<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt,reqno]{article}

\usepackage[colorlinks=true,
linkcolor=webgreen, filecolor=webbrown,
citecolor=webgreen]{hyperref}
\usepackage{epsf,psfig}
\usepackage{amsmath,amssymb}
\usepackage{color}


\def\Ei{\mbox{\rm Ei\,}}
\def\NN{{\mathbb N}}
\def\RR{{\mathbb R}}
\def\Re{\mbox{\rm Re}\,}
\def\Im{\mbox{\rm Im}\,}
\def\dj{d\kern-0.4em\char"16\kern-0.1em}
\def\Dj{\mbox{\raise0.3ex\hbox{-}\kern-0.4em D}}
\def\ra{\; \; \;}


\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{.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}}}


\newtheorem{theorem}{Theorem}
\newtheorem{tabel}{Table}
\newtheorem{definition}{Definition}
\newtheorem{lemma}{Lemma}
\newtheorem{corollary}{Corollary}
\newtheorem{remark}{Remark}

\renewcommand{\thefootnote}{}

\begin{document}

\begin{center}
\epsfxsize=4in
\leavevmode\epsffile{logo129.eps}
\end{center}

\vskip .5in

\begin{center}


 {\LARGE \bf The function $_{\lowercase{v}}
 M_{\lowercase{m}}(\lowercase{s};\lowercase{a},\lowercase{z})$
 and some well-known sequences}\vskip 1.5cm
 {\large Aleksandar Petojevi\'{c}} \smallskip \\
 Aleja Mar\v{s}ala Tita 4/2\\
 24000 Subotica\\
 Yugoslavia\medskip \\
 Email address:\; \href{mailto:apetoje@ptt.yu}{apetoje@ptt.yu}\\
 \vskip2cm
 {\bf Abstract}
\end{center}

\footnotetext{This work was supported in part by the Serbian
 Ministry of  Science, Technology and Development under Grant \# 2002:
 {\it Applied Orthogonal Systems, Constructive Approximation and Numerical
 Methods}.}

{\em In this paper we define the function $_vM_m(s;a,z)$, and we study 
 the special cases $_1M_m(s;a,z)$ and
 $_nM_{-1}(1;1,n+1)$.  We prove
  some new equivalents of Kurepa's hypothesis for the left
  factorial. Also, we present a generalization
  of the alternating factorial numbers.}

\section{Introduction}
Studying the Kurepa function
\[K(z)= \ !z =\int_{0}^{+\infty}\frac{t^{z}-1}{t-1}\,e^{-t}\,dt \quad
(\Re z&gt;0),\] G. V. Milovanovi\'{c} gave a generalization of the
function
\[M_m(z)=
  \int_{0}^{+\infty}\frac{t^{z+m}-Q_{m}(t;z)}{(t-1)^{m+1}}e^{-t}dt\quad
   (\Re z &gt; -(m+1)),\]
where the polynomials $Q_{m}(t;z)$, $m = -1,0,1,2,...$ are given
by
\[Q_{-1}(t;z)=0 \quad Q_{m}(t;z)=\sum_{k=0}^{m} {m+z \choose
k}(t-1)^k.\]
The function $\{ M_{m}(z)\}_{m=-1}^{+\infty}$ has
the integral representation
\[M_m(z)=\frac{z(z+1)\cdots(z+m)}{m!}\int_0^1\xi^{z-1}(1-\xi)^m
  e^{(1-\xi)/\xi}\Gamma\Bigl(z,\frac{1-\xi}{\xi}\Bigr)\,d\xi,\]
where $\Gamma(z,x)$, the incomplete gamma function, is defined by
\begin{equation}\label{i1}
 \Gamma(z,x)=\int_x^{+\infty}t^{z-1}e^{-t}\,dt.
\end{equation}
Special cases include
\begin{equation}\label{po1}
 M_{-1}(z)=\Gamma(z) \qquad \mbox{and}\qquad M_0(z)=K(z),
\end{equation}
where $\Gamma(z)$ is the gamma function
\[\Gamma(z)=\int_{0}^{+\infty}t^{z-1}e^{-t}\,dt. \]

The numbers $M_m(n)$ were introduced by
Milovanovi\'c \cite{Mil2} and
Milovanovi\'c and Petojevi\'{c} \cite{MiPe}.
For non-negative integers $n,m\in \NN$ 
the following identities hold:
 \[ M_m(0)=0,\quad
 M_{m}(n)=\sum_{i=0}^{n-1}\frac{(-1)^i}{i!}\sum_{k=i}^{n-1}k!{m+n\choose
 k+m+1}. \]
For the numbers $M_m(n)$ the following relations hold:
\begin{eqnarray*}
 M_m(n+1)&amp;=&amp;n!+ \sum_{\nu=0}^m M_\nu(n),\\[3mm]
 \lim_{n\to+\infty}\frac{M_\nu(n)}{M_{\nu-1}(n)}&amp;=&amp;1,\\[3mm]
 \lim_{n\to+\infty}\frac{M_m(n)}{(n-1)!}&amp;=&amp;1.
\end{eqnarray*}
 The generating function of the numbers
$\{M_m(n)\}_{n=0}^{+\infty}$ is given by
\[\frac1{m!}\left(A_m(x)e^{x-1}\left(\Ei(1)-\Ei(1-x)+B_m(x)e^x-C_m(x)\right)\right)
=\sum_{n=0}^{+\infty}M_m(n)\,\frac{x^n}{n!}\,\] where $A_m(x)$,
$B_m(x)$, and $C_m(x)$ are polynomials defined as follows:
 \begin{eqnarray*}
 \frac{A_m(x)}{m!}&amp;=&amp;
 \sum_{k=0}^m{\binom{m}{k}}\frac{(x-1)^k}{k!}\,,\\ [3mm]
 \frac{B_m(x)}{m!}&amp;=&amp;
 \sum_{\nu=0}^{m-1}\left(\sum_{k=1}^{m-\nu}
 {\binom{m}{k+\nu}}\frac{(-1)^{k-1}}{k}\sum_{j=0}^{k-1}\frac{(-1)^j}{j!}\right)
 \frac{(x-1)^\nu}{\nu!}\,,\\ [3mm]
 \frac{C_m(x)}{m!}&amp;=&amp;
 \sum_{j=0}^{m-1}\left(\sum_{\nu=0}^{j}(-1)^\nu
 {\binom{j}{\nu}}\sum_{k=j+1}^{m}\frac{(-1)^{k-1}}{k-\nu}
 {\binom{m}{k}}\right)\frac{(x-1)^j}{j!}\,,
 \end{eqnarray*}
Here $\Ei(x)$ is the exponential integral defined by
\begin{equation}\label{ei}
  \Ei(x)=\mbox{\rm p.v.}\int_{-\infty}^{x}\frac{e^t}{t}\,dt
\qquad(x&gt;0).
\end{equation}

In this paper, we give a generalization of the function $M_m(z)$
which we denote as $_vM_m(s;a,z)$.  These generalization are of
interest because its special cases include:
\begin{eqnarray*}
 _1M_1(1;1,n+1)&amp;=&amp;n!,\\[3mm]
 _1M_0(1;1,n)&amp;=&amp;!n,\\[3mm]
_nM_{-1}(1;1,n+1)&amp;=&amp;A_n.
\end{eqnarray*}
where $n!$, $!n$, and $A_n$ are the {\it right factorial numbers}
(sequence A000142 in \cite{Slo}), the {\it left factorial numbers}
(sequence A003422 in \cite{Slo}) and the {\it alternating
factorial numbers} (sequence A005165 in \cite{Slo}), respectively.
They are defined as follows:
\begin{equation}\label{eq1.1}
0!=1,\; n!=n\cdot (n-1)!; \qquad \ !0=0, \; \ !n=\sum_{k=0}^{n-1}k!
\quad\mbox{and}\quad A_n=\sum_{k=1}^{n}(-1)^{n-k}k!.
\end{equation}

\section{Definitions}

We now introduce a generalization of the function $M_m(z)$.
\vspace{3mm}
\begin{definition}\label{d1} For $m=-1,0,1,2,...,$ and $\Re z&gt; v-m-2$
 the function $_vM_m(s;a,z)$ is defined by
  \begin{eqnarray*}
&amp;&amp;_vM_m(s;a,z)=\\[3mm]
&amp;&amp;=\sum_{k=1}^v\frac{(-1)^{2v+1-k}\Gamma(m+z+2-k)}{\Gamma(z+1-k)\Gamma(m+2)}\,
\mathcal{L}[s;\,{}_2F_1(a,k-z,m+2,1-t)],
  \end{eqnarray*}
where $v$ is a positive integer, and $s,\; a,\; z$ are complex
variables.
\end{definition}\vspace{3mm}
The hypergeometric function $_2F_1(a,b;c,x)$ is defined by the
series
\[ _2F_1(a,b,c;x)=\sum_{n=0}^\infty
\frac{(a)_n(b)_n}{(c)_n}\frac{x^n}{n!}\qquad (|x|&lt;1), \] and has
the integral representation
\[_2F_1(a,b,c;x)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}
  \int_0^1t^{b-1}(1-t)^{c-b-1}(1-tx)^{-a}dt,\] in the $x$
plane cut along the real axis from $1$ to $\infty$, if $\Re c&gt;\Re
b&gt;0$.\\ The symbols $(z)_n$ and $\mathcal{L}[s;F(t)]$ represent
the Pochhammer symbol
\[(z)_n=\frac{\Gamma(z+n)}{\Gamma(z)},\]
and Laplace transform
\[\mathcal{L}[s;F(t)]=\int_0^\infty e^{-st}F(t)dt.\]

\begin{table}[ht]\label{table1}
\caption{The numbers $_1M_m(1;a,n)$ for $m=1,2,3,4$ and
$a=0,1,2$}\vspace{3mm}
\begin{tabular*}{\textwidth}{@{\extracolsep{\fill}}llllll}
\hline $a=0$&amp;$\mbox{in \cite{Slo}}$&amp;$a=1$&amp;$\mbox{in
\cite{Slo}}$&amp;$a=2$&amp;$\mbox{in \cite{Slo}}$\cr \hline
 $_1M_{1}(1,0,n)$&amp;$A000217$&amp;$_1M_{1}(1,1,n)$&amp;$A014144$&amp;$_1M_{1}(1,2,n)$&amp;$A007489$\cr \\
 $_1M_{2}(1,0,n)$&amp;$A000292$&amp;$_1M_{2}(1,1,n)$&amp;$\mbox{unlisted}$&amp;$_1M_{2}(1,2,n)$&amp;$A014145$\cr \\
 $_1M_{3}(1,0,n)$&amp;$A000332$&amp;$_1M_{3}(1,1,n)$&amp;$\mbox{unlisted}$&amp;$_1M_{3}(1,2,n)$&amp;$\mbox{unlisted}$\cr \\
 $_1M_{4}(1,0,n)$&amp;$A000389$&amp;$_1M_{4}(1,1,n)$&amp;$\mbox{unlisted}$&amp;$_1M_{4}(1,2,n)$&amp;$\mbox{unlisted}$ \\
\hline
\end{tabular*}
\end{table}
The term ``unlisted" in Table 1 means that the sequence cannot currently be
found in Sloane's on-line encyclopedia of integer sequences\cite{Slo}.\vspace{3mm}

\begin{lemma}\label{l1} Let $m=-1,0,1,2,\ldots\; .$ Then
\[_1M_m(1;1,z)=M_m(z).\]
\end{lemma}\vspace{3mm}

\noindent {\it Proof.} \, The proof presented here is due to
Professor G. V. Milovanovi\'{c}.
\\ Since
\[(k+m+1)!=(m+1)!(m+2)_k\quad\mbox{and}\quad
 (1-z)_k=\frac{(-1)^k\Gamma(z)}{\Gamma(z-k)},\]
we have
\[{m+z\choose k+m+1}=\frac{\Gamma(m+z+1)}{\Gamma(z-k)(k+m+1)!}=
 \frac{\Gamma(m+z+1)}{\Gamma(z)(m+1)!}\cdot\frac{(1-z)_k(-1)^k}{(m+2)_k}\,,\]
so that
 \begin{eqnarray*}
 \frac{t^{z+m}-Q_{m}(t;z)}{(t-1)^{m+1}}&amp;=&amp;\sum_{k=0}^{+\infty}
{m+z\choose k+m+1}(t-1)^k\qquad (|t-1|&lt;1)\\[3mm]
 &amp;=&amp;\frac{\Gamma(m+z+1)}{\Gamma(z)(m+1)!}\sum_{k=0}^{+\infty}
 \frac{(1-z)_k(1)_k}{(m+2)_k}\cdot\frac{(1-t)^k}{k!}\\ [3mm]
 &amp;=&amp;\frac{\Gamma(m+z+1)}{\Gamma(z)(m+1)!}\,{}_2F_1(1,1-z,m+2;1-t).
 \end{eqnarray*}

\section{The function $_{\lowercase{1}}M_{\lowercase{m}}(\lowercase{s};\lowercase{a},
\lowercase{z})$}

\subsection{The numbers $\{_{\lowercase{1}}M_{\lowercase{m}}(1;-\lowercase{n},\lowercase{r})\}
_{r=0}^{+\infty}\, _{\lowercase{n=0}}^{+\infty}\,
_{\lowercase{m=-1}}^{+\infty}$}
We have
\[_1M_{m}(1;-n,z)=\frac{\Gamma(m+z+1)}{\Gamma(z)\Gamma(m+2)}\,
\mathcal{L}[1;\,{}_2F_1(-n,1-z,m+2;1-t)],\]
starting with the polynomials
\[ \sum_{k=0}^\infty {m+z\choose
k+m+1}{n\choose k}(1-t)^k, \quad n\in \NN.\] Since
\[(-n)_k=(-1)^k\frac{n!}{(n-k)!}\]
we have
\begin{equation}\label{s1}
  \sum_{k=0}^\infty {m+z\choose
k+m+1}{n\choose
k}(1-t)^k=\frac{\Gamma(m+z+1)}{\Gamma(z)\Gamma(m+2)}\,{}_2F_1(-n,1-z,m+2;1-t)
\end{equation}
or, by continuation,
\[ =\pi\; {\rm cosec}\;\pi z \int_0^1\xi^{1-z}(1-\xi)^{m+z+1}
 \bigl(1-(1-t)\xi\bigr)^{n}d\xi.\]
Hence, the following definition is reasonable.\vspace{3mm}

\begin{definition}\label{d2} For $n\in \NN$ and $m=-1,0,1,2,...,$ the
polynomials $z\mapsto {_mP_n(z)}$ are defined by
\[_mP_n(z)=\mathcal{L}[1;\,{}_2F_1(-n,1-z,m+2;1-t)].\]
\end{definition}

\begin{table}[ht]\label{table2}
\caption{The polynomials $_{m}P_2(z)$, $_{m}P_3(z)$ and
$_{m}P_4(z)$ }\vspace{3mm}
\begin{tabular*}{\textwidth}{@{\extracolsep{\fill}}llll}
\hline $m$&amp;$_{m}P_2(z)$&amp;$_{m}P_3(z)$&amp;$_{m}P_4(z)$\cr \hline
 $-1$&amp;$\frac{1}{2}z^2-\frac{3}{2}z+2$&amp;$-\frac{1}{3}z^3+\frac{7}{2}z^2-\frac{49}{6}z+6$&amp;
 $\frac{3}{8}z^4-\frac{61}{12}z^3+\frac{193}{8}z^2-$\cr \\
 $\;\;\;0$&amp;$\frac{1}{6}z^2-\frac{1}{2}z+\frac{4}{3}$&amp;
 $-\frac{1}{12}z^3+z^2-\frac{29}{12}z+\frac{5}{2}$&amp;$-\frac{509}{12}z+24
 $\cr \\
 $\;\;\;1$&amp;$\frac{1}{12}z^2-\frac{1}{4}z+\frac{7}{6}$&amp;
 $-\frac{1}{30}z^3+\frac{9}{20}z^2-\frac{67}{60}z+\frac{17}{10}$&amp;$ $\cr
 \hline
\end{tabular*}
\end{table}\vspace{3mm}

The polynomials $_mP_n(z)$ can be expressed in terms of the  {\it
derangement numbers\/} (sequence A000166 in \cite{Slo})
\[S_k=k!\sum_{\nu=0}^k\frac{(-1)^\nu}{\nu!}\qquad (k\ge0)\]
in the form

\begin{theorem}\label{t1}For $m=-1,0,1,2,...$ and $n\in \NN$ we
have
\[_mP_n(z)={n+m+1\choose n}^{-1}\sum_{k=0}^\infty{n+m+1\choose
k+m+1}{z-1\choose k}(-1)^kS_k.\]
\end{theorem}\vspace{3mm}

\noindent {\it Proof.} \, Using the relation (\ref{s1}) we have
\begin{eqnarray*}
 _mP_n(z)&amp;=&amp;\frac{\Gamma(z)\Gamma(m+2)}{\Gamma(m+z+1)}
 \int_0^\infty e^{-t}\sum_{k=0}^\infty {m+z\choose k+m+1}{n\choose
 k}(1-t)^kdt\\[3mm]
 &amp;=&amp;\frac{\Gamma(z)(m+1)!}{\Gamma(m+z+1)}
 \sum_{k=0}^\infty
 \frac{\Gamma(m+z+1)}{\Gamma(z-k)(k+m+1)!}{n\choose
 k}\int_0^\infty e^{-t}(1-t)^kdt\\[3mm]
 &amp;=&amp;{n+m+1\choose n}^{-1}\sum_{k=0}^\infty{n+m+1\choose
k+m+1}{z-1\choose k}\int_0^\infty e^{-t}(1-t)^kdt.
\end{eqnarray*}
Now use
 \[\mathcal{L}[s;(t+\alpha)^{z-1}]=\frac{e^{\alpha s}\Gamma(z,\alpha
 s)}{s^z}\qquad (\Re s&gt;0),\]
to obtain
\[_mP_n(z)={n+m+1\choose n}^{-1}\sum_{k=0}^\infty{n+m+1\choose k+m+1}{z-1\choose
k}(-1)^k\frac{\Gamma(k+1,-1)}{e}.\] Here $\Gamma(z,x)$ is the
incomplete gamma function. The result follows from
\[ \Gamma(k+1,-1)=eS_k. \]

\begin{lemma}\label{le1}For $n\in \NN$ we have
\[_{-1}P_n(z)=1+n!\sum_{k=1}^n\frac{(-1)^kS_k}{(n-k)!\,(k!)^2}\prod_{i=1}^k(z-i).\]
\end{lemma}\vspace{3mm}

\noindent {\it Proof.} \, Applying Theorem \ref{t1} for $m=-1$, we
have
\begin{eqnarray*}
_{-1}P_n(z)&amp;=&amp;\sum_{k=0}^n{n\choose k}{z-1\choose
k}(-1)^kS_k,\\[3mm]
&amp;=&amp;1+n!\sum_{k=1}^n\frac{(-1)^kS_k}{(n-k)!(k!)^2}\prod_{i=1}^k(z-i).
\end{eqnarray*}\vspace{3mm}

\begin{remark} Let the sequence $X_{n,k}$ be defined by
\begin{displaymath}
  X_{n,k} = \begin{cases}
  Y_n, &amp; \quad {\rm if}\ n=k, \\
  (n-k)X_{n-1,k}, &amp; \quad {\rm if}\ n&gt;k,
  \end{cases}
\end{displaymath}
\\ where $Y_n=(n!)^2.$ Since $X_{n,k}=(n-k)!\,(k!)^2$ we have
\[_{-1}P_n(z)=1+n!\sum_{k=1}^n\frac{(-1)^kS_k}{X_{n,k}}\prod_{i=1}^k(z-i).\]
\end{remark}\vspace{3mm}

\begin{theorem}\label{t2}For $m=-1,0,1,2,...$ and $n\in \NN$ we
have
\begin{eqnarray*}
  _mP_0(z)&amp;=&amp; _mP_1(z)=1\\[3mm]
 _mP_n(z)&amp;=&amp; \frac{1}{n+m+1}\,{}[(m+1)\,{}_{m-1}P_n(z)+n\,{}
 _mP_{n-1}(z)],\quad m&gt;-1.
\end{eqnarray*}
\end{theorem}\vspace{3mm}

\noindent {\it Proof.} \, According to Theorem \ref{t1} we
have
\[\frac{1}{n+m+1}\,{}[(m+1)\,{}_{m-1}P_n(z)+n\,{}
 _mP_{n-1}(z)]=\]
\begin{eqnarray*}
&amp;&amp;\frac{(m+1)!n!}{(n+m+1)!}\sum_{k=0}^\infty{n+m\choose
k+m}{z-1\choose k}(-1)^kS_k+\\[3mm]
&amp;+&amp;\frac{(m+1)!n!}{(n+m+1)!}\sum_{k=0}^\infty{n+m\choose
k+m+1}{z-1\choose k}(-1)^kS_k .
\end{eqnarray*}
The recurrence for $_mP_n(z)$ now follows from
\[{a\choose b}+{a\choose b+1}={a+1\choose b+1}.\]\vspace{3mm}

\begin{corollary}\label{co1} For $m=-1,0,1,2,...$ and $n\in \NN$
we have
\[_1M_m(1;-n,z)=\frac{n!\, z(z+1)\ldots (z+m)}{(n+m+1)!}\sum_{k=0}^\infty{n+m+1\choose
k+m+1}{z-1\choose k}(-1)^kS_k.\]
\end{corollary}\vspace{3mm}

\begin{corollary}\label{co2} For $n\in \NN$ the result is as
follows
\begin{eqnarray*}
_1M_{-1}(1;-n,z)&amp;=&amp;1+n!\sum_{k=1}^n\frac{(-1)^kS_k}{(n-k)!\,(k!)^2}\prod_{i=1}^k(z-i),\\[3mm]
_1M_{m}(1;0,z)&amp;=&amp;\,_1M_{m}(1;-1,z)=\frac{\Gamma(m+z+1)}{\Gamma(z)\Gamma(m+2)},\\[3mm]
_1M_{m}(1;-n,z)&amp;=&amp;\frac{1}{n+m+1}\cdot[\,(m+z)\cdot\,
_1M_{m-1}(1;-n,z)+\\[2mm]
 &amp;+&amp;n\cdot \, _1M_m(1;-n+1,z)],   \quad m&gt;-1.
 \end{eqnarray*}
\end{corollary}\vspace{3mm}
The numbers $_1M_m(1;-n,r)_{r=0}^{\infty}$ can now be evaluated
recursively. \vspace{3mm}

\begin{table}[ht]\label{table3}
\caption{The numbers $_1M_m(1;-n,r)$ for $m=0,1,2,3,4$ and
$n=1,2$}\vspace{3mm}
\begin{tabular*}{\textwidth}{@{\extracolsep{\fill}}llll}
\hline $_1M_{m}(1,-n,r)$&amp;$\mbox{sequence in
\cite{Slo}}$&amp;$_1M_{m}(1,-n,r)$&amp;$\mbox{sequence in \cite{Slo}}$\cr
\hline
 $_1M_{0}(1,-1,r)$&amp;$0,1,2,3,4...$&amp;$_1M_{0}(1,-2,r)$&amp;$A000125$\cr \\
 $_1M_{1}(1,-1,r)$&amp;$A000217$&amp;$_1M_{1}(1,-2,r)$&amp;$A055795$\cr \\
 $_1M_{2}(1,-1,r)$&amp;$A000292$&amp;$_1M_{2}(1,-2,r)$&amp;$A027660$\cr \\
 $_1M_{3}(1,-1,r)$&amp;$A000332$&amp;$_1M_{3}(1,-2,r)$&amp;$A055796$\cr \\
 $_1M_{4}(1,-1,r)$&amp;$A000389$&amp;$_1M_{4}(1,-2,r)$&amp;$A055797$ \\
\hline
\end{tabular*}
\end{table}

\subsection{The numbers $\{_{\lowercase{1}}M_{\lowercase{m}}
(1/\lowercase{n};\lowercase{m+2},\lowercase{r})
\}_{\lowercase{r=0}}^{+\infty}\, _{\lowercase{n=1}}^{+\infty}\,
_{\lowercase{m=-1}}^{+\infty}$} In Table 4 twelve well-known
sequences from \cite{Slo} are given. These sequences are special
cases of the function $_vM_m(s;a,r)$ for $v=1,\, s=1/n,$ and
$a=m+2$. The sequences have the following common
characteristic.\vspace{3mm}

\begin{lemma} For $m=-1,0,1,2,...,$ we have
\[_1M_{m}(1/n;m+2,z)=\frac{n^z\Gamma(z+m+1)}{(m+1)!}.\]
\end{lemma}\vspace{3mm}

\noindent {\it Proof.} \, Since
\[_2F_1(m+2,1-z,m+2,1-t)=t^{z-1}\] we have
\[\mathcal{L}[1/n;\,{}_2F_1(m+2,1-z,m+2,1-t)]=n^z\Gamma(z).\]

\begin{table}[htb]\label{table4}
\caption{The numbers $_1M_{m}(1/n;m+2,r)$ for $m=-1,0,1,2,3,4$ and
 $n=1,2,3,4$}\vspace{3mm}
\begin{tabular*}{\textwidth}{@{\extracolsep{\fill}}llll}
\hline $_1M_{m}(1/n,m+2,r)$&amp;$\mbox{in
\cite{Slo}}$&amp;$_1M_{m}(1/n,m+2,r)$&amp;$\mbox{in \cite{Slo}}$\cr \hline
 $_1M_{-1}(1,1,r)$&amp;$A000142$&amp;$_1M_{-1}(1/2,1,r)$&amp;$A066318$\cr \\
 $_1M_{-1}(1/3,1,r)$&amp;$A032179$&amp;$_1M_{-1}(1/4,1,r)$&amp;$\mbox{unlisted}$\cr \\
 $_1M_{0}(1,2,r)$&amp;$A000142$&amp;$_1M_{0}(1/2,2,r)$&amp;$A000165$\cr \\
 $_1M_{0}(1/3,2,r)$&amp;$A032031$&amp;$_1M_{0}(1/4,2,r)$&amp;$A047053$\cr \\
 $_1M_{1}(1,3,r)$&amp;$A001710$&amp;$_1M_{1}(1/2,3,r)$&amp;$A014297$\cr \\
 $_1M_{1}(1/3,3,r)$&amp;$\mbox{unlisted}$&amp;$_1M_{1}(1/4,3,r)$&amp;$\mbox{unlisted}$\cr \\
 $_1M_{2}(1,4,r)$&amp;$A001715$&amp;$_1M_{2}(1/2,4,r)$&amp;$\mbox{unlisted}$\cr \\
 $_1M_{2}(1/3,4,r)$&amp;$\mbox{unlisted}$&amp;$_1M_{2}(1/4,4,r)$&amp;$\mbox{unlisted}$\cr \\
 $_1M_{3}(1,5,r)$&amp;$A001720$&amp;$_1M_{3}(1/2,5,r)$&amp;$\mbox{unlisted}$\cr \\
 $_1M_{3}(1/3,5,r)$&amp;$\mbox{unlisted}$&amp;$_1M_{3}(1/4,5,r)$&amp;$\mbox{unlisted}$\cr \\
 $_1M_{4}(1,6,r)$&amp;$A001725$&amp;$_1M_{4}(1/2,6,r)$&amp;$\mbox{unlisted}$\cr \\
 $_1M_{4}(1/3,6,r)$&amp;$\mbox{unlisted}$&amp;$_1M_{4}(1/4,6,r)$&amp;$\mbox{unlisted}$ \\
\hline
\end{tabular*}
\end{table}

\subsection{Some equivalents of Kurepa's hypothesis}

The special values $M_{-1}(z)=\Gamma(z)$ and $M_0(z)=K(z)$ given
in (\ref{po1}) yield
\begin{equation}\label{kh1}
   _1M_{-1}(1,1,n+1)=n! \quad\mbox{and}\quad _1M_0(1;1,n)= !n
\end{equation}
where $n!$ and $!n$ are the right factorial numbers and the left
factorial numbers given in (\ref{eq1.1}).
The function $n!$ and
$!n$ are linked by Kurepa's hypothesis:\vspace{3mm}
\\{\bf KH hypothesis.} {\it For $n\in \NN \backslash \{1 \}$ we
have} \[ \gcd( \, !n,\, n!\,)=2\] {\it where $\gcd(a,b)$
denotes the greatest common divisor of integers $a$ and
$b$}.\vspace{3mm}
\\This is listed as Problem B44 of Guy's classic book \cite{Gu}.
In \cite{Du2}, it was proved that the KH is equivalent to the
following assertion
\[!p \not\equiv 0 \pmod p , \ra {\rm for \; all \; primes} \;
p&gt;2.\]
\\The sequences $a_{n},\, b_n,\, c_n,\, d_n$ and $e_n$
(sequences A052169, A051398, \\ A051403, A002467 and A002720 in
\cite{Slo}) are defined as follows:
\begin{eqnarray*}
a_{2}=1\quad a_{3}=2\quad
a_{n}&amp;=&amp;(n-2)a_{n-1}+(n-3)a_{n-2},\\[1mm] b_{3}=2\quad
b_{n}&amp;=&amp;-(n-3)b_{n-1}+2(n-2)^2,\\[1mm] c_{1}=3\quad c_{2}=8\quad
c_{n}&amp;=&amp;(n+2)(c_{n-1}-c_{n-2}),\\[1mm] d_{0}=0\quad d_{1}=1\quad
d_{n}&amp;=&amp;(n-1)(d_{n-1}+d_{n-2}),\\[1mm] e_{0}=1 \quad e_{1}=2\quad
e_{n}&amp;=&amp;2ne_{n-1}-(n-1)^2e_{n-2}.
\end{eqnarray*}
They are related to the left factorial function. For instance, let
$p&gt;3$ be a prime number.   Then
\[!p\equiv -3a_{p-2}\equiv -b_{p}\equiv -2c_{p-3}\equiv d_{p-2}\equiv e_{p-1} \pmod p.\]
We give the details for the last congruence.\vspace{3mm}

\noindent {\it Proof.} \, Let
\[L_{n}^{\nu}(x)=\sum_{k=0}^{n}\frac{\Gamma(\nu + n+1)}
{\Gamma(\nu +k+1)} \frac{(-x)^{k}}{k!(n-k)!}\] be the Laguerre
polynomials, and set $L_{n}^{0}(x)=L_{n}(x).$ Using the relation
${p-1 \choose k}\equiv (-1)^k \pmod p$, we have
\[L_{p-1}(x)\equiv -(p-1)!\sum_{k=0}^{p-1}\frac{x^k}{k!} \pmod p .\]
Wilson's theorem yields
\[!p\equiv -L_{p-1}(-1) \pmod p.\]
The recurrence for Laguerre polynomials
\[(n+1)L_{n+1}^{\nu} (x)=(\nu+2n+1-x)L_{n}^{\nu}(x)-(\nu +n)L_{n-1}^{\nu}(x),\] for
$\nu =0,\; x=-1$ produces
\[h_{p-1}(p-1)!\equiv \!p \pmod p,\]
where
\[h_{1}=2 \quad h_{2}=\frac{7}{2}\quad h_{n}=2h_{n-1}-\frac{n-1}{n}h_{n-2}.\]
The identity $e_n=h_{n}n!$ finally yields
\[!p\equiv e_{p-1} \pmod p .\]

\section{The numbers $_{\lowercase{n}}M_{\lowercase{-1}}(\lowercase{1};\lowercase{1},
\lowercase{n+1})$} The special values
\[_nM_{-1}(1;1,n+1)=A_n,\]
are the alternating factorial numbers given in (\ref{eq1.1}). This
sequence satisfies the recurrence relation
\[A_{0}=0,\; A_{1}=(-1)^{n-1},\; A_{n}=-(n-1)A_{n-1}+nA_{n-2}.\]

These numbers can be expressed in terms of the gamma function as
follows
\begin{eqnarray*}
  A_{n} &amp;=&amp;
 \sum_{k=1}^{n}(-1)^{n-k}\Gamma(k+1)=\int_0^\infty
e^{-x}\left(\sum_{k=1}^{n}(-1)^{n-k}x^{k}\right)dx \\
 [3mm]
 &amp;=&amp; \int_0^\infty e^{-x}\frac{x^{n+1}-(-1)^{n}x}{x+1}dx.
\end{eqnarray*}
The same relation is now used in order to define the function
$A_z$:\vspace{3mm}

\begin{definition}\label{de4.1}For every complex number $z, \; Re\;z&gt;0,$ the
function $A_z$ is defined by
\[A_{z}\stackrel{\rm
def}{=}\int_0^\infty
  e^{-x}\frac{x^{z+1}-(-1)^{z}x}{x+1}dx.\]
\end{definition}\vspace{3mm}
The identity
$\frac{x^{z+1}-(-1)^{z}x}{x+1}=x^{z}-\frac{x^{z}-(-1)^{z-1}x}{x+1}$
gives
\[\int_0^\infty
e^{-x}\frac{x^{z+1}-(-1)^{z}x}{x+1}dx=\int_0^\infty
e^{-x}x^zdx-\int_0^\infty e^{-x}\frac{x^z-(-1)^{z-1}x}{x+1}dx,\]
i.e.,
\begin{equation}\label{eq4.0}
  A_{z}=\Gamma(z+1)-A_{z-1}.
\end{equation}
This gives $A_{0}=\Gamma(2)-A_{1}=0$ and $
A_{-1}=\Gamma(1)-A_{0}=1.$ An inductive argument shows that
$A_{-n},$ the residue of $A_z$ at the pole $z=-n$, is given by
\[res\; A_{-n}=(-1)^{n}\sum_{k=0}^{n-2}\frac{1}{k!},\qquad
  n=2,3,4,...\]
The derivation employs the fact that $\Gamma(z)$ is meromorphic
with simple poles at $z=-n$ and residue $(-1)^n/n!$ there.

The function $A_z$ can be expressed in terms of the exponential
integral $\Ei(x)$ and the incomplete gamma function $\Gamma(z,x)$
by
\[ A_{z}= \mathcal{L}[1;\,{}\frac{t^{z+1}-(-1)^z}{t+1}]=
 e\Gamma(z+2)\Gamma(-z-1,1)-(-1)^ze\Ei(-1)-(-1)^z.\]

\subsection{The generating function for $A_{n-1}$}
The {\it total number of arrangements of a set with $n$ elements}
(sequence A000522 in \cite{Slo}) is defined (see \cite{Cam},
\cite{Ga}, \cite{Ri} and \cite{Si}) by:
\begin{equation}\label{eq4.1}
a_{0}=1,\ra a_{n}=na_{n-1}+1, \quad \mbox{or} \quad
a_{n}=n!\sum_{k=0}^n\frac{1}{k!}.
\end{equation}
The sequence $\{a_n\}$ satisfies:
\begin{equation}\label{eq4.2}
a_{0}=1,\ra a_{1}=2,\ra a_{n}=(n+1)a_{n-1}-(n-1)a_{n-2},
\end{equation}
and
\begin{equation}\label{eq4.3}
a_0=1,\ra a_{n}=\sum_{k=0}^{n-1}{n\choose
k}(-1)^{n+1-k}(n+1-k)a_k.
\end{equation}\medskip
\\Relation (\ref{eq4.2}) comes from the theory of continued
fractions and (\ref{eq4.3}) follows directly from (\ref{eq4.1}).

We now establish a connection between the sequence $\{a_n\}$ and
the alternating factorial numbers $A_n.$\vspace{3mm}

\begin{lemma}\label{le4.1}Let $\; a_{-1}=1$ and $n\in \NN\backslash \{1\}$. Then
$$ A_{n-1}=\sum_{k=0}^{n}{n \choose k}(-1)^{n-k}a_{k-1}. $$
\end{lemma}\vspace{3mm}

\noindent {\it Proof.} \, Using (\ref{eq4.3}) and induction on $n$
we have
\[ n!=\sum_{k=0}^n(-1)^{n-k}{n \choose k}a_k.\]
Inversion yields
\[a_n=n!-\sum_{k=1}^n{n \choose k-1}(-1)^{n+1-k}a_{k-1}.\]
The relation  ${n+1 \choose k}-{n\choose k}={n\choose k-1},\;
k\geq 1$ produces
\[\sum_{k=0}^{n+1}{n+1\choose
k}(-1)^{n+1-k}a_{k-1}+\sum_{k=0}^{n}{n\choose
k}(-1)^{n-k}a_{k-1}=n!.\] 
The result now follows from
(\ref{eq4.0}).\vspace{3mm}

\begin{theorem}\label{th4.1}The exponential generating function for $\{ A_{n-1}\}$
 is given by
\[g(x)=e^{1-x}[E_i(-1)-E_i(x-1)+e^{-1}]-1=\sum_{n=2}^\infty
  A_{n-1}\frac{x^n}{n!},\]
where $E_{i}(x)$ is the exponential integral (\ref{ei}).
\end{theorem}\vspace{3mm}

\noindent {\it Proof.} \,The expansion of the exponential integral
\[E_{i}(x)=\gamma +\ln(x)+\sum_{k=1}^\infty \frac{x^k}{k\cdot
k!},\] where $\gamma$ is Euler's constant, appears in
\cite[p.~57,~5.1.10.]{AS}. The statement of the theorem can be
written as
\[e[E_i(-1)-E_i(x-1)+e^{-1}]=e\left[-\ln(x-1)+\sum_{k=1}^\infty
  \frac{(-1)^k-(x-1)^k)}{k\cdot k!}\right]=\]
\begin{equation}\label{eq4.4}
 = \sum_{k=0}^\infty
  a_{k-1}\frac{x^k}{k!}.
\end{equation}
Expand $e^{-x}$ as a Taylor series to obtain
\[e^{-x}=\sum_{k=0}^\infty (-1)^k\frac{x^k}{k!}.\]
Then
\begin{eqnarray*}
e^{1-x}[E_i(-1)-E_i(x-1)+e^{-1}]-1&amp;=&amp;\sum_{k=0}^\infty
  a_{k-1}\frac{x^k}{k!}\sum_{k=0}^\infty
  (-1)^k\frac{x^k}{k!}-1\\[3mm]
 &amp;=&amp;\sum_{n=0}^\infty\sum_{k=0}^n
  a_{k-1}\frac{x^k}{k!}(-1)^{n-k}\frac{x^{n-k}}{(n-k)!}-1\\[3mm]
 &amp;=&amp;\sum_{n=0}^\infty\sum_{k=0}^n
  {n \choose k}a_{k-1}(-1)^{n-k}\frac{x^{n}}{n!}-1\\[3mm]
 &amp;=&amp;\sum_{n=0}^\infty A_{n-1}\frac{x^n}{n!}-1.
\end{eqnarray*}

By induction on $n$ we get\vspace{3mm}

\begin{lemma}Let $n\in N$. The function $g(x)$ in Theorem 4.1
satisfies
\[g^{(n)}(x)=\frac{d}{dx}g^{(n-1)}(x)=(-1)^n\left[
g(x)+1+\sum_{k=0}^{n-1}\frac{k!}{(x-1)^{k+1}}\right]. \]
\end{lemma}\vspace{3mm}
This identity gives the algorithm for computing the $n$-th
derivation of the function $g(x).$ \vspace{3mm}

\subsection{The AL hypothesis}
In \cite[p.~100]{Gu} the following problem is given:\vspace{3mm}
\\{\bf Problem B43.} Are there infinitely many numbers $n$ such
that $A_{n-1}$ is a prime?\vspace{3mm}

Here
\[A_n=\sum_{k=1}^{n}(-1)^{n-k}k!.\]

If there is a value of $n-1$ such that $n$ divides $A_{n-1}$, then
$n$ will divide $A_{m-1}$ for all $m&gt;n$, and there would be only a
finite number of prime values. The required condition for the
existence of infinitely many numbers $n$ such that $A_{n-1}$ is a
prime may be expressed as follows:\vspace{3mm}
\\{\bf AL hypothesis.} {\it For every prime number} $p$
\[A_{p-1}\not\equiv 0 \pmod p,\]
{\it holds.}\vspace{3mm}

Let $p$ be a prime number and $n,m\in \NN \backslash \{1\}$. It is
not difficult to prove the following results:

\[A_{n-1}\equiv -1-\sum_{k=2}^{n}[k-1-(-1)^{n-k}]\Gamma(k) \pmod n,\]

\begin{eqnarray*}
A_{n-1}&amp;=&amp;
 \frac{\Gamma(n+1)-1+\sum_{k=2}^{n-1}[(-1)^{n-k}n-k+1-(-1)^{n-k}]
\Gamma(k)}{n-1},\\[3mm]
 &amp;=&amp;1-!n+2\sum_{k=1}^{n-1}A_{k}\\[3mm]
 &amp;=&amp;3-!n-!(n-1)\cdot 2n+4\sum_{i=2}^{n}\sum_{k=1}^{i-1}A_{k}.
\end{eqnarray*}


\[\sum_{k=1}^{n}\sum_{i=0}^{m-1}(-1)^i(\Gamma(k+1))^{m-i}A_{k-1}^{i}=
\left\{\begin{array}{rl}\ra A_{n}^m,\ra\ra\ra\ra  m \; {\rm
even}\\[1mm]
\\A_{n}^m+2\sum_{j=1}^{n-1}A_{j}^m,\;  m \; {\rm odd}
\end{array}\right. ,\]

\[A_{p-1}=-\sum_{i=1}^{p}\frac{1}{\Gamma(i)}+p\sum_{i=1}^{p}\frac{n_i(-1)^{i-1}}
{\Gamma(i)}, \quad n_i\in \NN \quad (i=1,2,...p)\]

The first step in solving problem B43 is proving the AL hypothesis. Using
the previous identities, equivalents for the 
AL hypothesis can be given.

\section{Conclusions}

The main contribution is to define the function $_vM_m(s;a,z)$
by which problems B43 and B44 are connected. The Kurepa hypothesis is
an unsolved problem since 1971 and there seems to be no significant
progress in solving it, apart from numerous equivalents, such
as these in Section 3.3. Further details can be found in \cite{IvMi}.

However, apart from $n!, \ !n$, and $A_n$, twenty-five more well-known
sequences in \cite{Slo} are special cases of the function
$_vM_m(s;a,z).$ The first study of the function gave the author
the idea to find an algorithm for computing some special cases
(Corollary 2 and Lemma 3) before solving the above mentioned
problems.

The definition of the function $_vM_m(s;a,z)$ suggest another
method of studying the function by using the characteristics of
the inverse Laplace transform.

\section*{Acknowledgements}
I would like to thank the referee for the numerous comments and
suggestions.

\begin{thebibliography}{20}

\bibitem{AS}
M. Abramowitz, I. A. Stegun, {\it Handbook of Mathematical Functions},
Dover Publications, Inc. New York, 1965. 

\bibitem{Br}
C. Brezinski, {\it History of Continued Fractions and Pad\'e
Approximants}, Springer-Verlag, Berlin, 1991.

\bibitem{Cam}
P. J. Cameron, Sequences realized by oligomorphic permutation
groups, {\it J. Integer Sequences} {\bf 3} (1) (2000), Article 00.1.5.

\bibitem{Com}
L. Comtet, {\it Advanced Combinatorics,} Reidel, Dordrecht, 1974.

\bibitem{Ga}
J. M. Gandhi, On logarithmic numbers, {\it Math.\ Student} {\bf 31}
(1963), 73--83.

\bibitem{Gu}
R. Guy, {\it Unsolved Problems in Number Theory}, Springer-Verlag,
1994.

\bibitem{IvMi}
A. Ivi\'c and \v Z. Mijajlovi\'c, On Kurepa problems in number
theory, {\it Publ.\ Inst.\ Math.\/} (N.S.) {\bf 57 (71)} (1995),
19--28.

\bibitem{Du2}
\Dj. Kurepa, On the left factorial function $!n$,  {\it Math.\ 
Balkanica} {\bf 1} (1971), 147--153.

\bibitem{Du3}
\Dj. Kurepa, Left factorial function in complex domain,
  {\it Math.\ Balkanica} {\bf 3} (1973), 297--307.

\bibitem{Mil2}
G. V. Milovanovi\'c, A sequence of Kurepa's functions,
{\it Sci.\ Rev.\ Ser.\ Sci.\ Eng.}  No.~{\bf 19--20} (1996), 137--146.

\bibitem{MiPe}
G. V. Milovanovi\'c and A. Petojevi\'{c}, Generalized
factorial function, numbers and polynomials and related problems,
{\it Math.\ Balkanica}, to appear.

\bibitem{Pe}
O. Perron, {\it Die Lehren von den Kettenb\"{u}chen}, Chelsea
Publishing Company, New York, 1954.

\bibitem{Pet1} A. Petojevi\'c, On Kurepa's hypothesis for left
factorial, {\it Filomat (Nis)}, {\bf 12} (1) (1998), 29--37.

\bibitem{Pr}
A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev, {\it Integrals
and Series. Elementary Functions,} Nauka, Moscow, 1981. (Russian)

\bibitem{Ri} J. Riordan,
{\it An Introduction to Combinatorial Analysis,} Wiley,
1958.

\bibitem{Si} D. Singh, The numbers {\it L(m,n)} and their relations
with prepared Bernoulli and Eulerian numbers, {\it Math.\ Student} {\bf
20} (1952), 66--70.

\bibitem{Slo} N. J. A. Sloane, {\it The On-Line Encyclopedia of Integer
Sequences}, published electronically at\hfil\break
{\tt http://www.research.att.com/\char'176 njas/sequences/}

\bibitem{Wilf}
H. S. Wilf, {\it Generatingfunctionology,} Academic Press, New
York, 1990.

\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}: \ \ 
Primary 11B83; Secondary 33C05, 44A10.

\noindent \emph{Keywords: left factorial, alternating factorial,
hypergeometric function, Laplace transform}

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences 
\seqnum{A000142},
\seqnum{A003422},
\seqnum{A005165},
\seqnum{A000217},
\seqnum{A014144},
\seqnum{A007489},
\seqnum{A000292},
\seqnum{A000332},
\seqnum{A000389},
\seqnum{A014145},
\seqnum{A000166},
\seqnum{A000125},
\seqnum{A000217},
\seqnum{A000389},
\seqnum{A055795},
\seqnum{A027660},
\seqnum{A055796},
\seqnum{A055797},
\seqnum{A032179},
\seqnum{A032031},
\seqnum{A001710},
\seqnum{A001715},
\seqnum{A001720},
\seqnum{A001725},
\seqnum{A066318},
\seqnum{A000165},
\seqnum{A047053},
\seqnum{A014297},
\seqnum{A052169},
\seqnum{A051398},
\seqnum{A051403},
\seqnum{A002467},
\seqnum{A002720}, and
\seqnum{A000522}.)

\vspace*{+.1in}
\noindent
Received March 21, 2002;
revised version received August 14, 2002.
Published in Journal of Integer Sequences August 31, 2002.



\bigskip
\hrule
\bigskip

\noindent 
Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.math.uwaterloo.ca/JIS/}.





\end{document}

</pre></body></html>