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\input{paths.tex}
\renewcommand{\theenumi}{(\roman{enumi})}
\newtheorem{Proposition}[Theorem]{Proposition}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\title{ Vicious walkers and Young tableaux II: With a wall}
\author[A. J. Guttmann]{Anthony J. Guttmann{\dag }}
\author[C. Krattenthaler]{Christian Krattenthaler{\S}}
\author[X. G. Viennot]{\\ Xavier G. Viennot{\ddag }} 
\address{\dag Department of Mathematics and Statistics, The
University of Melbourne\\ Parkville, Victoria 3052, Australia}
\address{\S Institut f\"{u}r Mathematik, Universit\"{a}t Wien,\\
Strudlhofgasse 4, A-1090 Vienna, Austria}
\address{\ddag LaBRI, Universit\'{e} Bordeaux 1, 351 cours de la
Lib\'{e}ration, 33405 Talence Cedex, France.
}
\dedicatory{19th January, 2000}
\begin{abstract}
We derive new results for the number of star
and watermelon configurations of vicious walkers
in the presence of an impenetrable wall by showing 
that these follow from standard results in the
theory of Young tableaux, and combinatorial descriptions
of symmetric functions.
\end{abstract}
\maketitle

\section{Introduction}
In an earlier paper \cite{ESG} an expression for the number of
star configurations of vicious walkers on a $D$-dimensional
lattice was obtained,
and the result for the corresponding number of watermelon
configurations was conjectured.
Later, in \cite{GOV} it was shown how certain
results from the theory of Young tableaux, and related
results in algebraic combinatorics enabled one to readily
prove both results.

In this paper, we show how some results from the theory of
symmetric functions can be used to prove analogous results
for the more difficult case of walkers in the presence of
an impenetrable wall. We also give rigorous asymptotic results.

{\em Vicious walkers} 
describes the situation in which two or more walkers arriving
at the same lattice site annihilate one another. Accordingly,
the only configurations we consider are those in which such contacts
are forbidden. Alternatively expressed, we consider mutually 
self-avoiding networks of lattice walks which also model 
directed polymer
networks. The connection of these vicious walker problems to the
5 and  6 vertex model of statistical mechanics was also discussed in \cite{GOV}.

The problem, together with a number of physical applications,
was introduced by Fisher \cite{FIS}. The general model is one
of $p$ random walkers on a $D$-dimensional lattice who at 
regular time intervals simultaneously take one step with equal 
probability in the direction of one of the allowed lattice vectors.

The two standard topologies of interest are that of a \textit{star}
and a \textit{watermelon}. Consider a directed square lattice,
rotated $45^o,$ so that the unit vectors on the lattice are
$({\bf i}+{\bf j})/\sqrt(2)$ and $({\bf i}-{\bf j})/\sqrt(2).$
Both configurations consist of $p$ chains of length $n$ which
start at $(0,0),(0,2),(0,4),\cdots,(0,2p-2).$
The watermelon configurations end at
$(n,k),(n,2+k),(n,4+k),\cdots,(n,k+2p-2).$ For stars, the end-points
of the chains all have $x$-coordinate equal to $n,$ but the $y$
 coordinates
are unconstrained, apart from the ordering imposed by the non-crossing
condition. Thus if the end-points are
$(n,e_1),(n,e_2),(n,e_3),\cdots,(n,e_p),$ then 
$e_1 < e_2 < e_3 < \cdots < e_p \le 2p-2+n.$ In the 
problem considered
here, the additional constraint of an impenetrable wall
imposes the condition that at no stage
may any walker step to a point with negative $y$-coordinate.

In \cite{ESG} recurrence relations and the corresponding differential
equations for stars and watermelons on the directed square lattice were
obtained. In the case of watermelons, a determinental form was
evaluated by standard techniques applied to the determinant. In
the case of stars, the results obtained were conjectural, being
equivalent to an earlier conjecture \cite{AME}.
In \cite{GOV} it was shown  how a number of ``standard'' results in the
theory of Young tableaux and partitions lead to a much more intuitive
derivation of the above results, and provided a proof of the
conjectured results.

In \cite{GV} we introduced a generalisation of the vicious walker model, called
the {\em $n$-friendly walker model.} In that model, the ``vicious''
constraint is systematically relaxed, so that any two walks may stay together for
up to $n$ steps. Thus $n=0$ corresponds to the vicious walker model described
above, $n = 1$ to the
so-called {\em osculating walker} model in which walkers may touch at a
vertex, but then must part, and so on. Walkers that may share an
arbitrary number of steps we call the {\em $\infty$-friendly walker} model.
Only numerical conjectures for exponents were obtained in \cite{GV} for the
$\infty$-friendly walker model.
Only numerical conjectures for exponents were obtained in \cite{GV} for the
$\infty$-friendly walker model. Here we provide asymptotics for
this model, which both proves the earlier conjectures, and makes the
earlier results more precise.

In this paper we first rederive the results of \cite{GOV}  by making
use of Schur functions and {\em odd orthogonal characters}, which
will be needed in our derivation of the results for watermelons
and stars in the presence of a wall.
As well as rederiving results in the absence of a
wall, we also develop the asymptotics for the number of such configurations.

We then solve the problem in the presence of a wall, or impenetrable barrier
below which the walks may not go. We also give asymptotic results for
both stars and watermelons in this case. We then extend these
results to $m$-friendly stars and watermelons in the presence of a
wall.

Finally, we treat the case of walkers confined in a slit. That is to say,
the walkers may not go below one wall, nor above a parallel wall. They
are confined to lie between the two. Asymptotics for the number of 
stars and watermelons is given, and then generalised to the $m$-friendly
walker case.


\section{Stars with fixed end-points, without a wall.}
A typical star configuration is shown in Figure~\ref{F1}.
Let $e_1<e_2<\dots<e_p$ with $e_i\equiv m\pmod2$, $i=1,2,\dots,p$.
We seek the number of stars with $p$ branches,
the $i$-th branch running from
$A_i=(0,2i-2)$ to $E_i=(m,e_i)$, $i=1,2,\dots,p$. In
Figure~\ref{F1}.a  an example with $p=4$, $m=6$, $e_1=0$, $e_2=2$,
$e_3=6$, $e_4=10$ is given.

\begin{Theorem} \label{T1}
The number of such stars is given by:
\begin{equation} \label{e1}
2^{-\binom p2}\prod _{i=1} ^{p}\frac {(m-i+p+1)!} {\(\frac {m+e_i} {2}\)!\,\(\frac {m-e_i}
{2}+p-1\)!}\prod _{1\le i<j\le p} ^{}(e_j-e_i).
\end{equation}
\end{Theorem}
\begin{pf}
The Schur function
$s_\la(x_1,x_2,\dots,x_m)$ is defined by (see \cite[I, (3.1)]{MacdAC}
or \cite[p.~403, (A.4)]{FuHaAA})
\begin{equation} \label{e4}
s_\lambda (x_1, x_2,\dots ,x_m)=\frac {
 \det\limits _{1\le i,j\le m}(x_j^{\lambda _i+m-i})} {
\det\limits_{1\le i,j\le m}(x_j^{m-i})}.
\end{equation}
It is well-known that a combinatorial description of Schur
functions may be given in terms of ({\em semistandard}) {\em tableaux}.
Let $\lambda$ be a partition. A filling
of the cells of the Ferrers board of $\lambda$ with elements of
the set $\{1,2,\dots\}$
which is {\em weakly increasing\/} along rows and {\em strictly increasing\/}
along columns
is called a ({\em semistandard}) {\em tableau of shape $\lambda$\/}.
Figure~\ref{F1}.b shows such a semistandard tableau of shape
$(4,3,2)$.

\begin{figure}[t]
\begin{Figure}\label{F1}
        \Gitter(7,13)(0,-1)
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        \Pfad(0,6),333334\endPfad
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        \DickPunkt(6,0)
        \DickPunkt(6,2)
        \DickPunkt(6,6)
        \DickPunkt(6,10)
        \Label\r{\mathit 2}(1,1)
        \Label\r{\mathit 4}(3,1)
        \Label\r{\mathit 5}(4,0)
        \Label\r{\mathit 3}(2,4)
        \Label\r{\mathit 4}(3,3)
        \Label\r{\mathit 6}(5,3)
        \Label\r{\mathit 4}(3,7)
        \Label\r{\mathit 6}(5,7)
        \Label\r{\mathit 6}(5,11)
        \Label\l{A_1}(0,0)
        \Label\l{A_2}(0,2)
        \Label\l{A_3}(0,4)
        \Label\l{A_4}(0,6)
        \Label\r{E_1}(6,0)
        \Label\r{E_2}(6,2)
        \Label\r{E_3}(6,6)
        \Label\r{E_4}(6,10)
\begin{picture}(15,5)(0,0)
\put(12,4){
        \begintableau{3}{4}
        \row{2346}
        \row{446}
        \row{56}
        \endtableau
}
\put(13,2){\small b. A tableau}
\put(3,-3){\small a. A star}
\end{picture}
$$
\vskip.3cm
$$
\end{Figure}
\end{figure}

The {\em weight} $\mathbf x^T$
of a tableau $T$ is defined as
\begin{equation}\label{e5}
 \mathbf{x}^{T}:=\prod
x_{T_{i,j}} ,
\end{equation}
where the product is over all entries $T_{ij}$ of $T$.
Given this terminology, the Schur function $s_\la(x_1,x_2,\dots,x_m)$
is also given by (see
\cite[I, (5.12) with $\mu=\emptyset$]{MacdAC}),
\begin{equation}\label{e6}
 s_\lambda(x_1,x_2,\dots,x_m) = \sum_{T}\mathbf{x}^{T},
\end{equation}
where the sum is over all
tableaux $T$ of shape $\lambda$ with entries $\le m$.

In \cite{GOV} it was proved that the number of stars with $p$
branches, as described above, follows from
a standard bijection between stars and tableaux, see
Figure~\ref{F1}. First label down-steps by the $x$-coordinate of their
end point, so that a step from $(a-1,b)$ to $(a,b-1)$ is labelled by
$a$, see Figure~\ref{F1}.a. Then, out of the labels of the $j$-th
branch, form the $j$-th column of the corresponding tableau. The
resulting array of numbers is indeed a tableau. This can
be readily seen, since the entries are
trivially strictly increasing along columns, and they are weakly
increasing along rows because the branches do not touch each other.

Thus, given a star with $p$ branches, the $i$-th branch running from
$A_i=(0,2i-2)$ to $E_i=(m,e_i)$, $i=1,2,\dots,p$, one obtains a tableau
with column lengths $\frac {m-e_1}2,\frac {m-e_2}2\dots,\break
\frac {m-e_p}2$.
The {\em shape} (the vector
of row lengths) can be easily extracted from the column lengths.
This correspondence between stars and tableaux is a bijection between
stars with $p$ branches, the $i$-th branch running from
$A_i=(0,2i-2)$ to $E_i=(m,e_i)$, $i=1,2,\dots,p$, and tableaux with
entries at most $m$ and column lengths
$\frac {m-e_1}2,\frac {m-e_2}2\dots,\break\frac {m-e_p}2$.

Further, it is known \cite{Sta, GeViAB} that the number
of semistandard tableaux is given by the
product formula, $$ \prod _{\rh\in\la} ^{}\frac {m+c_\rh} {h_\rh}. $$
where $c_\rh$ and $h_\rh$ are the {\em content\/} and the {\em hook
length\/} of the cell $\rh$. The content $c_\rh$ of a cell $\rh=(i,j)$
is $j-i$, whereas the hook-length $h_\rh$ of a cell $\rh$ is the
number of cells in the same row to the right of $\rh$ plus the number
of cells in the same column below $\rh$ plus 1.

However this product is given directly in terms of Schur functions
by the formula (see \cite[I, Sec.~3,  Ex.~1 and 4]{MacdAC},
 \cite[Ex.~A.30, (ii)]{FuHaAA})
\begin{equation} \label{e7}
s_\la(\underbrace{1,1,\dots,1}_m)=\prod _{1\le i<j\le m} ^{}\frac {\la_i-i-\la_j+j}
{j-i}=
\prod _{\rh\in\la} ^{}\frac {m+c_\rh} {h_\rh}.
\end{equation}

Because of \eqref{e6}, with $x_1=x_2=\cdots=x_m=1$,
the number of these tableaux is then given by
\eqref{e7}. The expression obtained can, with some work, be converted
into \eqref{e1}.
\end{pf}

\section{Enumeration of stars with arbitrary end points, without
wall restriction}

The following result was proved in \cite{GOV} using the Bender-Knuth
formula, but no proof of the Bender-Knuth formula was given. As
we require some of the concepts of the proof in subsequent
sections, we briefly repeat the proof given in \cite{GOV}
but include a proof of the Bender-Knuth conjecture.

\begin{Theorem} \label{T2}
The number of stars of length $m$ with $p$ branches equals
\begin{equation} \label{e10}
\prod _{1\le i\le j\le m} ^{}\frac {p+i+j-1} {i+j-1}.
\end{equation}
\end{Theorem}
\begin{pf}Using the correspondence between stars and tableaux
described in the  proof of Theorem~\ref{T1}, we see that we
must count tableaux with entries at most $m$ having at most $p$
columns. This enumeration problem (actually the corresponding
``$q$-enumeration" problem) is known under the name {\em
Bender--Knuth conjecture\/}, and was first proved by B.~Gordon around
1970 (but appeared only much later as \cite{GordAC}). Since then,
many further proofs have been given. See \cite{AndrAJ},
\cite[Th\'eor\'eme~1.1, first identity]{DesaAB},
\cite[I, Sec.~5, Ex.~19]{MacdAC},
\cite[Prop.~7.2]{ProcAD}, \cite{ProcAB},
\cite[Sec.~7]{StemAE} for a selection.
What all these proofs share more or less
explicitly is the following identity, which relates Schur functions
and {\em odd orthogonal characters\/} of the symmetric group 
of rectangular shape,
\begin{equation} \label{e11}
\sum _{\mu_1\le p} ^{}s_\mu(x_1,x_2,\dots,x_m)=(x_1x_2\cdots
x_m)^{p/2}\so_{\big((p/2)^m\big)}
(x_1^{\pm1},x_2^{\pm1},\dots,x_m^{\pm1},1).
\end{equation}
The odd orthogonal characters $\so_\la
(x_1^{\pm1},x_2^{\pm1},\dots,x_m^{\pm1},1)$ where $x_1^{\pm1}$ is 
a shorthand notation
for $x_1,x_1^{-1}$, etc.), and $\la$ is an
$m$-tuple $(\la_1,\la_2,\dots,\la_m)$ of integers, or of
half-integers, is defined by
\begin{equation} \label{e12}
\so_\la(x_1^{\pm1},x_2^{\pm1},\dots,x_m^{\pm1},1)=\frac {\det\limits_{1\le i,j\le
m}(x_j^{\la_i+m-i+1/2}-x_j^{-(\la_i+m-i+1/2)})}
{\det\limits_{1\le i,j\le
m}(x_j^{m-i+1/2}-x_j^{-(m-i+1/2)})}.
\end{equation}
Recall that the Schur functions $s_\la(x_1,x_2,\dots,x_m)$ are defined by
\eqref{e4}. While Schur functions are polynomials in
$x_1,x_2,\dots,x_m$ (cf\@. \eqref{e6}), odd orthogonal characters\break
$\so_\la(x_1^{\pm1},x_2^{\pm1},\dots,x_m^{\pm1},1)$ are polynomials
in $x_1,x_1^{-1},x_2,x_2^{-1},\dots,x_m,x_m^{-1},1$. They have a
combinatorial descriptions in terms of certain tableaux as well, see
\cite[Sec.~2]{FuKrAA}, \cite[Sec.~6--8]{ProcAK}, \cite[Theorem~2.3]{SunaAE}.

A variety of different proofs of \eqref{e11} has been given. There
are proofs by a combination of combinatorial and manipulatory
arguments (cf\@.
\cite{GordAC}, \cite[Sec.~7, Cor.~7.4.(a)]{StemAE},
\cite[Theorem~2.3.(1)]{OkadAI}), by use of the theory of
Hall--Littlewood functions (cf\@. \cite[I, Sec.~5, Ex.~16]{MacdAC}),
by use of combinatorial descriptions of
orthogonal characters coming from
algebraic geometry, due to
DeConcini, Procesi, Lakshimibai, Musili and Seshadri
(cf\@. \cite[Sec.~7]{ProcAD}, \cite[Theorem~3]{ProcAB},
\cite[Proof of (3.8) in the case $p=0$, stated separately as
(3.12)]{KratBC}). Eventually, a
completely elementary proof was found by Bressoud \cite{BresAN}.

However, what we now require is the evaluation of the left-hand side of
\eqref{e11} at $x_1=x_2=\cdots=x_m=1$. We may use well-known
formulae for the evaluation of odd orthogonal characters at
$x_1=x_2=\cdots=x_m=1$, namely (see \cite[(24.29)]{FuHaAA},
\cite[Theorem~4.5.(2)]{SunaAE}),
\begin{multline} \label{e13}
\so_\la(1,1,\dots,1)=\prod _{1\le i<j\le m} ^{}\frac {\la_i-i-\la_j+j}
{j-i}\prod _{1\le i\le j\le m}\frac {\la_i+\la_j+2m+1-i-j}
{2m+1-i-j}\\
=\prod _{\rh\in\la} ^{}\frac {2m+1+e_\rh} {h_\rh},
\end{multline}
where, again, $h_\rh$ is the hook length of cell $\rh$, and $e_\rh$
is given by
$$e_\rh=e_{(i,j)}=\begin{cases} \la_i+\la_j-i-j&i\le j\\
i+j-\la_i'-\la_j'-2&i>j.\end{cases}$$
Here, $\la'$ denotes the partition conjugate to $\la$ (see
\cite[p.~2]{MacdAC} for the definition of conjugate partition).

Using this formula for $\la=\big((p/2)^n\big)$ in \eqref{e11} with
$x_1=x_2=\cdots=x_n=1$, finally leads to \eqref{e10}.
\end{pf}

\begin{Theorem} \label{thm:star-without}
The number of stars with $p$ branches of length $m$
is asymptotically
\begin{equation} \label{eq:star-without}
\begin{cases}
\displaystyle
2^{mp+p^2/4}m^{-p^2/4+p/4}\pi^{-p/4}\bigg(\prod _{\ell=1} ^{p/2}(2\ell-2)!\bigg)(1+O(m^{-1}))&
\text {if $p$ is even,}\\
\displaystyle
2^{mp+p^2/4-1/4}m^{-p^2/4+p/4}\pi^{-p/4+1/4}
\bigg(\prod _{\ell=1} ^{(p-1)/2}(2\ell-1)!\bigg)(1+O(m^{-1}))&
\text {if $p$ is odd,}\end{cases}
\end{equation}
as $m$ tends to infinity.
\end{Theorem}

\begin{proof}We know that the number of stars with $p$ branches of length $m$
is given by the product formula
\begin{equation} \label{eq:num-star}
\prod _{1\le i\le j\le m}
\frac {p+i+j-1} {i+j-1}.
\end{equation}
Therefore, proving \eqref{eq:star-without} amounts to appropriately
rewriting \eqref{eq:num-star} and then applying Stirling's formula.

For convenience, let us introduce the notations $H(n):=\prod _{\ell=1}
^{n}(i-1)!=\prod _{\ell=1} ^{n}(n-i)!$
and $H_2(n):=\prod _{\ell=1} ^{\fl{n/2}}(n-2i)!$. Then the product
\eqref{eq:num-star} can be rewritten as follows,
\begin{align} \notag
\prod _{1\le i\le j\le m}&
\frac {p+i+j-1} {i+j-1}=\prod _{i=1} ^{m}\frac {(p+i+m-1)!\,(2i-2)!}
{(p+2i-2)!\,(i+m-1)!}\\
\label{eq:num-star-H}
&=\frac {H(p+2m)\,H_2(p)\,H_2(2m)\,H(m)}
{H(p+m)\,H_2(p+2m)\,H_2(0)\,H(2m)}.
\end{align}
Our aim is to write this as a product whose range depends only on
$p$. To do so, we need to distinguish between the cases of $p$ being
even or odd.

If $p$ is even, then \eqref{eq:num-star-H} can be written as
$$\prod _{\ell=1} ^{p/2}\frac {(2\ell-2)!} {(2m+2\ell-2)!}
\prod _{\ell=1} ^{p}\frac {(2m+\ell-1)!} {(m+\ell-1)!}.$$
Application of Stirling's formula, and some simplification, yields
the first line of \eqref{eq:star-without}.

If $p$ is odd, then \eqref{eq:num-star-H} can be written as
$$\frac {(p+2m-1)!!} {(p-1)!!}
\prod _{\ell=1} ^{(p+1)/2}\frac {(2\ell-2)!} {(2m+2\ell-2)!}
\prod _{\ell=1} ^{p}\frac {(2m+\ell-1)!} {(m+\ell-1)!}.$$
Renewed application of Stirling's formula, and some simplification, yields
the second line of \eqref{eq:star-without}.
\end{proof}

Next we consider the $\infty$-friendly model for stars.

\begin{Theorem} \label{thm:star-inf-without}
The number of $\infty$-friendly stars with $p$ branches of length $m$
is asymptotically
\begin{equation} \label{eq:star-inf-without}
\begin{cases}\displaystyle
2^{mp+p^2/2-p/2+1}m^{-p^2/4+p/4}\pi^{-p/4}
\bigg(\prod _{\ell=1} ^{p/2}\frac {(\ell-1)!\,(4\ell-4)!}
{(\ell+p/2-2)!}\bigg)(1+O(m^{-1}))&
\text {if $p$ is even,}\\
\displaystyle
2^{mp+p^2/2-p/2+1}m^{-p^2/4+p/4}\pi^{-p/4+1/4}\\
\displaystyle
\kern3cm\times
\bigg(\prod _{\ell=1} ^{(p-1)/2}\frac {(\ell-1)!\,(4\ell-1)!}
{(\ell+p/2-5/2)!}\bigg)(1+O(m^{-1}))&
\text {if $p$ is odd,}\end{cases}
\end{equation}
as $m$ tends to infinity.
\end{Theorem}

\begin{proof}
The situation is more difficult here, as we do not have a nice closed
product formula (such as \eqref{eq:num-star}) for $\infty$-stand
stars.

The first step is the same as in the proof of
Theorem~\ref{thm:watermelon-inf-without}. We transform $\infty$-stand
star into families of non-intersecting lattice paths by shifting the
$i$-th path up by $2(i-1)$ units. Thus,
$\infty$-friendly stars with $p$ branches of length $m$ are in bijection
with families $(P_1,P_2,\dots,P_p)$ of
non-intersecting lattice paths, $P_i$ running from $A_i=(0,4(i-1))$ to
some point on the line $x=m$, $i=1,2,\dots,p$.
To enumerate these families of non-intersecting lattice paths, we
resort to Proposition~\ref{prop:Stem}.
Apparently, we must again distinguish between the cases of $p$ being
even or odd.

Let first $p$ be even. Then we may directly apply Proposition~\ref{prop:Stem} with $A_i=(0,4(i-1))$, $i=1,2,\dots,p$,
and $\mathbf E=\{(m,y):y\equiv m$ (mod $2)\}$. The number $Q_{\mathbf
E}(A_i,A_j)$ of all pairs of
non-intersecting lattice paths, one connecting $A_i$ to $\mathbf E$,
the other connecting $A_j$ to $\mathbf E$, which has to be computed
in order to apply \eqref{eq:Stem}, is easily determined by using
$2\times 2$ Gessel--Viennot determinants. With $E_i=(m,-m+2(i-1))$,
so that $\mathbf E=\{\dots,E_1,E_2,\dots\}$, we have
\begin{align} \notag
Q_{\mathbf E}&(A_i,A_j)=\sum _{s<t} ^{}\big(\v{\mathcal P(A_i\to E_s)}
\cdot \v{\mathcal P(A_j\to E_t)}-\v{\mathcal P(A_i\to E_t)}\cdot \v{\mathcal P(A_j\to E_s)}\big)\\
\notag
&=\sum _{s<t} ^{}\(\binom {m} {s-2i+1}\binom {m}{t-2j+1}-
\binom {m} {t-2i+1}\binom {m}{s-2j+1}\)\\
\notag
&=\sum _{r\ge0} ^{}\sum _{s} ^{}\(\binom {m} {s-2i+1}\binom {m}{s+r-2j+1}-
\binom {m} {s+r-2i+1}\binom {m}{s-2j+1}\)\\
\notag
&=\sum _{r\ge0} ^{}\(\binom {2m}{m-2i+2j-r}-\binom {2m}{m+r-2i+2j}\)\\
\notag
&=\sum _{-2(j-i)<r\le 2(j-i)} ^{}\binom {2m}{m+r}\\
\notag
&=\sum _{-(j-i)<r\le (j-i)} ^{}\(\binom {2m}{m+2r-1}+\binom
{2m}{m+2r}\)\\
\label{eq:Q}
&=\sum _{-(j-i)<r\le (j-i)} ^{}\binom {2m+1}{m+2r}.
\end{align}
(From the third to the fourth line we used the Vandermonde summation.)

If we substitute this into \eqref{eq:Stem}, we obtain
\begin{equation*}
\Pf\limits_{1\le i<j\le p}\(\sum _{-(j-i)<r\le (j-i)} ^{}\binom
{2m+1}{m+2r}\)
\end{equation*}
for the number of $\infty$-friendly stars with $p$ branches of length $m$.
To this Pfaffian we may apply Gordon's reduction
\eqref{eq:Gordon-even} with $g_\al=\binom {2m+1}{m+2\al}$, to get
\begin{equation} \label{eq:Pfaff-star-inf}
\det\limits_{1\le i,j\le p/2}\(\binom {2m+1}{m+2i-2j}+\binom
{2m+1}{m+2i+2j-2}\)
\end{equation}
for the number of $\infty$-friendly stars under consideration.

We have to estimate the determinant in \eqref{eq:Pfaff-star-inf} as
$m$ becomes large. We already encountered a similar situation in the proof of Theorem~\ref{thm:watermelon-inf-without},
where we had to estimate \eqref{eq:GV-det}. Also here, we start by taking
some factors out of the determinant,
\begin{multline} \label{eq:det-factors2}
\det\limits_{1\le i,j\le p/2}\(\binom {2m+1}{m+2i-2j}+\binom
{2m+1}{m+2i+2j-2}\)\\
\kern-4cm=\prod _{i=1} ^{p/2}\frac {(2m+1)!}
{(m+2i+p-2)!\,(m-2i+p+1)!}\\
\times
\det\limits_{1\le i,j\le p/2}\big((m+2i-2j+1)_{p+2j-2}\,(m-2i+2j+2)_{p-2j}\\
+(m+2i+2j-1)_{p-2j}\, (m-2i-2j+4)_{p+2j-2}\big).
\end{multline}
The determinant on the right-hand side is a polynomial in $m$. As in
\eqref{eq:det-factors}, it suffices to find the leading term of this
polynomial as lower terms give negligible contributions to the asymptotics.

In order to do so, we use again the ``trick" of considering a more
general determinant, namely
\begin{multline} \label{eq:gen-det2}
\det\limits_{1\le i,j\le p/2}\big((m+X_i-2j+5/2)_{p+2j-2}\,
(m-X_i+2j+1/2)_{p-2j}\\
+(m+X_i+2j+1/2)_{p-2j}\, (m-X_i-2j+5/2)_{p+2j-2}\big).
\end{multline}
Clearly, we get back the determinant in \eqref{eq:det-factors2} for
$X_i=2i-3/2$.

The determinant in \eqref{eq:gen-det2} is a polynomial in $m$ and the
$X_i$'s of degree $p(p-1)$. It is divisible by $\prod _{1\le
i<j\le p/2} ^{}(X_j^2-X_i^2)$. Therefore the degree in $m$
of the determinant in \eqref{eq:det-factors2} is at most $3p^2/4-p/2$.

The leading term of \eqref{eq:gen-det2} is
\begin{align*}
&\det\limits_{1\le i,j\le p/2}\big((m+X_i)^{p+2j-2}(m-X_i)^{p-2j}+
(m+X_i)^{p-2j}(m-X_i)^{p+2j-2}\big)\\
&\kern10pt=\prod _{i=1} ^{p/2}(m+X_i)^{p-2}(m-X_i)^{p}
\det\limits_{1\le i,j\le p/2}\(\(\frac {m+X_i} {m-X_i}\)^{2j}+
\(\frac {m+X_i} {m-X_i}\)^{-2j+2}\)\\
&\kern10pt=\prod _{i=1} ^{p/2}(m-X_i)^{2p-2}
\prod _{i=1}
^{p/2}\(\(\frac {m+X_i} {m-X_i}\)^2+1\)
\\
&\kern10pt\kern1cm
\times\prod _{1\le i<j\le p/2} ^{}\(\(\frac {m+X_i} {m-X_i}\)^2
-\(\frac {m+X_j} {m-X_j}\)^2\)\(1-\(\frac {(m+X_i)} {(m-X_i)}\frac {(m+X_j)} {(m-X_j)}\)^2\)\\
&\kern10pt=\prod _{1\le i<j\le p/2} ^{}\big((m+X_j)^2(m-X_i)^2-
(m+X_i)^2(m-X_j)^2\big)\\
&\kern10pt\kern3cm
\cdot
\big((m+X_j)^2(m+X_i)^2-(m-X_i)^2(m-X_j)^2\big)\\
&\kern10pt\kern3cm
\times
\prod _{i=1} ^{p/2}\big((m+X_i)^2+(m-X_i)^2\big)\\&\kern10pt=\prod _{1\le i<j\le p/2} ^{}2m(X_j-X_i)(2m^2+\cdots)
2m(X_j+X_i)(2m^2+\cdots)
\prod _{i=1} ^{p/2}(2m^2+\cdots).
\end{align*}
Here we used \eqref{eq:ortho1} to evaluate the determinant in the
second line.
The leading term of the determinant in \eqref{eq:det-factors2} is this
expression under the
substitution of $X_i=2i-3/2$, $i=1,2,\dots,p$, which equals
\begin{multline*}
m^{6\binom {p/2}2+p}2^{4\binom {p/2}2+p/2}
\prod _{1\le i<j\le p/2} ^{}(2j-2i)(2j+2i-3)\\
=m^{3p^2/4-p/2}2^{p^2/2-p+1}
\prod _{\ell=1} ^{p/2}
\frac {(\ell-1)!\,(4\ell-4)!} {(\ell+p/2-2)!}.
\end{multline*}
Hence, the asymptotics of the
number of $\infty$-friendly stars with $p$ branches of length $m$
is this expression multiplied by the asymptotics of the
product on the right-hand side of \eqref{eq:det-factors2}, the latter
being easily found by Stirling's formula. After some simplification,
the first line of \eqref{eq:star-inf-without} (the
case of even $p$) is found.

\smallskip
If $p$ is odd, then Proposition~\ref{prop:Stem} cannot be applied
directly. As described in \cite{StemAE} (cf\@. the Remark after Theorem~3.1
in \cite{StemAE}), there is a simple trick to
make Proposition~\ref{prop:Stem} work also in this case. This
consists of adding a ``dummy path" $P_{p+1}$ of length 0. More
precisely, a ``phantom" starting point, $A_{p+1}$ say, which is not
connected to any other lattice point, is added, and
it is also added as a ``phantom"end point to the set $\mathbf E$.
Then we consider families of non-intersecting lattice paths connecting $A_1,A_2,\dots,A_{p+1}$ to $\mathbf E$.
Since $p+1$ is even, we may apply Proposition~\ref{prop:Stem} to this
situation. Conveniently, the only possible way to connect $A_{p+1}$ to a
point of $\mathbf E$ is to connect it to itself. Therefore, as we may
simply disregard this ``dummy path" from $A_{p+1}$ to itself, the number
of these families of non-intersecting lattice paths is exactly equal
to the number of families of non-intersecting lattice paths connecting $A_1,A_2,\dots,A_{p}$ to $\mathbf E$, which
is the number we want to find and approximate.

So we apply Proposition~\ref{prop:Stem} with $A_i=(0,4(i-1))$,
$i=1,2,\dots,p$, $A_{p+1}$ the ``phantom" vertex, and with $\mathbf
E=\{(m,y):y\equiv m$ (mod $2)\}\cup\{A_{p+1}\}$. Clearly, for $1\le i<j\le
p$, the quantity $Q_{\mathbf
E}(A_i,A_j)$ is given by formula \eqref{eq:Q} as
before. On the other hand, $Q_{\mathbf E}(A_i,A_{p+1})$ is simply equal 
to $2^m$, because, once again, the only way to connect $A_{p+1}$ to a
point in $\mathbf E$ is to connect it with itself, and {\em any} path from
$A_i$ to the line $x=m$ will not intersect the ``dummy path" from
$A_{p+1}$ to itself, of which there are exactly $2^m$.

Hence, the Pfaffian that we obtain from \eqref{eq:Stem} is
\begin{equation*}
\Pf\limits_{1\le i<j\le p}\(\begin{cases}
\sum _{-(j-i)<r\le (j-i)} ^{}\binom {2m+1}{m+2r}&j\le p\\
2^m&j=p+1
\end{cases}\).
\end{equation*}
To this Pfaffian we may apply Gordon's reduction
\eqref{eq:Gordon-odd} with $g_\al=\binom {2m+1}{m+2\al}$ and $X=2^m$, to get
\begin{equation} \label{eq:Pfaff-star-inf-odd}2^m\det\limits_{1\le i,j\le (p-1)/2}\(\binom {2m+1}{m+2i-2j}-\binom
{2m+1}{m+2i+2j}\)
\end{equation}
for the number of $\infty$-friendly stars under consideration.

Again, to get the determinant into a tractable form, we take some
factors out of it, to obtain
\begin{multline} \label{eq:det-factors2-odd}2^m\det\limits_{1\le i,j\le (p-1)/2}\(\binom {2m+1}{m+2i-2j}-\binom
{2m+1}{m+2i+2j}\)\\=2^m\prod _{i=1} ^{(p-1)/2}\frac {(2m+1)!} {(m+2i+p-1)!\,(m-2i+p)!}\\
\times
\det\limits_{1\le i,j\le (p-1)/2}\big((m+2i-2j+1)_{p+2j-1}\,(m-2i+2j+2)_{p-2j-1}\\
-(m+2i+2j+1)_{p-2j-1}\, (m-2i-2j+2)_{p+2j-1}\big).
\end{multline}As before, it suffices to find the leading term of the determinant,
as a polynomial in $m$. To do so, we consider the more general
determinant\begin{multline} \label{eq:gen-det2-odd}
\det\limits_{1\le i,j\le (p-1)/2}\big((m+X_i-2j+3/2)_{p+2j-1}\,
(m-X_i+2j+3/2)_{p-2j-1}\\
-(m+X_i+2j+3/2)_{p-2j-1}\, (m-X_i-2j+3/2)_{p+2j-1}\big).
\end{multline}
Clearly, we get back the determinant in \eqref{eq:det-factors2} for$X_i=2i-1/2$.

The determinant in \eqref{eq:gen-det2-odd} is a polynomial in $m$ and the$X_i$'s of degree $(p-1)^2$. It is divisible by
$$\prod _{1\le
i<j\le (p-1)/2} ^{}(X_j^2-X_i^2)\prod _{i=1} ^{(p-1)/2}X_i.$$Therefore the degree in $m$
of the determinant in \eqref{eq:det-factors2-odd} is at most
$3p^2/4-3p/2+3/4$.

The leading term of \eqref{eq:gen-det2-odd} is
\begin{align*}
&\det\limits_{1\le i,j\le (p-1)/2}\big((m+X_i)^{p+2j-1}\,
(m-X_i)^{p-2j-1}-(m+X_i)^{p-2j-1}\, (m-X_i)^{p+2j-1}\big)\\
&\kern1pt=\prod _{i=1} ^{(p-1)/2}(m+X_i)^{p-1}(m-X_i)^{p-1}
\det\limits_{1\le i,j\le (p-1)/2}\(\(\frac {m+X_i} {m-X_i}\)^{2j}+
\(\frac {m+X_i} {m-X_i}\)^{-2j}\)\\
&\kern1pt=\prod _{i=1} ^{(p-1)/2}(m-X_i)^{2p-2}
\prod _{i=1}^{(p-1)/2}\(\(\frac {m+X_i} {m-X_i}\)^4-1\)
\\&\kern1pt\kern1cm
\times
\prod _{1\le i<j\le (p-1)/2} ^{}\(\(\frac {m+X_i} {m-X_i}\)^2
-\(\frac {m+X_j} {m-X_j}\)^2\)
\(1-\(\frac {(m+X_i)} {(m-X_i)}\frac {(m+X_j)} {(m-X_j)}\)^2\)\\
&\kern1pt=\prod _{1\le i<j\le (p-1)/2} ^{}\big((m+X_j)^2(m-X_i)^2-
(m+X_i)^2(m-X_j)^2\big)\\
&\kern1pt\kern3cm\cdot
\big((m+X_j)^2(m+X_i)^2-(m-X_i)^2(m-X_j)^2\big)\\&\kern1pt\kern3cm
\times
\prod _{i=1} ^{(p-1)/2}\big((m+X_i)^2-(m-X_i)^2\big)\big((m+X_i)^2+(m-X_i)^2\big)\\
&\kern1pt=\prod _{1\le i<j\le (p-1)/2} ^{}\kern-14pt
2m(X_j-X_i)(2m^2+\cdots)2m(X_j+X_i)(2m^2+\cdots)
\prod _{i=1} ^{(p-1)/2}\kern-4pt 4mX_i(2m^2+\cdots).
\end{align*}Here we used \eqref{eq:sympl} to evaluate the determinant in the
second line.

The leading term of the determinant in \eqref{eq:det-factors2-odd} is this
expression under the substitution of $X_i=2i-1/2$, $i=1,2,\dots,p$, which equals
\begin{multline*}
m^{6\binom {(p-1)/2}2+3(p-1)/2}2^{4\binom {(p-1)/2}2+3(p-1)/2}\prod _{1\le i<j\le (p-1)/2} ^{}(2j-2i)(2j+2i-1)\prod _{i=1} ^{(p-1)/2}(2i-1/2)\\
=m^{3p^2/4-3p/2+3/4}2^{p^2/2-p+3/2}
\prod _{\ell=1} ^{(p-1)/2}\frac {(\ell-1)!\,(4\ell-1)!} {(\ell+p/2-5/2)!}.
\end{multline*}
Hence, the asymptotics of the
number of $\infty$-friendly stars with $p$ branches of length $m$
is this expression multiplied by the asymptotics of the
product on the right-hand side of \eqref{eq:det-factors2-odd}, the latter
being easily found by Stirling's formula. After some simplification,
the second line of \eqref{eq:star-inf-without} (the
case of odd $p$) is found.

This completes the proof of the theorem.
\end{proof}
Since the expressions \eqref{eq:star-without} and\eqref{eq:star-inf-without} are identical except for a
multiplicative constant, we obtain as a consequence
the following result for $n$-friendly models.

\begin{Corollary} \label{cor:n-stand-star}
As $m$ tends to infinity, $n$-friendly stars with $p$ branches of
length $m$ have, up to a multiplicative
constant, the same asymptotic behaviour. More precisely, the number of$n$-friendly stars with $p$ branches of length $m$ is $\asymp
2^{mp}m^{-p^2/4+p/4}$, i.e.,there are positive constants $d_1$ and $d_2$
such that for large enough $m$ this number is between
$d_12^{mp}m^{-p^2/4+p/4}$ and $d_22^{mp}m^{-p^2/4+p/4}$.Under the assumption that there is a constant $d(n)$ such that this
number is asymptotically exactly equal to $d(n)2^{mp}m^{-p^2/4+p/4}$,
then we must have $d(0)<d(1)<d(2)<\cdots$, i.e., for any $n$ there are,asymptotically, strictly less $n$-friendly stars with $p$ branches of
length $m$ than $(n+1)$-friendly stars with $p$ branches of
length $m$.\end{Corollary}
This follows without difficulty from Theorems~\ref{thm:star-without}
and \ref{thm:star-inf-without}, in the same way as
Corollary~\ref{cor:n-stand-watermelon} followed from Theorems~\ref{thm:watermelon-without} and \ref{thm:watermelon-inf-without}.

\section{Enumeration of stars with fixed end points, with
wall restriction}
The previous sections rederive known results, albeit by new
techniques. We now show how these techniques may be used to
derive new results, for the case of vicious walkers in the
presence of a wall.

\begin{Theorem} \label{T3}
Let $e_1<e_2<\dots<e_p$ with $e_i\equiv m\pmod2$, $i=1,2,\dots,p$.
The number of stars with $p$ branches, the $i$-th branch running from
$A_i=(0,2i-2)$ to $E_i=(m,e_i)$, $i=1,2,\dots,p$, and never going
below the $x$-axis, equals
\begin{equation} \label{e20}
2^{-p^2+p}\prod _{i=1} ^{p}\frac {(e_i+1)\,(m+2i-2)!}
{\(\frac {m+e_i} {2}+p\)!\,\(\frac {m-e_i} {2}+p-1\)!}
\prod _{1\le i<j\le p}
^{}(e_j-e_i)(e_i+e_j+2).
\end{equation}
\end{Theorem}
\begin{pf}
(1) {\it By first principles}. As in \cite{GOV},
we could directly use the main theorem of
non-intersecting lattice paths
\cite[Cor.~2]{GeViAB} (see \cite[Theorem~1.2]{StemAE}, to write the
number of stars in question in the form
\begin{equation} \label{e21}
\det_{1\le i,j\le p}\big(\v{\Pp{A_j\to E_i}}\big),
\end{equation}
where $\Pp{A\to E}$ denotes the set of all lattice paths from $A$ to
$E$ that do not go below the $x$-axis. By the ``reflection principle"
(see e.g\@. \cite[p.~22]{ComtAA}), each path number $\v{\Pp{A_j\to E_i}}$
could then be written as a difference of two binomials. It was shown
in \cite[Proof of Theorem~7]{KratAP}, how to evaluate the resulting
determinant (actually, a $q$-analogue was evaluated there). The
evaluation relies on the determinant lemma \cite[Lemma~34]{KratAP}.

However, since we are only interested in plain enumeration there is a
simpler way. As a first step, we may freely attach $2i-2$ up-steps at the
beginning of the $i$-th branch, $i=1,2,\dots,p$, see Figure~\ref{F2}.
It is obvious that the number of stars with starting points
$A'_i=(-2i+2,0)$ (instead of $A_i=(0,2i-2)$) and end points as in the
statement of the Theorem, each branch not going below the $x$-axis (see
Figure~\ref{F2}.b),
is exactly the same as the number of stars
in the statement of the Theorem (see Figure~\ref{F2}.a).

\begin{figure}[t]
\begin{Figure}\label{F2}
\begin{picture}(0,5)(0,0)
\put(2,-3){\small a. A star}
\put(13,-3){\small b. Star with attached ``up-pieces"}
\end{picture}
        \Gitter(8,14)(0,-1)
        \Koordinatenachsen(8,14)(0,-1)
        \Pfad(0,0),3343443\endPfad
        \Pfad(0,2),3334434\endPfad
        \Pfad(0,4),3333434\endPfad
        \Pfad(0,6),3333334\endPfad
        \DickPunkt(0,0)
        \DickPunkt(0,2)
        \DickPunkt(0,4)
        \DickPunkt(0,6)
        \DickPunkt(7,1)
        \DickPunkt(7,3)
        \DickPunkt(7,7)
        \DickPunkt(7,11)
        \Label\l{A_1}(0,0)
        \Label\l{A_2}(0,2)
        \Label\l{A_3}(0,4)
        \Label\l{A_4}(0,6)
        \Label\r{E_1}(7,1)
        \Label\r{E_2}(7,3)
        \Label\r{E_3}(7,7)
        \Label\r{E_4}(7,11)
\hbox{\hskip10cm}
        \Gitter(8,14)(-6,-1)
        \Koordinatenachsen(8,14)(-6,-1)
        \Pfad(0,0),3343443\endPfad
        \Pfad(-2,0),333334434\endPfad
        \Pfad(-4,0),33333333434\endPfad
        \Pfad(-6,0),3333333333334\endPfad
        \DickPunkt(0,0)
        \DickPunkt(-2,0)
        \DickPunkt(-4,0)
        \DickPunkt(-6,0)
        \DickPunkt(7,1)
        \DickPunkt(7,3)
        \DickPunkt(7,7)
        \DickPunkt(7,11)
        \Label\u{A'_1}(0,0)
        \Label\u{A'_2}(-2,0)
        \Label\u{A'_3}(-4,0)
        \Label\u{A'_4}(-6,0)
        \Label\r{E_1}(7,1)
        \Label\r{E_2}(7,3)
        \Label\r{E_3}(7,7)
        \Label\r{E_4}(7,11)
\hskip4cm
$$
\vskip.3cm
$$
\end{Figure}
\end{figure}

If we apply the main theorem of
non-intersecting lattice paths now, then we again obtain a determinant
for the number in question, namely the determinant \eqref{e21} with
$A_j$ replaced by $A_j'$,
\begin{equation} \label{e22}
\det_{1\le i,j\le p}\big(\v{\Pp{A_j'\to E_i}}\big),
\end{equation}
$\Pp{A\to E}$ denoting the set of all lattice paths from $A$ to
$E$ that do not go below the $x$-axis.

Again, by reflection principle, the path number $\v{\Pp{A_j'\to E_i}}$
can be easily computed, so that the determinant \eqref{e22} equals
\begin{equation} \notag
\det_{1\le i,j\le p}\(\frac {e_i+1} {m+2j-1}\binom {m+2j-1}{\frac
{m-e_i} {2}+j-1}\).
\end{equation}
Now we remove as many factors from the determinant as possible. In
that way we obtain
\begin{multline} \label{e23}
(-1)^{\binom p2}
\prod _{i=1} ^{p}\frac {(e_i+1)\,(m+2i-2)!}
{\(\frac {m+e_i} {2}+p\)!\,\(\frac {m-e_i} {2}+p-1\)!}\\
\times
\det_{1\le i,j\le p}\Big(\big(\frac {-m+e_i} {2}-p+1\big)\cdots
\big(\frac {-m+e_i} {2}-j\big)\big(\frac {m+e_i} {2}+p\big)\cdots
\big(\frac {m+e_i} {2}+j-1\big)\Big).\\
\end{multline}
The determinant in \eqref{e23} can clearly be rewritten as
\begin{multline} \notag
\det_{1\le i,j\le p}\Big(\big(\frac {e_i+1} {2}-\frac {m-1} {2}-p\big)\cdots
\big(\frac {e_i+1} {2}-\frac {m-1} {2}-j-1\big)\\
\cdot\big(\frac {e_i+1} {2}
+\frac {m-1} {2}+p\big)\cdots
\big(\frac {e_i+1} {2}-\frac {m-1} {2}+j+1\big)\Big)\\
=\det_{1\le i,j\le p}\(\Big(\Big(\frac {e_i+1} {2}\Big)^2-
\Big(\frac {m-1} {2}-p\Big)^2\Big)\cdots
\Big(\Big(\frac {e_i+1} {2}\Big)^2-\Big(\frac {m-1}
{2}-j-1\Big)^2\Big)\).
\end{multline}
This determinant can be reduced by elementary row manipulations to
$$\det_{1\le i,j\le p}\(\Big(\frac {e_i+1} {2}\Big)^{2(p-j)}\),
$$
which is apparently a Vandermonde determinant and therefore equals
\begin{equation} \notag
\prod _{1\le i<j\le p} ^{}\(\Big(\frac {e_i+1} {2}\Big)^2-\Big(\frac {e_j+1}
{2}\Big)^2\)=2^{-p^2+p}\prod _{1\le i<j\le p}
^{}(e_i-e_j)(e_i+e_j+2).
\end{equation}
Substituting this in \eqref{e23} gives \eqref{e20}.

\medskip
(2) {\it Using knowledge about symplectic characters}. The {\em (even) symplectic
character} $\sp_\la(x_1^{\pm1},x_2^{\pm1},\dots,x_n^{\pm1})$ is
defined by (see \cite[(24.18)]{FuHaAA})
\begin{equation} \label{e24}
\sp_\la(x_1^{\pm1},x_2^{\pm1},\dots,x_n^{\pm1})
=\frac {\det\limits_{1\le i,j\le
n}(x_j^{\la_i+n-i+1}-x_j^{-(\la_i+n-i+1)})}
{\det\limits_{1\le i,j\le
n}(x_j^{n-i+1}-x_j^{-(n-i+1)})}.
\end{equation}
Proctor \cite{ProcAF} defined also {\em odd\/} symplectic characters
$\sp_\la(x_1^{\pm1},x_2^{\pm1},\dots,x_n^{\pm1},1)$,
which are for example defined by
\begin{multline} \label{e24o}
\sp_\la(x_1^{\pm1},x_2^{\pm1},\dots,x_n^{\pm1},1)\\
=\frac {1} {2}\det_{1\le i,j\le n}\big(h_{\la_i-i+j}
(x_1^{\pm1},x_2^{\pm1},\dots,x_n^{\pm1},1)+h_{\la_i-i-j+2}
(x_1^{\pm1},x_2^{\pm1},\dots,x_n^{\pm1},1)\big),
\end{multline}
where $h_k(z_1,z_2,\dots,z_r)=\sum _{1\le i_1\le \dots\le i_m\le r}
^{}z_{i_1}\cdots z_{i_k}$ denotes the {\em $k$-th complete homogeneous
symmetric function\/}.


It is well-known that a combinatorial description of symplectic
characters is given in terms of {\em symplectic tableaux}.
Let $\lambda$ be a partition. A symplectic tableau of shape $\la$ is
a semistandard tableau of shape $\la$ with the additional property
that
\begin{equation} \label{e25}
\text {entries in row $i$ are at least $2i-1$.}
\end{equation}
It is obvious that because of weak increase along rows this condition
may be restricted to the entries in the first column.

Let $n$ be fixed, and let $T$ be a symplectic tableau with entries at
most $2n+1$.
The weight $\mathbf x^T$
of the symplectic tableau $T$ is defined by
\begin{equation}\label{e26}
\mathbf{x}^T = \prod_{l=1}^{n}x_l^{\v{\{T_{i,j} = 2l-1\}} -
        \v{\{T_{i,j} = 2l\}}}.
\end{equation}
where $T_{ij}$ denotes the entry in cell $(i,j)$ of $T$. Note that
entries $2n+1$ do not contribute to the weight.
Given this terminology, the (even) symplectic character
$\sp_\la(x_1^{\pm1},x_2^{\pm1},\dots,x_n^{\pm1})$
is also given by (see
\cite[Theorem~2.3]{SunaAE}, \cite[Theorem~4.2]{ProcAK}),
\begin{equation}\label{e27}
 \sp_\lambda(x_1^{\pm1},x_2^{\pm1},\dots,x_n^{\pm1}) = \sum_{T}\mathbf{x}^{T},
\end{equation}
where the sum is over all symplectic
tableaux $T$ of shape $\lambda$ with entries $\le 2n$, whereas the
odd symplectic character
$\sp_\la(x_1^{\pm1},x_2^{\pm1},\dots,x_n^{\pm1},1)$
is also given by (see
\cite[Theorem~4.2]{ProcAK}),
\begin{equation}\label{e27o}
 \sp_\lambda(x_1^{\pm1},x_2^{\pm1},\dots,x_n^{\pm1},1) = \sum_{T}\mathbf{x}^{T},
\end{equation}
where the sum is over all symplectic
tableaux $T$ of shape $\lambda$ with entries $\le 2n+1$.

The formula for symplectic characters needed here is (see
\cite[(3.27)]{ESKiAA},
\cite[Theorem~4.5.(1)]{SunaAE} and \cite[Prop.~3.2]{ProcAF})
\begin{multline} \label{e28}
\sp_\la(\underbrace{1,1,\dots,1}_m)=\prod _{1\le i<j\le m}
^{}\frac {\la_i-i-\la_j+j}
{j-i}\prod _{1\le i\le j\le m}\frac {\la_i+\la_j+m-i-j+2} {m+2-i-j}\\
=\prod _{\rh\in\la} ^{}\frac {m-f_\rh} {h_\rh},
\end{multline}
where $m$ may be even or odd,
where, again, $h_\rh$ is the hook length of cell $\rh$, and $f_\rh$
is given by
$$f_\rh=f_{(i,j)}=\begin{cases} \la_i'+\la_j'-i-j&i\le j\\
i+j-\la_i-\la_j-2&i>j.\end{cases}$$
Here again, $\la'$ denotes the partition conjugate to $\la$.

\begin{figure}[t]
 \begin{Figure}\label{F3}
\begin{picture}(0,5)(0,0)
\put(2,-3){\small a. A star with wall}
\put(13,0){\small b. A symplectic tableau}
\end{picture}
        \Gitter(8,14)(0,-1)
        \Koordinatenachsen(8,14)(0,-1)
        \Pfad(0,0),3343443\endPfad
        \Pfad(0,2),3334434\endPfad
        \Pfad(0,4),3333434\endPfad
        \Pfad(0,6),3333334\endPfad
        \DickPunkt(0,0)
        \DickPunkt(0,2)
        \DickPunkt(0,4)
        \DickPunkt(0,6)
        \DickPunkt(7,1)
        \DickPunkt(7,3)
        \DickPunkt(7,7)
        \DickPunkt(7,11)
        \Label\r{\mathit 2}(2,2)
        \Label\r{\mathit 4}(4,2)
        \Label\r{\mathit 5}(5,1)
        \Label\r{\mathit 3}(3,5)
        \Label\r{\mathit 4}(4,4)
        \Label\r{\mathit 6}(6,4)
        \Label\r{\mathit 4}(4,8)
        \Label\r{\mathit 6}(6,8)
        \Label\r{\mathit 6}(6,12)
        \Label\l{A_1}(0,0)
        \Label\l{A_2}(0,2)
        \Label\l{A_3}(0,4)
        \Label\l{A_4}(0,6)
        \Label\r{E_1}(7,1)
        \Label\r{E_2}(7,3)
        \Label\r{E_3}(7,7)
        \Label\r{E_4}(7,11)
\begin{picture}(16,5)(0,0)
\put(15,4){
        \begintableau{3}{4}
        \row{2346}
        \row{446}
        \row{56}
        \put(-1.8,2.3){$1\le$}
        \put(-1.8,1.3){$3\le$}
        \put(-1.8,0.3){$5\le$}
        \endtableau
}
\end{picture}
$$

\vskip.3cm
$$
\end{Figure}
\end{figure}

Now we use a slight variant of the correspondence described in the proof of
Theorem~\ref{T1}. Given a star, we label down-steps by the
$x$-coordinate of their
{\em starting} point, i.e., a step from $(a,b)$ to $(a+1,b-1)$ is labelled by
$a$, see Figure~\ref{F3}.a. Then, again, out of the labels of the $j$-th
branch, we form the $j$-th column of the corresponding tableau. It is
evident that the condition that the branches do not go below the
$x$-axis, under this correspondence translates exactly into
condition \eqref{e25}. Therefore, in that manner we obtain a
bijection between stars with $p$ branches, the $i$-th branch running from
$A_i=(0,2i-2)$ to $E_i=(m,e_i)$, $i=1,2,\dots,p$, and never going
below the $x$-axis, with symplectic tableaux with entries at most
$m-1$ and with column lengths $\frac {m-e_1}2,\frac {m-e_2}2\dots,\frac
{m-e_p}2$. So, formula
\eqref{e28} solves this enumeration problem and gives \eqref{e20}
upon some manipulation.
\end{pf}

\section{Enumeration of stars with arbitrary end points, with
wall restriction}

\begin{Theorem} \label{T4}
The number of stars of length $m$ with $p$ branches which do not go
below the $x$-axis, and whose end points have $y$-coordinates at
least $s$, $s\equiv m\pmod2$, equals
\begin{equation} \label{e30}
\prod _{i=1} ^{m-s+1}\prod _{j=1} ^{p}\prod _{k=1} ^{s}\frac {i+j+k-1} {i+j+k-2}.
\end{equation}
\end{Theorem}

\begin{pf} Using the correspondence between stars adjacent to a wall and
symplectic tableaux
described in the second proof of Theorem~\ref{T3}, we see that we
want to count symplectic tableaux with entries at most $m-1$ having at most
$(m-s)/2$ rows and at most $p$ columns. This problem was encountered before
by Proctor \cite{ProcAE}. (He is actually interested in enumerating
plane partitions of trapezoidal shape. However, he demonstrates in
\cite{ProcAE} that these are in bijection with the symplectic
tableaux we are considering here.) The solution of the problem lies
in the following identity which relates symplectic characters and
Schur functions of rectangular shape:
\begin{equation} \label{e31}
s_{(c^r)}(\underbrace{1,1,\dots,1}_{N+1})=
\sum _{\nu\subseteq (c^r)} ^{}\sp_\nu(\underbrace{1,1,\dots,1}_N).
\end{equation}
Here $(c^r)$ is short for $(c,c,\dots,c)$, with $r$ occurrences of
$c$. Recall, that in
the argument of a symplectic character
$\sp_\la(x_1^{\pm1},x_2^{\pm1},\dots)$ the term $x_i^{\pm1}$ denotes
the {\em two} arguments $x_i,x_i^{-1}$.

Actually, an identity for ``universal" characters is true, see
\cite[(3.1) summed over all $p$]{KratBC}. This underlying ``universal"
character identity is proved by a combinatorial rule due to
Littlewood \cite{LittAA}, see \cite[Proof of Lemma~4]{ProcAE} and
\cite[Sec.~4]{KratBC}.

Clearly, the formula \eqref{e7}, used in \eqref{e31} with $c=p$,
$r=(m-s)/2$, $N=m-1$, gives immediately
what we want. With some work the resulting expression can be
transformed into \eqref{e30}.
\end{pf}

Formula \eqref{e31} also implies that there should be a bijection
between symplectic tableaux with entries at most $m$ having at most
$s$ rows and at most $p$ columns, on the one hand, and (ordinary)
tableaux with entries at most $m+1$ having $s$ rows and $p$ columns,
on the other hand. Such a bijection has been constructed by Haiman
\cite[Prop.~8.11]{HaimAD}, it is based on Sch\"utzenberger's
\cite{SchuAA} jeu de taquin. Actually, his bijection is between
tableaux of trapezoidal shape (which however are in bijection with
symplectic tableaux as shown by Proctor) and tableaux of rectangular
shape. This bijection could easily be converted into a bijection
between stars of length $m$ with $p$ branches which do not go
below the $x$-axis, and whose end points have $y$-coordinates at
least $m-s$ and watermelons of length $m$ with $p$ branches of deviation
$s$.

\begin{Theorem} \label{thm:star-with}
The number of stars with $p$ branches of length $m$which do not go below the $x$-axis, and whose end pointshave $y$-coordinates at least $s$, $s\equiv m\pmod2$,
is asymptotically
\begin{equation} \label{eq:star-with}2^{mp+p^2-p/2}m^{-p^2/2}\pi^{-p/2}
\bigg(\prod _{\ell=1} ^{p}(\ell-1)!\bigg)(1+O(1/m))
\end{equation}as $m$ tends to infinity.
\end{Theorem}
\begin{proof}
We know that the number of stars with $p$ branches of length $m$
which do not go below the $x$-axis, and whose end points
have $y$-coordinates at least $s$, $s\equiv m\pmod2$,
is given by the product formula\begin{equation} \label{e30}
\prod _{i=1} ^{(m+s)/2}\prod _{j=1} ^{(m-s)/2}\prod _{\ell=1} ^{p}\frac {i+j+\ell-1} {i+j+\ell-2}=
\prod _{\ell=1} ^{p}\frac {(\ell-1)!\,(m+\ell-1)!} {(\tfrac {m+s}{2}+\ell-1)!\,(\tfrac {m-s} {2}+\ell-1)!}.
\end{equation}
Application of Stirling's formula yields \eqref{eq:star-with} after
a short calculation.
\end{proof}


\section{Enumeration of watermelons without
wall restriction}

As discussed in the introduction, watermelons are a proper subset of
stars, their end points being $(n,k),(n,2+k),(n,4+k),\cdots,(n,k+2p-2).$
The parameter $k$ we refer to as {\em the deviation}.

In \cite{ESG} it was shown that the number of watermelons of length $m$ with
$p$ branches and deviation $k$ 
(where $k\equiv m\pmod 2$) is given by
\begin{equation} \label{eq:num-watermelon}
\prod _{\ell=1} ^{p}\prod _{i=1} ^{(m+k)/2}\prod _{j=1} ^{(m-k)/2}
\frac {i+j+\ell-1} {i+j+\ell-2}=\prod _{\ell=1} ^{p}
\frac {(m+\ell-1)!\,(\ell-1)!} {(\frac {m+k} {2}+\ell-1)!\, (\frac
{m-k} {2}+\ell-1)!}.
\end{equation}

We were not able to give a closed form expression for the total number of
such watermelons, but we can give an asymptotic result.
\begin{Theorem} \label{thm:watermelon-without}
The number of watermelons with $p$ branches of length $m$ (and
arbitrary deviation)
is asymptotically
\begin{equation} \label{eq:watermelon-without}
2^{mp+p^2-p/2-1/2}m^{-p^2/2+1/2}\pi^{-p/2+1/2}p^{-1/2}
\bigg(\prod _{\ell=1} ^{p}(\ell-1)!\bigg)(1+O(m^{-1/2}\log^3m))
\end{equation}
as $m$ tends to infinity.
\end{Theorem}
\begin{proof}
We know that the number of watermelons with $p$ branches of length $m$
and with deviation $k$ (where $k\equiv m\pmod 2$) is given by
\begin{equation} \label{eq:num-watermelon}
\prod _{\ell=1} ^{p}\prod _{i=1} ^{(m+k)/2}\prod _{j=1} ^{(m-k)/2}
\frac {i+j+\ell-1} {i+j+\ell-2}=\prod _{\ell=1} ^{p}
\frac {(m+\ell-1)!\,(\ell-1)!} {(\frac {m+k} {2}+\ell-1)!\, (\frac
{m-k} {2}+\ell-1)!}.
\end{equation}
Therefore our task is to estimate the sum
\begin{equation} \label{eq:sum-watermelon}
\sum _{k\equiv m \text { (mod } 2)} ^{}\prod _{\ell=1} ^{p}
\frac {(m+\ell-1)!\,(\ell-1)!} {(\frac {m+k} {2}+\ell-1)!\, (\frac
{m-k} {2}+\ell-1)!}.
\end{equation}
We follow the standard way of carrying out such estimations, as
described in \cite[Ex.~5.4]{OdlyAA}.

The dominant terms in the sum on the right-hand side
are those corresponding to $k$'s which
are near $0$. Consequently, we split the sum into three parts,
the terms with ``small" $k$, the terms with ``large" $k$, and the
terms with $k\sim 0$. Let $S(k)$ denote the summand on the
right-hand side of \eqref{eq:sum-watermelon},
$$S(k)=\prod _{\ell=1} ^{p}
\frac {\Gamma(m+\ell)\,\Gamma(\ell)} {\Gamma(\frac {m+k} {2}+\ell)\,
\Gamma(\frac {m-k} {2}+\ell)}.
$$
Then the precise way we split the sum
in \eqref{eq:sum-watermelon} is
\begin{equation} \label{eq:sumsplit}
\sum _{-m\le k\le m} ^{}S(k)=\sum _{-m\le k\le - \sqrt m\log m} S(k)+
\sum _{\sqrt m\log m\le k\le m} S(k)+
\sum _{\v{k}<\sqrt m\log m} ^{}S(k),
\end{equation}
where it is understood without saying that all sums are only over
those $k$ which are of the same parity as $m$.
We will show that the contributions of the first and second term in
\eqref{eq:sumsplit} are negligible, and we will compute the
contribution which the third term provides.

To see that the first term in \eqref{eq:sumsplit} is negligible, it
is enough to observe that every summand $S(k)$ with $-m\le k\le
-\sqrt m\log m$ is bounded above by $S(-\sqrt m\log m)$, and to
compute the asymptotics of $S(-\sqrt m\log m)$ by means of
Stirling's formula,
\begin{align*}
(-\sqrt m&\log m)=(1+O(1/m))\prod _{\ell=1} ^{p}\Gamma(\ell)
\frac {\(\frac {m} {e}\)^{m}m^{\ell-1}\sqrt{2\pi m}}
{\(\frac {m+\sqrt m\log m} {2e}\)^{\frac {1} {2}(m+\sqrt m\log m)}
\(\frac {m+\sqrt m\log m} {2}\)^{\ell-1}}\\
&\kern1cm
\cdot\frac {1}
{\(\frac {m-\sqrt m\log m} {2e}\)^{\frac {1} {2}(m-\sqrt m\log m)}
\(\frac {m-\sqrt m\log m} {2}\)^{\ell-1}
\sqrt{\frac {1} {2}\pi^2(m^2-m\log^2m)}}\\
&=(1+O(\log^2 m/m))\prod _{\ell=1} ^{p}(\ell-1)!
\frac {2^{m+2\ell-1}} {m^{\ell-1}\sqrt{\pi m}}\\
&\kern1cm
\cdot
\frac {1} {\(1+\frac {\log m} {\sqrt m}\)^{\frac {1} {2}(m+\sqrt m\log
m)+\ell-1}
\(1-\frac {\log m} {\sqrt m}\)^{\frac {1} {2}(m-\sqrt m\log m)+\ell-1}}\\
&=(1+O(\log^2 m/m))\prod _{\ell=1} ^{p}(\ell-1)!
\frac {2^{m+2\ell-1}} {m^{\ell-1}\sqrt{\pi m}}\\
&\kern1cm
\cdot
\frac {1} {\exp\(\(\frac {\log m} {\sqrt m}-\frac {\log^2m} {2m}+O\(\frac
{\log^3 m} {m\sqrt m}\)\)(\frac {1} {2}(m+\sqrt m\log m)+\ell-1)\)}\\
&\kern1cm
\cdot
\frac {1} {\exp\(\(-\frac {\log m} {\sqrt m}-\frac {\log^2m} {2m}+O\(\frac
{\log^3 m} {m\sqrt m}\)\)(\frac {1} {2}(m-\sqrt m\log m)+\ell-1)\)}\\
&=(1+O(\log^3 m/\sqrt m))\prod _{\ell=1} ^{p}(\ell-1)!
\frac {2^{m+2\ell-1}} {m^{\ell-1}\sqrt{\pi m}}e^{-\frac {1} {2}\log^2m}\\
&=(1+O(\log^3 m/\sqrt m))\frac {2^{pm+p^2}}
{m^{p^2/2}\pi^{p/2} }m^{-\frac {p} {2}\log m}
\prod _{\ell=1} ^{p}(\ell-1)!\\
&=O(1/m^3) ,
\end{align*}
because the term $m^{\frac {1} {2}p\log m}$ which appears in the denominator
of the expression in the next-to-last line grows super exponentially.
Therefore, the first term in \eqref{eq:sumsplit} is
$$O\big(\tfrac {1} {2}(m-\sqrt m-\log m)/m^3\big)=O(1/m).$$
The second term in \eqref{eq:sumsplit} is equal to the first term,
hence has the same order of magnitude.

\smallskip
Now we turn to the third term in \eqref{eq:sumsplit}. To carry out
our computations, we would have to distinguish between two cases, depending
on whether $m$ is even or odd. The computations in both cases are
however rather parallel. Therefore we carry them out in detail just
for the case that $m$ is even and leave it to the reader to complete
the computations for the other case.

In the case that $m$ is even, the third term in
\eqref{eq:sumsplit}, after replacement of $k$ by $2k$, becomes
\begin{equation} \label{eq:term3}
\sum _{\v k\le \frac {1} {2}\sqrt m\log m} ^{}\prod _{\ell=1} ^{p}
\frac {(m+\ell-1)!\,(\ell-1)!} {(m/2+k+\ell-1)!\, (m/2-k+\ell-1)!}.
\end{equation}
For nonnegative $k$, we may rewrite the summand as
\begin{align} \notag
\prod _{\ell=1} ^{p}&
\frac {(m+\ell-1)!\,(\ell-1)!} {(\frac {m} {2}+\ell-1)!^2}
\cdot \frac {(\frac {m} {2}-k+\ell)(\frac {m} {2}-k+\ell+1)\cdot
(\frac {m} {2}+\ell-1)}
{(\frac {m} {2}+\ell)(\frac {m} {2}+\ell+1)\cdot
(\frac {m} {2}+k+\ell-1)}\\
\notag
&=\prod _{\ell=1} ^{p}
\frac {(m+\ell-1)!\,(\ell-1)!} {(\frac {m} {2}+\ell-1)!^2}
\cdot \frac {(1+\frac {-k+\ell} {m/2})(1+\frac {-k+\ell+1} {m/2})\cdot
(1+\frac {\ell-1} {m/2})}
{(1+\frac {\ell} {m/2})(1+\frac {\ell+1} {m/2})\cdot
(1+\frac {k+\ell-1} {m/2})}\\
\label{eq:S(k)-asy}
&=\prod _{\ell=1} ^{p}
\frac {(m+\ell-1)!\,(\ell-1)!} {(\frac {m} {2}+\ell-1)!^2}
\cdot e^{-\frac {2} {m}k^2+O(k^3/m^2)}.
\end{align}
For nonpositive $k$ there is a similar computation which leads to the
same result.

We have to sum the expression \eqref{eq:S(k)-asy}
over $k$ between $-\frac {1} {2}\sqrt m\log m$ and
$\frac {1} {2}\sqrt m\log m$. In that range, the $O(.)$ term is at worst
$O(m^{-1/2}\log^3m)$. Thus, the sum \eqref{eq:term3} turns into
\begin{equation} \label{eq:term3a}
\bigg(\sum _{\v k\le \frac {1} {2}\sqrt m\log m} ^{}
e^{-\frac {2p} {m}k^2}\bigg)
(1+O(m^{-1/2}\log^3m))
\prod _{\ell=1} ^{p}
\frac {(m+\ell-1)!\,(\ell-1)!} {(\frac {m} {2}+\ell-1)!^2}.
\end{equation}
If we extend the sum to run over all integers $k$ then we make an
error which is bounded by $O(1/m)$. The complete sum
$\sum _{k=-\infty} ^{\infty}
e^{-\frac {2p} {m}k^2}$ can be approximated by
Lemma~\ref{lem:expsum-asy} with $N=0$ and $\al=-2p/m$.
The asymptotics of the product
in \eqref{eq:term3a} is easily determined by using Stirling's
formula. Thus we obtain \eqref{eq:watermelon-without}.
\end{proof}

\begin{Theorem} \label{thm:watermelon-inf-without}
The number of $\infty$-friendly watermelons with $p$ branches of length $m$ (and
arbitrary deviation)
is asymptotically
\begin{equation} \label{eq:watermelon-inf-without}
2^{mp+2p^2-3p/2-1/2}m^{-p^2/2+1/2}\pi^{-p/2+1/2}p^{-1/2}
\bigg(\prod _{\ell=1} ^{p}(\ell-1)!\bigg)(1+O(m^{-1/2}\log^3m))
\end{equation}
as $m$ tends to infinity.
\end{Theorem}
\begin{proof}Consider an $\infty$-friendly watermelon with $p$ branches
of length $m$ and with deviation $k$. It consists of a family $(P_1,P_2,\dots,P_p)$ of
non-crossing lattice paths, $P_i$ running from $A_i=(0,2(i-1))$ to
$E_i=(m,k+2(i-1))$, $i=1,2,\dots,p$. Figure~\ref{Fig1}.a displays an
example for $p=4$, $m=6$, and $k=0$.

\begin{figure}[t]
$$
        \Gitter(7,13)(0,-1)
        \Koordinatenachsen(7,13)(0,-1)
\raise-2.25pt\hbox{$\Pfad(0,0),343443\endPfad$}
\raise-.75pt\hbox{$\Pfad(0,2),433443\endPfad$}
\raise.75pt\hbox{$\Pfad(0,4),344433\endPfad$}
\raise2.25pt\hbox{$\Pfad(0,6),443334\endPfad$}
        \DickPunkt(0,0)
        \DickPunkt(0,2)
        \DickPunkt(0,4)
        \DickPunkt(0,6)
        \DickPunkt(6,0)
        \DickPunkt(6,2)
        \DickPunkt(6,4)
        \DickPunkt(6,6)
        \Label\l{A_1}(0,0)
        \Label\l{A_2}(0,2)
        \Label\l{A_3}(0,4)
        \Label\l{A_4}(0,6)
        \Label\r{E_1}(6,0)
        \Label\r{E_2}(6,2)
        \Label\r{E_3}(6,4)
        \Label\r{E_4}(6,6)
\hbox{\hskip7cm}
        \Gitter(7,13)(0,-1)
        \Koordinatenachsen(7,13)(0,-1)
        \Pfad(0,0),343443\endPfad
        \Pfad(0,4),433443\endPfad
        \Pfad(0,8),344433\endPfad
        \Pfad(0,12),443334\endPfad
        \DickPunkt(0,0)
        \DickPunkt(0,4)
        \DickPunkt(0,8)
        \DickPunkt(0,12)
        \DickPunkt(6,0)
        \DickPunkt(6,4)
        \DickPunkt(6,8)
        \DickPunkt(6,12)
        \Label\l{\tilde A_1}(0,0)
        \Label\l{\tilde A_2}(0,4)
        \Label\l{\tilde A_3}(0,8)
        \Label\l{\tilde A_4}(0,12)
        \Label\r{\tilde E_1}(6,0)
        \Label\r{\tilde E_2}(6,4)
        \Label\r{\tilde E_3}(6,8)
        \Label\r{\tilde E_4}(6,12)
\hskip6cm$$
\centerline{\small a. An $\infty$-friendly watermelon
\hskip1cm
b. A corresponding set of non-intersecting lattice paths}
\vskip.3cm
\caption{}
\label{Fig1}
\end{figure}
Shifting the $i$-th path, $P_i$, by $2(i-1)$ units up,
we obtain a set $(\tilde P_1,\tilde P_2,\dots,\tilde P_p)$ of
{\em non-intersecting} paths, $\tilde P_i$ running from $(0,4(i-1))$
to $(m,k+4(i-1))$, $i=1,2,\dots,p$. Clearly, this correspondence is a
bijection. The number of such sets of non-intersecting lattice paths
is given by the corresponding Gessel-Viennot determinant,
\begin{equation} \label{eq:GV-det}
\det\limits_{1\le i,j\le p}\(\binom m{\frac {m+k} {2}+2i-2j}\).
\end{equation}

Consequently,
our task is to sum \eqref{eq:GV-det} over all $k\equiv m\pmod 2$,
$-m\le k\le m$, and approximate
the sum as $m$ tends to infinity.
The procedure is quite similar to the proof of
Theorem~\ref{thm:watermelon-without}, with the complication that
\eqref{eq:GV-det} cannot be written in closed form.

To begin with, we bring the determinant \eqref{eq:GV-det} into a more
convenient form, by taking out some factors,
\begin{multline} \label{eq:det-factors}
\det\limits_{1\le i,j\le p}\(\binom m{\frac {m+k} {2}+2i-2j}\)=
\prod _{\ell=1} ^{p}\frac {m!} {(\frac {m+k} {2}+2\ell-2)!\,(\frac
{m-k} {2}-2\ell+2p)!}\\
\times\det\limits_{1\le i,j\le p}\big((\tfrac {m+k} {2}+2i-2j+1)_{2j-2}\, (\tfrac
{m-k} {2}-2i+2j+1)_{2p-2j}\big),
\end{multline}
where $(a)_k$ denotes the standard shifted factorial,
$(a)_k:=a(a+1)\cdots(a+k-1)$, $k\ge1$, $(a)_0:=1$.

The determinant in \eqref{eq:det-factors} is a polynomial in $m$ and
$k$, of degree at most $4\binom p2$,
\begin{equation} \label{eq:det-expansion}
\det\limits_{1\le i,j\le p}\big((\tfrac {m+k} {2}+2i-2j+1)_{2j-2}\, (\tfrac
{m-k} {2}-2i+2j+1)_{2p-2j}\big)
=\sum _{0\le s+t\le 4\binom p2}
^{}A(s,t)m^sk^t
\end{equation}
say. We claim that, for our asymptotic
considerations, it is sufficient to just consider the leading terms
on the right-hand side of \eqref{eq:det-expansion}. For, consider a
single term $A(s,t)m^sk^t$ on the right-hand side of
\eqref{eq:det-expansion}. It has to be multiplied by the product on
the right-hand side of \eqref{eq:det-factors}, and the resulting
expression is summed over all $k\equiv m\pmod2$, $-m\le k\le m$, so
that one obtains
\begin{equation*}
\sum _{k\equiv m\text { (mod }2)} ^{}A(s,t)m^sk^t
\prod _{\ell=1} ^{p}\frac {m!} {(\frac {m+k} {2}+2\ell-2)!\,(\frac
{m-k} {2}-2\ell+2p)}.
\end{equation*}
This is now handled in the same way as the sum
\eqref{eq:sum-watermelon}. In essence, it is the expression (compare
\eqref{eq:term3a})
\begin{equation*}
\sum _{k} ^{}A(s,t)m^s(2k)^te^{-\frac {2p} {m}k^2}
\prod _{\ell=1} ^{p}\frac {m!} {(\frac {m} {2}+2\ell-2)!\,(\frac
{m} {2}-2\ell+2p)}
\end{equation*}
that needs to be estimated. By \eqref{eq:expsum-asy} with $N=t$ and
$\al=2p/m$, this
is
\begin{multline} \label{eq:single-term}
A(s,t)m^{s+ {t}/ {2}+ {1}/ {2}}
2^{t/2-3/2}p^{-t/2-1/2}\Gamma(\tfrac {t} {2}+\tfrac {1} {2})
(1+O(m^{-1/2}\log^3m))\\
\times
\prod _{\ell=1} ^{p}\frac {m!} {(\frac {m} {2}+2\ell-2)!\,(\frac
{m} {2}-2\ell+2p)}.
\end{multline}
Obviously, the larger $s$ and $t$ are, the larger will be the
contribution of the corresponding term, the largest coming from those
for which $s+t/2$ is maximal.

As it turns out,
the actual degree in $m$ and $k$ of the leading terms of the
determinant in \eqref{eq:det-expansion}
is significantly smaller than $4\binom p2$.
To see what it is, and what the
leading terms are, we consider a more general determinant,
\begin{equation} \label{eq:gen-det1}
\det\limits_{1\le i,j\le p}\big((\tfrac {m} {2}+X_i-2j+1)_{2j-2}\,
(\tfrac {m} {2}-X_i+2j+1)_{2p-2j}\big).
\end{equation}
Clearly, we get back the determinant in \eqref{eq:det-expansion} for
$X_i=2i+k/2$.

The determinant in \eqref{eq:gen-det1} is a polynomial in $m$ and the
$X_i$'s of degree $4\binom p2$. It is divisible by $\prod _{1\le
i<j\le p} ^{}(X_j-X_i)$. Therefore the degree in $m$ and $k$, after
substitution of $X_i=2i+k/2$, $i=1,2,\dots,p$,
of the determinant in \eqref{eq:det-expansion} is at most $3\binom p2$.

The leading term of \eqref{eq:gen-det1} is
\begin{align*}
\det\limits_{1\le i,j\le p}&\big((\tfrac {m} {2}+X_i)^{2j-2}\,
(\tfrac {m} {2}-X_i)^{2p-2j}\big)=(\tfrac {m} {2}-X_i)^{p(2p-2)}
\det\limits_{1\le i,j\le p}\(\(\frac {\tfrac {m} {2}+X_i}
{\tfrac {m} {2}-X_i}\)^{2j-2}\)\\
&=\prod _{1\le i<j\le p} ^{}\((\tfrac {m} {2}+X_j)^2(\tfrac {m}
{2}-X_i)^2-(\tfrac {m} {2}+X_i)^2(\tfrac {m}
{2}-X_j)^2\)\\
&=\prod _{1\le i<j\le p} ^{}(X_j-X_i)\,m\((\tfrac {m} {2}+X_j)(\tfrac {m}
{2}-X_i)+(\tfrac {m} {2}+X_i)(\tfrac {m}
{2}-X_j)\).
\end{align*}
The leading term of the determinant in \eqref{eq:det-expansion} is this
expression under the
substitution of $X_i=2i+k/2$, $i=1,2,\dots,p$, which equals
$$\prod _{1\le i<j\le p} ^{}(2j-2i)\,m\(\tfrac {m^2} {2}-\tfrac {k^2}
{2}-2ik-2jk-4ij\)=m^{3\binom p2}\prod _{\ell=1}
^{p}(\ell-1)!(1+O(k^2/m^2)).
$$
Hence, in order to compute the asymptotics of the sum of
\eqref{eq:GV-det} over all $k\equiv m\pmod 2$,
it suffices to determine the asymptotics
of the sum over all $k\equiv m\pmod2$ of (recall \eqref{eq:det-factors}
and the remark after \eqref{eq:single-term})
$$
m^{3\binom p2}\prod _{\ell=1} ^{p}\frac {(\ell-1)!\,m!} {(\frac {m+k} {2}+2\ell-2)!\,(
\frac
{m-k} {2}-2\ell+2p)}.
$$
As the considerations leading to \eqref{eq:single-term} showed,
this can be handled completely analogously to the computation
of the asymptotics of the sum \eqref{eq:sum-watermelon}. The result
is exactly \eqref{eq:watermelon-inf-without}.
\end{proof}

Since the expressions \eqref{eq:watermelon-without} and
\eqref{eq:watermelon-inf-without} are identical except for a
multiplicative constant of $2^{p^2-p}$, we obtain as a consequence
the following result for $n$-friendly models.

\begin{Corollary} \label{cor:n-stand-watermelon}
As $m$ tends to infinity, $n$-friendly watermelons with $p$ branches of
length $m$ (and arbitrary deviation) have, up to a multiplicative
constant, the same asymptotic behaviour. More precisely, the number of
$n$-friendly watermelons with $p$ branches of length $m$ is $\asymp
2^{mp}m^{-p^2/2+1/2}$, i.e.,
there are positive constants $c_1$ and $c_2$
such that for large enough $m$ this number is between
$c_12^{mp}m^{-p^2/2+1/2}$ and $c_22^{mp}m^{-p^2/2+1/2}$.
Under the assumption that there is a constant $c(n)$ such that this
number is asymptotically exactly equal to $c(n)2^{mp}m^{-p^2/2+1/2}$,
then we must have $c(0)<c(1)<c(2)<\cdots$, i.e., for any $n$ there are,
asymptotically, strictly less $n$-friendly watermelons with $p$ branches of
length $m$ than $(n+1)$-friendly watermelons with $p$ branches of
length $m$.
\end{Corollary}
\begin{proof}The first assertion follows immediately from
Theorems~\ref{thm:watermelon-without} and
\ref{thm:watermelon-inf-without} since the number of
$n$-friendly watermelons is bounded below by the number of ``genuine"
watermelons, and is bounded above by the number of $\infty$-stand
watermelons. So, explicitly, we may choose
$$c_1=2^{p^2-p/2-1/2}\pi^{-p/2+1/2}p^{-1/2}
\bigg(\prod _{\ell=1} ^{p}(\ell-1)!\bigg)$$
and
$$c_2=2^{2p^2-3p/2-1/2}\pi^{-p/2+1/2}p^{-1/2}
\bigg(\prod _{\ell=1} ^{p}(\ell-1)!\bigg).$$

The second assertion can be proved as follows. Clearly, for any $n$ we have
$c(n)\le c(n+1)$. To see that in fact strict inequality holds, we
identify a set of $(n+1)$-friendly watermelons which are not $n$-stand
watermelons, the cardinality of which is
$\asymp (2^{mp}m^{-p^2/2+1/2})$ (as is the cardinality of $n$-friendly watermelons).
As this set of $(n+1)$-friendly watermelons we may choose families
$(P_1,P_2,\dots,P_p)$ of paths, such that $P_i$ runs from
$(0,2(i-1))$ through $(2\fl{n/2}+4,2(i-1))$ to $(m,k+2(i-1))$,
$i=1,2,\dots,p$, $k$ arbitrary integral, and $P_1$ and $P_2$
touch each other along $n+1$ consecutive
edges. (This is indeed possible. Let $P_1$ start with an up-step and
$P_2$ start with a down step, then let $P_1$ and $P_2$ go up and down
in parallel for $2\fl{n/2}+2$ steps, then let $P_1$ continue with a
down-step, thus reaching $(2\fl{n/2}+4,0)$,
and $P_2$ continue with an up-step, thus reaching $(2\fl{n/2}+4,2)$.
As $2\fl{n/2}+2\ge n+1$, such paths $P_1$ and
$P_2$ do indeed touch each other along $n+1$ consecutive edges.)

If we disregard the portion of the paths between $x=0$ and
$x=2\fl{n/2}+4$, then what remains is a $(n+1)$-friendly watermelon with
$p$ branches of length $m-2\fl{n/2}-4$. The cardinality of these is
at least the cardinality of ``genuine" watermelons with
$p$ branches of length $m-2\fl{n/2}-4$, which,
asymptotically, is given by \eqref{eq:watermelon-without} with $m$
replaced by $m-2\fl{n/2}-4$. This is
$$2^{mp-2\fl{n/2}p+p^2-9p/2-1/2}m^{-p^2/2+1/2}\pi^{-p/2+1/2}p^{-1/2}
\bigg(\prod _{\ell=1} ^{p}(\ell-1)!\bigg)(1+O(m^{-1/2}\log^3m)),
$$
which is $\asymp (2^{mp}m^{-p^2/2+1/2})$,
as desired.
\end{proof}


\section{Enumeration of watermelons with given deviation, with
wall restriction}

The number of
such watermelons follows immediately from Theorem \cite{T3} upon setting
$m = n$ and $e_i = k + 2(i-1).$ After a little simplification we arrive at
\begin{Corollary} \label{C1}
The number of watermelons of length $m$ with $p$ branches which do not go
below the $x$-axis, and with deviation $k \ge 0$
 equals
\begin{equation} \label{e60}
W^{(p)}_{m,k} =\prod _{i=0} ^{\frac{p}{2}-1}\frac{(k+2p-1-2i)!}{(k+2i)!}\prod _{j=0} ^{p-1}
\frac {(m+2j)!j!}{(\frac{m-k}{2}+j)!(\frac{m+k}{2}+j+p)!}
\end{equation}
\end{Corollary}

\section{The enumeration of vicious walkers, between two walls}
The Gessel--Viennot determinant in the case of the presence of two
walls yields the following result.
\begin{Theorem} \label{thm:vicious-with-2}
Let $0\le a_1<a_2<\dots<a_p\le h$, all $a_i$'s of the same parity, and$0\le e_1<e_2<\dots<e_p\le h$, all $e_i$'s of the same parity, such that$a_i+e_i\equiv m\pmod2$, $i=1,2,\dots,p$.The number of vicious walkers with $p$ branches of length $m$, the
$i$-th branch running from $A_i=(0,a_i)$ to $E_i=(m,e_i)$,$i=1,2,\dots,p$,which do not go below the $x$-axis and not above the line $y=h$,
is given by
\begin{equation} \label{eq:vicious-with-2}
\det_{1\le s,t\le p}\(\sum _{k=-\infty} ^{\infty}\(\binom {m} {\frac {m+e_t-a_s}{2}+k(h+2)}-\binom {m} {\frac {m+e_t+a_s} {2}+k(h+2)+1}\)\).\end{equation}
\end{Theorem}
\begin{proof}According to the main theorem of non-intersecting lattice paths
\cite[Cor.~2]{GeViAB} (see \cite[Theorem~1.2]{StemAE}), the number of
vicious walkers in question equals
\begin{equation} \label{e2}
\det_{1\le s,t\le p}\big(\v{\mathcal P^{++}\big(A_t\to E_s\big)}\big),\end{equation}
where $\mathcal P^{++}\big(A\to E\big)$ denotes the set of all lattice paths from $A$ to$E$ which do not go below the $x$-axis and not above the line $y=h$.
There is a well-known formula (see \cite[(1.7)]{MohaAE}) for the latter
number, which is obtained by iterated reflection principle.
Substitution of this formula into \eqref{e2}
gives \eqref{eq:vicious-with-2} immediately.\end{proof}
By bringing the sums in \eqref{eq:vicious-with-2} outside the
determinant (using the multilinearity of the determinant), the number of these vicious walkers can be described as a multiple sum of
determinants. In some cases, such as for certain watermelons and stars, 
the determinants can be evaluated. Thus, a multiple hypergeometric
sum is obtained.

\begin{Theorem} \label{thm:vicious-with-2a}Let $0\le e_1<e_2<\dots<e_p\le h$ with $e_i\equiv m\pmod2$, $i=1,2,\dots,p$.The number of vicious walkers with $p$ branches of length $m$, the
$i$-th branch running from $A_i=(0,2i-2)$ to $E_i=(m,e_i)$,
which do not go below the $x$-axis and not above the line $y=h$,$i=1,2,\dots,p$,is given by\begin{multline} \label{eq:vicious-with-2a}\sum _{k_1,\dots,k_p=-\infty} ^{\infty}
2^{p-p^2}\prod _{s=1} ^{p}\frac{(e_s+2k_s(h+2)+1)\,(m+2s-2)!} {\(\frac {m+e_s}{2}+k_s(h+2)+p\)!\,\(\frac {m-e_s} {2}-k_s(h+2)+p-1\)!}\\
\times
\prod _{1\le s<t\le p} ^{}(e_t-e_s+2(h+2)(k_t-k_s))
(e_t+e_s+2(h+2)(k_t+k_s)+2).\end{multline}\end{Theorem}\begin{proof}As described above the statement of the theorem, we
write the expression \eqref{eq:vicious-with-2}, with $a_i=2i-2$, as a
sum of determinants,\begin{equation} \label{eq:sum-of-dets}\sum _{k_1,\dots,k_p=-\infty} ^{\infty}\det_{1\le s,t\le p}
\(\binom {m} {\frac {m+e_t}
{2}+k_t(h+2)-s+1}-\binom {m} {\frac {m+e_t} {2}+k_t(h+2)+s}\).\end{equation}
Suppose that we disregard the terms $k_t(h+2)$ in the determinant,
\begin{equation} \label{eq:sum-of-dets2}\det_{1\le s,t\le p}
\(\binom {m} {\frac {m+e_t}{2}-s+1}-\binom {m} {\frac {m+e_t} {2}+s}\).\end{equation}
Gessel--Viennot theory (again) says that this determinant counts
vicious walkers with $p$ branches of length $m$, the
$i$-th branch running from $A_i=(0,2i-2)$ to $E_i=(m,e_i)$,
which do not go below the $x$-axis. By Theorem~4?? (I am not sure because I lost the copy
of the draft of our paper that you gave me in a train in Japan) from\cite{GuKVAA}, the number of these vicious walkers is given by
\begin{equation} \label{e20}2^{-p^2+p}\prod _{s=1} ^{p}\frac {(e_s+1)\,(m+2s-2)!}
{\(\frac {m+e_s} {2}+p\)!\,\(\frac {m-e_s} {2}+p-1\)!}
\prod _{1\le s<t\le p}^{}(e_t-e_s)(e_s+e_t+2).
\end{equation}
Hence, the determinant in \eqref{eq:sum-of-dets2} must equal the
expression \eqref{e20}. In fact, as the equality between
\eqref{eq:sum-of-dets2} and \eqref{e20} can be reduced to an equation which is polynomial in $e_1,e_2\dots,e_p$, the equality is true for{\em any} choice of $e_1,e_2,\dots,e_p$. In particular, it remains true if we replace $e_i$ by $e_i+2k_i(h+2)$, $i=1,2,\dots,p$.
However, the
determinant in \eqref{eq:sum-of-dets2} under these replacements becomes the determinant in \eqref{eq:sum-of-dets}. Thus, if we substitute the expression \eqref{e20}, with these replacements, into\eqref{eq:sum-of-dets}, we obtain \eqref{eq:vicious-with-2a}immediately.
\end{proof}For the subsequent asymptotic calculations, however, we need a
different type of expression for the number of vicious walkers under
consideration. It is valid for arbitrary starting and end points of
the vicious walkers. It generalizes a well-known formula for the number of (single) paths between two parallel boundaries (see \cite[(2.42)]{MohaAE}).(There must be a way to go from the formula below to the
formula which expressed the number of vicious walkers in terms of a
Schur function, at least in the case of watermelons, that you and Xavier had derived. This could be tedious, however.)
\begin{Theorem} \label{thm:vicious-with-2b}
Let $0\le a_1<a_2<\dots<a_p\le h$, all $a_i$'s of the same parity, and$0\le e_1<e_2<\dots<e_p\le h$, all $e_i$'s of the same parity, such that
$a_i+e_i\equiv m\pmod2$, $i=1,2,\dots,p$.The number of vicious walkers with $p$ branches of length $m$, the
$i$-th branch running from $A_i=(0,a_i)$ to $E_i=(m,e_i)$,$i=1,2,\dots,p$,which do not go below the $x$-axis and not above the line $y=h$,
is given by
\begin{multline} \label{eq:vicious-with-2b}
\sum _{k_1,\dots,k_p=1} ^{h+1}
 \frac {1} {(h+2)^p} \bigg(2^p \prod _{s=1}^{p}\cos\frac {k_s\pi} {h+2}\bigg)^m\\
\times e^{-\frac {\pi i} {h+2}\(\sum _{t=1} ^{p}k_t(e_t+1)\)}
\det_{1\le s,t\le p}\( e^{\frac {\pi ik_t} {h+2}(a_s+1)}-
e^{-\frac {\pi ik_t} {h+2}(a_s+1)}\).
\end{multline}
\end{Theorem}

\begin{proof}We wish to rewrite \eqref{eq:vicious-with-2}, which
gives an expression for the number of vicious walkers in the
statement of the theorem.Trivially, the binomial coefficient $\binom nk$ is the coefficient of $z^{-1}$ in the Laurent series$$\frac {(1+z)^n} {z^{k+1}}.$$
Thus, the $(s,t)$-entry in the determinant \eqref{eq:vicious-with-2} equals the coefficient of $z^{-1}$ in\begin{align} \notag\sum _{k=0} ^{\infty}\bigg(\frac {(1+z)^m\,z^{k(h+2)}} {z^{\frac {m+e_t-a_s}
{2}+m(h+2)+1}}&-\frac {(1+z)^m\,z^{k(h+2)}} {z^{\frac {m+e_t+a_s} {2}+m(h+2)+2}}\bigg)\\
\notag
&=\frac {(1+z)^m} {z^{\frac {m+e_t-a_s}
{2}+m(h+2)+1}(1-z^{h+2})}-\frac {(1+z)^m} {z^{\frac {m+e_t+a_s} {2}+m(h+2)+2}(1-z^{h+2})}\\&=\frac {(1+z)^m\(z^{(a_s-e_t)/2}-z^{-(a_s+e_t+2)/2}\)} {z^{\frac {m}{2}+m(h+2)+1}(1-z^{h+2})}. \label{eq:coef}
\end{align}
Equivalently, it equals the residuum of the Laurent series\eqref{eq:coef} at $z=0$. Now consider the contour integral of \eqref{eq:coef}
(with respect to $z$, of course)
along a circle of radius $r$ around the origin. It is a standard fact that in the limit $r\to\infty$ this integral vanishes, because the integrand \eqref{eq:coef} is of the order $O(1/z^2)$. Therefore, by
the theorem of residues, the sum of the residues of \eqref{eq:coef} must be 0, or, equivalently,
the residuum at $z=0$, which we are interested in, equals the negative of the sum of the other residues. As the other poles of \eqref{eq:coef}
are the $(h+2)$-th roots of unity different from 1, we obtain
\begin{multline*}
-\sum _{k=1} ^{h+1}
\frac {\(1+e^{\frac {2\pi ik} {h+2}}\)^m\(e^{\frac {\pi ik} {h+2}(a_s-e_t)}-
e^{-\frac {\pi ik} {h+2}(a_s+e_t+2)}\)} {e^{\frac {2\pi ik} {h+2}(\frac{m} {2}+1)}\(-(h+2)e^{\frac {2\pi ik} {h+2}(h+1)}\)}\\
=\sum _{k=1} ^{h+1}\frac {1} {h+2}\(2\cos\frac {\pi k} {h+2}\)^m {\(e^{\frac {\pi ik} {h+2}(a_s-e_t)}-
e^{-\frac {\pi ik} {h+2}(a_s+e_t+2)}\)}
\end{multline*}for the $(s,t)$-entry in \eqref{eq:vicious-with-2}. Substituting this
into \eqref{eq:vicious-with-2}, and bringing the summations and a few
factors outside
of the determinant, by multilinearity of the determinant,
we get \eqref{eq:vicious-with-2b}.\end{proof}\section{The asymptotics of vicious walkers, between two wall}
Theorem~\ref{thm:vicious-with-2b} allows us to derive asymptotic formulas for the number of vicious walkers between two walls, for arbitrary starting and end points.\begin{Theorem} \label{thm:vicious-with-2-asy}
Let $0\le a_1<a_2<\dots<a_p\le h$, all $a_i$'s of the same parity, and$0\le e_1<e_2<\dots<e_p\le h$, all $e_i$'s of the same parity, such that $a_i+e_i\equiv m\pmod2$,
$i=1,2,\dots,p$.
The number of vicious walkers with $p$ branches of length $m$, the
$i$-th branch running from $A_i=(0,a_i)$ to $E_i=(m,e_i)$,$i=1,2,\dots,p$,which do not go below the $x$-axis and not above the line $y=h$,is asymptotically
\begin{multline} \label{eq:vicious-with-2-asy}
 \frac {4^{p^2}} {(h+2)^p} \bigg(2^p \prod _{s=1}^{p}\cos\frac {s\pi} {h+2}\bigg)^m \prod _{1\le s<t\le p} ^{}\sin\frac {\pi (a_t-a_s)} {2(h+2)} \cdot
   \sin\frac {\pi (e_t-e_s)} {2(h+2)}\\
\times \prod _{1\le s\le t\le p}\sin\frac {\pi (a_t+a_s+2)} {2(h+2)} \cdot   \sin\frac {\pi (e_t+e_s+2)} {2(h+2)}.
\end{multline}\end{Theorem}
\begin{proof}Theorem~\ref{thm:vicious-with-2b} tells us that the number of vicious walkers that we want to estimate can be written in
form of a finite sum $\sum _{\ell} ^{}c_\ell b_\ell^m$, where the
$c_\ell$'s and $b_\ell$'s are independent of $m$. Hence,
what we have to find is $b=\max_\ell b_\ell$. Then, asymptotically,
the number of vicious walkers is $b^m\sum _{\ell:\ b_\ell=b}^{}c_\ell$.
The $b_\ell$'s have the form
\begin{equation} \label{eq:b_l}2^p \prod _{s=1}
^{p}\cos {k_s\pi} / (h+2).
\end{equation}To find the largest among these, we have
to choose $k_s$ close to the lower limit of the summation in
\eqref{eq:vicious-with-2b}, 1, or
close to the upper limit, $h+1$. However, we are not allowed to
make a choice such that $k_s=k_t$ for some $s\ne t$, because in that case the determinant in \eqref{eq:vicious-with-2b} vanishes. For the same reason a choice where $k_s=h+2-k_t$ for some $s$ and $t$ is forbidden.
Therefore the expression \eqref{eq:b_l} will be maximal if the set$\{k_1,k_2\dots,k_p\}$ is chosen from$\{1,2,\dots,p,h+2-p,\dots,h,h+1\}$, subject to the two restrictions mentioned above. These are exactly $2^p$ different choices. As a short calculation shows, the sum of the corresponding summands in\eqref{eq:vicious-with-2b} equals
\begin{multline} \label{eq:prod-det} \frac {1} {(h+2)^p} \bigg(2^p \prod _{s=1}^{p}\cos\frac {s\pi} {h+2}\bigg)^m
\det_{1\le s,t\le p}
\( e^{\frac {\pi it} {h+2}(a_s+1)}-
e^{-\frac {\pi it} {h+2}(a_s+1)}\)\\\times\det_{1\le s,t\le p}\( e^{-\frac {\pi it} {h+2}(e_s+1)}-
e^{\frac {\pi it} {h+2}(e_s+1)}\).
\end{multline}Both determinants are easily evaluated by means of \eqref{eq:sympl}.Little manipulation then leads to \eqref{eq:vicious-with-2-asy}.
\end{proof}
If we specialize $a_i$ to $a+2i-2$ and $e_i$ to $e+2i-2$ in Theorem~\ref{thm:vicious-with-2-asy}, then we obtain the asymptotics
for watermelons between two walls.

\begin{Corollary} \label{thm:watermelons-with-2-asy}
Let $a$ and $e$ be integers with $0\le a,e\le h-2p+2$ and $a+e\equiv m\pmod2$.
The number of watermelons with $p$ branches of length $m$, in which
the lowest branch starts at height $a$ and terminates at height $e$,
which do not go below the $x$-axis and not above the line $y=h$,
is asymptotically\begin{multline} \label{eq:watermelons-with-2-asy}
 \frac {4^{p^2}} {(h+2)^p} \bigg(2^p \prod _{s=1}^{p}\cos\frac {s\pi} {h+2}\bigg)^m
 \prod _{1\le s<t\le p} ^{}\sin^2\frac {\pi (t-s)} {(h+2)} \\\times
 \prod _{1\le s\le t\le p}\sin\frac {\pi (a+t+s-1)} {(h+2)} \cdot
   \sin\frac {\pi (e+t+s-1)} {(h+2)}.\end{multline}

Let $a$ and $e$ be integers with $0\le a,e\le h-2p+2$ and $a+e\equiv
m\pmod2$.
The number of $\infty$-stand watermelons with $p$ branches of length $m$, in which the lowest branch starts at height $a$ and terminates at height $e$,which do not go below the $x$-axis and not above the line $y=h$,
is asymptotically\begin{multline} \label{eq:infty-watermelons-with-2-asy} \frac {4^{p^2}} {(h+2p)^p} \bigg(2^p \prod _{s=1}^{p}\cos\frac {s\pi} {h+2p}\bigg)^m \prod _{1\le s<t\le p} ^{}\sin^2\frac {2\pi (t-s)} {(h+2p)} \\\times
 \prod _{1\le s\le t\le p}\sin\frac {\pi (a+2t+2s-3)} {(h+2p)} \cdot   \sin\frac {\pi (e+2t+2s-3)} {(h+2p)}.\end{multline}
\end{Corollary}

\begin{proof}There is nothing to say about the first claim.To establish the second claim, we shift the $i$-th branch of the$\infty$-friendly watermelon by $2(i-1)$ units up,
as in the proof of Theorem~\ref{thm:watermelon-inf-without}.
Thus we obtain a set of vicious walkers with $p$ branches of length $m$, the$i$-th branch starting from $A_i=(0,a+4i-4)$ and terminating at
$E_i=(m,e+4i-4)$, $i=1,2,\dots,p$,
which do not go below the $x$-axis and not above the line $y=h+2p-2$~(!).Hence, Theorem~\ref{thm:vicious-with-2-asy} with $a_i=a+4i-4$, $e_i=e+4i-4$,$i=1,2,\dots,p$, and $h$ replaced by $h+2p-2$ gives immediately the
desired asymptotics.\end{proof}
Clearly, by performing the obvious summations, we could also obtain
the asymptotics for watermelons of arbitrary deviation. The resulting
sums apparently do not simplify, however.
It must be noted, however, that in contrast to watermelons
without restriction and with the restriction of {\em one} wall, here, i.e., in the presence
of the restriction of two walls, the asymptotics of (ordinary)watermelons and $\infty$-friendly watermelons (compare the bases of the
exponentials in the two
statements in Corollary~\ref{thm:watermelons-with-2-asy})is of different order of
magnitude (except in the case of single branch watermelons, of course). 
Hence, from these considerations, it is impossible to
conclude whether $n$-friendly watermelons restricted by two walls will have the same order of magnitude as (ordinary) watermelons, or not.
\smallskip
Let us now turn to the asymptotics for stars. Again, we begin with a general theorem.\begin{Theorem} \label{thm:starvicious-with-2}Let $0\le a_1<a_2<\dots<a_p\le h$ be integers, all of the same parity.The number of vicious walkers with $p$ branches of length $m$, the
$i$-th branch starting from $A_i=(0,a_i)$,$i=1,2,\dots,p$,which do not go below the $x$-axis and not above the line $y=h$,
is asymptotically
\begin{multline} \label{eq:starvicious-with-2even}
\frac {4^{p^2}} {(h+2)^p}  \bigg(2^p \prod _{s=1} ^{p}\cos \frac {s \pi}{h+2}\bigg)^m \prod _{1\le s<t\le p} ^{}\sin \frac {\pi  (a_t-a_s)} {2(h+2)} \cdot
    \sin \frac {\pi  (t-s)} {h+2}\\
\times \prod _{1\le s\le t\le p} ^{}\sin \frac {\pi  (a_t+a_s+2)} {2(h+2)} \cdot
  \sin \frac {\pi  (t+s)} {h+2}
 \prod _{s=0} ^{p}\prod _{t=1} ^{p}\frac {\sin \frac   {\pi  (t-s+\fl{(h+2)/2})}   {h+2}} {\sin \frac {\pi  (t-s+p)} {h+2}}\\\times \prod _{s=1} ^{p}\frac {\sin \frac {(s+\fl{(h+2)/2}-p) \pi} {h+2}}
   {\sin \frac {(2s+\fl{(h+2)/2}-p) \pi} {h+2}},\end{multline}if $m+a_i$ is even, and
\begin{multline} \label{eq:starvicious-with-2odd}
\frac {4^{p^2}} {(h+2)^p}  \bigg(2^p \prod _{s=1} ^{p}\cos \frac {s \pi}
{h+2}\bigg)^m \prod _{1\le s<t\le p} ^{}\sin \frac {\pi  (a_t-a_s)} {2(h+2)} \cdot
    \sin \frac {\pi  (t-s)} {h+2}\\\times
 \prod _{1\le s\le t\le p} ^{}\sin \frac {\pi  (a_t+a_s+2)} {2(h+2)}\cdot   \sin \frac {\pi  (t+s)} {h+2}
 \prod _{s=0} ^{p}\prod _{t=1} ^{p}\frac {\sin \frac
   {\pi  (t-s+\fl{(h+1)/2})}   {h+2}} {\sin \frac {\pi  (t-s+p)} {h+2}},
\end{multline}if $m+a_i$ is odd.
\end{Theorem}
\begin{proof}It is obvious that, in view of Theorem~\ref{thm:vicious-with-2-asy}, we have to compute the sum of\eqref{eq:vicious-with-2-asy} over all possible choices of
$e_1<e_2<\dots<e_p$. Here we have to distinguish between two cases,depending on whether $m+a_i$ is even or odd.Let first $m+a_i$ be odd. This implies that all the $e_i$'s are odd as well, so that we have to compute the sum of\eqref{eq:vicious-with-2-asy} over all possible choices of
$1\le e_1<e_2<\dots<e_p\le h$, $e_i=2e'_i-1$ for some integer $e'_i$,$i=1,2,\dots,p$. If we remember that expression \eqref{eq:vicious-with-2-asy} came from
\eqref{eq:prod-det}, we see that this is
\begin{multline} \label{eq:vicious-with-2-asy-sum}
 \frac {4^{p^2}} {(h+2)^p} \bigg(2^p \prod _{s=1}^{p}\cos\frac {s\pi} {h+2}\bigg)^m \prod _{1\le s<t\le p} ^{}\sin\frac {\pi (a_t-a_s)} {2(h+2)} \prod _{1\le s\le t\le p}\sin\frac {\pi (a_t+a_s+2)} {2(h+2)} \\
\times (-i)^p
\sum _{1\le e'_1<\dots<e'_p\le \fl{(h+1)/2}} ^{}\det_{1\le s,t\le p}\( e^{\frac {2\pi it} {h+2}e'_s}-
e^{-\frac {2\pi it} {h+2}e'_s}\).
\end{multline}In order to evaluate the sum in the last line, we rewrite it in terms of symplectic characters$\sp_\la(x_1^{\pm1},x_2^{\pm1},\dots,x_p^{\pm1})$,which are
defined by (see \cite[(24.18)]{FuHaAA})\begin{equation} \label{e24}\sp_\la(x_1^{\pm1},x_2^{\pm1},\dots,x_p^{\pm1})
=\frac {\det\limits_{1\le s,t\le
p}(x_t^{\la_s+p-s+1}-x_t^{-(\la_s+p-s+1)})}{\det\limits_{1\le s,t\le
p}(x_t^{p-s+1}-x_t^{-(p-s+1)})}.\end{equation}
Therefore, writing $q$ for $e^{2\pi i/(h+2)}$ and $H$ for$\fl{(h+1)/2}$, the sum in the last
line of \eqref{eq:vicious-with-2-asy-sum} equals\begin{multline} \label{eq:sum-sympl}
\det_{1\le s,t\le p}
\( q^{st}-q^{-st}\)\sum _{1\le e'_1<\dots<e'_p\le H} ^{}
\sp_{(e'_p-p,\dots,e'_1-1)}(q^{\pm1},q^{\pm2},\dots,q^{\pm p})\\=\det_{1\le s,t\le p}
\( q^{st}-q^{-st}\)
\sum _{\nu\subseteq ((H-p)^p)} ^{}\sp_\nu(q^{\pm1},q^{\pm2},\dots,q^{\pm p}).\end{multline}Now we appeal to the formula (see
\cite[(3.4)]{KratBC}),\begin{equation} \label{e31}s_{(c^r)}(x_1,x_1^{-1},\dots,x_p,x_p^{-1},1)=\sum _{\nu\subseteq (c^r)} ^{}\sp_\nu(x_1^{\pm1},\dots,x_p^{\pm1}).\end{equation}Use of this formula in \eqref{eq:sum-sympl} gives
\begin{equation} \label{eq:sum-schur}\det_{1\le s,t\le p}\( q^{st}-q^{-st}\)
s_{((H-p)^p)}(q^{-p},\dots,q^{-1},1,q,\dots,q^p).
\end{equation}
Clearly, the determinant is easily evaluated by means of\eqref{eq:sympl}, whereas the specialized Schur function can be evaluated by means of the hook-content formula (see\cite[I, Sec.~3,  Ex.~1]{MacdAC}, \cite[Ex.~A.30, (ii)]{FuHaAA})
\begin{equation} \label{e7}
s_\la(q^{-L},q^{-L+1},\dots,q^{-L+P})=q^{\sum _{l\ge1} ^{}(l-L-1)\la_l}\prod _{\rh\in\la} ^{}\frac {1-q^{P+c_\rh}} {1-q^{h_\rh}},\end{equation}
where $c_\rh$ and $h_\rh$ are the {\em content\/} and the {\em hook
length\/} of the cell $\rh$. Substitution of all this in\eqref{eq:vicious-with-2-asy-sum} and some manipulation then leads to\eqref{eq:star vicious-with-2odd}.Now let $m+a_i$ be even. This implies that all the $e_i$'s are even
as well, so that we have to compute the sum of\eqref{eq:vicious-with-2-asy} over all possible choices of$0\le e_1<e_2<\dots<e_p\le h$, $e_i=2e'_i-2$ for some integer $e'_i$,
$i=1,2,\dots,p$. Again, if we remember that
expression \eqref{eq:vicious-with-2-asy} came from\eqref{eq:prod-det}, we see that this is
\begin{multline} \label{eq:vicious-with-2-asy-sum-even} \frac {4^{p^2}} {(h+2)^p} \bigg(2^p \prod _{s=1}
^{p}\cos\frac {s\pi} {h+2}\bigg)^m \prod _{1\le s<t\le p} ^{}\sin\frac {\pi (a_t-a_s)} {2(h+2)}
 \prod _{1\le s\le t\le p}\sin\frac {\pi (a_t+a_s+2)} {2(h+2)} \\\times (-i)^p
\sum _{1\le e'_1<\dots<e'_p\le \fl{(h+2)/2}} ^{}
\det_{1\le s,t\le p}\( e^{\frac {2\pi it} {h+2}(e'_s-\frac {1} {2})}-
e^{-\frac {2\pi it} {h+2}(e'_s-\frac {1} {2})}\).\end{multline}
This time, it is possible to rewrite the sum in the last line in
terms of odd orthogonal characters (cf.~\eqref{e12}).Namely, writing $q$ for $e^{2\pi i/(h+2)}$ and $H$ for$\fl{(h+2)/2}$, the sum in the last line of \eqref{eq:vicious-with-2-asy-sum-even} equals
\begin{multline} \label{eq:sum-ortho}\det_{1\le s,t\le p}\( q^{st}-q^{-st}\)\sum _{1\le e'_1<\dots<e'_p\le H} ^{}\so_{(e'_p-p,\dots,e'_1-1)}(q^{\pm1},q^{\pm2},\dots,q^{\pm p},1)\\=\det_{1\le s,t\le p}
\( q^{st}-q^{-st}\)\sum _{\nu\subseteq ((H-p)^p)} ^{}\so_\nu(q^{\pm1},q^{\pm2},\dots,q^{\pm p},1).
\end{multline}
Again there is a formula which allows to evaluate the sum in the last
line (see \cite[(3.2)]{KratBC}),\begin{equation} \label{eq:Schur-orthosum}s_{((c^{r-\ell},(c-1)^\ell)}(x_1,x_1^{-1},\dots,x_p,x_p^{-1},1)=\underset
{\oddrows\!\big((c^r)/\nu\big)=\ell}
{\sum _{\nu\subseteq (c^r)} ^{}}\so_\nu(x_1^{\pm1},\dots,x_p^{\pm1},1),\end{equation}
where $\oddrows\!\big((c^r)/\nu\big)=\ell$ means that the number of
rows of odd length in the skew shape $(c^r)/\nu$ equals exactly$\ell$.Use of this formula in \eqref{eq:sum-ortho} gives\begin{equation} \label{eq:sum-schur2}\det_{1\le s,t\le p}
\( q^{st}-q^{-st}\)\sum _{\ell=0} ^{p}s_{((H-p)^{p-\ell},(H-p-1)^\ell)}(q^{-p},\dots,q^{-1},1,q,\dots,q^p).
\end{equation}
Again, the hook-content formula \eqref{e7} applies and yields\begin{multline*}
\det_{1\le s,t\le p}\( q^{st}-q^{-st}\)
\sum _{\ell=0} ^{p}q^{-(H-p)\binom {p+1}2+\binom {\ell+1}2}\frac {\prod _{s=1}
^{H-p-1}(q^{p+s+1};q)_p} {\prod _{s=1} ^{H-p}(q^{s};q)_p}\\\times\frac
{(q^{H+\ell+1};q)_{p-\ell}\,(q^{\ell+1};q)_{p-\ell}\,(q^{H-p};q)_\ell}
{(q;q)_{p-\ell}}\end{multline*}
for the expression \eqref{eq:sum-schur2}. Here we used the standard notation for shifted $q$-factorials,
$(a;q)_k:=(1-a)(1-aq)\cdots(1-aq^{k-1})$, $k\ge1$, $(a;q)_0:=1$.
In standard basic hypergeometric notation$${}_r\phi_s\!\left[\matrix a_1,\dots,a_r\\ b_1,\dots,b_s\endmatrix; q,z\right]=\sum _{\ell=0} ^{\infty}\frac {\poq{a_1}{\ell}\cdots\poq{a_r}{\ell}}
{\poq{q}{\ell}\poq{b_1}{\ell}\cdots\poq{b_s}{\ell}}\left((-1)^\ell q^{\binom\ell2}\right)^{s-r+1}z^\ell\ ,$$this can be written in the form\begin{equation*}\det_{1\le s,t\le p}
\( q^{st}-q^{-st}\)q^{-(H-p)\binom {p+1}2}\bigg(\prod _{s=1} ^{H-p}\frac{(q^{p+s+1};q)_p} {(q^{s};q)_p}\bigg)
{}_2\phi_1\!\[\matrix q^{H-p},q^{-p}\\q^{H+1}\endmatrix; q,
-q^{p+1}\].
\end{equation*}The $_2\phi_1$-series in this expression can be summed with the help of the $q$-analogue of Kummer's summation (see \cite[Appendix (II.9)]{GaRaAA}),
$$_2\phi_1\!\left[\matrix a,b\\aq/b\endmatrix;q,-q/b\right]=
\frac {(-q;q)_\infty\,(aq;q^2)_\infty\,(aq^2/b^2;q^2)_\infty}{(-q/b;q)_\infty\,(aq/b;q)_\infty}\ .$$Thus we have finally evaluated the sum in
\eqref{eq:vicious-with-2-asy-sum-even}. From there, it is then routine
to work out the expression \eqref{eq:starvicious-with-2even}.\end{proof}Clearly, if we specialize Theorem~\ref{thm:starvicious-with-2} to$a_i=a+2i-2$, $i=1,2,\dots,p$, we obtain the asymptotics for stars
between two walls.\begin{Corollary} \label{thm:star-with-2}
Let $a$ be an integer with $0\le a\le h-2p+2$.
The number of stars with $p$ branches of length $m$, the$i$-th branch starting from $A_i=(0,a+2i-2)$,
$i=1,2,\dots,p$,which do not go below the $x$-axis and not above the line $y=h$,
is asymptotically\begin{multline} \label{eq:star-with-2even}
\frac {4^{p^2}} {(h+2)^p}  \bigg(2^p \prod _{s=1} ^{p}\cos \frac {s \pi}
{h+2}\bigg)^m
 \prod _{1\le s<t\le p} ^{}\sin^2 \frac {\pi  (t-s)} {(h+2)} \\
\times \prod _{1\le s\le t\le p} ^{}\sin \frac {\pi  (a+t+s-1)} {(h+2)} \cdot
  \sin \frac {\pi  (t+s)} {h+2}
 \prod _{s=0} ^{p}\prod _{t=1} ^{p}\frac {\sin \frac   {\pi  (t-s+\fl{(h+2)/2})}   {h+2}} {\sin \frac {\pi  (t-s+p)} {h+2}}\\\times
 \prod _{s=1} ^{p}\frac {\sin \frac {(s+\fl{(h+2)/2}-p) \pi} {h+2}}   {\sin \frac {(2s+\fl{(h+2)/2}-p) \pi} {h+2}},\end{multline}if $m+a$ is even, and\begin{multline} \label{eq:star-with-2odd}\frac {4^{p^2}} {(h+2)^p}  \bigg(2^p \prod _{s=1} ^{p}\cos \frac {s \pi}{h+2}\bigg)^m \prod _{1\le s<t\le p} ^{}\sin^2 \frac {\pi  (t-s)} {(h+2)} \\
\times
 \prod _{1\le s\le t\le p} ^{}\sin \frac {\pi  (a+t+s-1)} {(h+2)}
\cdot   \sin \frac {\pi  (t+s)} {h+2} \prod _{s=0} ^{p}\prod _{t=1} ^{p}\frac {\sin \frac   {\pi  (t-s+\fl{(h+1)/2})}   {h+2}} {\sin \frac {\pi  (t-s+p)} {h+2}},
\end{multline}
if $m+a$ is odd.\end{Corollary}
A different specialization
yields the asymptotics for $\infty$-friendly stars between two walls.\begin{Corollary} \label{thm:infty-star-with-2}Let $a$ be an integer with $0\le a\le h-2p+2$.The number of $\infty$-friendly stars with $p$ branches of length $m$, the
$i$-th branch starting from $A_i=(0,a+2i-2)$,$i=1,2,\dots,p$,which do not go below the $x$-axis and not above the line $y=h$,
is asymptotically
\begin{multline} \label{eq:infty-star-with-2even}\frac {4^{p^2}} {(h+2p)^p}  \bigg(2^p \prod _{s=1} ^{p}\cos \frac {s \pi}{h+2p}\bigg)^m
 \prod _{1\le s<t\le p} ^{}\sin \frac {2\pi  (t-s)} {(h+2p)} \cdot    \sin \frac {\pi  (t-s)} {h+2p}\\
\times \prod _{1\le s\le t\le p} ^{}\sin \frac {\pi  (a+2t+2s-3)} {(h+2p)} \cdot
 \sin \frac {\pi  (t+s)} {h+2p}
 \prod _{s=0} ^{p}\prod _{t=1} ^{p}\frac {\sin \frac   {\pi  (t-s+\fl{(h+2p)/2})}
   {h+2p}} {\sin \frac {\pi  (t-s+p)} {h+2p}}\\
\times \prod _{s=1} ^{p}\frac {\sin \frac {(s+\fl{(h+2p)/2}-p) \pi} {h+2p}}   {\sin \frac {(2s+\fl{(h+2p)/2}-p) \pi} {h+2p}},\end{multline}
if $m+a$ is even, and\begin{multline} \label{eq:infty-star-with-2odd}\frac {4^{p^2}} {(h+2p)^p}  \bigg(2^p \prod _{s=1} ^{p}\cos \frac {s \pi}{h+2p}\bigg)^m \prod _{1\le s<t\le p} ^{}\sin \frac {2\pi  (t-s)} {(h+2p)} \cdot    \sin \frac {\pi  (t-s)} {h+2p}\\\times \prod _{1\le s\le t\le p} ^{}\sin \frac {\pi  (a+2t+2s-3)} {(h+2p)}
\cdot   \sin \frac {\pi  (t+s)} {h+2p}
 \prod _{s=0} ^{p}\prod _{t=1} ^{p}\frac {\sin \frac
   {\pi  (t-s+\fl{(h+1)/2})}   {h+2p}} {\sin \frac {\pi  (t-s+p)} {h+2p}},\end{multline}if $m+a$ is odd.
\end{Corollary}
\begin{proof}As in the proof of Theorem~\ref{thm:watermelon-inf-without}, we shift the $i$-th branch by $2(i-1)$ units up. Thus we obtain a set of vicious walkers
with $p$ branches of length $m$, the
$i$-th branch starting from $A_i=(0,a+4i-4)$,$i=1,2,\dots,p$,which do not go below the $x$-axis and not above the line $y=h+2p-2$.Hence, Theorem~\ref{thm:starvicious-with-2} with $a_i=a+4i-4$,$i=1,2,\dots,p$, and $h$ replaced by $h+2p-2$ gives immediately the
desired asymptotics.\end{proof}
As before for watermelons,
it must be noted that in contrast to stars without restriction and with the restriction of {\em one} wall, here, i.e., in the presence of the restriction of two walls, the asymptotics of (ordinary) stars and $\infty$-friendly stars (compare the bases of the exponentials in
Corollaries~\ref{thm:star-with-2} and \ref{thm:infty-star-with-2})is of different order of
magnitude (except in the case of single branch stars, of course).Hence, again, from these considerations, it is impossible to conclude whether $n$-friendly stars restricted by two walls will have
the same order of magnitude as (ordinary) stars, or not.







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\section{Appendix}
In this appendix we collect together the tools used in the 
analysis carried out in the body of the paper.
These comprise certain basic asymptotic results, a formulation
of lattice paths with arbitrary endpoints as Pfaffians, and
the evaluation of certain determinants and Pfaffians.

\subsection{Asymptotics}
Aside from Stirling's formula,
the approximation that we use extensively throughout is the following:
\begin{Lemma} \label{lem:expsum-asy}
Let $N$ be e nonnegative integer. Then
\begin{equation} \label{eq:expsum-asy}
\sum _{k=0} ^{\infty}k^Ne^{-\al k^2}=
\int _{0} ^{\infty}y^Ne^{-\al y^2}\,dy+O(1)=\frac 
{\Gamma\(\tfrac {N} {2}+\tfrac {1} {2}\)}
{2\,\al^{{N} /{2}+ {1}/ {2}}}
+O(1)
\end{equation}
as $\al$ tends to\/ $0^+$.
\end{Lemma}
\begin{proof} The Poisson summation theorem (see
\cite[(5.75) with $a=0$]{OdlyAA}, \cite[(2.8.1)]{Titch})
says that
\begin{equation} \label{eq:Poisson}
\sum _{k=-\infty} ^{\infty}f(k)=\sum _{m=-\infty} ^{\infty}
\int _{-\infty} ^{\infty}f(y)e^{-2\pi imy}\,dy,
\end{equation}
for suitable functions $f$. It is for example valid for continuous,
absolutely integrable functions $f$
of bounded variation. The choice of $f(x)=e^{-\al x^2}$ in
\eqref{eq:Poisson} gives \eqref{eq:expsum-asy} for $N=0$ upon little
manipulation. From now on let $N\ge1$.
In \eqref{eq:Poisson} we choose 
$$f(x)=\begin{cases} x^N\cdot e^{-\al x^2}&x\ge0,\\
0&x<0.\end{cases}$$
This function does satisfy the above mentioned requirements.
Thus we obtain
\begin{align} \notag
\sum _{k=0} ^{\infty}k^Ne^{-\al k^2}
&=\sum _{m=-\infty} ^{\infty}e^{-\pi^2 m^2/\al}
\int _{0} ^{\infty}y^Ne^{-\al(y+\pi im/\al)^2}\,dy\\
\notag
&=\int _{0} ^{\infty}y^Ne^{-\al y^2}\,dy+
\sum _{m=1} ^{\infty}e^{-\pi^2 m^2/\al}
\int _{-\infty} ^{\infty}{\v y}^Ne^{-\al(y+\pi im/\al)^2}\,dy\\
\label{eq:Poisson-appl}
&=\int _{0} ^{\infty}y^Ne^{-\al y^2}\,dy+
\sum _{m=1} ^{\infty}e^{-\pi^2 m^2/\al}
\int _{-\infty} ^{\infty}\v{y-\tfrac {\pi im}\al}^Ne^{-\al y^2}\,dy.
\end{align}
Next we approximate the integral which appears in the sum. 
We split the integral into two parts,
\begin{equation*} 
\int _{-\infty} ^{\infty}\v{y-\tfrac {\pi im}\al}^Ne^{-\al y^2}\,dy
=\int _{\v y\ge \frac {\pi m}{\al\sqrt 3}}
\v{y-\tfrac {\pi im}\al}^Ne^{-\al y^2}\,dy+
\int _{\v y <\frac {\pi m}{\al\sqrt 3}}
\v{y-\tfrac {\pi im}\al}^Ne^{-\al y^2}\,dy.
\end{equation*}
For the first part, i.e., for $\v y\ge \frac {\pi m}{\al\sqrt 3}$ and
$y$ real, we have
$\v{y-\tfrac {\pi im}\al}\le 2\v y$.
For the second part, i.e., for $\v y< \frac {\pi m}{\al\sqrt 3}$ and
$y$ real, we have
$\v{y-\tfrac {\pi im}\al}\le \tfrac {2\pi m} {\al \sqrt 3}$.
Thus we obtain
\begin{align*} 
\int _{-\infty} ^{\infty}\v{y-\tfrac {\pi im}\al}^Ne^{-\al
y^2}\,dy&\le
\int _{\v y\ge \frac {\pi m}{\al\sqrt 3}}
(2\v{y})^Ne^{-\al y^2}\,dy+
\int _{\v y <\frac {\pi m}{\al\sqrt 3}}
\(\tfrac {2\pi m} {\al\sqrt 3}\)^Ne^{-\al y^2}\,dy\\
&\le 2^N\int _{-\infty} ^{\infty}
\v{y}^Ne^{-\al y^2}\,dy+
\(\tfrac {2\pi m} {\al\sqrt 3}\)^N\int _{-\infty} ^{\infty}
e^{-\al y^2}\,dy.
\end{align*}
The integrals in the last line are easily evaluated by recalling one
of the definitions of the gamma function (see \cite[1.1(1)]{Erdelyi}),
\begin{equation} \label{eq:gamma}
\Gamma(x)=\int _{0} ^{\infty}t^{x-1}e^{-t}\,dt.
\end{equation}
For, substitution of $-\al y^2$ for $t$ and replacement of $x$ by
$(N+1)/2$ yields
\begin{equation} \label{eq:exp-int}
\int _{0} ^{\infty}y^Ne^{-\al y^2}\,dy
=\frac {\Gamma\(\tfrac {N} {2}+\tfrac {1} {2}\)} {2\,\al^{N/2+1/2}}.
\end{equation}
Combining all this and substituting back into
\eqref{eq:Poisson-appl}, we get
\begin{equation*} 
\sum _{k=0} ^{\infty}k^Ne^{-\al k^2}
=\frac {\Gamma\(\tfrac {N} {2}+\tfrac {1}
{2}\)} {2\,\al^{N/2+1/2}}
+O\(\sum _{m=1} ^{\infty}e^{-\pi^2 m^2/\al}
\(2^{N+1}
\frac {\Gamma\(\tfrac {N} {2}+\tfrac {1} {2}\)} {2\,\al^{N/2+1/2}}+
\(\tfrac {2\pi m} {\al\sqrt 3}\)^N
\frac {\sqrt\pi} {\sqrt\al}\)\).
\end{equation*}
The appearance of the exponential $e^{-\pi^2 m^2/\al}$ makes the $O(.)$
term ``arbitrarily" small. Thus, \eqref{eq:expsum-asy} follows
immediately.
\end{proof}

\subsection{Nonintersecting lattice paths and Pfaffians}
If one wants to enumerate nonintersecting lattice paths with given
starting and end points, then the solution is given by the
Gessel--Viennot determinant. If, however, we want to enumerate
nonintersecting lattice paths with given starting points, but where
the end points may be any points from a given {\em set\/} of points,
then the solution can be given in terms of a Pfaffian, as was shown
by Okada \cite[Theorem~3]{OkadAA} and Stembridge \cite[Theorem~3.1]{StemAE}.

\begin{Proposition} \label{prop:Stem}
Let $p$ be even. Let $A_1,A_2,\dots,A_p$ be lattice points, and let
$\mathbf E=\{E_i:i\in I\}$ be a set of lattice points, 
where $I$ is a linearly ordered set of
indices, with the property that if $1\le i<j\le p$ and $k,l\in I$,
$k<l$, then any path from $A_i$ to $E_l$ must intersect any path from
$A_j$ to $E_k$.
The number of families $(P_1,P_2,\dots,P_p)$ of nonintersecting
lattice paths, where $P_i$ runs from $A_i$ to some point in the set
$I$, is given by
\begin{equation} \label{eq:Stem}
\Pf\limits_{1\le i<j\le p}(Q_{\mathbf E}(A_i,A_j)),
\end{equation}
where $Q_{\mathbf E}(A_i,A_j)$ is the number of all pairs of
nonintersecting lattice paths, one connecting $A_i$ to $\mathbf E$,
the other connecting $A_j$ to $\mathbf E$.
\end{Proposition}

\subsection{Some determinants and Pfaffians}
In our computations we need the following determinant evaluations.
Both of them are readily proved by
the standard argument that proves Vandermonde-type determinant
evaluations.

\begin{Lemma} \label{lem:dets}
Let $N$ by a nonnegative integer. Then
\begin{align} \label{eq:ortho1}
\det\limits_{1\le i,j\le N}\(x_i^{j}+x_i^{1-j}\)&=
(x_1x_2\cdots x_N)^{1-N}
\prod _{1\le i<j\le N} ^{}(x_i-x_j)(1-x_ix_j)\prod _{i=1}
^{N}(x_i+1),\\
\label{eq:sympl}
\det\limits_{1\le i,j\le N}\(x_i^{j}-x_i^{-j}\)&=
(x_1x_2\cdots x_N)^{-N}
\prod _{1\le i<j\le N} ^{}(x_i-x_j)(1-x_ix_j)\prod _{i=1}
^{N}(x_i^2-1)\\
\label{eq:ortho2}
\det\limits_{1\le i,j\le N}\(x_i^{j-1/2}-x_i^{-j+1/2}\)&=
(x_1x_2\cdots x_N)^{-N+1/2}
\prod _{1\le i<j\le N} ^{}(x_i-x_j)(1-x_ix_j)\prod _{i=1}
^{N}(x_i-1).
\end{align}
\end{Lemma}

In the proof of Theorem~\ref{thm:watermelon-inf-with}
we need to express another, very similar determinant in terms of odd
orthogonal characters. The latter are defined by
\begin{equation} \label{e12}
\so_\la(x_1^{\pm1},x_2^{\pm1},\dots,x_N^{\pm1},1)=\frac {\det\limits_{1\le i,j\le
N}(x_i^{\la_j+N-j+1/2}-x_i^{-(\la_j+N-j+1/2)})} 
{\det\limits_{1\le i,j\le
N}(x_i^{N-j+1/2}-x_i^{-(N-j+1/2)})},
\end{equation}
where $\la$ is a partition of length $N$.
Then a combination of \eqref{e12} and \eqref{eq:ortho2} gives
\begin{multline} \label{eq:unortho}
\det\limits_{1\le i,j\le N}\(x_i^{2j-3/2}-x_i^{-2j+3/2}\)=
\det\limits_{1\le i,j\le N}\(x_i^{j-1/2}-x_i^{-j+1/2}\)\\
\kern4cm\times
\so_{(N-1,N-2,\dots,1,0)}(x_1^{\pm1},x_2^{\pm1},\dots,x_N^{\pm1},1)\\
=(x_1x_2\cdots x_N)^{-N+1/2}
\prod _{1\le i<j\le N} ^{}(x_i-x_j)(1-x_ix_j)\prod _{i=1}
^{N}(x_i-1)\\
\times
\so_{(N-1,N-2,\dots,1,0)}(x_1^{\pm1},x_2^{\pm1},\dots,x_N^{\pm1},1).
\end{multline}
It will also be important to know 
that the odd orthogonal character
$\so_\la(\mathbf x)$ is a Laurent polynomial in the $x_i$'s,
consisting of a sum of exactly
\begin{equation} \label{e13}
\so_\la(1,1,\dots,1)=\prod _{1\le i<j\le N} ^{}\frac {\la_i-i-\la_j+j}
{j-i}\prod _{1\le i\le j\le N}\frac {\la_i+\la_j+2N+1-i-j}
{2N+1-i-j}
\end{equation}
monomials (see \cite[(24.29)]{FuHaAA},
\cite[Theorem~4.5.(2)]{SunaAE}), 
in each of which the exponent of $x_i$ is between $-\la_1$
and $\la_1$, $i=1,2,\dots,N$.

For the proof of Theorem~\ref{thm:star-inf-without} we need
the following reductions of Pfaffians to determinants. These are
essentially due to Gordon \cite[implicitly in Sec.~4, 5]{GordAB},
see also \cite[proof of Theorem~7.1]{StemAE}.
\begin{Lemma} \label{lem:Gordon}
Let $(g_i)$ be a sequence with the property $g_{-i}=g_i$, and let $N$
be a positive integer. Then
\begin{equation} \label{eq:Gordon-even}
\Pf\limits_{1\le i<j\le 2N}\bigg(\sum _{-(j-i)<\al\le j-i} ^{}g_\al\bigg)=
\det\limits_{1\le i,j\le N}(g_{i-j}+g_{i+j-1}),
\end{equation}
and
\begin{equation} \label{eq:Gordon-odd}
\Pf\limits_{1\le i<j\le 2N+2}\(\left\{\begin{matrix} 
\sum _{-(j-i)<\al\le j-i} ^{}g_\al&j\le 2N+1\\
X&j=2N+2
\end{matrix}\right\}\)=
X\cdot\det\limits_{1\le i,j\le N}(g_{i-j}-g_{i+j}).
\end{equation}
\end{Lemma}


\end{document}

