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\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma} 
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem*{remark}{Remark}

\def\v#1{{\vert #1\vert}}
\def\Tr{\operatorname{Tr}}
\def\po#1#2{(#1)_#2}
\def\fl#1{\left\lfloor#1\right\rfloor}
\def\cl#1{\left\lceil#1\right\rceil}
\def\bX{{\bar X}}

\begin{document}

We consider the average of $\overset n\otimes \rho$, evaluated at
$y=0$, with respect
to the probability distribution
\begin{equation} \label{eq:1}
q(u)=\frac {(1-u)} {\pi\,(1-x^2-z^2)^u}, \quad
-\infty<u<1,
\end{equation}
taken over the unit disk $\{(x,z):x^2+z^2\le 1\}$.
This average can be described explicitly as follows.

\begin{theorem} \label{thm:1}
The average 
$$\int_{x^2+z^2\le 1}\big(\overset n\otimes \rho\big|_{y=0}\big)\,
q(u)\,dx\,dz$$ 
equals the matrix $\tau_n(u)=(T_{IJ})_{I,J\in[n]}$,
where
\begin{multline} \label{eq:2}
T_{IJ}=\chi(n_{\notin\in}\equiv n_{\in\notin}\text { \em mod } 2)
\frac {\big( {n-n_{\in\in}-
n_{\notin\notin}} -1\big)!!} {2^{n+(n-n_{\in\in}-
n_{\notin\notin})/2}}\\
\times
\frac {\Ga\( {2}-u\)\,\Ga\(\frac {3} {2}+\frac {n} {2}+
\frac{n_{\in\in}}2-\frac {n_{\notin\notin}}2-u\)\, 
\Ga\(\frac {3} {2}+\frac {n} {2}+\frac {n_{\notin\notin}}2-
\frac {n_{\in\in}}2-u\)} {\Ga\(\frac {3} {2}+\frac {n} {2}-u\)\,
\Ga\(2+\frac {n} {2}-u\)\, \Ga\(\frac {3} {2}+\frac {n} {2}-\frac
{n_{\in\in}}2- \frac {n_{\notin\notin}}2-u\)}.
\end{multline}
Here, $\chi(\mathcal A)$ denotes the truth function, $\chi(\mathcal A)=1$ if
$\mathcal A$ is true
and $\chi(\mathcal A)=0$ otherwise.
\end{theorem}

\begin{remark} \em It is important for later considerations to observe
that because of the term $\chi(n_{\notin\in}\equiv n_{\in\notin}\text { mod
}2)$ in
\eqref{eq:2}
the entry $T_{IJ}$ is nonzero if and only if the cardinalities of the 
sets $I$ and $J$ have the same parity. 
\end{remark}

Now, given integers $h,s$, $0\le h\le n$, $0\le s\le n/2-h$, and two
disjoint $h$-element subsets $A$ and $B$ of $[n]$, define the vector
 $v^{(1)}_{h,s}(A,B)$ by
\begin{multline} \label{eq:3}
v^{(1)}_{h,s}(A,B):=\\
\underset {Y\subseteq [n]\backslash (A\cup B),\ 
\v{Y}\equiv 0\text{ (mod 2)}}
{\sum _{X\subseteq A} ^{}}\kern -10pt e_{X\cup X'\cup
Y}\,(-1)^{\v{X}+\v{Y}/2}\sum _{i=0} ^{s}(-4)^i\binom {n-2h-2i}{s-i}
\binom {\v Y/2}i
,
\end{multline}
where $X'$ is again the ``{\it complement of $X$ in $B$}", and 
define the vector
 $v^{(2)}_{h,s}(A,B)$ by
\begin{multline} \label{eq:4}
v^{(2)}_{h,s}(A,B):=\\
\underset {Y\subseteq [n]\backslash (A\cup B),\ 
\v{Y}\equiv 1\text{ (mod 2)}}
{\sum _{X\subseteq A} ^{}}\kern-10pt e_{X\cup X'\cup
Y}\,(-1)^{\v{X}+(\v{Y}-1)/2}\sum _{i=0} ^{s}(-4)^i\binom {n-2h-2i}{s-i}
\binom {\v Y/2}i
.
\end{multline}
Then we have the following theorem:

\begin{theorem} \label{thm:2}
If $A$ and $B$ are two disjoint $h$-element
subsets of $[n]$, and if $h,s$ are integers with $0\le h\le n$, $0\le
s\le n/2-h$, then the vectors $v^{(j)}_{h,s}(A,B)$, $j=1,2$, are 
eigenvectors of
the averaged matrix $\tau_n(u)=(T_{IJ})_{I,J\in[n]}$ for the eigenvalue
\begin{equation} \label{eq:5}
\la_{h,s}:=
\frac {1} {2^n}\frac { \Ga(2+n-h-u)\,
\Ga(1+h-u)\, \Ga\( {2}-u\)} {\Ga\({2}+ {n} -h-s-u\)\, \Ga(2+h+s-u)\,\Ga(1-u)}.
\end{equation}
The vectors
\begin{multline} \label{eq:vect}
\{v^{(1)}_{h,s}(P):0\le h\le n,\ 
0\le s\le n/2-h,\ P\text { a ballot path from $(0,0)$
to $(n,n-2h)$}\}\\
\cup \{v^{(2)}_{h,s}(P):0\le h\le n,\ 
0\le s< n/2-h,\ P\text { a ballot path from $(0,0)$
to $(n,n-2h)$}\},
\end{multline}
where $v^{(j)}_{h,s}(P)=v^{(j)}_{h,s}(A_P,B_P)$, $j=1,2$, with $A_P$ and
$B_P$ as in \cite[Sec.~2.2]{SlatAC},
form a basis of the $2^n$-dimensional vector space.
\end{theorem}

\begin{proof}We first concentrate on the first assertion, i.e., 
that for any integers $h,s$, $0\le h\le n$, $0\le s\le n/2-h$, and two
disjoint $h$-element subsets $A$ and $B$ of $[n]$, we have
$$\tau_n(u)\cdot v^{(j)}_{h,s}(A,B)=\la_{h,s} \,v^{(j)}_{h,s}(A,B)\ ,$$
for $j=1,2$. We treat the case of $j=1$, the other case being
completely analogous.
Restricting our attention to the $I$-th component, we see 
from the
definition \eqref{eq:3} of $v^{(1)}_{h,s}(A,B)$ that we need to establish 
\begin{multline}\label{eq:8}
\underset {Y\subseteq [n]\backslash (A\cup B),\ 
\v{Y}\equiv 0\text{ (mod 2)}}
{\sum _{X\subseteq A} ^{}}\kern-1cm T_{I,X\cup X'\cup
Y}\,(-1)^{\v{X}+\v{Y}/2}
\sum _{i=0} ^{s}(-4)^i\binom {n-2h-2i}{s-i}
\binom {\v Y/2}i
\\=
\begin{cases} 
\la_{h,s} (-1)^{\v{U}+\v{V}/2}\sum _{i=0} ^{s}(-4)^i\binom {n-2h-2i}{s-i}
\binom {\v V/2}i
&\text {if $I$ is of the form $U\cup U'\cup V$}\\
&\text {for some $U$ and $V$, $U\subseteq A$,}\\
&\text {$V\subseteq [n]\backslash (A\cup B)$,
$\v{V}\equiv 0\text { (mod 2)}$}\\
0&\text {otherwise.}\end{cases}
\end{multline}
We prove \eqref{eq:8} by a case by case analysis. The first two cases cover
the case ``otherwise" in \eqref{eq:8}, the third case treats the first alternative
in \eqref{eq:8}.

\smallskip
{\it Case 1. The cardinality of $I$ is of parity different from the
parity of $h$}. Clearly, the cardinality of any set $X\cup X'\cup Y$ which
occurs as index at the left-hand side of \eqref{eq:8} has the same
parity as $h$. The parity of $I$ however is different from the
partity of $h$. 
As we observed in the Remark after Theorem~\ref{thm:1}, this implies that
any coefficient $T_{I,X\cup X'\cup Y}$ on the left-hand side
vanishes. Thus, \eqref{eq:8} is proved in this case.

\smallskip
{\it Case 2. The cardinality of $I$ is of the same parity as $h$, but
$I$ does not have the form $U\cup U'\cup V$
for any $U$ and $V$, $U\subseteq A$,
$V\subseteq [n]\backslash (A\cup B)$, $\v{V}\equiv 0\text{ (mod 2)}$}.
Now the sum on the left-hand side of \eqref{eq:8} contains nonzero
contributions. We have to show that they cancel each other. We do
this by grouping summands in pairs, the sum of each pair being 0.

Consider a set $X\cup X'\cup Y$ which occurs as index at the
left-hand side of \eqref{eq:8}. Let $e$ be minimal such that 
\begin{enumerate}
\item[]either: the $e$-th largest element of $A$ and the
$e$-th largest element of $B$ are both in $I$,
\item[]or: the $e$-th largest element of $A$ and the
$e$-th largest element of $B$ are both not in $I$.
\end{enumerate}
That such an $e$ must exist is guaranteed by our assumptions about
$I$. Now consider $X$ and $X'$. If the $e$-th largest element of $A$
is contained in $X$ then the $e$-th largest element of $B$ is not
contained in $X'$, and vice versa. Define a new set $\bX$ by adding
to $X$ the $e$-th largest element of $A$ if it is not already
contained in $X$, respectively by removing it from $X$ if it is
contained in $X$. Then, it is easily checked that 
$$T_{I,X\cup X'\cup Y}=T_{I,\bX\cup \bX'\cup Y}.$$
On the other hand, we have $(-1)^{\v{X}}=-(-1)^{\v\bX}$ since the
cardinalities of $X$ and $\bX$ differ by $\pm1$. Both facts combined
give
$$T_{I,X\cup X'\cup Y}\,(-1)^{\v{X}}+T_{I,\bX\cup \bX'\cup Y}\,
(-1)^{\v\bX}=0.$$
Hence, we have found two summands on the left-hand side of
\eqref{eq:8} which cancel each other. 

Summarizing,
this construction finds for any $X,Y$ sets $\bX,Y$ such that the
corresponding summands on the left-hand side of \eqref{eq:8} cancel each
other. Moreover, this construction applied to $\bX,Y$ gives back
$X,Y$. Hence, what the construction does is exactly what we claimed,
namely it groups the summands into pairs which contribute 0 to the
whole sum. Therefore the sum is 0, which establishes \eqref{eq:8} in this case
also.

\smallskip
{\it Case 3. $I$ has the form $U\cup U'\cup V$
for some $U$ and $V$, $U\subseteq A$,
$V\subseteq [n]\backslash (A\cup B)$, $\v{V}\equiv 0\text{ (mod 2)}$}.
This assumption implies in particular that the cardinality of $I$ is
of the same parity as $h$. If we write $N(j,k,l)$
for the number of sets $X\cup X'\cup Y$,
for some $X$ and $Y$, $X\subseteq A$, $\v{(U\cup U')\cap (X\cup
X')}=j$,
$Y\subseteq [n]\backslash (A\cup B)$, $\v{Y}=2k$, $\v{V\cap Y}=l$,
then the left-hand side
in \eqref{eq:8} reduces to
\begin{multline}\label{eq:9}
\sum _{j,k,l,i\ge0} ^{}N(j,k,l)\,(-1)^{\v{U}+h-j+k}
\frac {(\v{V}+2h-2j+2k-2l-1)!!} {2^{n+\v{V}/2+h-j+k-l}}\\
\cdot
\frac {\Ga\(2-u\)\, \Ga(\frac {3} {2}+\frac {\v{V}}
{2}+h+k-u)\,
\Ga(\frac {3} {2}+n-\frac {\v{V}} {2}-h-k-u)} 
{\Ga\(\frac {3} {2}+\frac {n} {2}-u\)\, \Ga\(2+\frac {n}
{2}-u\)\Ga(\frac {3} {2}+\frac {\v{V}} {2}+h-j+k-l-u)}\\
\cdot
(-4)^i\binom {n-2h-2i}{s-i}\binom ki.
\end{multline}
Clearly, we used expression \eqref{eq:2} with
$n_{\in\in}=j+l$, $n_{\in\notin}=h-j+\v{V}-l$,
$n_{\notin\in}=h-j+2k-l$, and $n_{\notin\notin}=n-2h-\v{V}-2k+l+j$.

To determine $N(j,k,l)$, note first that there are $\binom hj$
possible sets $X\cup X'$ which intersect $U\cup U'$ 
in exactly $j$ elements. Next, let us assume that we already made a
choice for $X\cup X'$.
In order to determine the number of possible sets $Y$ such that
$\v{Y}=2k$ and $\v{V\cap Y}=l$, we have to
choose $l$ elements from
$V$, for which we have $\binom {\v{V}}{l}$ possibilities, and
we have to choose $2k-l$ elements from $[n]\backslash
(I\cup A\cup B)$ to obtain a total number of $2k$ elements, for which
we have $\binom {n-2h-\v{V}}{2k-l}$ possibilities. Hence,
\begin{equation}\label{eq:10}
N(j,k,l)=\binom hj\binom {\v{V}}{l}\binom {n-2h-\v{V}}{2k-l}.
\end{equation}

So it remains to evaluate the double sum \eqref{eq:9}, using the
expression \eqref{eq:10}
for $N(j,k,l)$. 

We start by writing the sum over $j$ in \eqref{eq:9} in hypergeometric
notation \eqref{eq:hyp},
\begin{multline}\notag 
\sum _{k,l,i\ge0} ^{}(-1)^{\v{U}+h+k}
\binom {\v{V}}{l}\binom {n-2h-\v{V}}{2k-l}
\frac {(\v{V}+2h+2k-2l-1)!!} {2^{n+\v{V}/2+h+k-l}}\\
\cdot
\frac {\Ga\(2-u\)\, \Ga(\frac {3} {2}+\frac {\v{V}}
{2}+h+k-u)\,
\Ga(\frac {3} {2}+n-\frac {\v{V}} {2}-h-k-u)} 
{\Ga\(\frac {3} {2}+\frac {n} {2}-u\)\, \Ga\(2+\frac {n}
{2}-u\)\Ga(\frac {3} {2}+\frac {\v{V}} {2}+h+k-l-u)}\\
\cdot
(-4)^i\binom {n-2h-2i}{s-i}\binom ki\,
{}_2F_1\!\[\matrix -\frac {1} {2}-\frac {\v{V}} {2}-h-k+l+u,-h\\
\frac {1} {2}-\frac {\v{V}} {2}-h-k+l\endmatrix; 1\].
\end{multline}
We can evaluate the ${}_{2}F_{1}$-series by the Chu--Vandermonde summation 
formula (see \cite[(1.7.7), Appendix~(III.4)]{SlatAC}),
\begin{equation} \label{vand}
{} _{2} F _{1} \!\left [ \begin{matrix} { a, -n}\\ { c}\end{matrix} ;
   {\displaystyle 1}\right ]  = 
  {\frac{({ \textstyle c-a}) _{n} }{({ \textstyle c}) _{n} }},
\end{equation}
where $n$ is a nonnegative integer. Thus we get the expression
\begin{multline} \label{eq:11}
(-1)^{\v{U}}\,(1-u)_h
\sum _{k,l,i\ge0} ^{}(-1)^{k}
\binom {\v{V}}{l}\binom {n-2h-\v{V}}{2k-l}
\frac {(\v{V}+2k-2l-1)!!} {2^{n+\v{V}/2+k-l}}\\
\cdot
\frac {\Ga\(2-u\)\, \Ga(\frac {3} {2}+\frac {\v{V}}
{2}+h+k-u)\,
\Ga(\frac {3} {2}+n-\frac {\v{V}} {2}-h-k-u)} 
{\Ga\(\frac {3} {2}+\frac {n} {2}-u\)\, \Ga\(2+\frac {n}
{2}-u\)\Ga(\frac {3} {2}+\frac {\v{V}} {2}+h+k-l-u)}\\
\cdot
(-4)^i\binom {n-2h-2i}{s-i}\binom ki.
\end{multline}
Now we claim that
\begin{multline} \label{eq:triple}
\sum_{l = 0}^{{v}}\sum_{k = \cl{(v+l)/2}}^{\fl{(n+l)/2}}
      \sum_{i = 0}^{s}{{\left( -1 \right) }^{ k}}
\frac {\left( 2k-2l-1 \right) !!} {{2^{n+k-l}}}
             {\binom {{v}} l} 
            {\binom {n-v} {n+l-2k}}\\
\cdot
{\frac {    \Gamma({ \textstyle 2 - u}) \,
            \Gamma({ \textstyle {\frac 3 2} + k - u}) \,
            \Gamma({ \textstyle {\frac 3 2} - k + n - u}) } 
          {   \Gamma({ \textstyle {\frac 3 2} + {\frac n 2} - u}) \,
            \Gamma({ \textstyle 2 + {\frac n 2} - u}) \,
          \Gamma({ \textstyle {\frac 3 2} + k - l - u}) }}\,
(-4)^i            {\binom {n-2i} {s-i}} 
            {\binom {k-{\frac {{v}} 2} } i} 
 \\= \frac {1} {2^n}
  {\frac {\Gamma({ \textstyle 2 - u}) \,\Gamma({ \textstyle 2 + n - u}) \,
       } 
    {\Gamma({ \textstyle 2 + n - s - u}) \,
      \Gamma({ \textstyle 2 + s - u}) }}
 \sum_{i = 0}^{s}{{\left( -4 \right) }^i}\,
          {\binom {{\frac {{v}} 2}} i}\,{\binom {n-2i} {s-i}}.
\end{multline}
Use of this identity with $v=\v{V}$, with $u$ replaced $u-h$, $n$
replaced by $n-2h$, and with $k$ replaced by $k+\v{V}/2$, 
transforms \eqref{eq:11} into
\begin{multline*} 
(-1)^{\v{U}+\v{V}/2}
\frac {1} {2^n}\frac {\Gamma(2-u)\,\Gamma(2+n-h-u)\,\Gamma(1+h-u)} 
{\Gamma(1-u)\,\Gamma(2+n-h-s-u)\,\Gamma(2+h+s-u)}\\
\times
 \sum_{i = 0}^{s}{{\left( -4 \right) }^i}\,
{\binom {n-2h-2i} {s-i}}\,          {\binom {{\frac {\v{V}} 2}} i},
\end{multline*}
which, in view of \eqref{eq:5}, 
is exactly equal to the first alternative on the right-hand side of
\eqref{eq:8}, as desired.

It remains to establish \eqref{eq:triple}.
For fixed $n,s,u$ this identity is polynomial in $v$, because the
summation over $l$ can be extended to run from $0$ to $n$ (instead of
from $0$ to $v$), and the
summation over $k$ can be extended to run from $0$ to $\fl{(n+l)/2}$ (instead of
from $\cl{(v+l)/2}$ to $\fl{(n+l)/2}$). Therefore it suffices
to prove \eqref{eq:triple} for all nonnegative {\em even} integers
$v$.

From now on let $v$ be an even, nonnegative integer.
We start with the left-hand side of \eqref{eq:triple}, which we will,
step by step, transform into the right-hand side of \eqref{eq:triple}.
We make the sum over $l$ the innermost sum and write it in
hypergeometric notation, so that we obtain
\begin{multline} \notag
\frac {1} {2^n}\sum _{i=0} ^{s}\sum _{k=i+v/2} ^{\fl{(n+v)/2}}
{{\left( -1 \right) }^{k + i}}\,{4^i}\,\binom {n-2i}{s-i}
{\frac {     ({ \textstyle {\frac 1 2}}) _{k} \,
     ({ \textstyle 1 - {v} + 2 k}) _{n-2 k} \,
     ({ \textstyle 1 - {\frac {{v}} 2} + k - i}) _{i} } 
   {({ \textstyle 1}) _{i} \,
     ({ \textstyle 1}) _{n-2 k} }}\\
\cdot
\frac {  \Gamma({ \textstyle 2 - u}) \,
     \Gamma({ \textstyle {\frac 3 2} - k + n - u}) } 
  {\Gamma({ \textstyle {\frac 3 2} + {\frac n 2} - u}) \,
     \Gamma({ \textstyle 2 + {\frac n 2} - u}) }
     {} _{3} F _{2} \!\left [ \begin{matrix} { -{v}, {v} - 2 k,
-{\frac 1
      2} - k + u}\\ { {\frac 1 2} - k, 1 - 2 k + n}\end{matrix} ; {\displaystyle
      1}\right ]
\end{multline}
for the left-hand side of \eqref{eq:triple}. To the $_3F_2$-series we
apply the quadratic transformation (cf\@. \cite[Appendix (III.21), 
$q\uparrow1$]{GaRaAA})
$$
{} _{3} F _{2} \!\left [ \begin{matrix} { 2 a, 2 b, c}\\ { {\frac 1 2} + a + b,
   d}\end{matrix} ; {\displaystyle 1}\right ]  =
  {} _{4} F _{3} \!\left [ \begin{matrix} { a, b, c, d-c}\\ { {\frac 1 2} + a + b,
   {\frac d 2}, {\frac {1 + d} 2}}\end{matrix} ; {\displaystyle 1}\right ]
,
$$
provided both series terminate.
This gives
\begin{multline} \label{t1}
\frac {1} {2^n}\sum _{i=0} ^{s}\sum _{k=i+v/2} ^{\fl{(n+v)/2}}
{{\left( -1 \right) }^{k + i}}\,{4^i}\,\binom {n-2i}{s-i}
{\frac {      ({ \textstyle {\frac 1 2}}) _{k} \,
     ({ \textstyle 1 - {v} + 2 k}) _{n-2 k } \,
     ({ \textstyle 1 - {\frac {{v}} 2} + k - i}) _{i} } 
   {({ \textstyle 1}) _{i} \,
     ({ \textstyle 1}) _{n-2 k } }}\\
\cdot
\frac {\Gamma({ \textstyle 2 - u}) \,
     \Gamma({ \textstyle {\frac 3 2} - k + n - u}) } 
   {\Gamma({ \textstyle {\frac 3 2} + {\frac n 2} - u}) \,
     \Gamma({ \textstyle 2 + {\frac n 2} - u})}
     {} _{4} F _{3} \!\left [ \begin{matrix} { {\frac 3 2} - k + n - u,
{\frac {{
      v}} 2} - k, -{\frac 1 2} - k + u, -{\frac {{v}} 2}}\\ { {\frac 1
      2} - k, {\frac 1 2} - k + {\frac n 2}, 1 - k + {\frac n 2}}\end{matrix} ;
      {\displaystyle 1}\right ].
\end{multline}
(The $_4F_3$-series does indeed terminate as required, as it contains
$-v/2$ as an upper parameter, which is a nonpositive integer since,
by assumption, $v$ is even.)
Next we make use of Sears' balanced $_4F_3$-transformation (cf\@. \cite[(4.3.5.1)]{SlatAC})
\begin{multline}\notag
{} _{4} F _{3} \!\left [ \begin{matrix} { a, b, c, -N}\\ { e, f, 1 + a + b + c - e -
   f - N}\end{matrix} ; {\displaystyle 1}\right ] =
{\frac {( e-a )_N\,( f-a ) _{N}}  {( e)_N\,( f) _{N}}}\\
\times
  {} _{4} F _{3} \!\left [ \begin{matrix} { -N, a, 1 + a + c - e - f - N, 1 + a + b -
    e - f - N}\\ { 1 + a + b + c - e - f - N, 1 + a - e - N, 1 + a - f -
    N}\end{matrix} ; {\displaystyle 1}\right ]   ,
\end{multline}
where $N$ is a nonnegative integer.
We use this identity in \eqref{t1} (again, this is legal because 
$v/2$ is indeed a nonnegative integer), 
write the resulting
$_4F_3$-series as a sum over $l$, interchange the sums over $k$ and
$l$, and finally write the now innermost sum over $k$ in
hypergeometric notation. As a result we obtain the expression
\begin{multline} \label{t2}
\frac {1} {2^n}\sum _{i=0} ^{s} \sum_{l = 0}^{v/2}
{{\left( -1 \right) }^{{\frac {{v}} 2} }}\,
       {4^i} \,\binom {n-2i}{s-i}
{\frac {\Gamma({ \textstyle 2 - u}) \,
       \Gamma({ \textstyle {\frac 3 2} - {\frac {{v}} 2} + l + n - i - u})
        \,({ \textstyle {\frac 1 2}}) _{{\frac {{v}} 2} + i} \,
       ({ \textstyle -{\frac {{v}} 2}}) _{l} } 
     {\Gamma({ \textstyle {\frac 3 2} + {\frac n 2} - u}) \,
       \Gamma({ \textstyle 2 + {\frac n 2} - u}) \,({ \textstyle 1}) _{l} \,
       ({ \textstyle 1}) _{n-2i-{v} } }}\\
\cdot
\frac {       ({ \textstyle 1 - {\frac {{v}} 2} + {\frac n 2}}) _{l} \,
       ({ \textstyle 1 + 2 i}) _{n-2i-{v}} \,
       ({ \textstyle {\frac 3 2} + {\frac n 2} - u}) _{l} \,
       ({ \textstyle -1 - n + u}) _{{\frac {{v}} 2}} \,
       ({ \textstyle -1 - {\frac n 2} + u}) _{{\frac {{v}} 2}} } 
 {       ({ \textstyle {\frac 1 2} - {\frac {{v}} 2} - i}) _{{\frac {{v}}
        2}} \,({ \textstyle {\frac 1 2} - {\frac {{v}} 2} + {\frac n 2} -
        i}) _{{\frac {{v}} 2}} \,
       ({ \textstyle 1 - {\frac {{v}} 2} + {\frac n 2} - i}) _{l} \,
       ({ \textstyle 2 - {\frac {{v}} 2} + {\frac n 2} - u}) _{l} \,
       ({ \textstyle 2 - {\frac {{v}} 2} + n - u}) _{l} }\\
\cdot
{} _{2} F _{1} \!\left [ \begin{matrix} { {\frac 1 2} - {\frac n
        2} + i, {\frac {{v}} 2} - l - {\frac n 2} + i}\\ { -{\frac 1 2} +
        {\frac {{v}} 2} - l - n + i + u}\end{matrix} ; {\displaystyle
        1}\right ].
\end{multline}
The $_2F_1$-series is terminating. This is because either $1/2-n/2+i$
or $v/2-l-n/2+i$ is integral (again, remember that $v$ is even by
assumption), and because $1/2-n/2+i\le 1/2-n/2+s\le 1/2$ and
$v/2-l-n/2+i\le -l$. The latter inequality holds 
since we may restrict the sum over $i$ in
\eqref{t2} to $i\le n/2-v/2$ as the term $(1)_{n-2i-v}$ in the
denominator makes all
summands with $i>n/2-v/2$ vanish.
Hence, the $_2F_1$-series can be summed by means of Chu--Van\-der\-monde
summation \eqref{vand}. We simplify the result and write the sum over
$l$ in hypergeometric notation to get
\begin{multline} \notag
\frac {1} {2^n}\sum _{i=0} ^{s}
{{\left( -1 \right) }^{{\frac {{v}} 2} + i}}\,{4^i}\,\binom {n-2i}{s-i}
{\frac {     ({ \textstyle {\frac 1 2}}) _{i} \,
     ({ \textstyle 1 + 2 i}) _{n-2i-{v} } \,
     ({ \textstyle -1 - n + u}) _{{\frac {{v}} 2}} } 
   {({ \textstyle 1}) _{n-2i-{v} } \,
     ({ \textstyle {\frac 1 2} - {\frac {{v}} 2} + {\frac n 2} - i})
      _{{\frac {{v}} 2}} \,({ \textstyle -1 - i + u}) _{i} }}\\
\cdot
     {} _{3} F _{2} \!\left [ \begin{matrix} { {\frac 3 2} + {\frac n 2} - u, 1
      - {\frac {{v}} 2} + {\frac n 2}, -{\frac {{v}} 2}}\\ { 1 -
{\frac {{
      v}} 2} + {\frac n 2} - i, 2 - {\frac {{v}} 2} + n -
      u}\end{matrix} ; {\displaystyle 1}\right ].
\end{multline}
Now we apply the following transformation formula due to
Thomae (see e.g\@. \cite[(3.1.1)]{GaRaAA}),
\begin{equation} \label{eq:Thomae}
{} _{3} F _{2} \!\left [ \begin{matrix} { a, b, -N}\\ { d, e}\end{matrix} ;
   {\displaystyle 1}\right ]  =
{\frac { ({ \textstyle e-b}) _{N} }
      {({ \textstyle e}) _{N} }}  
{} _{3} F _{2} \!\left [ \begin{matrix} { -N, b, d-a}\\ { d, 1 + b - e -
       N}\end{matrix} ; {\displaystyle 1}\right ] ,
\end{equation}
where $N$ is a nonnegative integer. We may indeed apply
\eqref{eq:Thomae} because, by assumption, $v/2$ is a nonnegative
integer. This gives
\begin{multline} \notag
\frac {1} {2^n}\sum _{i=0} ^{s}
{{\left( -1 \right) }^{{\frac {{v}} 2} + i}}\,{4^i}\,\binom {n-2i}{s-i}
{\frac {     ({ \textstyle {\frac 1 2}}) _{i} \,
     ({ \textstyle 1 + 2 i}) _{n-2i-{v} } \,
     ({ \textstyle 1 + {\frac n 2} - u}) _{{\frac {{v}} 2}} \,
     ({ \textstyle -1 - n + u}) _{{\frac {{v}} 2}} } 
   {({ \textstyle 1}) _{n-2i-{v} } \,
     ({ \textstyle {\frac 1 2} - {\frac {{v}} 2} + {\frac n 2} - i})
      _{{\frac {{v}} 2}} \,({ \textstyle 2 - {\frac {{v}} 2} + n - u})
      _{{\frac {{v}} 2}} \,({ \textstyle -1 - i + u}) _{i} }}\\
\cdot
     {} _{3} F _{2} \!\left [ \begin{matrix} { -{\frac {{v}} 2}, 1 -
      {\frac {{v}} 2} + {\frac n 2}, -{\frac 1 2} - {\frac {{v}} 2} - i
      + u}\\ { 1 - {\frac {{v}} 2} + {\frac n 2} - i, -{\frac {{v}} 2}
      - {\frac n 2} + u}\end{matrix} ; {\displaystyle 1}\right ].
\end{multline}
Here we need another $_3F_2$-transformation (Thomae??)
(cf\@. \cite[Ex.~7, p.~98]{BailAA}),
$$
{} _{3} F _{2} \!\left [ \begin{matrix} { a, b, c}\\ { d, e}\end{matrix} ;
   {\displaystyle 1}\right ]=
\frac {\Ga(e)\, \Ga(d+e-a - b - c )} {\Ga(e-a)\, \Ga(d+e-b - c )}
  {} _{3} F _{2} \!\left [ \begin{matrix} { a, d-b, d-c}\\ { d, d+e-b - c
 }\end{matrix} ; {\displaystyle 1}\right ]  .
$$
It transforms the expression obtained so far into
\begin{multline} \notag
\frac {1} {2^n}\sum _{i=0} ^{s}{{\left( -1 \right) }^{{\frac {{v}} 2} + i}}\,{4^i}\,
\binom {n-2i}{s-i}
{\frac {      ({ \textstyle {\frac 1 2}}) _{i} \,
     ({ \textstyle {\frac 1 2} - {\frac n 2}}) _{{\frac {{v}} 2}} \,
     ({ \textstyle 1 + 2 i}) _{n-2i-{v} } } 
   {({ \textstyle 1}) _{n-2i-{v} } \,
     ({ \textstyle {\frac 1 2} - {\frac {{v}} 2} + {\frac n 2} - i})
      _{{\frac {{v}} 2}} \,({ \textstyle -1 - i + u}) _{i} }}\\
\cdot
     {} _{3} F _{2} \!\left [ \begin{matrix} { {\frac 3 2} + {\frac n 2} - u,
      -i, -{\frac {{v}} 2}}\\ { {\frac 1 2} - {\frac n 2}, 1 - {\frac {{
      v}} 2} + {\frac n 2} - i}\end{matrix} ; {\displaystyle 1}\right
].
\end{multline}
We apply the transformation formula \eqref{eq:Thomae} once again.
Then we write the resulting $_3F_2$-series as a sum over $k$ and
interchange sums over $i$ and $k$. Writing the now innermost
sum over $i$ in hypergeometric notation, we obtain
\begin{multline}
\frac {1} {2^n}\sum_{k = 0}^{s}{(-4)^{ k}}
{\frac {       ({ \textstyle 1 + {\frac {{v}} 2} - k}) _{k} \,
       ({ \textstyle 1 - k + n - s}) _{s-k } \,
       ({ \textstyle 2 - k + n - u}) _{k} } 
     {({ \textstyle 1}) _{k} \,({ \textstyle 1}) _{s-k } \,
       ({ \textstyle 2 - u}) _{k} }}\\
\cdot
       {} _{2} F _{1} \!\left [ \begin{matrix} { k - s, k - n + s}\\ { 2 + k -
        u}\end{matrix} ; {\displaystyle 1}\right ].
\end{multline}
Once again, the $_2F_1$-series can be evaluated by means of the
Chu--Vandermonde summation \eqref{vand}. Subsequently, little simplification
turns the resulting expression into the right-hand side of
\eqref{eq:triple}.

\smallskip
Now we turn to the second assertion of the theorem, that the vectors
\eqref{eq:vect} form a basis of the $2^n$-dimensional vector space.
We will resort to a result from \cite[Lemmas~4 and 5]{KrSlAA}, which says
that the set of vectors 
\begin{equation}\label{eq:12}
\{v_{h,s}(P):0\le h\le s\le n-h,\ P\text { a ballot path from $(0,0)$
to $(n,n-2h)$}\}
\end{equation}
forms a basis of the $2^n$-dimensional vector space, where the
vectors $v_{h,s}(P)=v_{h,s}(A_P,B_P)$ are defined in \cite[(2.14)]{KrSlAA}.
Let the ballot path $P$ be fixed. 
By definition \eqref{eq:3} of
$v^{(1)}_{h,s}(P)=v^{(1)}_{h,s}(A_P,B_P)$ we have
$$v^{(1)}_{h,s}(P)=\sum _{y=0} ^{\fl{n/2-h}}(-1)^y 
\bigg( \sum_{i = 0}^{s}{{\left( -4 \right) }^i}\,
{\binom {n-2h-2i} {s-i}}\,          {\binom {y}
i}\bigg)v_{h,h+2y}(P),$$
and a similar equation holds for $v^{(2)}_{h,s}(P)$.
This means that the vectors $v^{(1)}_{h,s}(P)$,
$s=0,1,\dots,\fl{n/2-h}$, are in the span of the vectors
\begin{equation} \label{eq:vhy1}
v_{h,h+2y}(P),\quad y=0,1,\dots,\fl{n/2-h},
\end{equation}
and the vectors $v^{(2)}_{h,s}(P)$,
$s=0,1,\dots,\cl{n/2-h-1}$, are in the span of the vectors
\begin{equation} \label{eq:vhy2}
v_{h,h+2y+1}(P),\quad y=0,1,\dots,\cl{n/2-h-1}.
\end{equation}
We know that the latter, i.e., the vectors $v_{h,h+y}(P)$,
$y=0,1,\dots,n-2h$,
are linearly independent. Therefore, if we can show that the
transition matrices between these two sets of vectors, the matrices
\begin{equation} \label{eq:matrix}
\bigg((-1)^y 
 \sum_{i = 0}^{s}{{\left( -4 \right) }^i}\,
{\binom {n-2h-2i} {s-i}}\,          {\binom {y}
i}\bigg)_{0\le s,y\le N},
\end{equation}
where $N=\fl{n/2-h}$ or $N=\cl{n/2-h-1}$,
are nonsingular, then we also know that the former, i.e., the vectors
$v^{(1)}_{h,s}(P)$, $s=0,1,\dots,\fl{n/2-h}$, and
$v^{(2)}_{h,s}(P)$, $s=0,1,\dots,\cl{n/2-h-1}$, are linearly
independent. This suffices to establish our claim, because for
different $h$ and $P$ the vectors \eqref{eq:vhy1} and \eqref{eq:vhy2} generate 
different subspaces, and altogether (compare
\eqref{eq:12}) they form a
basis of the complete $2^n$-dimensional vector space.
On the other hand, that the matrix \eqref{eq:matrix} is nonsingular
is trivial. For, it is of the form $(p_s(x_y))_{0\le s,y\le N}$
(with $x_y\equiv y$),
where $p_s(x)$ is a polynomial in $x$ of degree $s$. The determinant
of such a matrix is
\begin{equation} \label{eq:Vandermonde-allg}
\det_{0\le s,y\le N}\(p_s(x_y)\)=a_0a_1\cdots a_N\prod _{0\le i<j\le N}
^{}(x_j-x_i),
\end{equation}
with $a_s$ denoting the leading coefficient of $p_s(x)$,
as is easily seen (cf\@. e.g\@. \cite{KratBN}). In particular, the
determinant must be nonzero, and, hence, the corresponding matrix
must be nonsingular.

\smallskip
The proof of Theorem~\ref{thm:2} is now complete.
\end{proof}

In fact, I am already heading for the more general situation of a
spherically symmetric weight function. (That is, in retrospective, 
the above proof is useless, except for the part where it is argued
that the vectors are indeed a complete basis of the $2^n$-dimensional
vector space.) I have very precise conjectures.

\begin{conjecture} \label{conj:1}
Let $w$ be a spherically symmetric probability density on the unit
disk, i.e., $w=w(x,z)$ depends only on $r=\sqrt{x^2+z^2}$.
Furthermore, let $\tau_n(w)$ be the average 
$\int_{x^2+z^2\le 1} \big(\overset n  \otimes \rho\big\vert_{y=0}
\big)\,w(x,z) \,dx\,dz$.

If $A$ and $B$ are two disjoint $h$-element
subsets of $[n]$, and if $h,s$ are integers with $0\le h\le n$, $0\le
s\le n/2-h$, then the vectors $v^{(j)}_{h,s}(A,B)$, $j=1,2$, are 
eigenvectors of
the averaged matrix $\tau_n(w)$ for the eigenvalue
\begin{align} \label{eq:eigen}
\la_{h,s}&=\frac {\pi} {2^{n-1}} \int _{0} ^{1}s(1-s^2)^h \sum
_{\ell=0} ^{s}\binom s\ell \binom {n-2h-s}\ell s^{2\ell}\,w(s)\,ds\\
\label{eq:eigen1}
&=\frac {\pi} {2^{n-1}} \int _{0} ^{1}s \sum
_{\ell=0} ^{s}(-1)^\ell\binom s\ell \binom {n-2h-\ell}s
(1-s^2)^{h+\ell}\,w(s)\,ds.
\end{align}
\end{conjecture}
In view of what is proved in
Theorem~\ref{thm:2} (second part), this describes all the eigenvalues, together
with multiplicities, of $\tau_n(w)$.

That the expressions \eqref{eq:eigen} and \eqref{eq:eigen1} are
identical is no big deal to show, this follows from some
$_2F_1$-transformation formula.

Now let us consider the relative entropy of
$\overset n\otimes \rho$ with respect to $\tau_n(w)$. Regarding that
I have the
following conjecture.

\begin{conjecture} \label{conj:2}
Let $\tau_n(w)$ be the matrix as defined in Conjecture~\ref{conj:1}.
Then there holds
\begin{multline} \label{eq:6}
\Tr\(\overset n\otimes \rho\cdot\log \tau_n(w)\)\\
=\frac {1} {2^n}\sum _{0\le h+s\le n/2}
^{}(2-\chi(h+s=n/2)) \frac {(n-2h+1)} {(n+1)}\binom {n+1}h\\
\cdot 
\bigg(\sum _{\ell=0} ^{s}(-1)^\ell \binom s\ell 
\binom {n-2h-\ell}s(1-r^2)^{h+\ell} \bigg)
\log\la_{h,s}\\
=\frac {1} {2^n}\sum _{0\le h+s\le n/2}
^{}(2-\chi(h+s=n/2)) \frac {(n-2h+1)} {(n+1)}\binom {n+1}h\\
\cdot 
\bigg((1-r^2)^h\sum _{\ell=0} ^{s} \binom s\ell 
\binom {n-2h-s}\ell r^{2\ell} \bigg)
\log\la_{h,s},
\end{multline}
with $\la_{h,s}$ as given in \eqref{eq:eigen}.
\end{conjecture}

Clearly, this would give, as a corollary, the following expression for the 
relative entropy of
$\overset n\otimes \rho$ with respect to $\tau_n(w)$:
\begin{multline} \label{eq:7}
\frac {n} {2}(1-r)\log((1-r)/2)+\frac {n}
{2}(1+r)\log((1+r)/2)\\
-\frac {1} {2^n}\sum _{0\le h+s\le n/2}
^{}(2-\chi(h+s=n/2)) \frac {(n-2h+1)} {(n+1)}\binom {n+1}h\\
\cdot 
\bigg(\sum _{\ell=0} ^{s}(-1)^\ell \binom s\ell 
\binom {n-2h-\ell}s(1-r^2)^{h+\ell} \bigg)
\log\la_{h,s}.
\end{multline}

A remarkable feature is that the integrand in the expressions for the
eigenvalue (see \eqref{eq:eigen}) appears also in the expression for
the relative entropy (with $s$ replaced by $r$;
see \eqref{eq:6}). This phenomenon could
already be observed in the nondegenerate case (i.e., where $y$ is not
set equal to 0; see the Conjecture at
the end of our paper). There is something going on which we should
understand (but for which I do not have any explanation).

\begin{thebibliography}{10}

\bibitem{BailAA}
W. N. Bailey, {\it Generalized hypergeometric series},
Cambridge University Press, Cambridge, 1935.

\bibitem{GaRaAA} G.    Gasper and M. Rahman,  
{\em Basic Hypergeometric Series},
Encyclopedia of Mathematics And Its Applications~35, Cambridge
  University Press, Cambridge, 1990.

\bibitem{KratBN}
C.    Krattenthaler, {\em Advanced determinant calculus},
S\'eminaire Lotharingien Combin\@. {\bf 42} (``The Andrews
Festschrift"), paper~B42q, 67~pp.

\bibitem{KrSlAA}
C.    Krattenthaler and P. Slater, {\em Asymptotic redundancies for 
universal quantum coding}, preprint.

\bibitem{SlatAC} L. J. Slater,
{\em Generalized Hypergeometric Functions},
Cambridge University Press,
Cambridge, 1966.

\end{thebibliography}


\end{document}
