\documentclass[12pt,reqno]{amsart}

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\def\today{\ifcase\month\or
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 July\or August\or September\or October\or November\or December\fi
 \space\number\day, \number\year}
\def\nmb#1#2{#2} %zum Nummerieren
\def\totoc{} % f\"ur Inhaltsverzeichnung
%\def\dfrac#1#2{{\displaystyle{#1\over#2}}}
%\def\tfrac#1#2{{\textstyle{#1\over#2}}}
\def\iprod#1#2{\langle#1,#2\rangle}
\def\pder#1#2{\frac{\partial #1}{\partial #2}}
\def\iint{\int\!\!\int}
\def\({\left(}
\def\){\right)}
\def\[{\left[}
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\def\supp{\operatorname{supp}}
\def\Df{\operatorname{Df}}
\def\dom{\operatorname{dom}}
\def\Ker{\operatorname{Ker}}
\def\Tr{\operatorname{Tr}}
\def\Res{\operatorname{Res}}
\def\Aut{\operatorname{Aut}}
\def\kgV{\operatorname{kgV}}
\def\ggT{\operatorname{ggT}}
\def\diam{\operatorname{diam}}
\def\Im{\operatorname{Im}}
\def\Re{\operatorname{Re}}
\def\ord{\operatorname{ord}}
\def\rang{\operatorname{rang}}
\def\rng{\operatorname{rng}}
\def\grd{\operatorname{grd}}
\def\inv{\operatorname{inv}}
\def\maj{\operatorname{maj}}
\def\des{\operatorname{des}}
\def\varmaj{\operatorname{\overline{maj}}}
\def\vardes{\operatorname{\overline{des}}}
\def\pvarmaj{\operatorname{\overline{maj}'}}
\def\pmaj{\operatorname{maj'}}
\def\ln{\operatorname{ln}}
\def\der{\operatorname{der}}
\def\Hom{\operatorname{Hom}}
\def\tr{\operatorname{tr}}
\def\Span{\operatorname{Span}}
\def\grad{\operatorname{grad}}
\def\div{\operatorname{div}}
\def\rot{\operatorname{rot}}
\def\Sp{\operatorname{Sp}}
\def\sgn{\operatorname{sgn}}
\def\liml{\lim\limits}
\def\supl{\sup\limits}
\def\bigcupl{\bigcup\limits}
\def\bigcapl{\bigcap\limits}
\def\limsupl{\limsup\limits}
\def\liminfl{\liminf\limits}
\def\intl{\int\limits}
\def\suml{\sum\limits}
\def\maxl{\max\limits}
\def\minl{\min\limits}
\def\prodl{\prod\limits}
\def\tg{\operatorname{tan}}
\def\ctg{\operatorname{cot}}
\def\arctg{\operatorname{arctan}}
\def\arccot{\operatorname{arccot}}
\def\arcctg{\operatorname{arccot}}
\def\tgh{\operatorname{tanh}}
\def\ctgh{\operatorname{coth}}
\def\arcsinh{\operatorname{arcsinh}}
\def\arccosh{\operatorname{arccosh}}
\def\arctgh{\operatorname{arctanh}}
\def\arcctgh{\operatorname{arccoth}}
\def\3{\ss}

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%Input-Datei zum Erzeugen von Gitterpunktwegen.%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%LaTeX-Version%
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%
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%
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\def\so{\operatorname{\text {\it so}}}
\def\sp{\operatorname{\text {\it sp}}}
\def\oddrows{\operatorname{oddrows}}
\def\poq#1#2{(#1;q)_#2}


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\begin{document}

\section{Basic asymptotics}

Aside from Stirling's formula,
the approximation that we use 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}

\section{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}

\section{A few 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}

\section{The asymptotics of watermelons, without wall restriction} 


\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*} 
S(-\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 superexponentially.
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}

Next we consider the $\infty$-stand model for watermelons, i.e., we
allow the branches to touch arbitrarily often, but not cross each
other.

\begin{Theorem} \label{thm:watermelon-inf-without}
The number of $\infty$-stand 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 $\infty$-stand watermelon with $p$ branches
of length $m$ and with deviation $k$. It consists of a family $(P_1,P_2,\dots,P_p)$ of
noncrossing 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 $\infty$-stand watermelon
\hskip1cm
b. A corresponding set of nonintersecting 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 nonintersecting} 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 nonintersecting 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
contrubution 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 $n$-stand models.

\begin{Corollary} \label{cor:n-stand-watermelon}
As $m$ tends to infinity, $n$-stand 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 
$n$-stand 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 $n$-stand watermelons with $p$ branches of
length $m$ than $(n+1)$-stand 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
$n$-stand 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 $(n+1)$-stand 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 $n$-stand watermelons).
As this set of $(n+1)$-stand 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 $(n+1)$-stand 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{The asymptotics of stars, without wall restriction} 



\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 $\infty$-stand model for stars.

\begin{Theorem} \label{thm:star-inf-without}
The number of $\infty$-stand 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 nonintersecting lattice paths by shifting the
$i$-th path up by $2(i-1)$ units. Thus, 
$\infty$-stand stars with $p$ branches of length $m$ are in bijection
with families $(P_1,P_2,\dots,P_p)$ of
nonintersecting 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 nonintersecting 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
nonintersecting 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 $\infty$-stand 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 $\infty$-stand 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 $\infty$-stand 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 nonintersecting
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 nonintersecting lattice paths is exactly equal
to the number of families of nonintersecting
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 $\infty$-stand 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 $\infty$-stand 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 $n$-stand models. 

\begin{Corollary} \label{cor:n-stand-star}
As $m$ tends to infinity, $n$-stand stars with $p$ branches of
length $m$ have, up to a multiplicative
constant, the same asymptotic behaviour. More precisely, the number of 
$n$-stand 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 $n$-stand stars with $p$ branches of
length $m$ than $(n+1)$-stand 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{The asymptotics of watermelons, restricted by one wall} 

\begin{Theorem} \label{thm:watermelon-with}
The number of watermelons with $p$ branches of length $m$ (and
arbitrary deviation) which do not go below the $x$-axis
is asymptotically
\begin{multline} \label{eq:watermelon-with}
2^{mp+9p^2/4-p/4-3/2}m^{-3p^2/4-p/4+1/2}\pi^{-p/2}p^{-p^2/4-p/4-1/2}\\
\times
\Ga\(\tfrac {p^2} {4}+\tfrac {p} {4}+\tfrac {1} {2}\)
\bigg(\prod _{\ell=1} ^{p}(\ell-1)!\bigg)(1+O(m^{-1/2}\log^3m))
\end{multline}
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 $k\equiv m\pmod 2$) which do not go
below the $x$-axis is given by 
\begin{equation} \label{eq:num-watermelon-with}
\prod _{\ell=1} ^{p}
\frac {(\ell-1)!\,(k+2\ell-1)_{p-\ell+1}\,(m+2\ell-2)!} {(\frac {m+k}
{2}+\ell+p-1)!\,(\frac {m-k} {2}+\ell-1)!}.
\end{equation}
Our task is to sum this expression over all $k\ge0$ with $k\equiv
m\pmod 2$ and then approximate it. This is done completely
analogously to the proof of Theorem~\ref{thm:watermelon-without}.
The only difference is that here the sum does not extend to negative
$k$. Again, the dominant terms are those for $k\sim0$. Therefore we
concentrate on the terms for $0\le k\le \sqrt m\log m$. The other
terms are negligible, as is seen in the same way as in the proof of
Theorem~\ref{thm:watermelon-without}. For carrying out the
computations, we would again have to distinguish between the two cases
of $m$ being even or odd. Also here, the computations are rather
parallel. So let us assume in the following that $m$ is even.

If we carry out the computations parallel to those leading from
\eqref{eq:term3} to \eqref{eq:term3a}, then we see that the sum over
all even $k\ge0$ of \eqref{eq:num-watermelon-with} is asymptotically
\begin{multline} \label{eq:term-wat}
\bigg(\sum _{k=0} ^{\infty}
e^{-\frac {2p} {m}k^2}\prod _{\ell=1} ^{p}(2k+2\ell-1)_{p-\ell+1}\bigg)
\prod _{\ell=1} ^{p}
\frac {(\ell-1)!\,(m+2\ell-2)!} {(\frac {m} {2}+\ell+p-1)!\,
(\frac {m} {2}+\ell-1)!}\\
\times
(1+O(m^{-1/2}\log^3m)).
\end{multline}
The sum in this expression can be split into a linear combination of 
sums of the form 
$\sum _{k=0} ^{\infty}k^Ne^{-\frac {2p} {m} k^2}$. The asymptotics of
the latter is given by Lemma~\ref{lem:expsum-asy}. It implies
particularly that the largest contribution would come from the term
where the exponent $N$ of $k$ is maximal. This term is 
$$\sum _{k=0} ^{\infty}2^{\binom {p+1}2}
k^{\binom {p+1}2}e^{-\frac {2p} {m} k^2},$$
which by \eqref{eq:expsum-asy} gives a contribution of $\frac {1}
{2}(m/2p)^{p^2/4+p/4+1/2}\Ga(p^2/4+p/4+1/2)$.
The asymptotics of the product
in \eqref{eq:term-wat} is easily determined by using Stirling's
formula. Putting everything together we obtain \eqref{eq:watermelon-without}.
\end{proof}

Next we consider the $\infty$-stand model for watermelons with wall
restriction.

\begin{Theorem} \label{thm:watermelon-inf-with}
The number of $\infty$-stand watermelons with $p$ branches of length $m$ (and
arbitrary deviation) which do not go below the $x$-axis
is asymptotically
\begin{multline} \label{eq:watermelon-inf-with}
2^{mp+9p^2/4-9p/4-3/2}m^{-3p^2/4-p/4+1/2}\pi^{-p/2}p^{-p^2/4-p/4-1/2}\\
\times
\Ga\(\tfrac {p^2} {4}+\tfrac {p} {4}+\tfrac {1} {2}\)
\bigg(\prod _{\ell=1} ^{p}\frac {(\ell-1)!\,(2\ell-2)!\,(4\ell-2)!} 
{(\ell+p-1)!^2}\bigg)(1+O(m^{-1/2}\log^3m))
\end{multline}
as $m$ tends to infinity.
\end{Theorem}

\begin{proof}
As we are already used to, in the context of $\infty$-stand models
the situation is more difficult, here again because 
we do not have a nice closed
product formula (such as \eqref{eq:num-watermelon-with}) for the
number of our $\infty$-stand watermelons. 

The first step is the same as in the proof of
Theorem~\ref{thm:watermelon-inf-without}. We transform $\infty$-stand
watermelon into families of nonintersecting lattice paths by shifting the
$i$-th path up by $2(i-1)$ units. Thus, 
$\infty$-stand watermelons with $p$ branches of length $m$ and 
deviation $k$ which do not go below the $x$-axis are in bijection
with families $(P_1,P_2,\dots,P_p)$ of
nonintersecting lattice paths which do not go below the $x$-axis, 
$P_i$ running from $A_i=(0,4(i-1))$ to
$E_i=(m,k+4(i-1))$, $i=1,2,\dots,p$.

The number of these families of nonintersecting lattice paths is
given by the Gessel--Viennot determinant 
\begin{equation} \label{e21}
\det\limits_{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 $\v{\Pp{A_j\to E_i}}$
can then be written as a difference of two binomials. Thus we
obtain the determinant
\begin{equation} \label{eq:GV-det-with}
\det\limits_{1\le i,j\le p}\(\binom {m}{\frac {m-k} {2}+2j-2i}-
\binom {m}{\frac {m+k} {2}+2j+2i-3}\)
\end{equation}
for the number of watermelons under consideration.

We have to sum \eqref{eq:GV-det-with} over all $k\equiv m\pmod 2$,
$-m\le k\le m$, and approximate
the sum as $m$ tends to infinity. Having done the many similar proofs
before, in particular the proof of
Theorem~\ref{thm:watermelon-inf-without}, we are already experienced
of how to approach this problem. As in the proof of 
Theorem~\ref{thm:watermelon-inf-without}, we need to determine the
leading terms of \eqref{eq:GV-det-with}.

Once more, we bring the determinant \eqref{eq:GV-det-with} into a more
convenient form, by taking out some factors,
\begin{multline} \label{eq:det-factors3}
\det\limits_{1\le i,j\le p}\(\binom {m}{\frac {m-k} {2}+2j-2i}-
\binom {m}{\frac {m+k} {2}+2j+2i-3}\)\\
\kern-4.5cm
=
\prod _{\ell=1} ^{p}\frac {m!} {(\frac {m+k} {2}+2\ell+2p-3)!\,(\frac
{m-k} {2}-2\ell+2p)!}\\
\times\det\limits_{1\le i,j\le p}\big((\tfrac {m+k} {2}+2i-2j+1)_{2p+2j-3}\, 
(\tfrac {m-k} {2}-2i+2j+1)_{2p-2j}\\
(\tfrac {m+k} {2}+2i+2j-2)_{2p-2j}\, 
(\tfrac {m-k} {2}-2i-2j+4)_{2p+2j-3}\big).
\end{multline}
Again, we generalize and consider the determinant
\begin{multline} \label{eq:gen-det3}
\det\limits_{1\le i,j\le p}\big((\tfrac {m} {2}+X_i-2j+\tfrac 
{5} {2})_{2p+2j-3}\, 
(\tfrac {m} {2}-X_i+2j-\tfrac {1} {2})_{2p-2j}\\
(\tfrac {m} {2}+X_i+2j-\tfrac {1} {2})_{2p-2j}\, 
(\tfrac {m} {2}-X_i-2j+\tfrac {5} {2})_{2p+2j-3}\big).
\end{multline}
Clearly, we get back the determinant in \eqref{eq:det-factors3} for
$X_i=2i+k/2-3/2$.

The determinant in \eqref{eq:gen-det3} is a polynomial in $m$ and the
$X_i$'s of degree $p(4p-3)$. It is divisible by $\prod _{1\le
i<j\le p} ^{}(X_j^2-X_i^2)\prod _{i=1} ^{p}X_i$. 

The leading term of \eqref{eq:gen-det3} is
\begin{align*} 
&\det\limits_{1\le i,j\le p}\big((\tfrac m2+X_i)^{2p+2j-3}
(\tfrac m2-X_i)^{2p-2j}-
(\tfrac m2+X_i)^{2p-2j}(\tfrac m2-X_i)^{2p+2j-3}\big)\\
&\kern10pt=\prod _{i=1} ^{p}(\tfrac m2+X_i)^{2p-3/2}(\tfrac
m2-X_i)^{2p-3/2}
\det\limits_{1\le i,j\le p}\(\(\frac {\tfrac m2+X_i} {\tfrac
m2-X_i}\)^{2j-3/2}-
\(\frac {\tfrac m2+X_i} {\tfrac m2-X_i}\)^{-2j+3/2}\)\\
&\kern10pt=\prod _{i=1} ^{p}(\tfrac m2+X_i)^{p-1}(\tfrac m2-X_i)^{3p-2}
\prod _{i=1}
^{p}\(\(\frac {\tfrac m2+X_i} {\tfrac m2-X_i}\)-1\)
\\
&\kern10pt\kern1cm
\times
\prod _{1\le i<j\le p} ^{}\(\(\frac {\tfrac m2+X_i} {\tfrac m2-X_i}\)
-\(\frac {\tfrac m2+X_j} {\tfrac m2-X_j}\)\)
\(1-\(\frac {(\tfrac m2+X_i)} {(\tfrac m2-X_i)}\frac {(\tfrac m2+X_j)} 
{(\tfrac m2-X_j)}\)\)\\
&\kern10pt\kern1cm
\times
\so_{(p-1,p-2,\dots,1,0)}
\(\(\tfrac {\tfrac m2+X_1} {\tfrac m2-X_1}\)^{\pm1},\dots,
\(\tfrac {\tfrac m2+X_p} {\tfrac m2-X_p}\)^{\pm1},1\)\\
&\kern10pt=\prod _{1\le i<j\le p} ^{}
\Big(\big((\tfrac m2+X_j)(\tfrac m2-X_i)-
(\tfrac m2+X_i)(\tfrac m2-X_j)\big)\\
&\kern10pt\kern1cm
\cdot
\big((\tfrac m2+X_j)(\tfrac m2+X_i)-(\tfrac m2-X_i)(\tfrac m2-X_j)\big)\Big)
\prod _{i=1} ^{p}\big((\tfrac m2+X_i)-(\tfrac m2-X_i)\big)\\
&\kern.9cm
\times
\prod _{i=1} ^{p}(\tfrac m2+X_i)^{p-1}(\tfrac m2-X_i)^{p-1}
\so_{(p-1,p-2,\dots,1,0)}
\(\(\tfrac {\tfrac m2+X_1} {\tfrac m2-X_1}\)^{\pm1},\dots,
\(\tfrac {\tfrac m2+X_p} {\tfrac m2-X_p}\)^{\pm1},1\).
\end{align*} 
Here we used \eqref{eq:unortho} to evaluate the determinant in the
second line.
Now we recall the observation (see the paragraph containing
\eqref{e13}) that the odd orthogonal character is a certain Laurent
polynomial in its variables. In the present context it implies that
the very last line in our computation is a {\em polynomial\/} in the
quantities $(m/2+X_i)$, $(m/2-X_i)$, $i=1,2,\dots,p$, consisting of a
sum of exactly $\so_{(p-1,\dots,1,0)}(1,1,\dots,1)$ monomials, the
evaluation of the orthogonal character at all 1s being given by \eqref{e13}
itself. 

Hence, the determinant \eqref{eq:gen-det3} equals
\begin{multline} \label{eq:leading}
\Big(\prod _{1\le i<j\le p} ^{}m(X_j-X_i)m(X_j+X_i)\Big)
\Big(\prod _{i=1} ^{p}2X_i\Big)\\
\times
2^{-p}\bigg(\prod _{\ell=1} ^{p}\frac {(2\ell-2)!\,(4\ell-2)!}
{(\ell+p-1)!^2}\bigg)
\(\(\frac {m} {2}\)^{2p^2-2p}+\cdots\).
\end{multline}

The leading term of the determinant in \eqref{eq:det-factors3} is
obtained from this
expression under the 
substitution of $X_i=2i+k/2-3/2$, $i=1,2,\dots,p$. This substitution
turns \eqref{eq:leading} into
\begin{multline*} 
m^{3p^2-3p}2^{-p^2+p}\prod _{\ell=1} ^{p}\frac
{(\ell-1)!\,(2\ell-2)!\,(4\ell-4)!} {(\ell+p-1)!^2}
\prod _{\ell=1} ^{p}(2\ell+\tfrac {k} {2}-\tfrac {1} {2})_{p-\ell}
\prod _{\ell=1} ^{p}(2\ell+\tfrac {k} {2}-\tfrac {3} {2})\\
+\text { lower terms}.
\end{multline*}
When expanded, the term in this expression
which will give the largest contribution is
\begin{equation*} 
k^{\binom {p+1}2}m^{3p^2-3p}2^{-3p^2/2+p/2}\prod _{\ell=1} ^{p}\frac
{(\ell-1)!\,(2\ell-2)!\,(4\ell-4)!} {(\ell+p-1)!^2}.
\end{equation*}
What remains to do is to multiply this expression by the product on
the right-hand side of \eqref{eq:det-factors3}, then sum the
resulting expression over all $k\ge0$, manipulate the summand in a way
analogous to the manipulations in \eqref{eq:S(k)-asy}, and finally
estimate the result using Lemma~\ref{lem:expsum-asy}. After
simplification, the final result is \eqref{eq:watermelon-inf-with}.
\end{proof}


Again, since the expressions \eqref{eq:watermelon-with} and 
\eqref{eq:watermelon-inf-with} are identical except for a
multiplicative constant, we obtain as a consequence
the following result for $n$-stand models.

\begin{Corollary} \label{cor:n-stand-watermelon-with}
As $m$ tends to infinity, $n$-stand watermelons with $p$ branches of
length $m$ (and arbitrary deviation) which do not go below the
$x$-axis have, up to a multiplicative
constant, the same asymptotic behaviour. More precisely, the number of 
$n$-stand watermelons with $p$ branches of length $m$ which do not go
below the $x$-axis is $\asymp 
2^{mp}m^{-3p^2/4-p/4+1/2}$, i.e.,
there are positive constants $e_1$ and $e_2$
such that for large enough $m$ this number is between
$e_12^{mp}m^{-3p^2/4-p/4+1/2}$ and $e_22^{mp}m^{-3p^2/4-p/4+1/2}$.
Under the assumption that there is a constant $e(n)$ such that this
number is asymptotically exactly equal to $e(n)2^{mp}m^{-3p^2/4-p/4+1/2}$,
then we must have $e(0)<e(1)<e(2)<\cdots$, i.e., for any $n$ there are,
asymptotically, strictly less $n$-stand watermelons with $p$ branches of
length $m$ which do not go below the $x$-axis than 
$(n+1)$-stand watermelons with $p$ branches of
length $m$ which do not go below the $x$-axis.
\end{Corollary}

\section{The asymptotics of stars, restricted by one wall} 

\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 points 
have $y$-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{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
nonintersecting 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
$\infty$-stand 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 $\infty$-stand 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 $n$-stand 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:starvicious-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 $\infty$-stand 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 $\infty$-stand 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 $\infty$-stand 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 $n$-stand stars restricted by two walls will have
the same order of magnitude as (ordinary) stars, or not.

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\end{document}
