<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">%\documentstyle[aps,amsfonts,epsfig,preprint,pra]{revtex}
\documentclass{article}
\usepackage[colorlinks=true]{hyperref}

\usepackage{amsfonts,amssymb}
\usepackage{psfig,epsf}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{.1in}
\setlength{\topmargin}{-.5in}
\setlength{\textheight}{8.9in}

\renewcommand{\baselinestretch}{1.4}
\def\C{{\mathbb C}}
\begin{document} 
\begin{center}
\epsfxsize=4in
\leavevmode\epsffile{logo0114.eps}
\vskip 1cm
{\LARGE\bf Extended Bell and Stirling Numbers From Hypergeometric Exponentiation}
\vskip 1.5cm
\large J.-M. Sixdeniers \\
\large K. A. Penson \\ 
\large A. I. Solomon\footnotemark[1] \\ \smallskip
\footnotetext[1]{
Permanent address: Quantum Processes Group, Open University, Milton Keynes, MK7 6AA, United Kingdom.}
Universit\'{e} Pierre et Marie Curie, Laboratoire de Physique Th\'{e}orique des Liquides,
\linebreak
Tour 16, $5^{i\grave{e}me}$ \'{e}tage, 4 place Jussieu, 75252 Paris Cedex 05, France \\ \medskip

Email addresses:
\href{sixdeniers@lptl.jussieu.fr}{sixdeniers@lptl.jussieu.fr}, \href{penson@lptl.jussieu.fr}{penson@lptl.jussieu.fr} and \href{a.i.solomon@open.ac.uk}{a.i.solomon@open.ac.uk}

\vskip2.5cm
{\bf Abstract}
\end{center}

{\em
\noindent Exponentiating the hypergeometric series  $\:_0F_L(1,1,\ldots,1;z)$,
$L=0,1,2,\ldots,$ furnishes a recursion relation for the members of
certain
integer sequences $b_L(n)$, $n=0,1,2,\ldots$. For $L&gt;0$, the $b_L(n)$'s are
generalizations of the conventional Bell numbers, $b_0(n)$.
The corresponding associated Stirling numbers of the second kind are
also investigated. For $L=1$ one can give a
combinatorial interpretation of the numbers $b_1(n)$ and of some
Stirling numbers associated with them. We also consider  the $L\geq1$
analogues of Bell numbers for restricted partitions.
}

\bigskip

The conventional Bell numbers \cite{yablonsky}  $b_0(n)$, $n=0,1,2,\ldots$, have a well-known exponential generating function
\begin{equation}
B_0(z)\equiv e^{\textstyle{(e^z-1)}}=\sum_{n=0}^{\infty}b_0(n)\frac{z^n}{n!},\label{B}
\end{equation}
which can be derived by interpreting $b_0(n)$ as the number of partitions of a set of $n$ distinct elements.
In this note we obtain recursion relations for related sequences of
positive integers, called $b_L(n)$, $L=0,1,2,\ldots,$ obtained by exponentiating the hypergeometric
series $\:_0F_L(1,1,\ldots,1;z)$ defined by \cite{andrews}:
\begin{equation}
\:_0F_L(\underbrace{1,1,\ldots,1}_{L};z)=\sum_{n=0}^{\infty}\frac{z^n}{(n!)^{L+1}},\label{defF}
\end{equation}
(which we shall denote by
$\:_0F_L(z)$) and
which includes the special cases $\:_0F_0(z)\equiv e^z$ and $\:_0F_1(z)\equiv
I_0(2\sqrt{z})$, where $I_0(x)$ is the modified Bessel function of the
first kind. For $L&gt;1$, the functions 
$\:_0F_L(z)$ are related to the so-called hyper-Bessel functions
\cite{marichev}, \cite{kiryakova}, \cite{paris},  which have recently
found application in quantum mechanics \cite{witte},
\cite{klauder}.
Thus we are interested in $b_L(n)$ given by
\begin{equation}
e^{[\:_0F_L(z)-1]}=\sum_{n=0}^{\infty}b_L(n)\frac{z^n}{(n!)^{L+1}},\label{geneF}   
\end{equation}
thereby defining a {\em hypergeometric} generating function for
the numbers $b_L(n)$. 
From eq.~(\ref{geneF}) it follows formally that 
\begin{equation}
b_L(n)=(n!)^L\cdot\frac{d^n}{dz^n}\left.\left(e^{[\:_0F_L(z)-1]}\right)\right|_{z=0}.\label{deriv}
\end{equation}
For $L=0$ the r.h.s of eq.~(\ref{deriv}) can be evaluated in closed form:
\begin{equation}
b_0(n)=\frac{1}{e}\sum_{k=0}^{\infty}\frac{k^n}{k!}=\left\{\frac{1}{e^z}\left[\left(z\frac{d}{dz}\right)^ne^z\right]\right\}_{z=1}.\label{dob}
\end{equation}
The first equality in (\ref{dob}) is the celebrated Dobi\'{n}ski formula
\cite{yablonsky}, \cite{comtet}, \cite{wilf}. The second equality in
eq.~(\ref{dob}) follows from observing that for a power series
$R(z)=\sum_{k=0}^{\infty}A_kz^k$
we have
\begin{equation}
\left(z\frac{d}{dz}\right)^nR(z)=\sum_{k=0}^{\infty}A_k\:k^n\:z^k\label{opderiv}
\end{equation}
and applying eq.~(\ref{opderiv}) to the exponential series $(A_k=(k!)^{-1})$.

The reason for including the divisors $(n!)^{L+1}$ rather than $n!$ as in the usual exponential generating function arises from the fact that only by using eq.~(\ref{geneF}) are the numbers $b_L(n)$  actually integers.
This can be seen from general formulas for exponentiation of a power
series \cite{comtet}, which employ the (exponential) Bell polynomials,
complicated and rather unwieldy objects. It cannot however be
considered as a proof that the $b_L(n)$ are integers.  At
this stage we shall use  eq.~(\ref{geneF}) with $b_L(n)$ real and apply to it an
efficient method, described in \cite{wilf}, which will yield the recursion
relation for the $b_L(n)$. (For the proof that the $b_L(n)$ are integers, see
below eq.~(\ref{recur})).   To this end we first obtain a result for the
multiplication of two power-series of the type (\ref{geneF}). Suppose
we wish to multiply $f(x)=\sum_{n=0}^{\infty}a_L(n)\frac{x^n}{(n!)^{L+1}}$ and
$g(x)=\sum_{n=0}^{\infty}c_L(n)\frac{x^n}{(n!)^{L+1}}$. We get
$f(x)\cdot g(x)=\sum_{n=0}^{\infty}d_L(n)\frac{x^n}{(n!)^{L+1}}$, where
\begin{equation}
d_L(n)=(n!)^{L+1}\sum_{r+s=n}^{\infty}\frac{a_L(r)c_L(s)}{(r!)^{L+1}(s!)^{L+1}}=\sum_{r=0}^{n}\left(\scriptsize{\begin{array}{c} n \\ r \end{array}}\right)^{L+1}a_L(r)\:c_L(n-r).\label{dL}
\end{equation}
Substitute eq.~(\ref{defF}) into eq.~(\ref{geneF}) and take the logarithm of both
sides of eq.~(\ref{geneF}):
\begin{equation}
\sum_{n=1}^{\infty}\frac{z^n}{(n!)^{L+1}}= \ln\left(\sum_{n=0}^{\infty}b_L(n)\frac{z^n}{(n!)^{L+1}}\right).\label{log}
\end{equation}
Now differentiate both sides of eq.~(\ref{log}) and multiply  by $z$:
\begin{equation}
\left(\sum_{n=0}^{\infty}b_L(n)\frac{z^n}{(n!)^{L+1}}\right)\left(\sum_{n=0}^{\infty}n\:\frac{z^n}{(n!)^{L+1}}\right)=\sum_{n=0}^{\infty}n\:b_L(n)\frac{z^n}{(n!)^{L+1}},
\end{equation}
which with eq.~(\ref{dL}) yields the desired recurrence relation
\begin{eqnarray}
b_L(n+1) &amp; = &amp;
\frac{1}{n+1}\sum_{k=0}^{n}\scriptsize{\left(\!\!\begin{array}{c} n+1 \\ k
\end{array}\!\!\right)}^{L+1}(n+1-k)\:b_L(k),\hspace{1cm}n=0,1,\ldots\label{recurb}\\
&amp; = &amp; \sum_{k=0}^{n}\left(\scriptsize{\begin{array}{c} n \\ k\end{array}}\right)\scriptsize{\left(\begin{array}{c} n+1 \\ k\end{array}\right)}^{L}\:b_L(k),\label{recur}\\
b_L(0) &amp; = &amp; 1.\label{bLinit}
\end{eqnarray}
Since eq.~(\ref{recur}) involves only positive integers, it follows
that the $b_L(n)$ are indeed positive integers.
For  $L=0$ one gets the known recurrence relation for the Bell numbers \cite{wilf}:
\begin{equation}
b_0(n+1) =\sum_{k=0}^{n}\scriptsize{\left(\begin{array}{c} n \\ k \end{array}\right)}b_0(k).
\end{equation}
We have used eq.~(\ref{recur}) to calculate some of the $b_L(n)$'s,
listed in Table I, for $L=0,1,\ldots,6$. Eq.(\ref{recur}), for $n$ fixed, gives 
closed form expressions for the $b_L(n)$ directly as a function of $L$ (columns
in Table I):  $b_L(2)=1+2^L$,  $b_L(3)=1+3\cdot3^L+(3!)^L$,
$b_L(4)=1+4\cdot4^L+3\cdot6^L+6\cdot12^L+(4!)^L$, etc.

The sets of $b_L(n)$ have been checked against the
most complete source of integer sequences available \cite{sloane}. Apart from the case
$L=0$ (conventional Bell numbers) only the first non-trivial sequence
$L=1$ is listed:\footnote{(others have since been added)} it turns out that this sequence $b_1(n)$, listed under the heading
{A023998} in \cite{sloane}, can be given a combinatorial
interpretation as the number of block permutations on a set of $n$ objects which are
uniform, i.e. corresponding blocks have the same size
\cite{fitzgerald}.

Eq.(\ref{B}) can be generalized by including an additional
variable $x$, which will result in ``smearing out'' the conventional
Bell numbers $b_0(n)$ with a set of integers $S_0(n,k)$, such that for $k&gt;n$,
$S_0(n,k)=0$, and $S_0(0,0)=1$, $S_0(n,0)=0$. In particular,
\begin{equation}
B_0(z,x)\equiv e^{\textstyle x(e^z-1)}=\sum_{n=0}^{\infty}\left[\sum_{k=1}^{n}S_0(n,k)\:x^k\right]\frac{z^n}{n!},\label{B0}
\end{equation}
which leads to the (exponential) generating function of $S_0(n,l)$,
the conventional Stirling numbers of the second kind, (see \cite{yablonsky}, \cite{comtet}), in the form
\begin{equation}
\frac{(e^z-1)^l}{l!}=\sum_{n=l}^{\infty}\frac{S_0(n,l)}{n!}z^n,
\end{equation}
and defines the so-called exponential or Touchard polynomials $l_n^{(0)}(x)$ as
\begin{equation}
l_n^{(0)}(x)=\sum_{k=1}^{n}S_0(n,k)x^k.
\end{equation}
They satisfy
\begin{equation}
l_n^{(0)}(1)=b_0(n),
\end{equation}
justifying  the term ``smearing out'' used above. 

The appearance of integers in eq.~(\ref{geneF}) suggests a  natural
extension with an additional variable $x$:
\begin{equation}
B_L(z,x)\equiv e^{x[\:_0F_L(z)-1]}=\sum_{n=0}^{\infty}\left[\sum_{k=1}^{n}S_L(n,k)\:x^k\right]\frac{z^n}{(n!)^{L+1}},\label{expF}
\end{equation}
where we  include the right divisors $(n!)^{L+1}$
in the r.h.s of (\ref{expF}).

This in turn defines  ``hypergeometric''polynomials
of type $L$ and order $n$ through
\begin{equation}
l_n^{(L)}(x)=\sum_{k=1}^{n}S_L(n,k)x^k,\label{poly}
\end{equation}
which satisfy
\begin{equation}
l_n^{(L)}(1)=b_L(n),
\end{equation}
with the $b_L(n)$ of eq.~(\ref{recurb}).
Thus the polynomials of
eq.~(\ref{poly}) "smear out" the $b_L(n)$ with the generalized Stirling
numbers of the second kind, of type $L$, denoted by $S_L(n,k)$ (with
$S_L(n,k)=0$, if $k&gt;n$, $S_L(n,0)=0$ if $n&gt;0$ and $S_L(0,0)=1$), which have, from eq.~(\ref{expF}) the ``hypergeometric''generating function
\begin{equation}
\frac{(\:_0F_L(z)-1)^l}{l!}=\sum_{n=l}^{\infty}\frac{S_L(n,l)}{(n!)^{L+1}}\:z^n,\hspace{1cm}L=0,1,2,\ldots .\label{geneSL}
\end{equation}

Eq.(\ref{geneSL}) can be used to derive a recursion relation for
the numbers $S_L(n,k)$, in the same manner as eq.~(\ref{geneF})
yielded eq.~(\ref{bLinit}). Thus we take the logarithm of both sides
of eq.~(\ref{geneSL}), differentiate with respect to $z$, multiply by $z$
and obtain:
\begin{equation}
\left(\sum_{n=0}^{\infty}\frac{S_L(n,l-1)}{(n!)^{L+1}}\:z^n\right)\left(\sum_{n=0}^{\infty}\frac{n}{(n!)^{L+1}}\:z^n\right)=\sum_{n=0}^{\infty}\frac{n\:S_L(n,l)}{(n!)^{L+1}}\:z^n,\label{eqsnl}
\end{equation}
which, with the help of eq.~(\ref{dL}),  produces the required recursion
relation
\begin{eqnarray}
S_L(n+1,l) &amp; = &amp; \sum_{k=l-1}^{n}\left(\scriptsize{\begin{array}{c} n \\
k\end{array}}\right)\scriptsize{\left(\begin{array}{c} n+1 \\
k\end{array}\right)}^{L}\:S_L(k,l-1),\label{recurSL}\\
&amp; &amp;  \hspace{-1cm}S_L(0,0)=1,\hspace{1cm}S_L(n,0)=0,\label{initrecurSL}
\end{eqnarray}
which for $L=0$ is the recursion relation for the conventional
Stirling numbers of the second kind \cite{yablonsky}, \cite{comtet}, and
in eq.~(\ref{recurSL}) the appropriate summation range has been inserted.
Since the recursions of eq.~(\ref{recurSL}) and eq.~(\ref{initrecurSL}) involve
only integers we conclude that $S_L(n,l)$ are positive integers.

We have calculated some of the numbers $S_L(n,l)$ using
eq.~(\ref{geneSL}) and have listed them in Tables II and III,
for $L=1$ and $L=2$ respectively. Observe that $S_1(n,2)=\left(\scriptsize{\begin{array}{c} 2n+1 \\
n+1\end{array}}\right)-1$ and $S_L(n,n)=(n!)^L$, $L=1,2$. Also, by
fixing $n$ and $l$, the individual values of $S_L(n,l)$ have been
calculated as a function of $L$ with the help of eq.~(\ref{recurSL}),
see Table IV, from which we observe
\begin{equation}
S_L(n,n)=(n!)^L,\hspace{1cm} L=1,2,\ldots.\label{Snn}
\end{equation}
which is the lowest diagonal in Table IV. We now demonstrate that the
repetitive use of eq.~(\ref{recurSL}) permits one to establish
closed-form expressions for any supra-diagonal of order $p$, i.e. the sequence $S_L(n+p,n)$, for $p=1,2,3,\ldots$,
if one knows the expression for all $S_L(n+k,n)$ with $k&lt;p$.
We shall illustrate it here for $p=1,2$. To this end fix $l=n$ on both
sides of eq.~(\ref{recurSL}). It becomes, upon using eq.~(\ref{Snn}),
and defining $\alpha_L(n)\equiv S_L(n+1,n)$, a linear recursion relation
\begin{equation}
\alpha_L(n)=\frac{n[(n+1)!]^L}{2^L}+(n+1)^L\alpha_L(n-1),\hspace{1cm}\alpha_L(0)=0,
\end{equation}
with the solution
\begin{eqnarray}
\alpha_L(n)=S_L(n+1,n)&amp;=&amp;\frac{n(n+1)}{2}\left[\frac{(n+1)!}{2}\right]^L\label{alpha}\\
&amp;=&amp;\left[\frac{(n+1)!}{2}\right]^L\:S_0(n+1,n),\label{s1}
\end{eqnarray}
which gives the second lowest diagonal in Table IV. Observe that for
any $L$, $S_L(n+1,n)$ is proportional to $S_0(n+1,n)=n(n+1)/2$. The
sequence $S_1(n+1,n)=1,\:9,\:72,\:600,\:5400,8564480,\:\ldots$ is of
particular interest: it represents the sum of inversion numbers of all
permutations on $n$ letters \cite{sloane}. For more information
about this and related sequences see the entry {A001809} in \cite{sloane}.
The $S_L(n+1,n)$ for $L&gt;1$ do not appear to have a simple combinatorial
interpretation.
A recurrence equation for $\beta_L(n)\equiv S_L(n+2,n)$ is obtained upon
substituting eq.~(\ref{Snn}) and eq.~(\ref{alpha}) into eq.~(\ref{recurSL}):
\begin{equation}
\beta_L(n)=\frac{n(n+1)}{2!}\left[\frac{(n+2)!}{2!}\right]^L\left(\frac{n-1}{2^L}+\frac{1}{3^L}\right)+(n+2)^L\beta_L(n-1),\hspace{1cm}\beta_L(0)=0.
\end{equation}
It has the solution
\begin{equation}
S_L(n+2,n)=\frac{n(n+1)(n+2)}{3\cdot2^3}\left[\frac{(n+2)!}{2}\right]^L\left(\frac{3}{2^L}(n-1)+\frac{4}{3^L}\right)\label{SL}
\end{equation}
which is a closed form expression for the second lowest diagonal in
Table IV. Clearly, eq.~(\ref{SL}) for $L=0$ gives the combinatorial
form for the series of conventional Stirling numbers
\begin{equation}
S_0(n+2,n)=\frac{n(n+1)(n+2)(3n+1)}{4!}.\label{s2}
\end{equation}
In a similar way we obtain
\begin{eqnarray}
S_L(n+3,n)&amp;=&amp;\frac{n(n+1)(n+2)(n+3)}{3\cdot2^4}\left[\frac{(n+3)!}{3}\right]^L\nonumber\\
&amp; &amp;\times\left(n^2\left(\frac{3}{8}\right)^L+n\left(\frac{1}{4^{L-1}}-\frac{3^{L+1}}{8^L}\right)+\frac{2+2\cdot3^L}{8^L}-\frac{1}{4^{L-1}}\right)
\end{eqnarray}
which for $L=0$ reduces to
\begin{equation}
S_0(n+3,n)=\frac{1}{48}n^2(n+1)^2(n+2)(n+3).\label{s3}
\end{equation}
Combined with the standard definition \cite{comtet}, \cite{wilf}
\begin{equation}
S_0(n,l)=\frac{(-1)^l}{l!}\sum_{k=1}^{l}(-1)^k\left(\scriptsize{\begin{array}{c} l \\
k\end{array}}\right)\:k^n.\label{A1}
\end{equation}
eqs.(\ref{s1}), (\ref{s2}) and (\ref{s3}) give compact expressions for
the summation form of $S_0(n+p,n)$. Further, from eq.~(\ref{A1}), use
of eq.~(\ref{opderiv}) gives the following generating formula
\begin{eqnarray}
S_0(n,l) &amp; = &amp;\frac{(-1)^l}{l!}\left[\left(z\frac{d}{dz}\right)^n\left(\sum_{k=1}^{l}(-1)^k\left(\scriptsize{\begin{array}{c}l \\ k\end{array}}\right)\:z^k\right)\right]_{z=1} \\
&amp; = &amp; \frac{(-1)^l}{l!}\left[\left(z\frac{d}{dz}\right)^n[(1-z)^l-1]\right]_{z=1},\hspace{1cm}n\geq l.
\end{eqnarray}

The formula (\ref{B}) can be generalized by putting restrictions on
the type of resulting partitions. The generating function for the
number of partitions of a set of $n$ distinct  elements without
singleton blocks $b_0(1,n)$ is \cite{comtet}, \cite{ehrenborg}, \cite{suter},
\begin{equation}
B_0(1,z)=e^{e^{z}-1-z}=\sum_{n=0}^{\infty}b_0(1,n)\frac{z^n}{n!},
\end{equation}
or more generally, without singleton, doubleton $\ldots$, $p-$blocks
$(p=0,1,\ldots)$ is \cite{suter}
\begin{equation}
B_0(p,z)=e^{e^{z}-\sum_{k=0}^{p}\frac{z^k}{k!}}=\sum_{n=0}^{\infty}b_0(p,n)\frac{z^n}{n!},
\end{equation}
with the corresponding associated Stirling numbers defined by analogy with
eq.~(\ref{B0}) and eq.~(\ref{eqsnl}).
The numbers $b_0(1,n)$, $b_0(2,n)$, $b_0(3,n)$, $b_0(4,n)$ can be read
off from the sequences {A000296},  {A006505},
{A057837} and {A057814} in \cite{sloane},
respectively. For more properties of these numbers see \cite{bernstein}.

We carry over this type of extension to eq.~(\ref{geneF}) and define $b_L(p,n)$ through
\begin{equation}
B_L(p,z)\equiv e^{\:_0F_L(z)-\sum_{k=0}^{p}\frac{z^k}{(k!)^{L+1}}}=\sum_{n=0}^{\infty}b_L(p,n)\frac{z^n}{(n!)^{L+1}},\label{BLpz}
\end{equation}
where $b_L(0,n)=b_L(n)$ from eq.~(\ref{geneF}). (We know of no
combinatorial meaning of $b_L(p,n)$ for $L\geq1$, $p&gt;0$).
The $b_L(p,n)$ satisfy the following recursion relations:
\begin{eqnarray}
b_L(p,n)&amp; = &amp; \sum_{k=0}^{n-p}\left(\scriptsize{\begin{array}{c} n \\ k\end{array}}\right)\scriptsize{\left(\begin{array}{c} n+1 \\ k\end{array}\right)}^{L}\:b_L(p,k),\label{defblpn}\\
b_L(p,0)&amp; = &amp; 1,\\
b_L(p,1)&amp; = &amp; b_L(p,2)=\cdots=b_L(p,p)=0,\\
b_L(p,p+1)&amp; = &amp; 1.
\end{eqnarray}
That the $b_L(p,n)$ are integers follows from eq.~(\ref{defblpn}).
 Through eq.~(\ref{BLpz}) additional families of integer Stirling-like
numbers $S_{L,p}(n,k)$ can be readily defined and investigated. 

The numbers $b_0(p,n)$ are collected in Table V, and Tables VI and
VII contain the lowest values of $b_1(p,n)$ and $b_2(p,n)$, respectively.

Formula (\ref{B}) can be used to express $e$ in terms of $b_0(n)$
in various ways. Two such lowest order (in differentiation) forms are
\begin{eqnarray}
e &amp; = &amp; 1+\ln\left( \sum_{n=0}^{\infty}\frac{b_0(n)}{n!}\right)=\label{lnb0} \\
&amp; = &amp;\ln\left( \sum_{n=0}^{\infty}\frac{b_0(n+1)}{n!}\right).\label{lnb01}
\end{eqnarray}   
In the very same way, eq.~(\ref{geneF}) can be used to express the
values of $\:_0F_L(z)$ and its derivatives at $z=1$ in terms of certain
series of $b_L(n)$'s. For $L=1$, the analogues of eq.~(\ref{lnb0}) and
eq.~(\ref{lnb01}) are 
\begin{eqnarray}
I_0(2) &amp; = &amp; 1+\ln\left( \sum_{n=0}^{\infty}\frac{b_1(n)}{(n!)^2}\right), \\
I_0(2)+\ln(I_1(2)) &amp; = &amp; 1+\ln\left(\sum_{n=0}^{\infty}\frac{b_1(n+1)}{(n+1)(n!)^2}\right)
\end{eqnarray}
and for $L=2$ the corresponding formulas are
\begin{eqnarray}
\:_0F_2(1,1;1) &amp; = &amp; 1+\ln\left( \sum_{n=0}^{\infty}\frac{b_2(n)}{(n!)^3}\right), \\
\:_0F_2(1,1;1)+\ln\left(\:_0F_2(2,2;1)\right) &amp; = &amp; 1+\ln\left(\sum_{n=0}^{\infty}\frac{b_2(n+1)}{(n+1)^2(n!)^3}\right).
\end{eqnarray}

By fixing  $z_0$ at  values other than $z_0=1$, one can link the numerical
values of certain combinations of $\:_0F_L(1,1,\ldots;z_0)$ ,
$\:_0F_L(2,2,\ldots;z_0)$,\ldots and their logarithms, with other series
containing the $b_L(n)$'s.

The above considerations can be extended to the exponentiation of the more
general hypergeometric functions of type $\:_0F_L(k_1,k_2,\ldots,k_L;z)$
where $k_1,k_2,\ldots,k_L$ are positive integers. We conjecture that for every set of $k_n$'s
a different set of integers will be generated through an appropriate
adaptation of eq.~(\ref{geneF}). We quote one simple example of such a
series. For
\begin{equation}
\:_0F_2(1,2;z)=\sum_{n=0}^{\infty}\frac{z^n}{(n+1)(n!)^3}
\end{equation}
eq.~(\ref{geneF}) extends to
\begin{equation}
e^{[\:_0F_2(1,2;z)-1]}=\sum_{n=0}^{\infty}f_2(n)\frac{z^n}{(n+1)(n!)^3}
\end{equation}
where the numbers
\begin{equation}
f_2(n)=(n+1)(n!)^2\left[\frac{d^n}{dz^n}e^{[\:_0F_2(1,2;z)-1]}\right]_{z=0}
\end{equation}
turn out to be integers:
$f_2(n)$, $n=0,1,\ldots,8$ are: 1, 1, 4, 37, 641, 18276, 789377, 48681011, etc. (A061683).
The analogue of equations (\ref{recurSL}) and (\ref{lnb0}) is:
\begin{equation}
\:_0F_2(1,2;1)=1+\ln\left(\sum_{n=0}^{\infty}\frac{f_2(n)}{(n+1)(n!)^3}\right).
\end{equation}

\bigskip

\section*{Acknowledgements}
We thank L. Haddad for interesting discussions. We have used ${\rm
Maple}^{\copyright}$ to calculate most of the numbers discussed above.

\begin{table}
Table I: Table of $b_L(n)$: $L,n=0,1,\ldots,6.$  (The rows give sequences A000110, A023998, A061684--A061688.)
\begin{tabular}{cccccccc}
$L$ &amp; $b_L(0)$ &amp; $b_L(1)$ &amp; $b_L(2)$ &amp; $b_L(3)$ &amp; $b_L(4)$ &amp; $b_L(5)$ &amp; $b_L(6)$ \\ \hline
0 &amp;\hspace{0.3cm} 1 &amp; 1 &amp; 2 &amp; 5 &amp; 15 &amp; 52 &amp; 203 \\
1 &amp;\hspace{0.3cm} 1 &amp; 1 &amp; 3 &amp; 16 &amp; 131 &amp; 1 496 &amp; 22 482 \\
2 &amp;\hspace{0.3cm} 1 &amp; 1 &amp; 5 &amp; 64 &amp; 1 613 &amp; 69 026 &amp; 4 566 992 \\
3 &amp;\hspace{0.3cm} 1 &amp; 1 &amp; 9 &amp; 298 &amp; 25 097 &amp; 4 383 626 &amp; 1 394 519 922\\
4 &amp;\hspace{0.3cm} 1 &amp; 1 &amp; 17 &amp; 1 540 &amp; 461 105 &amp; 350 813 126 &amp; 573 843
627 152\\ 
5 &amp;\hspace{0.3cm} 1 &amp; 1 &amp; 33 &amp; 8 506 &amp; 9 483 041 &amp; 33 056 715 626 &amp;
293 327 384 637 282\\
6 &amp;\hspace{0.3cm} 1 &amp; 1 &amp; 65 &amp; 48 844 &amp; 209 175 233 &amp; 3 464 129 078
126 &amp; 173 566 857 025 139 312\\
\end{tabular}
\end{table}

\begin{table}
Table II: Table of $S_L(n,l)$:  for $L=1$ and $l,n=1,2,\ldots,8.$  (The triangle, read by columns, gives A061691, the rows and diagonals give A017063, A061690, A000142, A001809, A061689.)
\begin{tabular}{ccccccccc}
$l$ &amp; $S_1(1,l)$ &amp; $S_1(2,l)$ &amp; $S_1(3,l)$ &amp; $S_1(4,l)$ &amp; $S_1(5,l)$ &amp;
$S_1(6,l)$ &amp; $S_1(7,l)$ &amp; $S_1(8,l)$ \\ \hline
1 &amp;\hspace{0.3cm} 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1\\
2 &amp;\hspace{0.3cm}   &amp; 2 &amp; 9 &amp; 34 &amp; 125 &amp; 461 &amp; 1 715 &amp; 6 434 \\
3 &amp;\hspace{0.3cm}   &amp;   &amp; 6 &amp; 72 &amp; 650 &amp; 5 400 &amp; 43 757 &amp; 353 192 \\
4 &amp;\hspace{0.3cm}   &amp;   &amp;   &amp; 24 &amp; 600 &amp; 10 500 &amp; 161 700 &amp; 2 361 016 \\
5 &amp;\hspace{0.3cm}   &amp;   &amp;   &amp;    &amp; 120 &amp; 5 400 &amp; 161 700 &amp; 4 116 000  \\ 
6 &amp;\hspace{0.3cm}   &amp;   &amp;   &amp;    &amp;     &amp; 720  &amp; 52 920 &amp; 2 493 120 \\
7 &amp;\hspace{0.3cm}   &amp;   &amp;   &amp;    &amp;     &amp;      &amp; 5 040 &amp; 564 480 \\
8 &amp;\hspace{0.3cm}   &amp;   &amp;   &amp;    &amp;     &amp;      &amp;     &amp; 40 320 \\
\end{tabular}
\end{table}

\begin{table}
Table III: Table of $S_L(n,l)$: for $L=2$ and $l,n=1,2,\ldots,8.$  (The triangle, read by columns, gives A061692, the rows and diagonals give A061693, A061694, A001044, A061695.)
\begin{tabular}{ccccccccc}
$l$ &amp; $S_2(1,l)$ &amp; $S_2(2,l)$ &amp; $S_2(3,l)$ &amp; $S_2(4,l)$ &amp; $S_2(5,l)$ &amp;
$S_2(6,l)$ &amp; $S_2(7,l)$ &amp; $S_2(8,l)$ \\ \hline
1 &amp;\hspace{0.3cm} 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1\\
2 &amp;\hspace{0.3cm}   &amp; 4 &amp; 27 &amp; 172 &amp; 1 125 &amp; 7 591 &amp; 52 479 &amp; 369 580 \\
3 &amp;\hspace{0.3cm}   &amp;   &amp; 36 &amp; 864 &amp; 17 500 &amp; 351 000 &amp; 7 197 169 &amp; 151 633 440 \\
4 &amp;\hspace{0.3cm}   &amp;   &amp;   &amp; 576 &amp; 36 000 &amp; 1 746 000 &amp; 80 262 000 &amp;
3 691 514 176\\
5 &amp;\hspace{0.3cm}   &amp;   &amp;   &amp;    &amp; 14 400 &amp; 1 944 000 &amp; 191 394 000 &amp; 17 188 416 000\\ 
6 &amp;\hspace{0.3cm}   &amp;   &amp;   &amp;    &amp;     &amp; 518 400  &amp; 133 358 400 &amp; 23
866 214 400 \\
7 &amp;\hspace{0.3cm}   &amp;   &amp;   &amp;    &amp;     &amp;    &amp; 25 401 600   &amp; 11 379 916 800\\
8 &amp;\hspace{0.3cm}   &amp;   &amp;   &amp;    &amp;     &amp;    &amp;            &amp; 1 625 702 400\\
\end{tabular}
\end{table}

\begin{table}
\centerline{Table IV: Table of $S_L(n,l)$:  $l,n=1,2,\ldots,6.$}
\begin{tabular}{ccccccc}
$l$ &amp; $S_L(1,l)$ &amp; $S_L(2,l)$ &amp; $S_L(3,l)$ &amp; $S_L(4,l)$ &amp; $S_L(5,l)$ &amp;
$S_L(6,l)$ \\ \hline
1 &amp;\hspace{0.3cm} 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1\\
2 &amp;\hspace{0.3cm}   &amp; $(2!)^L$ &amp; $3\cdot3^L$ &amp; $4\cdot4^L+3\cdot6^L$ &amp; $5\cdot5^L+10\cdot10^L$ &amp; $6\cdot6^L+15\cdot15^L+10\cdot20^L$\\
3 &amp;\hspace{0.3cm}   &amp;   &amp; $(3!)^L$ &amp; $6\cdot12^L$ &amp; $10\cdot20^L$+$15\cdot30^L$
&amp; $15\cdot30^L+60\cdot60^L+15\cdot90^L$ \\
4 &amp;\hspace{0.3cm}   &amp;   &amp;   &amp; $(4!)^L$ &amp; $10\cdot60^L$ &amp; $20\cdot120^L+45\cdot180^L$\\
5 &amp;\hspace{0.3cm}   &amp;   &amp;   &amp;    &amp; $(5!)^L$ &amp; $15\cdot360^L$ \\ 
6 &amp;\hspace{0.3cm}   &amp;   &amp;   &amp;    &amp;  &amp; $(6!)^L$ \\ 
\end{tabular}
\end{table}

\begin{table}
Table V: Table of $b_0(p,n)$: $p=0,1,2,3;\:\:n=0,\ldots,10.$  (The columns give A000110, A000296, A006505, A057837.) \\
\begin{tabular}{ccccc}
$n$ &amp; $b_0(0,n)$ &amp; $b_0(1,n)$ &amp; $b_0(2,n)$ &amp; $b_0(3,n)$ \\ \hline
0 &amp;\hspace{0.3cm} 1 &amp; 1 &amp; 1 &amp; 1 \\
1 &amp;\hspace{0.3cm} 1 &amp; 0 &amp; 0 &amp; 0 \\
2 &amp;\hspace{0.3cm} 2 &amp; 1 &amp; 0 &amp; 0 \\
3 &amp;\hspace{0.3cm} 5 &amp; 1 &amp; 1 &amp; 0 \\
4 &amp;\hspace{0.3cm} 15 &amp; 4 &amp; 1 &amp; 1 \\ 
5 &amp;\hspace{0.3cm} 52 &amp; 11 &amp; 1 &amp; 1 \\
6 &amp;\hspace{0.3cm} 203 &amp; 41 &amp; 11 &amp; 1\\
7 &amp;\hspace{0.3cm} 877 &amp; 162 &amp; 36 &amp; 1 \\
8 &amp;\hspace{0.3cm} 4 140 &amp; 715 &amp; 92 &amp; 36 \\
9 &amp;\hspace{0.3cm} 21 147 &amp; 3 425 &amp; 491 &amp; 127 \\
10 &amp;\hspace{0.3cm} 115 975 &amp; 17 722 &amp; 2 557 &amp; 337 \\
\end{tabular}
\end{table}

\begin{table}
Table VI: Table of $b_1(p,n)$: $p=0,1,2;\:\:n=0,\ldots,9.$  (The columns give A023998, A061696, A061697.)
\begin{tabular}{cccc}
$n$ &amp; $b_1(0,n)$ &amp; $b_1(1,n)$ &amp; $b_1(2,n)$ \\ \hline
0 &amp;\hspace{0.3cm} 1 &amp; 1 &amp; 1 \\
1 &amp;\hspace{0.3cm} 1 &amp; 0 &amp; 0 \\
2 &amp;\hspace{0.3cm} 3 &amp; 1 &amp; 0 \\
3 &amp;\hspace{0.3cm} 16 &amp; 1 &amp; 1\\
4 &amp;\hspace{0.3cm} 131 &amp; 19 &amp; 1 \\ 
5 &amp;\hspace{0.3cm} 1 496  &amp; 101 &amp; 1 \\
6 &amp;\hspace{0.3cm} 22 482 &amp; 1 776 &amp;  201\\
7 &amp;\hspace{0.3cm} 426 833 &amp; 23 717 &amp; 1 226 \\
8 &amp;\hspace{0.3cm} 9 934 563 &amp; 515 971 &amp; 5 587 \\
9 &amp;\hspace{0.3cm} 277 006 192 &amp; 11 893 597 &amp; 493 333 \\
\end{tabular}
\end{table}

\begin{table}
Table VII: Table of $b_2(p,n)$: $p=0,1,2;\:\:n=0,\ldots,8.$  (The columns give A061698--A061700.)
\begin{tabular}{cccc}
$n$ &amp; $b_2(0,n)$ &amp; $b_2(1,n)$ &amp; $b_2(2,n)$ \\ \hline
0 &amp;\hspace{0.3cm} 1 &amp; 1 &amp; 1 \\
1 &amp;\hspace{0.3cm} 1 &amp; 0 &amp; 0 \\
2 &amp;\hspace{0.3cm} 5 &amp; 1 &amp; 0 \\
3 &amp;\hspace{0.3cm} 64 &amp; 1 &amp; 1\\
4 &amp;\hspace{0.3cm} 1 613 &amp; 109 &amp; 1 \\ 
5 &amp;\hspace{0.3cm} 69 026  &amp; 1 001 &amp; 1 \\
6 &amp;\hspace{0.3cm} 4 566 992 &amp; 128 876 &amp;  4 001\\
7 &amp;\hspace{0.3cm} 437 665 649 &amp; 4 682 637 &amp; 42 876 \\
8 &amp;\hspace{0.3cm} 57 903 766 800 &amp; 792 013 069 &amp; 347 117 \\
\end{tabular}
\end{table}
 
\begin{thebibliography}{99}      
\bibitem{yablonsky}S.V. Yablonsky, `` Introduction to Discrete
Mathematics'', Mir Publishers, Moscow, 1989.

\bibitem{andrews} G.E. Andrews, R. Askey and R. Roy, ``Special
Functions'', Encyclopedia of Mathematics and its Applications, vol.~71,
Cambridge University Press,  1999.

\bibitem{marichev} O.I. Marichev, \emph{Handbook of Integral
Transforms of Higher Transcendental Functions, Theory and Algorithmic
Tables}, Ellis Horwood Ltd, Chichester, 1983, Chap. 6.

\bibitem{kiryakova}V.S. Kiryakova and B.Al-Saqabi, ``Explicit solutions to
hyper-Bessel integral equations of second kind'', Comput. and
Math. with Appl. \textbf{37}, 75 (1999).

\bibitem{paris}R.B. Paris and A.D. Wood, ``Results old and new on the
hyper-Bessel equation'', Proc. Roy. Soc. Edinb. \textbf{106~A}, 259 (1987).

\bibitem{witte}N.S. Witte, ``Exact solution for the reflection and
diffraction of atomic de Broglie waves by a traveling evanescent laser
wave'', J. Phys. A \textbf{31}, 807 (1998).

\bibitem{klauder}J.R. Klauder, K.A. Penson and J.-M. Sixdeniers,
``Constructing coherent states through solutions of Stieltjes and
Hausdorff moment problems'', Physical Review A, \textbf{64}, 013817 (2001).

\bibitem{comtet}L. Comtet, "Advanced Combinatorics", D. Reidel,
Boston, 1984.

\bibitem{wilf} H.S. Wilf, ``Generatingfunctionology'', $2^{\rm nd}$
ed., Academic Press, New York, 1994.

\bibitem{sloane}N.J.A. Sloane, \htmladdnormallink{On-Line Encyclopedia of Integer Sequences}{http://www.research.att.com/~njas/sequences/}, published electronically at:
http://www.research.att.com/$\sim$/njas/sequences/.

\bibitem{bernstein}M. Bernstein and N.J.A. Sloane, ``\htmladdnormallink{Some canonical
sequences of integers}{http://www.research.att.com/~njas/doc/eigen.pdf}'', Linear Algebra Appl., \textbf{226/228}, 57 (1995).

\bibitem{fitzgerald}D.G. Fitzgerald and J. Leech, ``Dual symmetric
inverse monoids and representation theory'', J. Austr. Math. Soc., Series
A, \textbf{64}, 345 (1998).

\bibitem{delerue}P. Delerue, ``Sur le calcul symbolique \`{a} $n$
variables et fonctions hyperbess\'{e}liennes II'',
Ann. Soc. Sci. Brux. \textbf{67}, 229 (1953).

\bibitem{ehrenborg}R. Ehrenborg, "The Hankel Determinant of Exponential
Polynomials", Am. Math. Monthly,  \textbf{207}, 557 (2000).

\bibitem{suter} R. Suter, ``\htmladdnormallink{Two Analogues of a Classical Sequence}{http://www.research.att.com/~njas/sequences/JIS/index.html\#P001.8}'',
J. Integ. Seq.  \textbf{3}, Article 00.1.8 (2000).

\end{thebibliography}

\vspace*{+.5in}
\centerline{\rule{5.4in}{.01in}}

\noindent
(Mentions sequences
A000296
A001044
A001809
A006505
A010763
A023998
A057814
A057837
A061683
A061684
A061685
A061686
A061687
A061688
A061689
A061690
A061691
A061692
A061693
A061694
A061695
A061696
A061697
A061698
A061699
A061700
.)

\centerline{\rule{5.4in}{.01in}}

\vspace*{+.1in}
\noindent
Received April 5, 2001;
published in Journal of Integer Sequences, June 22, 2001.

\centerline{\rule{5.4in}{.01in}}

\vspace*{+.1in}
\noindent
Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.research.att.com/~njas/sequences/JIS/}.

\centerline{\rule{5.4in}{.01in}}


\end{document}

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