sum_{j=0}^k\frac{B_{n-j}}{2^j(n-j)!}{k\choose j}. \label{pf-2.4} \end{eqnarray} \end{thm} \begin{proof} From the generating functions of $S_3(n, k)$ and $B_n$, we have \begin{eqnarray*} \sum_{j=0}^n{n+k\choose j+k}S_3(j+k, k)B_{n-j}&=&(n+k)!\sum_{j=0}^n\frac{S_3(j+k, k)}{(j+k)!} \frac{B_{n-j}}{(n-j)!}\\ &=&(n+k)!\sum_{j=0}^n[t^{j+k}]\frac{(e^t-1-t-t^2/2)^k}{k!}[t^{n-j}]\frac{t}{e^t-1}\\ &=&(n+k)!\sum_{j=0}^n[t^j]\frac{(e^t-1-t-t^2/2)^k}{t^kk!}[t^{n-j}]\frac{t}{e^t-1}\\ &=&(n+k)![t^n]\frac{(e^t-1-t-t^2/2)^kt}{k!t^k(e^t-1)}\\ &=&(n+k)![t^n]\bigg(\sum_{j=0}^k(-1)^{k-j}{k\choose j}\frac{(e^t-1)^j(t+t^2/2)^{k-j}t}{k!(e^t-1)t^k}\bigg)\\ &=&(n+k)![t^n]\frac{(-1)^k(1+t/2)^kt}{k!(e^t-1)}\\ &&+[t^n]\sum_{j=1}^k(-1)^{k-j}{k\choose j}\frac{(e^t-1)^{j-1}(t+t^2/2)^{k-j}t}{k!t^k}\\ &=&(-1)^k(n+k)!\sum_{j=0}^k{k\choose j}\frac{B_{n-j}}{2^j(n-j)!k!}\\ &&+\frac{(n+k)!}{k!}[t^n]\sum_{j=1}^k(-1)^{k-j}{k\choose j}(e^t-1)^{j-1}\sum_{j_1=0}^{k-j}{k-j\choose j_1}\frac{t^{j_1-j+1}}{2^{j_1}}\\ &=&\frac{(-1)^k(n+k)!}{k!}\sum_{j=0}^k{k\choose j}\frac{B_{n-j}}{2^j(n-j)!}\\ &&+(n+k)!\sum_{j=1}^k\frac{(-1)^{k-j}}{j(k-j)!}\sum_{j_1=0}^{k-j}{k-j\choose j_1}\frac{S(n-j_1+j-1, j-1)}{2^{j_1}(n-j_1+j-1)!} \end{eqnarray*} Then (\ref{pf-2.4}) holds. \end{proof} There are many identities relating Stirling numbers of the first kind and harmonic numbers in \cite{ref3}. For example, \begin{eqnarray*} (-1)^{n+1}s(n+1, 2)&=&n!H_n, \\ (-1)^ns(n+1, 3)&=&\frac{n!}{2}(H^2_n-H_n^{(2)}),\\ (-1)^{n+1}s(n+1, 4)&=&\frac{n!}{6}(H^2_n-3H_nH_n^{(2)}+2H_n^{(3)}), \end{eqnarray*} where $H_n^{(s)}=1+2^{-s}+3^{-s}+\cdots+n^{-s}$. For associated Stirling numbers of the first kind and harmonic numbers, we can prove \begin{thm} For $n\geq 1$ and $k\geq 1$, we have \begin{eqnarray} \sum_{j=0}^n\frac{(-1)^jH_{j+1}s_2(n-j+k, k)}{(j+2)(n-j+k)!}&=&\frac{(-1)^k}{2}\sum_{j=0}^k\frac{(-1)^j(j+1)(j+2)s(n+j+2, j+2)}{(k-j)!(n+j+2)!}.\nonumber \\ \label{pf-2.5} \end{eqnarray} \end{thm} \begin{proof} By integrating the generating function for $H_n$ we have \begin{eqnarray*} \sum_{n=0}^{\infty}\frac{H_{n+1}t^n}{n+2}=\frac{\ln^2(1-t)}{2t^2}. \end{eqnarray*} One can verify that \begin{eqnarray*} \frac{[\ln(1-t)+t]^k\ln^2(1-t)}{2(-1)^kk!t^{k+2}}=\frac{(-1)^k}{2k!}\sum_{j=0}^k{k\choose j}\frac{\ln^{j+2}(1-t)}{t^{j+2}}. \end{eqnarray*} Then \begin{eqnarray*} [t^n]\frac{[\ln(1-t)+t]^k\ln^2(1-t)}{2(-1)^kk!t^{k+2}}&=&\sum_{j=0}^n[t^{n-j+k}] \frac{[\ln(1-t)+t]^k}{(-1)^kk!}[t^j]\frac{\ln^2(1-t)}{2t^2}\\ &=&\sum_{j=0}^n\frac{(-1)^{n-j}s_2(n-j+k, k)H_{j+1}}{(n-j+k)!(j+2)}\\ &=&\frac{(-1)^k}{2k!}\sum_{j=0}^k{k\choose j}[t^n]\frac{\ln^{j+2}(1-t)}{t^{j+2}}, \end{eqnarray*} \begin{eqnarray*} \sum_{j=0}^n\frac{(-1)^{n-j}H_{j+1}s_2(n-j+k, k)}{(j+2)(n-j+k)!}=\frac{(-1)^{n+k}}{2}\sum_{j=0}^k\frac{(-1)^j(j+1)(j+2)s(n+j+2, j+2)}{(k-j)!(n+j+2)!}. \end{eqnarray*} Hence (\ref{pf-2.5}) holds. \end{proof} There are some identities involving Stirling numbers and Cauchy numbers of the first kind. For example \begin{eqnarray*} \sum_{j=0}^na_jS(n, j)=\frac{1}{n+1}, \ \ a_n=\sum_{j=0}^n\frac{s(n, j)}{j+1}. \end{eqnarray*} See \cite{ref3, ref6} for more details. For associated Stirling numbers of the first kind and the Cauchy numbers of the first kind, we have \begin{thm} For $n\geq 1$ and $k\geq 1$, $s_2(n, k)$ and $a_n$ satisfy \begin{eqnarray} \sum_{j=0}^ns_2(n-j+k, k){n+k\choose j}a_j&=&(n+k)\sum_{j=1}^k\frac{(-1)^{k-j}}{j}{n+k-1\choose k-j}s(n+j-1, j-1)\nonumber\\ &&+(-1)^k{n+k\choose k}a_n. \label{pf-2.6} \end{eqnarray} \end{thm} The proof of (\ref{pf-2.6}) is similar to that of (\ref{pf-2.1}) and is omitted here. Formula (\ref{pf-2.6}) relates associated Stirling numbers and Cauchy numbers. %\mbox{}\\[4pt] %\begin{flushleft} \section{Asymptotic Expansion of Certain Sums Involving $2$-associated Stirling numbers of the second kind, Bernoulli numbers, and Euler Numbers } %\end{flushleft} We know that it is difficult to compute the accurate values of certain sums involving $r$-associated Stirling numbers. However, sometimes we can give their asymptotic expansion. In this section, we give asymptotic expansion of certain sums for $2$-associated Stirling numbers of the second kind, Bernoulli numbers, and Euler numbers by Darboux's method. We first recall a lemma (see\cite{ref8}):\\ {\bf Lemma:} Assume that $f(t)=\sum_{n\geq 0}a_nt^n$ is an analytic function for $|t|