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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\title{ A Ramanujan sequence that refines the Cayley formula for trees}
\author{\normalsize Jiang Zeng\\
\normalsize D\'epartement de Math\'ematique, Universit\'e Louis Pasteur,\\
\normalsize 7, rue Ren\'e Descartes, 67084 Strasbourg C\'edex, France}
\begin{document}
\maketitle
\begin{abstract}
A further combinatorial study of Entries 16 and 17 in Chapter~3 of Ramanujan's 
second notebook \cite{BEW} leads to a refinement of Cayley's formula for
counting trees and generalizes the recent results derived 
by Shor~\cite{Sho}, and
Dumont and Ramamonjisoa~\cite{DR}. Our approach also yields a new proof of
Clarke's extension of Cayley's formula~\cite{Cla}.
\end{abstract}
\section{Introduction}
In his edition of Ramanujan's notebook~\cite{BEW,Ber}
Berndt made the following remarks:
\begin{quote}
``Although no combinatorial problems are mentioned in Chapter 3, much of the
content of this chapter belongs under the umbrella of combinatorial analysis. 
[\ldots] Entries 16 and 17 do not seem to have been expanded upon in the 
literature and would appear to be a basis for further fruitful research.''
\end{quote}
In the years following the publication of Berndt's work 
several authors have made arithmatical and combinatorial studies 
of a sequence in Entry 17~\cite{How,HH,DR}. 
However nothing seems to be done yet 
regarding Entry 16. 
It's the purpose of this paper to 
show that the aforementioned sequence is actually a special case of another 
one in Entry 16 and give a combinatorial interpretation 
of the latter sequence.

In \cite{Ram}, for intergers $k,\, r\geq 0$,
Ramanujan defined a sequence $\psi_k(r,\, x)$ by
\begin{equation}
\sum_{k=0}^{+\infty}\frac{(x+k)^{r+k}e^{-u(x+k)}u^k}{k!}=
\sum_{k=1}^{r+1}\frac{\psi_{k}(r,\, x)}{(1-u)^{r+k}},
\end{equation}
and gave the following recurrence :
\begin{equation}
\psi_{k}(r+1,\, x)=(x-1)\psi_{k}(r,\, x-1)+\psi_{k-1}(r+1,\, x)
-\psi_{k-1}(r+1,\, x-1).
\end{equation}
 Berndt et al~\cite{BEW,Ber} have found the following simpler recurrence:
\begin{equation}
\psi_{k}(r,\, x)=(x-r-k+1)\psi_{k}(r-1,\, x)
+(r+k-2)\psi_{k-1}(r-1,\,x),
\end{equation}
where $ k\in[ r+1]:=\{1,\, 2,\ldots, r+1\}$, $\psi_{1}(0,\,x)=1$ and 
 $\psi_{k}(r,\, x)=0$ if $k\notin [r+1]$.
Making the substitution $Q_{n,\, k}(x)=\psi_{k+1}(n-1,\, x+n)$
in the above recurrence yields
\begin{equation}
Q_{n,\, k}(x)=(x+n-1)Q_{n-1,\, k}(x)+(n+k-2)Q_{n-1,\, k-1}(x),\label{jiang}
\end{equation}
where $Q_{1,\, 0}(x)=1$ and $Q_{n,\, k}(x)=0$ if $k\geq n$ or $k<0$.
The first terms of this sequence are given in Table 1.

\begin{table}[h]
\begin{center}
{\footnotesize
\begin{tabular}{c|lllll} 
$k\backslash n$& 1& 2& 3&4&5\\
\hline\\
0&  1&$1+x$&$2+3x+x^2 $&$6+11x+6x^2+x^3$&$24+50x+35x^2+10x^3+x^4$\\
1&   &1   &$4+3x $      &$18+22x+6x^2 $    &$96+150x+70x^2+10x^3$     \\
2&   &    &3           &$25+15x $           &$190+195x+45x^2 $           \\
3&   &    &            &15                  &$210+105x$                              \\
4&   &    &            &                    &105                 \\
\hline\\
$\sum_{k}$& 1&$2+x $&$(3+x)^2$ &$(4+x)^3$ &$(5+x)^4$\\
\end{tabular}
\caption{$Q_{n,\, k}(x)$ for $0\leq k<n\leq 5$.}}
\end{center}
\end{table}

The sequence $Q_{n,\, k}(x)$ has the advantage of providing a common
algebraic set-up for recent combinatorial results obtained by Shor~\cite{Sho}
on the one hand, and Dumont and Ramamonjisoa~\cite{DR} on the other hand. 
As will be shown, Shor~\cite{Sho} found a combinatorial interpretation of the
value of the polynomial $Q_{n,\, k}(x)$ at $x=1$, while 
Dumont and Ramamonjisoa~\cite{DR} made 
a combinatorial study of the sequences $Q_{n,\,k}(-1)$, 
$Q_{n,\,k}(0)$ and $Q_{n,\,k}(1)$.

In section 2 we shall give two combinatorial interpretations of 
 $Q_{n,\,k}(x)$ and as an application we derive a new proof of 
 Clarke's extension of Caylay's formula. 
 In section 3 we show how to derive (1) from Lambert's
 formula (11) and compute the generating function of  
 $Q_{n,\,k}(x)$.
 In section 4 we derive the results of Shor~\cite{Sho} and 
 Dumont-Ramamonjisoa~\cite{DR} by specializing $Q_{n,\,k}(x)$. 
 We end this paper with some further
 remarks and open questions.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
\section{Combinatorial interpretations}
A {\sl labeled tree} on $[n]:=\{1,\, 2,\ldots, n\}$ is a connected graph 
without cycles
on $[n]$. It's well-known (see \cite{Mo}) that the number of 
labeled trees on $[n+1]$ is $(n+1)^{n-1}$. A {\sl rooted labeled
tree} is a labeled tree with a {\sl distinguished} vertex. 
Note respectively ${\cal T}_n$ and ${\cal T}_n(\rightarrow \cdot)$ the set of
labeled trees and rooted labeled trees on $[n]$.

Given a tree $T\in {\cal T}_n(\rightarrow \cdot)$ and two nodes $j,\, k$ 
of $T$, we say that $k$ belongs to the descendance of $j$ if the path from the root to $k$
passes through $j$. In particular, each node is the descendant of itself.
If $k$ is a descendant of $j$ and $e=(j,\, k)$ an edge of $T$, 
we say that $k$ is a child of $j$,
and $j$ and $k$ are respectively
the \emph{father} node and the \emph{child} node of $e$.
An edge $e=(j,\, k)$ of $T$ is \emph{propre} if all the descendants of the
child node $k$ are large than the $j$, otherwise it is \emph{impropre}.
The \emph{degree} of a node $x$ in a rooted tree $T$ is the number of 
children of $x$. Note $\deg_T(x)$ the degree of $x$ and $\beta(x)$ the smallest 
descendant of $x$.

Throughout this paper we shall take the convention 
that an unrooted labeled tree is rooted at the smallest node. Then we can
generalize the above notions to unrooted labeled trees.
Note respectively ${\cal T}_{n,\,k}$ and 
${\cal T}_{n,\,k}(\rightarrow \cdot)$ the set
of labeled trees and rooted labeled trees on $[n]$ with 
$k$ impropre edges.
 Applying the same argument 
used by Shor~\cite{Sho} or Dumont-Ramamonjisoa~\cite{DR} 
we can prove the following more general result.

\begin{prop} There holds
\begin{equation}
Q_{n,\,k}(x)=\sum_{T\in {\cal T}_{n+1,\,k}}x^{\deg_T(1)-1}.\label{com1}
\end{equation}
\end{prop}
\pf 
We shall prove formula (\ref{com1}) 
by showing that the right-hand side satisfies recurrence (\ref{jiang}).
Formula (\ref{com1}) is obviously true for $n=1$. Suppose that (\ref{com1}) is
true for $n=m-1$.
We define a surjection $\varphi$ from ${\cal T}_{m+1,\,k}$ to 
${\cal T}_{m,\,k}\cup {\cal T}_{m,\,k-1}$ as follows. 
For any $T\in {\cal T}_{m+1,\,k}$, if $m+1$ is a leaf, delete it to obtain a tree 
$T'\in {\cal T}_{m,\,k}$; otherwise replace $m+1$ by the child node $x$ which has the 
largest $\beta (x)$ among all children of $m+1$ and contract
the edge joining $m+1$ and $x$ to obtain a tree $T'\in {\cal T}_{m,\,k-1}$.
Now let us consider the inverse of $\varphi$. 
To recover a tree $T\in {\cal T}_{m+1,\,k}$ we can proceed 
in two different ways :
\begin{itemize}
\item start from a tree $T'\in {\cal T}_{m,\,k}$ and
add a node $m+1$ as a leaf in $m$ ways. Clearly this will not
creat any improper edges but increase the 
degree of 1 by one if $m+1$ is joined
directly to 1. Hence the generating function for 
${\cal T}_{m+1,\,k}$ such that $m+1$
is a leaf is equal to $(x+m-1)Q_{m-1,\,k}(x)$.
\item strart from a tree $T'\in {\cal T}_{m,\,k-1}$ and
add $m+1$ as an interior node:
pick any node $x\not= 1$ of $T'$, replace $x$ 
by $m+1$ and join $x$ to $m+1$ by an
edge, make all proper children of $x$ 
in $T'$ to be children of $x$ in $T$, it
remains to partition the improper children of 
$x$ in $T'$, say 
$x_1,\,x_2,\ldots, x_b$ such that 
$\beta(x_1)<\beta(x_2)<\cdots <\beta(x_b)$, then it's
easy to see that the only possible partition is to make $x_1,\, \ldots, x_a$ to 
be the children of $m+1$ and $x_{a+1},\, \ldots, x_b$ to be the children of $x$, 
where $0\leq a\leq b$. That means there are $b+1$ possibilities to partition
the improper children of $x$.
Hence there are $m+k-2$ total such preimages. Moreover
this operation will increase one improper edge between $m+1$ and $x$, but not the degree of 1.
Hence the generating function 
for ${\cal T}_{m+1,\,k}$ with $m+1$ being an interior node is equal to 
$(m+k-2)Q_{m-1,\,k-1}(x)$. 
\end{itemize}
Summarizing the above two cases yields the desired recurrence.
\qed\medskip

The following result shows that $Q_{n,\,k}(x)$ is actually a polynomial
of $(x+1)$ with \emph{non negative integer} coefficients. 

\begin{prop} There holds
\begin{equation}
Q_{n,\,k}(x)=\sum_{T\in {\cal T}_{n,\,k}(\rightarrow\cdot)}(x+1)^{\deg_T(1)}.
\label{new}
\end{equation}
\end{prop}
\pf
As in the proof of proposition 1 we show that the right-hand side of (\ref{new})
satisfies the recurrence (\ref{jiang}).
We define a similar surjection $\varphi$ 
from ${\cal T}_{m+1,\,k}(\rightarrow\cdot)$ to 
${\cal T}_{m,\,k}(\rightarrow\cdot)\cup {\cal T}_{m,\,k-1}(\rightarrow\cdot)$ 
as follows.
For any $T\in {\cal T}_{m+1,\,k}(\rightarrow\cdot)$, if
$m+1$ is a leaf (then $m+1$ cannot be a root), delete it to obtain a tree
$T'\in {\cal T}_{m,\,k}(\rightarrow\cdot)$; 
otherwise replace $m+1$ by the child node $x$ 
which has the largest $\beta(x)$ among all children of $m+1$ and contract the
edge joining $m+1$ and $x$ to obtain a 
tree $T'\in {\cal T}_{m,\,k-1}(\rightarrow\cdot)$, where 
if $m+1$ had been a root, we take $x$ as the root of $T'$, otherwise $T'$ has
the same root as that of $T$. 
Since the rest of the proof is similar to that of proposition 1 we omit it.
\qed\medskip

\remark Propositions 1 and 2 can also be derived respectively 
from (\ref{equa}) and (\ref{diff}) by
applying the combinatorial theory of generating functions (see \cite{Fo,GJ,DR}). 
It would be interesting to find a direct derivation
of proposition 2 from proposition~1.

The first polynomials
of $Q_{n,\,k}(x-1)$ are listed in Table 2.

\begin{table}[h]
\begin{center}
{\footnotesize
\begin{tabular}{c|lllll} 
$k\backslash n$& 1& 2& 3&4&5\\
\hline\\
0&  1&$x$&$x+x^2 $&$2x+3x^2+x^3$&$6x+11x^2+6x^3+x^4$\\
1&   &1   &$1+3x $      &$2+10x+6x^2 $    &$6+40x+40x^2+10x^3$     \\
2&   &    &3           &$10+15x $           &$40+105x+45x^2 $           \\
3&   &    &            &15                  &$105+105x$                              \\
4&   &    &            &                    &105                 \\
\hline\\
$\sum_{k}$& 1&$1+x $&$(2+x)^2$ &$(3+x)^3$ &$(4+x)^4$\\
\end{tabular}
\caption{$Q_{n,\, k}(x-1)$ for $0\leq k<n\leq 5$.}}
\end{center}
\end{table}

\begin{cor} For integer $n\geq 1$ we have
\begin{equation}
Q_n(x):=\sum_{k=0}^{n-1}Q_{n,\,k}(x)=(x+n)^{n-1}.\label{cor1}
\end{equation}
\end{cor}
\pf 
Given a tree $T\in {\cal T}_{n}$ we can root $T$ by picking any 
of the $n$ nodes. Hence it is clear that 
\begin{equation}
\sum_{T\in {\cal T}_{n}(\rightarrow\cdot)}x^{\deg_T(1)}
=(x+n-1)\sum_{T\in{\cal T}_{n}}x^{\deg_T(1)-1}.
\end{equation}
Replacing $x$ by $x+1$ and applying propositions 1 and 2 we obtain
$$
Q_n(x)=(x+n)Q_{n-1}(x+1).
$$
Since $Q_1(x)=1$ iterating the above recurrence yields then (\ref{cor1}).
\qed\medskip

\begin{cor} The number of labeled trees on $[n+1]$ such that the 
node $1$ has degree $l$ is equal to
\begin{equation}
{n-1\choose l-1}n^{n-l}, \qquad l=1,\,2,\ldots, n.
\end{equation}
\end{cor}
\pf Applying corollary 3 we get
$$
\sum_{T\in{\cal T}_{n+1}}x^{\deg_T(1)-1}=(x+n)^{n-1}=
\sum_{l=0}^n{n-1\choose l-1}n^{n-l}x^{l-1}.
$$
Extracting the coefficients of $x^{l-1}$ yields the desired formula. \qed
\medskip

\remark The above corollary was first proved by Clarke~\cite{Cla}, 
although it's equivalent \cite[p. 25]{Mo}
to another result of
Cayley, that is, the number of forets with $n$ labelled nodes 
that consist of $k$ trees such
that $k$ specified nodes belong to distinct trees is 
equal to $kn^{n-k-1}$.
\medskip
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Generating functions}
We shall give several formulae for
the generating functions of $Q_{n,\,k}(x)$.
In contrast to the treatment in \cite{Ber},
all the series in this section are {\sl formal power series}.
The following result was already known to Lambert and Euler in
the 18th century (see \cite[p. 72]{Ber}), but
 the simple proof that we propose below seems new. 
\begin{lem} Let $u\in {\bf C}[[t]]$ be the solution of $t=ue^{-u}$. Then
\begin{eqnarray}
u&=&\sum_{n\geq 1}n^{n-1}\frac{t^n}{n!},\label{lem0}\\
\frac{e^{xu}}{1-u}&=&\sum_{n\geq 0}(x+n)^n\frac{t^n}{n!},\label{lem1}\\
\frac{e^{xu}-1}{x}&=&\sum_{n\geq 1}(x+n)^{n-1}\frac{t^n}{n!}.\label{lem2}
\end{eqnarray}
\end{lem}
\pf Note that (\ref{lem0}) is the special case $x=0$ of (\ref{lem2}).
Dividing the two sides of (\ref{lem1}) by $e^{xu}$ and
equating the coefficient of $u^m/m!$ ($m\geq 0$) yields
\begin{equation}
m!=\sum_{i=0}^m(-1)^{m-i}{m\choose i}(x+i)^m.\label{eul2}
\end{equation}
Hence it suffices to prove (\ref{eul2}). Let $x\notin [m]$ and
consider the applications $\varphi : [m]\longrightarrow [m]\cup\{x\}$
such that $\varphi([m])=[m]$. Clearly there are $m!$ such applications. On
the other hand, if we weight each application 
$\varphi : [m]\longrightarrow [m]\cup\{x\}$ by $x^{|\varphi^{-1}(x)|}$, then
the enumerator of the applications such that $|\varphi([m])\cap [m]|\leq i$
($0\leq i\leq m$)
is ${m\choose i}(x+i)^m$. Hence by the principle of inclusion-exclusion
we get (\ref{eul2}). Since $dt/du=(1-u)e^{-u}$ we obtain (\ref{lem2}) by 
integrating (\ref{lem1}) with respect to $t$ 
and then replacing $x$ by $x+1$.
\qed \medskip

\begin{lem} Let $n\geq 1$ be integer. Then
$$
\left(\frac{e^u}{1-u}\frac{\partial}{\partial u}\right)^n\left(
\frac{e^{xu}-1}{x}\right)=
\sum_{k=0}^{n-1}\frac{Q_{n,\,k}(x)}{(1-u)^{n+k}}e^{(x+n)u}.
$$
\end{lem}
\pf By induction on $n$ and using (\ref{jiang}).\qed\medskip

The following result is equivalent to (1), which is the starting point 
of Ramanujan for the sequence $\psi_{k}(r,\, x)$.
 Here we show how to derive it from Lambert's formula.
\begin{prop} Let $n\geq 1$ be integer. Then
\begin{equation}
\sum_{k=0}^{n-1}\frac{Q_{n,\,k}(x)}{(1-u)^{n+k}}=\sum_{k=0}^\infty
\frac{(x+n+k)^{n+k-1}e^{-u(x+n+k)}u^k}{k!}.\label{gene}
\end{equation}
\end{prop}
\pf Let $t=ue^{-u}$, then $\frac{du}{d t}=\frac{e^{u}}{1-u}$.
Deriving $n$ times the both sides of (\ref{lem2}) 
with respect to $t$ we get
$$
\left(\frac{e^{u}}{1-u}\frac{\partial}{\partial u}\right)^{n}
\left(\frac{e^{xu}-1}{x}\right)=\sum_{k=0}^{\infty}
(x+n+k)^{n+k-1}\frac{t^k}{k!}.
$$
Replacing $t$ by $ue^{-u}$ we deduce (\ref{gene}) from lemma 6.\qed
\medskip
\remark If $u=0$ the formula (\ref{gene}) reduces to (5). 
\medskip

Define the generating function
\begin{equation}
Y(t,\, u)=\sum_{n\geq 1}
\sum_{k=0}^{n-1}\frac{Q_{n,\, k}(x)}{(1-u)^k}\frac{t^n}{n!}.
\end{equation}
\begin{prop} We have
\begin{equation}
Y(t,\,u)=\sum_{n\geq 1}\frac{(x+n)^{n-1}e^{-u(x+n)}}{n!}
\left[(u+(1-u)t)^n-u^n\right].\label{gf}
\end{equation}
\end{prop}
\pf
 It follows from the above proposition that
\begin{eqnarray*}
Y(t,\,u)&=& \sum_{n\geq 1}\frac{(1-u)^nt^n}{n!}
\sum_{k\geq 0}
\frac{(x+n+k)^{n+k-1}e^{-u(x+n+k)}u^k}{k!}\\
&=& \sum_{l\geq 1}\frac{(x+l)^{l-1}e^{-u(x+l)}}{l!}
\sum_{k=0}^{l-1}{l\choose k}\left[(1-u)t\right]^{l-k}u^k.
\end{eqnarray*}
Replacing $l$ by $n$ and applying 
the binomial formula to the interior summation
in the last line we deduce the desired formula. \qed\medskip

The following result gives a generalization of 
Lambert's formula (\ref{lem2}). We shall give two proofs of this 
generalization.

\begin{prop}
Let $y\in {\bf C}[[t,\,u]]$ be the solution of the equation 
\begin{equation}
(1-u)t=ye^{-y}+u(e^{-y}-1).\label{newRam}
\end{equation}
Then
\begin{equation}
Y(t,\, u)=\frac{e^{xy}-1}{x}.\label{equa}
\end{equation}
\end{prop}
\fpf Write (\ref{newRam}) as
\begin{equation}
e^{-u}[u+(1-u)t]=(y+u)e^{-(y+u)}.\label{newRam'}
\end{equation}
Since the left-hand side of (\ref{newRam'}) is a series of $u$ and $t$ without
constant, by applying (\ref{lem2}) of Lemma 3, we obtain
$$
\frac{e^{x(y+u)}-1}{x}=\sum_{n\geq 1}(x+n)^{n-1}\frac{e^{-nu}[u+(1-u)t]^n}{n!},
$$
which can be rewritten as
\begin{equation}
\frac{e^{xy}-1}{x}=-\frac{e^{xu}-1}{x} e^{-xu}+
\sum_{n\geq 1}(x+n)^{n-1}\frac{e^{-(x+n)u}[u+(1-u)t]^n}{n!}.\label{step}
\end{equation}
Now, by (\ref{lem2}) we have 
$$
\frac{e^{xu}-1}{x}=\sum_{n\geq 1}(x+n)^{n-1}\frac{(ue^{-u})^n}{n!}.
$$
Substituting the above formula in the left-hand side of (\ref{step}) 
we get the left-hand side of (\ref{gf}). \qed\medskip

\spf 
Let $a=\frac{1}{1-u}$, then the equation (\ref{newRam}) can be written as
\begin{equation}
t=aye^{-y}+(a-1)(e^{-y}-1).\label{Ram}
\end{equation}
Equivalently  $y$ is the solution of the differential equation:
\begin{equation}
y'=\frac{e^y}{1-ay},\qquad y(0)=0.\label{DR}
\end{equation} 
Denote by $Z(t,\,u)$ the right-hand side of (\ref{equa}).
Then $Z$ is characterized by the differential equation
\begin{equation}
Z'=\frac{e^{(x+1)y}}{1-ay},\qquad Z(0,\,u)=0.\label{diff}
\end{equation}
Define $A=\frac{1}{1-ay}$ and $S=e^{y}$.
Then $A$ and $S$ are characterized by
\begin{eqnarray}
A'&=& aA^3S,\quad A(0)=1;\\
S'&=& AS^2,\quad S(0)=1.
\end{eqnarray}
Note that $Z'=AS^{x+1}$ and $Z''=(x+1+aA)A^2S^{x+2}$.
It's routine to show by induction that for $n\geq 1$,
\begin{equation}
Z^{(n)}=A^nS^{x+n}\sum_{k=0}^{n-1}Q_{n,\,k}(x)(aA)^k.
\end{equation}
It follows then $Z(t,\,u)=Y(t,\,u)$ by comparing with (15). \qed

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Three specializations}
Set $Q_{0,\,0}(1)=1$ and define the following 
three sequences :
\begin{eqnarray}
a(n,\,k)&=&  Q_{n+1,\, k}(-1),\quad n\geq k\geq 1;\\
b(n,\,k)&=&Q_{n,\,k}(0),\quad n> k\geq 0;\\
c(n,\,k)&=&Q_{n-1,\,k}(1),\quad n> k\geq 0.
\end{eqnarray} 
It then follows from (\ref{jiang}) that
\begin{eqnarray}
a(n,\, k)&=&(n-1)a(n-1,\, k)+(n+k-1)a(n-1,\, k-1),\label{rdr}\\
b(n,\, k)&=&(n-1)b(n-1,\, k)+(n+k-2)b(n-1,\, k-1),\\
c(n,\, k)&=&(n-1)c(n-1,\, k)+(n+k-3)c(n-1,\, k-1),\label{Rdr}
\end{eqnarray}
with $a(0,\, 0)=b(1,\, 0)=c(1,\, 0)=1$.
The following result is du to Ramanujan~\cite{Ber}
and Dumont-Ramamonjisoa~\cite{DR} relating to Entry 17 of \cite{Ber}.
\begin{cor} 
The three sequences $a(n,\, k),\, b(n,\, k)$ and $c(n,\, k)$ 
satisfy 
\begin{equation}
\sum_{k=1}^na(n,\, k)=n^n,\quad
 \sum_{k=0}^{n-1}b(n,\, k)=n^{n-1},\quad
\sum_{k=0}^{n-1}c(n,\, k)=n^{n-2}.\label{RDR}
\end{equation}
Moreover, if $y$ is defined as in (21), then
\begin{eqnarray}
y&=&\sum_{n\geq 1}\left(\sum_{k=0}^{n-1}b(n,\, k)a^k\right)\frac{t^n}{n!},
\label{dr1}\\
1-e^{-y}&=&\sum_{n\geq 0}
\left(\sum_{k=0}^na(n,\, k)a^k\right)\frac{t^{n+1}}{(n+1)!},\label{dr2}\\
e^y&=&1+\sum_{n\geq 1}\left(\sum_{k=0}^{n-1}c(n+1,\, k)a^k\right)\frac{t^n}{n!}.
\label{dr3}
\end{eqnarray}
\end{cor}
\pf 
Formulae (\ref{RDR}) follow from corollary 3. 
In proposition 9 substituting
$\frac{1}{1-u}$ by $a$ we then obtain 
formulae (\ref{dr1})-(\ref{dr3}) respectively 
by setting $x=0$, $x=-1$ and $x=1$. 
\qed\medskip

By propositions 1 and 2 we obtain 
immediately the following combinatorial interpretations for the
three sequences.
\begin{cor}
For $n>j\geq 0$ we have
\begin{eqnarray}
a(n,\,k)&=& |\{T\in {\cal T}_{n+1,\, k}(\rightarrow\cdot)\,:\, \deg_T(1)=0\}|,
\label{inter1}\\
b(n,\,k)&=&|{\cal T}_{n,\, k}(\rightarrow\cdot)|,\label{inter2}\\
c(n,\,k)&=&|{\cal T}_{n,\, k}|.\label{inter3}
\end{eqnarray}
\end{cor}

The combinatorial
 interpretations (\ref{inter2}) and (\ref{inter3}) are
 respectively due to Shor \cite{Sho} and Dumont-Ramamonjisoa \cite{DR}.
 In \cite{DR} Dumont and Ramamonjisoa also give
 a combinatorial interpretation for $a(n,\, k)$ in terms
of functions from $[n]$ to $[n]$. To see the connexion of the latter 
interpretation and (\ref{inter1}) we first note that 
any tree $T\in {\cal T}_{n+1,\, k}(\rightarrow\cdot)$ such that $\deg_T(1)=0$
can be identified with a tree $T'\in {\cal T}_{n,\, k}(\rightarrow \cdot)$
with a single distinguished vertex by cutting the node 1 and rescaling any
other node $i$ to $i-1$ for $2\leq i\leq n+1$.
Therefore it remains to find an adequate bijection between
 the set of all functions from $[n]$ to $[n]$ and 
the set of all rooted labeled trees on 
$n$ vertices with a single distinguished vertex or vertebrates.
We note that such a bijection has been found independently by
 Labelle \cite{Lab} and Goulden-Jackson \cite[p. 179-180]{GJ}. 

\section{Further remarks and open questions}
From combinatorial point of view there are also some 
less obvious recurrences of $Q_{n,\, k}(x)$.  For example,
the formula (2) of Ramanujan states, in terms of $Q_{n,\, k}(x)$, that
\begin{equation}
Q_{n,\, k}(x)= (x+n-1)Q_{n-1,\, k}(x)+Q_{n,\, k-1}(x)-Q_{n,\, k-1}(x-1).
\end{equation}
Comparing with (\ref{jiang}) this is equivalent to the following:
\begin{equation}
Q_{n,\, k}(x)= Q_{n,\, k}(x-1)+(n+k-1)Q_{n-1,\, k}(x).\label{Ramanu}
\end{equation}
Now comparing the last formula with (\ref{jiang}) and replacing $x$ by $x+1$ we deduce
another formula for $Q_{n,\, k}(x)$ given in \cite{Sho} :
\begin{equation}
Q_{n,\, k}(x)=(x-k+1)Q_{n-1,\, k}(x+1)+(n+k-2)Q_{n-1,\, k-1}(x+1).
\end{equation}
We can of course verify the above identities by induction on $n$ using (\ref{jiang}).
It would be interesting to find a combinatorial explanation
 of these recurrences. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%\newpage
\begin{thebibliography}{99}
\bibitem{BEW} B. C. \textsc{Berndt}, R.J. \textsc{Evens} and B.M. 
\textsc{Wilson}: 
\emph{Chapiter 3 of
Ramanujan's Second Notebooks}, Adv. in Math., {\bf 49}(1983),
123--169.

\bibitem{Ber} B. C. \textsc{Berndt}: 
\emph{Ramanujan's Second Notebooks, Part I, chap. 3: Combinatorial analysis
and series inversions}, Springer-Verlag, 1985.

\bibitem{Cla} L.C. \textsc{Clarke}: \emph{On Cayley's formula
for counting trees}, J. London Math. Soc., {\bf 33} (1958), 471--474.

\bibitem{DR} D. \textsc{Dumont} et A. \textsc{Ramamonjisoa}: 
\emph{Grammaire de Ramanujan et Arbres de Cayley},
The Electronic J. Combinatorics, {\bf 3}(2) (1996), \#R17.

\bibitem{Fo} D. \textsc{Foata}: \emph{La s\'erie g\'en\'eratrice expo\-nen\-tielle 
dans les prob\-l\`emes d'\'enu\-m\'eration}, 
Presses Universitaires de Montr\'eal, 1974.

\bibitem{GJ} I. \textsc{Goulden} and D. \textsc{Jackson}: 
\emph{Combinatorial Enumeration},
John Wiley, New York, 1983.

\bibitem{How} F. T. \textsc{Howard}: Explicit formulas for numbers of Ramanujan, 
Fibonacci Quartly {\bf 24}(2) (1986), 168-175.

\bibitem{HH} F. T. \textsc{Howard} and E. K. \textsc{Hayashi}: 
\emph{Congruences for numbers of Ramanujan}, 
Fibonacci Quartly {\bf 27}(1) (1989), 61-69.

\bibitem{Lab} G. \textsc{Labelle}: \emph{Une nouvelle d\'emonstration combinatoire
des formules d'inversion de Lagrange}, 
Adv. Math., {\bf 42}(1981), 217-247.


\bibitem{Mo} J. W. \textsc{Moon: Counting Labelled Trees: A Survey of methods
and results}, Canadian Mathematical Congress, 1970.

\bibitem{Ram} S. \textsc{Ramanujan}: \emph{Notebooks, Vol. 1,
chap. II}, P. 35-36,
Tata Institute of Fundamental Research, Bombay, 1957.

\bibitem{Sho} P. \textsc{Shor}: \emph{A New Proof of Cayley's Formula for
Counting Labelled Trees}, J. C. T., Series A, {\bf 71} (1995), 154--158.\end{thebibliography}

\end{document}


\begin{table}
{\scriptsize
\begin{tabular}{c|llll} 
$j\backslash n$& 1& 2& 3&4\\
\hline
0&  1&$1+x$&$2+3x+x^2 $&$6+11x+6x^2+x^3$\\
1&   &1   &$4+3x $      &$18+22x+6x^2 $      \\
2&   &    &3           &$28+15x $                 \\
3&   &    &            &15                                                \\
\hline
$\sum_{j}Q_{n,\, j}(x)$& 1&$2+x $&$(3+x)^2$ &$(4+x)^3$ \\
\end{tabular}
\caption{$Q_{n,\, j}(x)$ pour $0\leq j<n\leq 5$.}}
\end{table}
\begin{remark} Given $t\in {\cal T}_{n,\,k}(\rightarrow\cdot)$ we denote by
$\son(t)$ the set of sons of 1. Hence
$$
\sum_{t\in {\cal T}_{n,\,k}(\rightarrow\cdot)}(x+1)^{\deg(t,\,1)}=
\sum_{t\in {\cal T}_{n,\,k}(\rightarrow\cdot)}\sum_{S\subseteq \son(t)}x^{|S|}.
$$
Let $\tilde {\cal T}_{n,\,k}(\rightarrow\cdot)$ be the set of trees $t$
in ${\cal T}_{n,\,k}(\rightarrow\cdot)$ with a distinguished subset $S$ of 
$\son(t)$. Hence in view of the proceeding 
proposition it suffices to establish a
bijection $\varphi\,:\, \tilde {\cal T}_{n,\,k}(\rightarrow\cdot)\rightarrow {\cal
T}_{n+1,\,k}$ such that $\deg(\varphi(t,\, S))=|S|+1$.
\end{remark}


%%%%%%%%%%%%%%%%%%%Appendice%%%%%%%%%%%%%%%%%%%%
\begin{lem} Let $y(t)\in {\bf K}[[t]]$ be the solution
of the equation $t=ye^{-y}$, then
\begin{equation}
\sum_{n\geq 0}(x+n)^n\frac{t^n}{n!}=\frac{e^{xy}}{1-y}.
\end{equation}
\end{lem}
\pf
Define the {\sl diff\'erence operator} $\Delta$:
$$
\Delta f(x)=f(x+1)-f(x),\qquad f(x)\in {\bf K}[x].
$$ 
It's then easy to verify by induction on $m$ that
$$
\Delta^mf(x)=\sum_{i=0}^m{m\choose i}(-1)^{m-i}f(x+i),\qquad m\geq 1.
$$
and $\Delta^m(x^m/m!)=1$. It follows that
$$
\sum_{i=0}^m{m\choose i}
(-1)^{m-i}\frac{(x+i)^m}{m!}=1.
$$
Multipling the above equation by $y^m$ and summing over all $m\geq 0$ we deduce
that
$$
\sum_{n\geq 0}(x+n)^n\frac{y^ne^{-(x+n)y}}{n!}=\frac{1}{1-y},
$$
which is equivalent to the desired formula. \qed\medskip

The following result generalize corrollary 3 which correponds to the
$u=0$ case.

\begin{cor} We have
\begin{equation}
\sum_{k=0}^{n-1}\frac{Q_{n,\,k}(x)}{(1-u)^{n+k}}=\sum_{k=0}^\infty
\frac{(x+n+k)^{n+k-1}e^{-u(x+n+k)}u^k}{k!}.\label{gene}
\end{equation}
\end{cor}
\pf Let
$$
Q_n(a,\,x)=\sum_{k=0}^{n-1}Q_{n,\, k}(x)a^k.
$$
Then recurrence (\ref{jiang}) implies that
$$
Q_{n+1}(a,\,x)=(x+n+an)Q_n(a,\,x)+a^2\frac{\partial}{\partial a}Q_n(a,\,x).
$$
The above equation can be rewritten as
$$
a^{n+1}e^{-(x+n+1)/a}Q_{n+1}(a,\,x)=a^3e^{-1/a}
\frac{\partial}{\partial a}\left[
a^ne^{-(x+n)/a}Q_n(a,\, x)\right].
$$
Note that $Q_1(a,\, x)=1$, hence
$$
Q_n(a,\,x)=a^{-n}e^{(x+n)/a}\left(a^3e^{-1/a}
\frac{\partial}{\partial a}\right)^{n-1}\left[
a^ne^{-(x+n)/a}Q_n(a,\,x)\right].
$$
Now make the substitution $a=\frac{1}{1-t}$ and $u=te^{-t}$ to obtain
\begin{eqnarray*}
Q_n(a,\, x)&=&(1-t)^ne^{-(x+n)t}\left(\frac{\partial }{\partial u}\right)^{n-1}
\left(\frac{e^{(x+1)t}}{1-t}\right)\\
&=& (1-t)^ne^{-(x+n)t}\sum_{k=0}^\infty(x+n+k)^{n+k-1}\frac{(te^{-t})^k}{k!}.
\end{eqnarray*}
The last equality follows from lemma 3.\qed
\medskip

\begin{cor} The generating function $Y(t)$ defined in 
(\ref{zeng}) has the following expression:
\begin{equation}
Y(t)=\sum_{n\geq 1}\frac{(x+n)^{n-1}e^{(1/a-1)(x+n)}}{n!a^n}
\left[(t+a-1)^n-(a-1)^n\right].
\end{equation}
\end{cor}
\pf
Let $a=1/(1-u)$. It follows from corollary 9 that
\begin{eqnarray}
Y(t)&=& \sum_{n\geq 1}\frac{t^n}{n!a^n}\sum_ {k\geq 0}
\frac{(x+n+k)^{n+k-1}e^{(1/a-1)(x+n+k)}(1-1/a)^k}{k!}\\
&=& \sum_{l\geq 1}\frac{(x+l)^{l-1}e^{(1/a-1)(x+l)}}{l!}
\sum_{k=0}^{l-1}
{l\choose k}\left(\frac{t}{a}\right)^{l-k}(1-\frac{1}{a})^k.
\end{eqnarray}
Replacing $l$ by $n$ and applying the binomial formula to the interior summation
in the last line we deduce the desired formula. \qed\medskip

For example, in view of propositions 1 and 2 
formula (\ref{Ramanu}) is equivalent to 
$$
\sum_{T\in{\cal T}_{n+1,\, k}}x^{\deg_T(1)-1}=
\sum_{T\in {\cal T}_{n,\,k}(\rightarrow\cdot)}x^{\deg_T(1)}
+(n+k-1)
\sum_{T\in{\cal T}_{n,\, k}}x^{\deg_T(1)-1}.
$$
Hence there should be an operation on trees to
explain the above identity.






