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\begin{document}

\title{$2$-adic
valuations of certain
ratios of products \\ of
factorials and applications}

\author
{Shmuel Friedland$^1$ and Christian Krattenthaler$^{2}$%
\thanks{Research partially supported by
EC's IHRP Programme, grant HPRN-CT-2001-00272,
``Algebraic Combinatorics in Europe,"
and by the Austrian Science Foundation FWF, grant S9607-N13,
in the framework of the National Research Network
``Analytic Combinatorics and Probabilistic Number Theory."\newline
\indent\kern3pt}
\\\\
$^1$\it Department of Mathematics, Statistics and Computer Science,\\
\it University of Illinois at Chicago \\
\it Chicago, Illinois 60607-7045, USA\\
 e-mail: {\tt friedlan@uic.edu}\\
\\
$^2$\it Fakult\"at f\"ur Mathematik\\
\it Universit\"at Wien,\\
\it  Nordbergstra{\ss}e~15, A-1090 Vienna, Austria\\
WWW: {\tt http://www.mat.univie.ac.at/\~{}kratt}
}
%  $^2$\it Institut Camille Jordan,
%  \it Universit\'e Claude Bernard Lyon-I\\
%  \it 21, avenue Claude Bernard,\\\it F-69622 Villeurbanne Cedex,
%  France\\
%  \centerline{WWW: \tt
%  http://igd.univ-lyon1.fr/\~{}kratt}
%  }


\date{}
\maketitle


\begin{abstract}
We prove
the conjecture of Falikman--Friedland--Loewy
on the parity of the degrees of projective varieties of $n\times
n$ complex symmetric matrices of
rank at most $k$.
We also characterize the parity of the degrees of projective varieties of $n\times
n$ complex skew symmetric matrices of
rank at most $2p$.
We give recursive relations which determine the parity of the
degrees of projective varieties of $m\times n$ complex matrices
of
rank at most $k$.
In the case the degrees of these varieties are odd, we characterize
the minimal dimensions of subspaces of $n\times
n$ skew symmetric real matrices and of $m\times n$
real matrices containing a nonzero matrix of rank
at most $k$.
The parity questions studied here are also of combinatorial interest
since they concern the parity of the number of plane partitions
contained in a given box, on the one hand, and the parity of
the number of symplectic tableaux of rectangular shape, on the
other hand.
\end{abstract}

\bigskip
\noindent {\bf 2000
Mathematics Subject Classification:}
{\it Primary}: 11A07, 15A03; \\{\it Secondary}: 14M12, 14P25,  15A30.
\par\noindent {\bf Keywords:}  2-adic
valuations of ratio of products
of factorials, parity of degrees of determinantal varieties,
subspaces of real skew symmetric matrices, subspaces of real rectangular matrices,
parity of number of plane partitions, parity of number of symplectic
tableaux.




\section{Introduction}
 Consider the polynomial equation $z^{d}+a_1z^{d-1}+\dots+a_d=0$
 over the field of complex numbers $\C$.  The fundamental theorem
 of algebra
 says that this polynomial system has always a
 nontrivial complex solution $\zeta$.  Assume that $a_1,\ldots,a_d$ are
 real valued.  Clearly, this does not imply that the polynomial
 equation is solvable over the field of real numbers $\R$.
 Since the complex solutions come in complex pairs,
 it is well-known that
 a sufficient condition for an existence
 of a real solution $\zeta$ is that the degree $d$ is odd.
 A similar situation holds in
 a more general setting.

 Let $\P\R^n$ and $\P^n$ be the real and
 the complex projective space of dimension $n$, respectively. Let $V
 \subset \P\R^n$ be an algebraic variety, such that
 its
 complexification $V_{\C} \subset \P^n $ is irreducible and has
 codimension $m \ge 1$.  Hence $B(V_{\C})=V_{\C}$, where $B:\P^n \to \P^n$ is
 the involution $z \mapsto \bar z$.  Recall that any linear space
 $M \subset \P^n$ of dimension $m$ intersects $V_{\C}$. Furthermore, for
 a generic $M$, the set $V_{\C} \cap M$ consists of
 a finite number, $d$ say, of points.
 The positive integer $d$ is called the degree of
 $V_{\C}$ and is denoted by $\deg V_{\C}$. As in the previous
 simple case, it is well-known that if $\deg V_{\C}$ is odd then any linear
 space $L \subset \P\R^n$ of dimension $m$ intersects $V$. Indeed,
 for a generic $L \subset \P\R^n$, the set $V_{\C}\cap L_{\C}$
 consists of $\deg V_{\C}$ points.
 As this set is invariant under the involution $B$,
 we deduce that there exists $\zeta \in V_{\C} \cap L_{\C}$ such
 that $B(\zeta)=\zeta$,
 that is, $\zeta \in \P\R^n$.
 A continuity
 argument yields that $V\cap L \ne \emptyset$ for any $L$ in the
 Gra\ss mannian $\Gr(m+1,n+1,\R)$.  We refer to this result as the
 \emph{odd degree theorem}.

Assume now that $\deg V_{\C}$ is even.  Then it is not difficult
to find nontrivial examples where $V \cap L' =\emptyset$ for some
$L'\in \Gr(m+1,n+1,\R)$. Moreover examples can be found among
determinantal varieties, such that for any integer $k \in [0,p)$
there exists $L' \in \Gr(m+k+1,n+1,\R)$ satisfying $V\cap L'
=\emptyset$, while $V\cap L \ne\emptyset$ for any $L \in
\Gr(m+p+1,n+1,\R)$ (see \cite[\S2]{FL}).

The following generalization of the odd degree theorem is
proved
in \cite{FL}: Assume that $\deg V_{\C}$ is even and let $r$ be a
positive integer. Suppose that the codimension of the variety of
the singular points of $V_{\C}$ in $V_{\C}$ is at least $2r+1$.
Suppose furthermore that for a generic $M \in \Gr(m+2r+1,n+1,\C)$
the Euler characteristic of $V_{\C}\cap M$ is odd.  Then $V\cap L
\ne\emptyset$ for any $L \in \Gr(m+2r+1,n+1,\R)$.

The degree of $V_{\C}$ and the Euler characteristic of $V_{\C}\cap
M$ can be often computed using the projectivized complex bundles
and the corresponding Chern classes of their tangent bundles, see
for example \cite{HT2} and \cite{FL}. It turns out that the
degrees of determinantal varieties (see
the special cases
discussed below) are quotients of products of certain binomial
coefficients (see \cite{HT1} and \cite[Ch. II]{ACGH}). Binomial
coefficients appear also in
Stiefel--Whitney classes \cite{MS}.

Let $\rM_{m,n}(\F)$, $\rM_n(\F)$, $\rS_n(\F)$, $\rA_n(\F)$ be the spaces of
$m\times n$ matrices,
$n\times n$ matrices, $n \times n$ symmetric matrices,
and $n\times n$ skew (antisymmetric) matrices
with entries in $\F=\R,\C$, respectively. Let
$\rU_{k,m,n}(\F)$, $\rV_{k,n}(\F)$, $\rW_{k,n}(\F)$ be the
varieties of all matrices in
$\rM_{m,n}(\F)$, $\rS_n(\F)$, $\rA_n(\F)$ of rank $k \le \min(m,n)$ or less, respectively.
It is known that any $A\in \rA_n(\F)$ has an even rank,
see e.g.\
\cite[\S11.4]{Gan} or \S5.  Hence it is enough to consider
$\rW_{2p,n}(\F)$, where $1\le p\le \lfloor \frac{n}{2}\rfloor$.
Then the
projectivizations $\P\rU_{k,m,n}$, $\P \rV_{k,n}(\F)$, $\P \rW_{2p,n}(\F)$
are irreducible
varieties of
codimension $(m-k)(n-k)$, $\binom {n-k+1} 2$, $\binom {n-2p} 2$
in the projective
spaces
$\P\rM_{m,n}(\F)$, $\P \rS_n(\F)$, $\P \rA_n(\F)$, respectively.  Note that
$\P \rU_{k-1,m,n}(\F)$, $\P \rV_{k-1,n}(\F)$,
$\P \rW_{2(p-1),n}(\F)$ are the
varieties of the
singular points of
$\P \rU_{k,m,n}(\F)$, $\P \rV_{k,n}(\F)$, $\P \rW_{2p,n}(\F)$, respectively.
For $\P\rU_{k,m,n}(\F)$, $\P \rV_{k,n}(\F)$ see for example \cite[II]{ACGH},
and for $\P \rW_{2p,n}(\F)$ see \S5.
Let $d(m,n,k,\F)$, $d_s(n,k,\F)$, $d_a(n,2p,\F)$ be the smallest integer
$\ell$ such that every $\ell$-dimensional subspace of $\rM_{m,n}(\F)$, $\rS_n(\F)$, $\rA_n(\F)$
contains a nonzero matrix
whose rank is at most $k, 2p$, respectively.  Then
\begin{multline}\label{dnkc}
\kern-2pt
d(m,n,k,\C)=(m-k)(n-k)+1, \;
d_s(n,k,\C)=\binom {n-k+1} 2 +1, \;\\ d_a(n,2p,\C)=\binom {n-2p} 2 +1,
\end{multline}
and the problem is to determine $d(m,n,k,\R)$, $d_s(n,k,\R)$, $d_a(n,2p,\R)$.
The degrees of
$\P\rU_{k,m,n}(\C)$, $\P\rV_{k,n}(\C)$, $\P \rW_{2p,n}(\C)$
were computed by Harris and Tu in \cite{HT1},
\begin{eqnarray}
&&\gamma_{k,m,n}:=\deg
\P\rU_{k,m,n}(\C)=
\prod _{j=0} ^{n-k-1}\frac {\binom {m+j}{m-k}} {\binom {m-k+j}{m-k}}=
\prod_{j=0}^{n-k-1}\frac{(m+j)!\,j!}{(k+j)!\,(m-k+j)!},
\label{degkmn}\\
&&\delta_{k,n}:={\rm deg\;} \P \rV_{k,n}(\C)=\prod_{j=0}^{n-k-1} \frac{\binom {n+j}
{n-k-j}}{\binom {2j+1} j}, \quad \varepsilon_{2p,n}:= \deg \P
\rW_{2p,n}=\frac{\delta_{2p+1,n}}{2^{n-2p-1}}.
\label{degvk}
\end{eqnarray}
For the curiosity of the reader we remark that these quantities
have also combinatorial interpretations. The quantity $\gamma_{k,m,n}$
counts plane partitions which are contained in an
$(n-k)\times(m-k)\times k$ box (see \cite{Bressoud} and \S6). On the other
hand, the quantity $\varepsilon_{2p,n}$ counts symplectic tableaux
(see \cite{King}) of a rectangular shape of size $n\times p$, and
thus several other sets of combinatorial objects
(see \cite{Sundaram} and \cite{GhorKrat}
for more information on these topics).

It was shown in \cite{FFL} that $\delta_{n-q,n}$ is odd if
\begin{equation}\label{spc1}
n\equiv \pm q \;({\rm mod\;}2^{\lceil \log_2 2q\rceil}) .
\end{equation}
For values of $q$ and $n$ which satisfy this condition,
\begin{equation}\label{spc}
d_s(n,n-q,\C)=d_s(n,n-q,\R)= \binom {q+1} 2 +1.
\end{equation}
It was
furthermore shown in \cite{FFL} that this equality implies the following
interesting result.  Assume that $n \ge q$ and
that $n$ satisfies
(\ref{spc1}), then any
$\binom {q+1} 2$-dimensional subspace of $\rS_n(\R)$
contains a nonzero matrix with an eigenvalue of multiplicity at
least $q$. This statement for $q=2$ yields
Lax's
result \cite{Lax} that any $3$-dimensional subspace of $\rS_n(\R)$
contains a nonzero matrix with a multiple eigenvalue for $n\equiv
2 \;({\rm mod\;} 4)$.  (This result and its generalization in
\cite{FRS} is of importance in the study of singularities of
hyperbolic systems.)

On the other hand,
it was conjectured in \cite{FFL} that
also the converse holds, that is, that if $\delta_{n-q,n}$ is odd then
(\ref{spc1}) holds.  In particular, for $n$ and $q$ which do not
satisfy (\ref{spc1}), we do not have a \emph{simple} way to
compute $d_s(n,n-q,\R)$.
One of the main purposes of this
paper is to prove this conjecture,
see \S\S3,4.  Our results yield
in addition that
$\varepsilon_{2p,n}$ is odd if and only
if (\ref{spc1}) holds with $q=n-2p$,
see \S5. Hence for these values of $p,n$
we have
\begin{equation}\label{asrc}
d_a(n,2p,\C)=d_a(n,2p,\R)= \binom {n-2p} 2 +1.
\end{equation}
In particular, for $n\equiv
2 \;({\rm mod\;} 4)$ any two-dimensional subspace of real
$n\times n$ skew symmetric matrices contains a nonzero singular
matrix.
For $n$ and $q=n-2p$ which do not satisfy the condition
(\ref{spc1}), we do not have a \emph{simple} way to
compute $d_a(n,2p,\R)$.

We also consider the problem
of characterizing the values of $k,m,n$ for which
$\gamma_{k,m,n}$ is odd.
This problem seems to have a rather intricate solution.
We give some partial results on this problem
in \S6. In particular, we provide an algorithm for computing the
parity of $\gamma_{k,m,n}$ from the binary expansions of $k,m,n$
directly, without having to actually compute $\gamma_{k,m,n}$
(see Remark~\ref{rem2} and Proposition~\ref{oddconb}). As above,
if $\gamma_{k,m,n}$ is odd then
\begin{equation}\label{dmnkcr}
d(m,n,k,\C)=d(m,n,k,\R)=(m-k)(n-k)+1.
\end{equation}
See Corollary~\ref{cordkmn} for the corresponding geometric results
which we are able to prove.

Another purpose of this paper
is to estimate the
$2$-adic valuation of $\delta_{k,n}$, i.e., the
largest power of $2$ that divides $\delta_{k,n}$.
There are two reasons for these estimations.
First, we show that our methods can give good estimates
for the complex behavior of the $2$-adic valuation of
$\delta_{k,n}$.  Second, we recall the classical results
of Radon \cite{Rad} and Hurwitz \cite{Hur} on the maximal dimension of the spaces of
$n\times n$ scaled orthogonal matrices,
and the Adams result \cite{Ada} on maximal number of vectors fields on the tangent bundle of
$n-1$-dimensional sphere,
which are functions of the $2$-adic valuation of $n$.
We believe that the $2$-adic
valuation of $\deg \P V_{k,n}(\C)$
is related to a lower bound for the problem raised in
Friedland--Libgober \cite{FL}.

Consider the variety of $n\times n$ singular
matrices in $\rM_n(\F)$.  Clearly the degree
of this variety is
$n$ and
its codimension is $1$.  Hence any two-dimensional complex
subspace $L\subset \rM_n(\C)$ contains a nonzero singular matrix.
For $n$ odd, any two-dimensional real subspace of $\rM_n(\R)$
contains a nonzero singular matrix.  For $n$ even, the situation
is much more complicated.
For $n\in\N$, let $c+4d$ be the $2$-adic valuation of $n$. Thus
$n=(2a+1)2^{c+4d}$, where $a$ and $d$ are nonnegative integers and
$c\in\{0,1,2,3\}$. Then the Radon--Hurwitz number $\rho(n)$ is
defined by $\rho(n)=2^c+ 8d$.
The classical results of Radon \cite{Rad} and Hurwitz \cite{Hur}
state that $\rho(n)$ is the maximal dimension of an
$n$-dimensional subspace
$U$ of $\rM_n(\R)$
such that each
nonzero $A \in U$ is an orthogonal
matrix times $r \in \R^*$. In his famous work \cite{Ada}, Adams
gave a nonlinear version of the Radon--Hurwitz
result by showing
that $\rho(n)-1$ is the maximal number of linearly independent
vector fields on the $(n-1)$-dimensional sphere in $\R^n$.
In particular,
Adams's result implies that any $(\rho(n)+1)$-dimensional subspace of
$\rM_n(\R)$ contains a nonzero singular matrix.

The parity of binomial coefficients plays
a role in generalized Radon--Hurwitz
numbers \cite[Prop. I (f)]{BF}.
Similarly, we believe that the answer to the following problem
raised in \cite{FL} significantly depends on
the $2$-adic valuation of $\delta_{k,n}$:

\begin{problem}\label{mprob}  Assume that $\delta_{k,n}$ is even.  Find
an integer $r \ge 1$, preferably the smallest possible, such that
$ 2r < \binom {n-k+2} 2 - \binom {n-k+1} 2$,
and
such that the Euler characteristic of $\P \rV_{n,k}(\C) \cap M$ is odd for
a generic $M \in \Gr\!\left(\binom {n-k+1}
2+2r+1,\binom {n+1} 2,\C\right)$.
\end{problem}
For the above minimal value of $r$,
we have $\P \rV_{n,k}(\R) \cap
L\ne\emptyset$ for any
$$L \in \Gr\!\left(\binom {n-k+1}
2+2r+1,\binom {n+1} 2,\R\right).$$

We now briefly survey the contents of this paper.  In \S2 we give
some
auxiliary results on the 2-adic
valuation of $\delta_{n-q,n}$.
In \S3 we prove the conjecture
from \cite{FFL} characterizing the values of $q$ and $n$ for which
 $\delta_{n-q,n}$ is odd.  In \S4 we discuss
properties of the 2-adic
valuation of $\delta_{n-q,n}$ when the condition
(\ref{spc1}) is not satisfied. In particular, we characterize the
values of $q$ and $n$ for which the 2-adic valuation of
$\delta_{n-q,n}$ is $1$, and we show that, for fixed $q$,
the $2$-adic valuations of $\delta_{n-q,n}$ have a wave-like
behavior as $n$ increases. In \S5 we show that
$\varepsilon_{2p,n}$ is odd if and only if
(\ref{spc1}) holds with $q=n-2p$.
Hence for these values of $n$ and $p$
the equality (\ref{asrc}) holds.  Finally, in \S6 we study
the parity of the number of plane partitions contained in an $a\times
b\times c$ box, and thus the parity of $\gamma_{k,m,n}$. Some partial
results are given, as well as the above-mentioned algorithm for
computing this parity.




\section{Preliminary results
on the $2$-adic valuation of $\delta_{n-q,n}$}
For a nonzero integer $i$ we write $\nu_2(i)$ for the $2$-adic
valuation of $i$.  That is $i=(2j+1)2^{\nu_2(i)}$ for some
$j\in\Z$.
For positive integers $q$ and $n$ define
\begin{equation}\label{deftqn}
\th_{q,n}:=\prod _{j=0} ^{q-1}\frac {\binom {n+j} {q-j}} {\binom
{2j+1}{j}}. %\quad \big(=\delta_{n-q,n}\big).
\end{equation}
The reader should note that $\th_{q,n}=\delta_{n-q,n}$
(compare (\ref{degvk})).
It will be convenient later to extend the definition of $\th_{q,n}$
to all {\it nonnegative} integers $q$ and $n$, that is, to allow
$n=0$, respectively $q=0$, in (\ref{deftqn}) also. In particular, for
$q=0$ we set $\th_{0,n}=1$ by interpreting the empty product as $1$.

Clearly, we have $\theta_{q,n}=0$ for $q>n$.
We want to study the behavior of the $2$-adic
valuation of
$\th_{q,n}$ for $n\ge q$. The following proposition simplifies this study
as it exhibits a simple relationship
between the $2$-adic
valuation of $\th_{q,n}$ when $n-q$ is even
and those when $n-q$ is odd. In particular, this result allows one
to concentrate on the analysis of just one case, which will be the
case where $n-q$ is even.



\begin{prop} \label{prop:1}
Let $n$ and $q$ be
nonnegative integers, $n\ge q+1$. Then
\begin{equation}\label{recrnt}
\nu_2(\th_{q,n})=\nu_2(\th_{q+1,n})+q -\sum_{j=0}^q\nu_2(n-q+2j).
\end{equation}
In particular, if $n-q$ is odd then
$\nu_2(\th_{q,n})=\nu_2(\th_{q+1,n})+q\ge q$.
Hence, if $n$ and $q$ are both positive, and if $n-q$ is positive and
odd, then $\th_{q,n}$ is even.
\end{prop}
\begin{proof}
The ratio of $\th_{q+1,n}$ and $\th_{q,n}$ is
\begin{align*}
\frac {\th_{q+1,n}} {\th_{q,n}}&=\frac{1}{\binom {2q+1} q}
\prod_{j=0}^q \frac{(n+j)!}{(q+1-j)!\,(n-q-1+2j)!}
\prod_{j=0}^{q-1}\frac{(q-j)!\,(n-q+2j)!}{(n+j)!}\\
&=\frac{(q)!\,(q+1)!}{(2q+1)!}\frac{(n+q)!}{1!\,(n+q-1)!}\prod_{j=0}^{q-1}
\frac{(q-j)!\,(n-q+2j)!}{(q+1-j)!\,(n-q-1+2j)!}\\
&=\frac{(q+1)!}{2^q(2q+1)!!} (n+q)
\prod_{j=0}^{q-1}
\frac{(n-q+2j)}{(q+1-j)}=\frac { \prod _{j=0} ^{q}(n-q+2j)}
{2^q\,(2q+1)!!}.
\end{align*}
(Here $(2q+1)!!:=(2q+1)\cdot(2q-1)\cdots 3\cdot 1$.)
As $(2q+1)!!$ is odd, we deduce (\ref{recrnt}).
Assume that $n-q$ is odd.  Then
$n-q+2j$ is odd for $j=0,\ldots,q$, and the last part of the proposition follows.
\end{proof}


We now concentrate on the case where $n-q$ is even.

\begin{prop}\label{eqfthm1}
Let $n$ and $q$ be
nonnegative integers, $n\ge q$, such that the difference
of $n$ and $q$ is even, say $n-q= 2p$.  Then
\begin{equation}\label{forthqn}
\nu_2(\th_{q,n})=(n-1-p)p-\nu_2\left(\prod_{k=1}^p\frac{(n-k)!}{(k-1)!}\right).
\end{equation}
Here
again, in case that $p=0$, the empty product has to be interpreted as $1$.
In particular, $(n-1-p)p\ge \nu_2\left(\prod_{k=1}^p\frac{(n-k)!}{(k-1)!}\right)$.
\end{prop}
\begin{proof}

We prove (\ref{forthqn}) by a reverse induction on $q$. By the
definition (\ref{deftqn}) of $\th_{q,n}$ we have $\th_{n,n}=1$.
Hence $\nu_2(\th_{n,n})=0$, which confirms (\ref{forthqn}) in this case.

Proposition~\ref{prop:1} implies that for any positive integer $k$
we have
$$\nu_2(\th_{n-2k+1,n})=\nu_2(\th_{n-2k+2,n})+n-2k+1.$$
We now use
(\ref{recrnt}) for $q=n-2k$ to obtain the recursive formula
\begin{align*}
\nu_2(\th_{n-2k,n})&=\nu_2(\th_{n-2k+2,n})+ n-2k+1 +n-2k -
\sum_{j=0}^{n-2k}\nu_2(2k+2j)\\
&=\nu_2(\th_{n-2k+2,n})+n-2k -\sum_{j=0}^{n-2k}\nu_2(k+j)\\
&=
\nu_2(\th_{n-2k+2,n})+n-2k -\nu_2\left(\frac{(n-k)!}{(k-1)!}\right).
\end{align*}
Use this recursive relation for $k=p,p-1,\ldots,1$ to obtain
(\ref{forthqn}).  Since $\nu_2(\theta_{q,n})\ge 0$ we obtain
that the right-hand side of (\ref{forthqn}) is nonnegative.
\end{proof}

Our next goal is to give an explicit expression of the $2$-adic
valuation of $\th_{q,n}$ in terms of binary digit sums. More
precisely, for a nonnegative integer $m$ let $s(m)$ denote the sum of the digits of $m$
when written in binary notation.  Then $s(0)=0$, $s(2m)=s(m)$,
$s(2m+1)=1+s(m)$, and
\begin{equation}\label{AI}
s(2^e-1-k)=e-s(k) \quad \text{for any integers } e\ge 0,\; k\in [0,2^e-1].
\end{equation}
The basic result which ties together the $2$-adic valuation of
factorials and binary digit sums is the following one due to Legendre
(cf.\ \cite[Sec.~4.4]{GrKnPa} and \cite{Legendre}).
We bring its proof for completeness.

\begin{prop}\label{2adn!}  Let $n$ be a
nonnegative integer.
Then $\nu_2(n!)=n-s(n)$.
\end{prop}
\begin{proof}
We prove the proposition by induction.
Clearly $\nu_2(1)=0
=0-s(0)=1-s(1)$.  Assume that the proposition holds
for $n\le m-1$.  Let $n=m$.  Suppose first that $m=2l$.
Then $\nu_2((2l)!)=\nu_2((2l)!!)=
\nu_2(2^ll!)=l+\nu_2(l!)=l+l-s(l)=m-s(m).$
Assume now that $m=2l+1$.  Then $\nu_2(m!)=\nu_2((2l)!)=2l-
s(2l)=m-s(m)$.
\end{proof}

In what follows we are going to make
extensive use of the following lemma
and particularly of its corollary.
\begin{lemma}\label{coreqfthm1}
Let $p$ and $q$ be nonnegative integers,
and assume
that $n-q= 2p$.  Then
\begin{equation}\label{forthqn1}
\nu_2(\th_{q,n})=-p + \sum_{k=1}^p s(n-k)-\sum_{k=1}^p s(k-1).
\end{equation}
If $p=0$, we have to interpret empty sums as $0$, as before.
The equation holds also for $q=0$ if we interpret, as earlier,
$\th_{0,2p}$ as $1$
for any $p$.
\end{lemma}

\begin{proof}
Combine
(\ref{forthqn}) and Proposition~\ref{2adn!}.
\end{proof}

Since the $2$-adic valuation in (\ref{forthqn1}) must always be nonnegative,
we obtain the following corollary.
\begin{corol}
For all nonnegative integers $l$ and $p$,
we have
\begin{equation}
\sum_{j=l}^{l+p-1} s(j) \ge p +\sum_{j=0}^{p-1}s(j) \quad \text{if\/ } l\ge
p,
\label{basins1}\\
\end{equation}
and
\begin{equation}
\sum_{j=l}^{l+p-1} s(j)
\ge l +\sum_{j=0}^{p-1}s(j) \quad \text{if\/ }
l\le p. \label{basins2}
\end{equation}
For $l=p+1,p$, there holds equality,
\begin{equation}\label{fundeq}
0= -\sum _{j=0} ^{p-1}s(j)+\sum _{j=p+1} ^{2p}s(j)-p
=-\sum _{j=0} ^{p-1}s(j)+\sum _{j=p} ^{2p-1}s(j)-p.
\end{equation}
In particular,
\begin{equation} \label{eq:Ungl}
\sum_{j=l}^{l+p-1} s(j) \ge \sum_{j=0}^{p-1}s(j),
\end{equation}
and equality holds if and only if either $p=0$ or $l=0$.
\end{corol}
\begin{proof}
Use (\ref{forthqn1}) with $l=n-p=p+q$ to deduce
(\ref{basins1}).
Assume that $0\le l \le p$.  Then by cancelling out
the common terms in
(\ref{basins2}) we deduce that
(\ref{basins2}) follows from
(\ref{basins1}) with the roles of $l$ and $p$ being interchanged.

Put $q=0$ in (\ref{forthqn1}) and recall that $\th_{0,n}=1$. This
implies the second part of (\ref{fundeq}).
Use the equality $s(2p)=s(p)$ to deduce the first part of
(\ref{fundeq}).

The inequality (\ref{eq:Ungl}) follows trivially from
 (\ref{basins1}) and  (\ref{basins2}).
\end{proof}

\section{Proof of the Falikman--Friedland--Loewy Conjecture}

In this section, we use the results from the previous section to prove
the conjecture from \cite{FFL} characterizing the values of $q$ and
$n$ for which $\delta_{n-q,n}=\th_{q,n}$ is odd. For the sake of
convenience, we state the result in form of the following theorem.
The ``if" direction was already shown in \cite{FFL}. Our proof will
not only show the ``only if" direction, but, in passing, it will also
provide an independent proof of the ``if" direction.

\begin{theo} \label{thm:2}
For positive integers $q$ and $n$, the quantity
$\th_{q,n}$ is odd if and only if $n\ge q\ge 1$ and
$$n\equiv \pm q\pmod {2^{\cl{\log_2 2q}}}.$$
\end{theo}



\begin{proof}

For $n-q$ odd, the theorem follows
from Proposition~\ref{prop:1}. Therefore, for the rest of the
proof, let $n-q$ be even.

We repeatedly use
subsequently the following observation.
Let $r,t,j$ be nonnegative integers such that $2^t> j$.
Then $s(r2^t+j)=s(r)+s(j)$.  We divide the proof into the two
following cases.

\medskip\noindent
\textsc{Case 1}.  $n=2n_1$.  It is enough to assume that $q=2q_1$
and $n_1\ge q_1$.
Write  $p=n_1-q_1$ and substitute in (\ref{forthqn1}) to
obtain
\begin{equation} \label{AE}
\nu_2(\th_{q,n})=-(n_1-q_1) -\sum _{j=0} ^{n_1-q_1-1}s(j)
+\sum _{j=n_1+q_1} ^{2n_1-1}s(j).
\end{equation}
We now show that the right-hand side of \eqref{AE} is zero if and only if
$$n_1\equiv \pm q_1\pmod {2^{\cl{\log_2 (2q_1)}}},$$
which is obviously equivalent to the theorem in this case.

In \textsc{Case 1} we always use the abbreviation $Q=2^{\cl{\log_2 2q_1}}$.
Write $n_1=cQ+q_1+d$, where $0\le d<Q$.
We know that the
quantity from (\ref{AE}),
\begin{equation} \label{AK}
-\sum _{j=0} ^{n_1-q_1-1}s(j)
+\sum _{j=n_1+q_1} ^{2n_1-1}s(j)-(n_1-q_1)=
-\sum _{j=0} ^{cQ+d-1}s(j)
+\sum _{j=cQ+2q_1+d} ^{2cQ+2q_1+2d-1}s(j)-(cQ+d),
\end{equation}
is nonnegative.  We have to show that it is zero only if $d=0$ or $d=Q-2q_1$. To do so,
we distinguish various subcases, depending on the size of $d$.

\medskip\noindent
\textsc{Case 1a}: $2q_1+2d\le Q$. In this case, the quantity \eqref{AK} becomes
\begin{align*}
-&\sum _{r=0} ^{c-1}\sum _{j=0} ^{Q-1}s(rQ+j)
-\sum _{j=cQ} ^{cQ+d-1}s(j)
+\sum _{j=cQ+2q_1+d} ^{(c+1)Q-1}s(j)\\
&\kern2cm
+\sum _{r=c+1} ^{2c-1}\sum _{j=0} ^{Q-1}s(rQ+j)
+\sum _{j=2cQ} ^{2cQ+2q_1+2d-1}s(j)
-(cQ+d)\\
&= -Q\sum _{r=0} ^{c-1}s(r)-c\sum _{j=0} ^{Q-1}s(j)
-ds(c)-\sum _{j=0} ^{d-1}s(j)
+(Q-2q_1-d)s(c)+\sum _{j=2q_1+d} ^{Q-1}s(j)\\
&\kern.5cm
+Q\sum _{r=c+1} ^{2c-1}s(r)+(c-1)\sum _{j=0} ^{Q-1}s(j)
+(2q_1+2d)s(2c)+\sum _{j=0} ^{2q_1+2d-1}s(j)
-(cQ+d)\\
&=
Q\left(-\sum _{r=0} ^{c-1}s(r)+\sum _{r=c+1} ^{2c}s(r)-c\right)
-\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d} ^{2q_1+2d-1}s(j) -d.
\end{align*}
Using (\ref{fundeq}), we see that the above expression is equal to
\begin{equation} \label{AL}
-\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d} ^{2q_1+2d-1}s(j) -d.
\end{equation}
By the definitions of $Q$ and $d$, we have $Q/2< 2q_1<
 2q_1+2d-1< Q$. Thus, we have
\begin{equation} \label{AM}
-\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d} ^{2q_1+2d-1}s(j)-d
= -\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d-\frac {Q} {2}} ^{2q_1+2d-\frac {Q} {2}-1}s(j).
\end{equation}
It should be noted that, by the definitions of
$Q$ and $d$, we have $Q/2<2q_1+d$.
By
(\ref{eq:Ungl}), the quantity on the right-hand side is
zero only if the sums on the right-hand side are empty,
i.e., if $d=0$.

\medskip\noindent
\textsc{Case 1b}: $2q_1+d\le Q<2q_1+2d$. In this case, the quantity \eqref{AK} becomes
\begin{align*}
-&\sum _{r=0} ^{c-1} \sum _{j=0} ^{Q-1}s(rQ+j)
-\sum _{j=cQ} ^{cQ+d-1}s(j)
+\sum _{j=cQ+2q_1+d} ^{(c+1)Q-1}s(j)\\
&\kern2cm
+\sum _{r=c+1} ^{2c}\sum _{j=0} ^{Q-1}s(rQ+j)
+\sum _{j=(2c+1)Q} ^{2cQ+2q_1+2d-1}s(j)
-(cQ+d)\\
&= -Q\sum _{r=0} ^{c-1}s(r)-c\sum _{j=0} ^{Q-1}s(j)
-ds(c)-\sum _{j=0} ^{d-1}s(j)
+(Q-2q_1-d)s(c)+\sum _{j=2q_1+d} ^{Q-1}s(j)\\
&\kern.5cm
+Q\sum _{r=c+1} ^{2c}s(r)+c\sum _{j=0} ^{Q-1}s(j)
+(2q_1+2d-Q)s(2c)+\sum _{j=Q} ^{2q_1+2d-1}s(j)
-(cQ+d)\\
&=
Q\left(-\sum _{r=0} ^{c-1}s(r)+\sum _{r=c+1} ^{2c}s(r)-c\right)
-\sum _{j=0} ^{d-1}s(j) +\sum _{j=2q_1+d} ^{2q_1+2d-1}s(j) -d.
\end{align*}
Using (\ref{fundeq}) again, we deduce that the
above expression is equal to
\begin{equation} \label{AN}
-\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d} ^{2q_1+2d-1}s(j) -d.
\end{equation}
By the definitions of $Q$ and $d$, we have $Q/2< 2q_1<
2q_1+d\le Q\le 2q_1+2d-1<2Q$. Thus, the above quantity can be further
modified to
\begin{align*}
-\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d} ^{Q-1}s(j)
+&\sum _{j=Q} ^{2q_1+2d-1}s(j) -d\\
&=-\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d-\frac {Q} {2}} ^{\frac {Q} {2}-1}s(j)
+\sum _{j=0} ^{2q_1+2d-Q-1}s(j)
\\
&=-\sum _{j=2q_1+2d-Q} ^{d-1}s(j)
+\sum _{j=2q_1+d-\frac {Q} {2}} ^{\frac {Q} {2}-1}s(j).
\end{align*}
From $Q/2<2q_1$ and $2q_1+d\le Q$, we infer that $d<Q/2$.
Now we use identity \eqref{AI} with $2^e=\frac{Q}{2}$ for all the digit sums
in the last expression. This leads to the expression
$$\sum _{j=\frac {Q} {2}-d} ^{\frac {3} {2}Q-2q_1-2d-1}s(j)
-\sum _{j=0} ^{Q-2q_1-d-1}s(j). $$
We recall that $Q-2q_1-d\ge 0$.
Apply
(\ref{eq:Ungl}) again to conclude
that the last expression, and hence, $\nu_2(\th_{q,n})$, is
zero only if the sums in the last line are empty,
that is, if $d=Q-2q_1$.

\medskip\noindent
\textsc{Case 1c}: $Q<2q_1+d$ and $2q_1+2d\le 2Q$.
In this case, the quantity \eqref{AK} becomes
\begin{align} \notag
-\sum _{r=0} ^{c-1}&\sum _{j=0} ^{Q-1}s(rQ+j)
-\sum _{j=cQ} ^{cQ+d-1}s(j)
+\sum _{j=cQ+2q_1+d} ^{(c+2)Q-1}s(j)\\
\notag
&\kern2cm
+\sum _{r=c+2} ^{2c}\sum _{j=0} ^{Q-1}s(rQ+j)
+\sum _{j=(2c+1)Q} ^{2cQ+2q_1+2d-1}s(j)
-(cQ+d)\\
\notag
&= -Q\sum _{r=0} ^{c-1}s(r)-c\sum _{j=0} ^{Q-1}s(j)
-ds(c)-\sum _{j=0} ^{d-1}s(j)
+(2Q-2q_1-d)s(c+1)\\
\notag
&\kern2cm
 +\sum _{j=2q_1+d-Q} ^{Q-1}s(j)
+Q\sum _{r=c+2} ^{2c}s(r)+(c-1)\sum _{j=0} ^{Q-1}s(j)\\
\notag
&\kern2cm
+(2q_1+2d-Q)s(2c+1)+\sum _{j=0} ^{2q_1+2d-Q-1}s(j) -(cQ+d)\\
\notag
&= Q\left(-\sum _{r=0} ^{c-1}s(r)+\sum _{r=c+1} ^{2c}s(r)-c\right)
-\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d-Q} ^{2q_1+2d-Q-1}s(j)\\
\notag
&\kern2cm
+(2q_1+d-Q)(s(c)-s(c+1)+1)\\
&= -\sum _{j=0} ^{d-1}s(j)+\sum _{j=2q_1+d-Q} ^{2q_1+2d-Q-1}s(j)
+(2q_1+d-Q)(s(c)-s(c+1)+1).
\label{ANa}
\end{align}
In the second step we used the equality $s(2c+1)=s(c)+1$, and
in the last step we used again (\ref{fundeq}).
Since $Q<2q_1+d$, we have $2q_1+d-Q>0$. As
$s(c)-s(c+1)+1\ge0$ for any $c\ge 0$, the third term in the last line is
nonnegative.   As $d>0$,
the inequality (\ref{eq:Ungl}) implies that the sum of the first two terms is
strictly positive.

\medskip\noindent
\textsc{Case 1d}: $Q<2q_1+d$ and $2q_1+2d> 2Q$.
In this case, the quantity \eqref{AK} becomes
\begin{align*}
-\sum _{r=0} ^{c-1}&\sum _{j=0} ^{Q-1}s(rQ+j)
-\sum _{j=cQ} ^{cQ+d-1}s(j)
+\sum _{j=cQ+2q_1+d} ^{(c+2)Q-1}s(j)\\
&\kern2cm
+\sum _{r=c+2} ^{2c+1}\sum _{j=0} ^{Q-1}s(rQ+j)
+\sum _{j=(2c+2)Q} ^{2cQ+2q_1+2d-1}s(j)
-(cQ+d)\\
&=
-Q\sum _{r=0} ^{c-1}s(r)-c\sum _{j=0} ^{Q-1}s(j)
-ds(c)-\sum _{j=0} ^{d-1}s(j)
+(2Q-2q_1-d)s(c+1)\\
&\kern2cm
+\sum _{j=2q_1+d-Q} ^{Q-1}s(j)
+Q\sum _{r=c+2} ^{2c+1}s(r)+c\sum _{j=0} ^{Q-1}s(j)\\
&\kern2cm
+(2q_1+2d-2Q)s(2c+2)+\sum _{j=0} ^{2q_1+2d-2Q-1}s(j)
-(cQ+d).
\end{align*}
We now do the following substitutions.  First, $s(2(c+1))=s(c+1)$.
Second, in the sum
over $j=0,\ldots,2q_1+2d-2Q-1$
(where $2q_1+2d-2Q-1\le
Q+2(Q-1)-2Q-1=Q-1$), we let $s(j)=s(Q+j)-1$.
Third,
$$\sum _{r=c+2} ^{2c+1}s(r)=s(2c+1)-s(c+1)+\sum _{r=c+1}
^{2c}s(r)=s(c)+1-s(c+1) +\sum _{r=c+1} ^{2c}s(r).$$
Hence, the above expression is equal to
\begin{align}
\notag
&Q\left(-\sum _{r=0} ^{c-1}s(r)+\sum _{r=c+1} ^{2c}s(r)-c\right)
-\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d-Q} ^{Q-1}s(j)\\
\notag
&\kern2cm
+\sum _{j=Q} ^{2q_1+2d-Q-1}s(j)-(2q_1+2d-2Q)+
(Q-d)(s(c)-s(c+1)+1)\\
&=-\sum _{j=0} ^{d-1}s(j)+\sum _{j=2q_1+d-Q} ^{2q_1+2d-Q-1}s(j)
-(2q_1+2d-2Q)+(Q-d)(s(c)-s(c+1)+1),
\label{ANb}
\end{align}
where we used again (\ref{fundeq}).
Since $Q>d$ and $s(c)-s(c+1)+1\ge 0$ for any $c\ge0$, the fourth term in
the last line is nonnegative.
We have $\frac{Q}{2}<d<Q$ and $d\ge 2q_1+d-Q$.
Thus, we may apply (\ref{basins2})
to deduce that the sum of the first two terms is at least
$2q_1+d-Q$.  Hence the expression (\ref{ANb}) is not less than
$Q-d>0$.

The proof of \textsc{Case~1} is completed.

\medskip\noindent
\textsc{Case 2}.  Let $n=2n_1+1$, $q=2q_1+1$, where $n_1\ge q_1\ge 0$.
Write $p=n_1-q_1$ and substitute in (\ref{forthqn1}) to obtain
\begin{equation} \label{AF}
\nu_2(\th_{q,n})=-(n_1-q_1)
-\sum _{j=0} ^{n_1-q_1-1}s(j)
+\sum _{j=n_1+q_1+1} ^{2n_1}s(j).
\end{equation}
We now show that the right-hand side of \eqref{AF} is zero if and only if
$$n_1\equiv  q_1\pmod {2^{\cl{\log_2 (2q_1+1)}}}$$
or
$$n_1\equiv  -q_1-1\pmod {2^{\cl{\log_2 (2q_1+1)}}},$$
which is obviously equivalent to the theorem in this case.

In \textsc{Case~2} we always use the abbreviation
$Q=2^{\cl{\log_2 (2q_1+1)}}$.
Write $n_1=cQ+q_1+d$, where $0\le d<Q$.
We have to show that
the expression (\ref{AF}), that is,
\begin{multline}
-\sum _{j=0} ^{n_1-q_1-1}s(j)
+\sum _{j=n_1+q_1+1} ^{2n_1}s(j)-(n_1-q_1)\\
= -\sum _{j=0} ^{cQ+d-1}s(j)
+\sum _{j=cQ+2q_1+d+1} ^{2cQ+2q_1+2d}s(j)-(cQ+d),
\label{AO}
\end{multline}
is zero only if $d=0$ or $d=Q-2q_1-1$. To do so,
we distinguish again various cases, depending on the size of $d$.

\medskip\noindent
\textsc{Case 2a}: $2q_1+2d< Q$.
In this case, the quantity \eqref{AO} becomes
\begin{align*}
-{}&\sum _{r=0} ^{c-1}\sum _{j=0} ^{Q-1}s(rQ+j)
-\sum _{j=cQ} ^{cQ+d-1}s(j)
+\sum _{j=cQ+2q_1+d+1} ^{(c+1)Q-1}s(j)\\
&\kern2cm
+\sum _{r=c+1} ^{2c-1}\sum _{j=0} ^{Q-1}s(rQ+j)
+\sum _{j=2cQ} ^{2cQ+2q_1+2d}s(j)
-(cQ+d)\\
&=
-Q\sum _{r=0} ^{c-1}s(r)-c\sum _{j=0} ^{Q-1}s(j)
-ds(c)-\sum _{j=0} ^{d-1}s(j)\\
&\kern.45cm
+(Q-2q_1-d-1)s(c)+\sum _{j=2q_1+d+1} ^{Q-1}s(j)\\
&\kern.45cm
+Q\sum _{r=c+1} ^{2c-1}s(r)+(c-1)\sum _{j=0} ^{Q-1}s(j)
+(2q_1+2d+1)s(2c)+\sum _{j=0} ^{2q_1+2d}s(j)
-(cQ+d)\\
&=
Q\left(-\sum _{r=0} ^{c-1}s(r)+\sum _{r=c+1} ^{2c}s(r)-c\right)
-\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d+1} ^{2q_1+2d}s(j) -d\\
&=
-\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d+1} ^{2q_1+2d}s(j) -d,
\end{align*}
where in the last line we used again (\ref{fundeq}).
The reader should compare this expression with the one in \eqref{AL}.
Indeed, the remaining arguments are completely analogous to those
after \eqref{AL} in \textsc{Case~1a} of the current proof, which are
therefore left to the reader.

\medskip\noindent
\textsc{Case 2b}: $2q_1+d< Q\le 2q_1+2d$.
In this case, the quantity \eqref{AO} becomes
\begin{align*}
-{}&\sum _{r=0} ^{c-1}\sum _{j=0} ^{Q-1}s(rQ+j)
-\sum _{j=cQ} ^{cQ+d-1}s(j)
+\sum _{j=cQ+2q_1+d+1} ^{(c+1)Q-1}s(j)\\
&\kern2cm
+\sum _{r=c+1} ^{2c}\sum _{j=0} ^{Q-1}s(rQ+j)
+\sum _{j=(2c+1)Q} ^{2cQ+2q_1+2d}s(j)
-(cQ+d)\\
&=
-Q\sum _{r=0} ^{c-1}s(r)-c\sum _{j=0} ^{Q-1}s(j)
-ds(c)-\sum _{j=0} ^{d-1}s(j)\\
&\kern.5cm
+(Q-2q_1-d-1)s(c)+\sum _{j=2q_1+d+1} ^{Q-1}s(j)\\
&\kern.5cm
+Q\sum _{r=c+1} ^{2c}s(r)+c\sum _{j=0} ^{Q-1}s(j)
+(2q_1+2d+1-Q)s(2c)+\sum _{j=Q} ^{2q_1+2d}s(j)
-(cQ+d)\\
&=
Q\left(-\sum _{r=0} ^{c-1}s(r)+\sum _{r=c+1} ^{2c}s(r)-c\right)
-\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d+1} ^{2q_1+2d}s(j) -d\\
&= -\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d+1} ^{2q_1+2d}s(j) -d,
\end{align*}
where in the last line we used again (\ref{fundeq}).
The reader should compare this expression with the one in \eqref{AN}.
Indeed, the remaining arguments are completely analogous to those
after \eqref{AN} in \textsc{Case~1b} of the current proof, which are
therefore left to the reader.

\medskip\noindent
\textsc{Case 2c}. $Q\le 2q_1+d$ and $2q_1+2d<2Q$.
In this case, the quantity \eqref{AO} becomes
\begin{align*}
-\sum _{r=0} ^{c-1}&\sum _{j=0} ^{Q-1}s(rQ+j)
-\sum _{j=cQ} ^{cQ+d-1}s(j)
+\sum _{j=cQ+2q_1+d+1} ^{(c+2)Q-1}s(j)\\
&\kern2cm
+\sum _{r=c+2} ^{2c}\sum _{j=0} ^{Q-1}s(rQ+j)
+\sum _{j=(2c+1)Q} ^{2cQ+2q_1+2d}s(j)
-(cQ+d)\\
&=
-Q\sum _{r=0} ^{c-1}s(r)-c\sum _{j=0} ^{Q-1}s(j)
-ds(c)-\sum _{j=0} ^{d-1}s(j)
+(2Q-2q_1-d-1)s(c+1)\\
&\kern2cm
+\sum _{j=2q_1+d+1-Q} ^{Q-1}s(j)
+Q\sum _{r=c+2} ^{2c}s(r)+(c-1)\sum _{j=0} ^{Q-1}s(j)\\
&\kern2cm
+(2q_1+2d+1-Q)s(2c+1)+\sum _{j=0} ^{2q_1+2d-Q}s(j)
-(cQ+d)\\
&= Q\left(-\sum _{r=0} ^{c-1}s(r)+\sum _{r=c+1} ^{2c}s(r)-c\right)
-\sum _{j=0} ^{d-1}s(j) +\sum _{j=2q_1+d+1-Q} ^{2q_1+2d-Q}s(j)\\
&\kern2cm
+(2q_1+d+1-Q)(s(c)-s(c+1)+1)\\
&= -\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d+1-Q} ^{2q_1+2d-Q}s(j)
+(2q_1+d+1-Q)(s(c)-s(c+1)+1),
\end{align*}
where in the last step we used again (\ref{fundeq}).
Since $Q<2q_1+d+1$, we have $2q_1+d+1-Q>0$. In particular, since also
$s(c)-s(c+1)+1\ge0$ for any $c$, the third term in the last line is
nonnegative, and, because $d>0$,
the inequality (\ref{eq:Ungl}) says that the sum of the first two terms is
strictly positive.

\medskip\noindent
\textsc{Case 2d}. $Q\le 2q_1+d$ and $2q_1+2d\ge 2Q$.
In this case, the quantity \eqref{AO} becomes
\begin{align*}
-&\sum _{r=0} ^{c-1}\sum _{j=0} ^{Q-1}s(rQ+j)
-\sum _{j=cQ} ^{cQ+d-1}s(j)
+\sum _{j=cQ+2q_1+d+1} ^{(c+2)Q-1}s(j)\\
&\kern2cm
+\sum _{r=c+2} ^{2c+1}\sum _{j=0} ^{Q-1}s(rQ+j)
+\sum _{j=(2c+2)Q} ^{2cQ+2q_1+2d}s(j)
-(cQ+d)\\
&=
-Q\sum _{r=0} ^{c-1}s(r)-c\sum _{j=0} ^{Q-1}s(j)
-ds(c)-\sum _{j=0} ^{d-1}s(j)
+(2Q-2q_1-d-1)s(c+1)\\
&\kern2cm
+\sum _{j=2q_1+d+1-Q} ^{Q-1}s(j)
+Q\sum _{r=c+2} ^{2c+1}s(r)+c\sum _{j=0} ^{Q-1}s(j)\\
&\kern2cm
+(2q_1+2d+1-2Q)s(2c+2)+\sum _{j=0} ^{2q_1+2d-2Q}s(j)
-(cQ+d)\\
&=
Q\left(-\sum _{r=0} ^{c-1}s(r)+\sum _{r=c+1} ^{2c}s(r)-c\right)
-\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d+1-Q} ^{Q-1}s(j)\\
&\kern2cm
+\sum _{j=Q} ^{2q_1+2d-Q}s(j)
-(2q_1+2d-2Q+1)
+(Q-d)(s(c)-s(c+1)+1)\\
&= -\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d+1-Q} ^{2q_1+2d-Q}s(j)\\
&\kern2cm
-(2q_1+2d-2Q+1)
+(Q-d)(s(c)-s(c+1)+1),
\end{align*}
where in the last step we used again (\ref{fundeq}).
We have $d<Q$. In particular, since also
$s(c)-s(c+1)+1\ge0$ for any $c$, the fourth term in the last line is
nonnegative. Moreover, because $2q_1+d+1-Q\le d$,
the inequality (\ref{basins2})
says that the sum of the first two terms is
at least $2q_1+d+1-Q=(2q_1+2d-2Q+1)+(Q-d)$. Thus, the sum of the first
three terms in the last line is strictly positive. Combining both findings,
we infer that the whole last line is strictly positive
(in fact, at least $Q-d$).

This finishes the proof of \textsc{Case~2} and, thus, of the theorem.
\end{proof}

\section{Additional results on the 2-adic
valuation of $\delta_{n-q,n}=\th_{q,n}$}

In this section, we examine the $2$-adic behavior of
$\th_{q,n}=\delta_{n-q,n}$ in more detail. Keeping in mind
Proposition~\ref{prop:1}, we concentrate throughout this section on
the case that $n$ and $q$ have the same parity.

If we fix $q$ and let $n=q+2i$, $i=0,1,2,\dots$, vary, then we know
that whenever (\ref{spc1}) is satisfied, the $2$-adic valuation of
$\th_{q,n}=\th_{q,q+2i}$ will be zero. However, what happens in between?
By looking at some random values of $q$, one might get the impression
that, between two successive occurrences of zero, the $2$-adic valuations
$\nu_2(\th_{q,q+2i})$ are unimodal, that is, they
first grow (weakly) monotone until they reach their maximum value
half-way, and then they drop (weakly) monotone until they come back to zero in the
end. Moreover, one is led to guess that the $2$-adic valuations are
symmetric around the place where the maximum is attained. As it turns
out (see Example~\ref{Ex1} below), the unimodality conjecture is not
true, while the symmetry conjecture is indeed true. However, in some
sense, unimodality is ``almost" true. As we demonstrate in
Theorems~\ref{thm:8}, \ref{thm:9}, \ref{thm:10}, \ref{thm:11} below,
between two successive values of zero, the $2$-adic valuations
$\nu_2(\th_{q,q+2i})$ stay above a linear function of slope $1$ which
is tight at the opening zero, and at the same time they stay below
another linear function of slope 1 which is tight at the maximum
(see (\ref{AP}) and (\ref{AQ}) in Theorem~\ref{thm:8} and the analogous
inequalities in the subsequent theorems),
until they reach the maximum value, which is attained exactly
half-way, and the $2$-adic valuations beyond are the symmetric
images of those before. In the theorems, we determine in addition
the maximal value for each of these intervals.

The results which are found on the way to prove these theorems allow
one also to address the following question: characterize all values of $q$
and $n$ for which the $2$-adic valuation of $\th_{q,n}$ has a certain
fixed (small) value.
Clearly, Theorem~\ref{thm:2} does this if we fix this value to
$0$. In Corollaries~\ref{cor:8}, \ref{cor:9}, \ref{cor:10},
\ref{cor:11} we work out the analogous
characterization if we fix this value to
$1$. We could move on to $2$, $3$, etc., but at the cost of
a considerable increase of complication the further we go.

\begin{example} \label{Ex1}
The sequence $(\nu_2(\th_{q,q+2i}))_{i\ge0}$ is not unimodal in the
intervals discussed in Theorems~\ref{thm:8}--\ref{thm:11}
(although the theorems show that
it comes very close).
Here we give two counter-examples.

Consider $q=39$. In this case, the first
200 terms of the sequence  $(\nu_2(\th_{q,q+2i}))_{i\ge0}$ are:
\begin{align*}
0, &1, 3, 5, 7, 8, 8, 8, 8, 9, 12, 15, 18, 18, 15, 12, 9, 8, 8, 8, 8,
    7, 5, 3, 1, 0, 3, 6, 9, 10, 9, 8, 7, 9,\\
& 12, 15, 18, 20, 21, 22, 23, 25,
    29, 33, 37, 37, 33, 29, 25, 23, 22, 21, 20, 18, 15, 12, 9, 7, 8, 9, \\
&10, 9, 6, 3, 0, 1, 3, 5, 7, 8, 8, 8, 8, 9, 12, 15, 18, 18, 15, 12, 9, 8, 8,
    8, 8, 7, 5, 3, 1, 0, 4, 8, 12,\\
& 14, 14, 14, 14, 17, 21, 25, 29, 32, 34,
    36, 38, 41, 46, 51, 56, 56, 51, 46, 41, 38, 36, 34, 32,\\
& 29, 25, 21, 17, 14, 14, 14, 14, 12, 8, 4, 0, 1, 3, 5, 7, 8, 8, 8, 8, 9, 12, 15, 18, 18,
    15, 12, 9, 8,\\
& 8, 8, 8, 7, 5, 3, 1, 0, 3, 6, 9, 10, 9, 8, 7, 9, 12, 15,
    18, 20, 21, 22, 23, 25, 29, 33, 37, 37, 33,\\
& 29, 25, 23, 22, 21, 20, 18,
    15, 12, 9, 7, 8,9, 10, 9, 6, 3, 0, 1, 3, 5, 7, 8, 8, 8, 8.
\end{align*}
As predicted by Theorem~\ref{thm:2}, zeros occur for
all multiples of $\frac {1} {2}2^{\cl{\log_22\cdot 39}}=64$ and for
multiples of $64$ reduced by $39$. However, while the first
interval (the interval comprising the first 25 values) is unimodal,
the second one (comprising the next 39 values) is not, as can be seen
from the subsequence $0, 3, 6, 9, 10, 9, 8, 7, 9, 12, 15,\dots, 37,\dots$,
which rises first to $10$, then drops
to $7$, to rise again beyond $10$ until it reaches the maximum of
$37$ in this interval.

Consider now $q=46$. In this case, the first 200 terms are:
\begin{align*} 0,& 4, 2, 5, 6, 10, 10, 13, 14, 19, 14, 13, 10, 10, 6, 5, 2, 4, 0, 3,
    4, 8, 8, 11, 12, 17, 14, 15, 14, 16,\\
& 14, 15, 14, 19, 18, 22, 24, 29, 30,
    34, 36, 42, 36, 34, 30, 29, 24, 22, 18, 19, 14, 15, 14, 16,\\
& 14, 15, 14, 17, 12, 11, 8, 8, 4, 3, 0, 4, 2, 5, 6, 10, 10, 13, 14, 19, 14, 13, 10,
    10, 6, 5, 2, 4, 0,\\
& 4, 6, 11, 12, 16, 18, 24, 22, 24, 24, 27, 26, 28, 28,
    34, 34, 39, 42, 48, 50, 55, 58, 65, 58, 55,\\
& 50, 48, 42, 39, 34, 34, 28, 28, 26, 27, 24, 24, 22, 24, 18, 16, 12, 11, 6, 4, 0, 4, 2, 5, 6, 10, 10,\\
& 13, 14, 19, 14, 13, 10, 10, 6, 5, 2, 4, 0, 3, 4, 8, 8, 11, 12, 17, 14,
    15, 14, 16, 14, 15, 14, 19,\\
& 18, 22, 24, 29, 30, 34, 36, 42, 36, 34, 30, 29, 24, 22, 18, 19, 14, 15, 14, 16, 14, 15, 14, 17,\\
& 12, 11, 8, 8, 4, 3,
    0, 4, 2, 5, 6, 10, 10, 13, 14.
\end{align*}
%Christian: correction
This is an example where {\it both\/} types of intervals are not
unimodular. The first interval (comprising the first $64-46=18$ values)
begins $ 0, 4, 2, 5, 6, 10, 10, 13, 14, 19, \dots$,
that is, rises to $4$, drops
to $2$, before rising again up to the maximum of $19$ in this interval.
The second one (comprising the next $46$ values) starts by
$0, 3,  4, 8, 8, 11, 12,\break 17, 14, 15,\dots,42, \dots$, that is, rises to
$17$, drops to $14$, before rising again up to the maximum of $42$
in this interval. Similar remarks apply to the subsequent intervals.
\end{example}

Nevertheless, between two successive zeros, although the $2$-adic valuations\break
$\nu_2(\th_{q,q+2i})$ are not unimodal in general, the $2$-adic valuations still seem
to exhibit an overall increase until a maximum halfway and then
a decrease which is the symmetric image of the increasing values.
In the theorems below, we quantify this statement.
We split our results into four separate theorems.
On the one hand, we have to distinguish between the two possible
parities of $q$, and, on the other hand, for each integer $q$, there
are two types of intervals to be considered.

\begin{theo} \label{thm:8}  Let $q$ be a fixed positive even integer,
$Q=2^{\cl{\log_2 q}}$, and let $c$ be any nonnegative integer.
Then the values of the sequence
$(\nu_2(\th_{q,q+2i}))_{i\ge0}$ for
$i\in [cQ,(c+1)Q-q]$ are symmetric around its center $i_1=cQ+(Q-q)/2$.
The values at the extreme points of the interval
$i=cQ$ and $i=(c+1)Q-q$ are $0$, and
the value at the center is
$$\nu_2(\th_{q,q+2i_1})=
\frac {Q-q} {2}\log_2\frac {Q} {2}
-2\sum _{j=0} ^{\frac {Q-q-2} {2}}s(j),
$$
independent of $c$.  Furthermore,
for $i\in [cQ,i_1]$ the inequalities
\begin{equation} \label{AP}
\nu_2(\th_{q,q+2i})\ge i-cQ
\end{equation}
and
\begin{equation} \label{AQ}
\nu_2(\th_{q,q+2i})\le \nu_2(\th_{q,q+2i_1})-\(i_1-i\)
\end{equation}
hold.
\end{theo}

\begin{proof}
In this proof, we use again the notation $q_1=q/2$.
To show the symmetry  write
$ i=cQ+d$ with $0\le d\le Q-2q_1$. In particular, we have $2q_1+d\le
Q$, and, hence (cf.\ \eqref{AL} and \eqref{AN}), the 2-adic
valuation of
$\th_{q,q+2i}$ is given by
\begin{equation} \label{AR}
\nu_2(\th_{q,q+2i})=\nu_2(\th_{q,q+2(cQ+d)})
=-\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d} ^{2q_1+2d-1}s(j) -d.
\end{equation}
On the other hand, for the same reason, the 2-adic
valuation of
$\th_{q,q+2(cQ+Q-q-d)}$ is given by
\begin{align*}
\nu_2(&\th_{q,q+2(cQ+Q-2q_1-d)})=-\sum _{j=0} ^{Q-2q_1-d-1}s(j)
+\sum _{j=2q_1+(Q-2q_1-d)} ^{2q_1+2(Q-2q_1-d)-1}s(j)
-(Q-2q_1-d)\\
&=-\sum _{j=0} ^{Q-2q_1-d-1}s(j)
+\sum _{j=Q-d} ^{2Q-2q_1-2d-1}s(j)
-(Q-2q_1-d)\\
&=-\sum _{j=0} ^{Q-1}s(j)+\sum _{j=Q-2q_1-d} ^{Q-1}s(j)\\
&\kern2cm
+\sum _{j=0} ^{2Q-1}s(j)
-\sum _{j=0} ^{Q-d-1}s(j)
-\sum _{j=2Q-2q_1-2d} ^{2Q-1}s(j)
-(Q-2q_1-d)\\
&=\sum _{j=Q-2q_1-d} ^{Q-1}s(j)
+\sum _{j=Q} ^{2Q-1}s(j)
-\sum _{j=0} ^{Q-d-1}s(j)
-\sum _{j=2Q-2q_1-2d} ^{2Q-1}s(j)
-(Q-2q_1-d)\\
&=\sum _{j=Q-2q_1-d} ^{Q-1}s(j)
+\sum _{j=Q-d} ^{Q-1}s(j)
-\sum _{j=2Q-2q_1-2d} ^{2Q-1}s(j)
+(2q_1+d).
\end{align*}

Now we apply the reflection identity \eqref{AI} to all the sum of digit
functions. Thus, we obtain
\begin{align*}
\nu_2(\th_{q,q+2(cQ+Q-2q_1-d)})
&=(2q_1+2d)\log_2Q-\sum _{j=0} ^{2q_1+d-1}s(j)
-\sum _{j=0} ^{d-1}s(j)\\
&\kern2cm
-(2q_1+2d)\log_2(2Q)+\sum _{j=0} ^{2q_1+2d-1}s(j)
+(2q_1+d)\\
&=
-\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d} ^{2q_1+2d-1}s(j)
-d\\
&=\nu_2(\th_{q,q+2(cQ+d)}),
\end{align*}
proving the symmetry.
That the values of the extreme points of the interval $i=cQ$ and
$i=(c+1)Q-q$ are $0$, was already shown in Theorem~\ref{thm:2}, \textsc{Case~1}.

Next we determine the 2-adic value of $\th_{q,q+2i}$ at the
center $i_1=cQ+(Q-q)/2$ of the interval. By \eqref{AR}, we have
\begin{align*}
\nu_2(\th_{q,q+2i_1})&=
\nu_2(\th_{q,q+2(cQ+Q/2-q_1)})\\
&=
-\sum _{j=0} ^{\frac {Q} {2}-q_1-1}s(j)
+\sum _{j=\frac {Q} {2}+q_1} ^{Q-1}s(j)
-\left(\frac {Q} {2}-q_1\right)\\
&=
-\sum _{j=0} ^{\frac {Q} {2}-q_1-1}s(j)
+\sum _{j=q_1} ^{\frac {Q} {2}-1}s(j)\\
&=
\(\frac {Q} {2}-q_1\)\log_2\frac {Q} {2}
-2\sum _{j=0} ^{\frac {Q} {2}-q_1-1}s(j).
\end{align*}

To prove the inequality \eqref{AP}, we write again $i=cQ+d$,
with $d\le Q/2-q_1$. Under this condition, we
showed in the proof of Theorem~\ref{thm:2}, \textsc{Case~1a} (see \eqref{AM}),
that $\nu_2(\th_{q,q+2i})$ can be written in the form
\begin{equation} \label{AP1}
\nu_2(\th_{q,q+2i})=\nu_2(\th_{q,q+2(cQ+d)})=-\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d-\frac {Q} {2}} ^{2q_1+2d-\frac {Q} {2}-1}s(j).
\end{equation}
Since $d\le 2q_1+d-Q$,
the inequality (\ref{basins1}) implies that this expression is at
least $d=i-cQ$.

To prove inequality \eqref{AQ}, we compute the difference of the 2-adic
valuations of $\th_{q,q+2i_1}$ and $\th_{q,q+2i}$ for
$i\in[cQ,i_1]$: let again $i=cQ+d$,
$0\le d\le (Q-q)/2$. Then, using \eqref{AR}, we have
\begin{align*}
\nu_2(\th_{q,q+2i_1})-\nu_2&(\th_{q,q+2i})=
\nu_2(\th_{q,q+2(cQ+Q/2-q_1)})-\nu_2(\th_{q,q+2(cQ+d)})\\
&=-\sum _{j=0} ^{\frac {Q} {2}-q_1-1}s(j)
+\sum _{j=\frac {Q} {2}+q_1} ^{Q-1}s(j)
-\left(\frac {Q} {2}-q_1\right)
\\
&\kern2cm
+\sum _{j=0} ^{d-1}s(j)
-\sum _{j=2q_1+d} ^{2q_1+2d-1}s(j)
+d\\
&=-\sum _{j=d} ^{\frac {Q} {2}-q_1-1}s(j)
-\sum _{j=2q_1+d} ^{\frac {Q} {2}+q_1-1}s(j)
+\sum _{j=2q_1+2d} ^{Q-1}s(j)
-\left(\frac {Q} {2}-q_1-d\right).
\end{align*}
Now we apply the reflection formula \eqref{AI} to all the sum of digits
functions. This yields
\begin{align*}
\nu_2(\th_{q,q+2i_1})-\nu_2&(\th_{q,q+2i})
=\sum _{j=q_1} ^{\frac {Q} {2}-d-1}s(j)
+\sum _{j=\frac {Q} {2}-q_1} ^{Q-2q_1-d-1}s(j)
-\sum _{j=0} ^{Q-2q_1-2d-1}s(j)\\
&=\sum _{j=q_1} ^{\frac {Q} {2}-d-1}s(j)
+\sum _{j=\frac {Q} {2}-q_1} ^{Q-2q_1-d-1}s(j)
-\sum _{j=0} ^{\frac {Q} {2}-q_1-d-1}(s(2j)+s(2j+1))\\
&=\sum _{j=q_1} ^{\frac {Q} {2}-d-1}s(j)
+\sum _{j=\frac {Q} {2}-q_1} ^{Q-2q_1-d-1}s(j)
-2\sum _{j=0} ^{\frac {Q} {2}-q_1-d-1}s(j)
-\(\frac {Q} {2}-q_1-d\).
\end{align*}
Since $q_1\ge \frac {Q} {2}-q_1\ge \frac {Q} {2}-q_1-d$,
we may apply
(\ref{basins1}) twice to obtain finally
$$
\nu_2(\th_{q,q+2i_1})-\nu_2(\th_{q,q+2i})
\ge\(\frac {Q} {2}-q_1-d\)=i_1-i,
$$
as desired.
\end{proof}

\begin{corol} \label{cor:8}
For a fixed even $q$, the values of the sequence
$(\nu_2(\th_{q,q+2i}))_{i\ge0}$ are never $1$ for
$i\in [cQ,(c+1)Q-q]$.
\end{corol}
\begin{proof}  By the symmetry of the values around the center of the
interval, and by the inequality \eqref{AP}, the only possible
candidates are $i=cQ+1$ and $i=(c+1)Q-q-1$. Formula~\eqref{AP1} for
$\nu_2(\th_{q,q+2i})$ implies that in that case we must have
$$s\(q+1-\frac {Q} {2}\)=1.$$
This is only possible if $q+1-\frac {Q} {2}$ is a power of $2$, and
this, in its turn, implies that $q$ must be odd, a contradiction.
\end{proof}

\begin{theo} \label{thm:9}
Let $q$ be a fixed positive even integer,
$Q=2^{\cl{\log_2 q}}$, and let $c$ be any nonnegative integer.
Then the values of the sequence
$(\nu_2(\th_{q,q+2i}))_{i\ge0}$ for
$i\in [(c+1)Q-q,(c+1)Q]$ are symmetric around its center $i_2=(c+1)Q-q/2$.
The values at the extreme points of the interval $i=(c+1)Q-q$ and
$i=(c+1)Q$ are $0$, and the value at the center is
$$\nu_2(\th_{q,q+2i_2})=
\frac {q} {2}\log_2Q
-2\sum _{j=0} ^{\frac {q-2} {2}}s(j)+\frac {q} {2}(s(c)-s(c+1)+1).
$$
Furthermore,
for $i\in [i_2,(c+1)Q]$ the inequalities
\begin{equation} \label{AT}
\nu_2(\th_{q,q+2i})\ge (c+1)Q-i
\end{equation}
and
\begin{equation} \label{AS}
\nu_2(\th_{q,q+2i})\le \nu_2(\th_{q,q+2i_2})-\(i-i_2\)
\end{equation}
hold.  \qed
\end{theo}

\begin{proof}
We use again the notation $q_1=q/2$.
To show the symmetry write
$i=cQ+d$ with $Q-2q_1\le d\le Q$. In particular, we have $2q_1+d\ge
Q$, and, hence, if $d\le Q-q_1$, the 2-adic
valuation of
$\th_{q,q+2i}$ is given by
(cf.\ \eqref{ANa}, which also holds if $2q_1+d=Q$ as the
quantity vanishes in this case)
\begin{multline}
\nu_2(\th_{q,q+2i})=\nu_2(\th_{q,q+2(cQ+d)})\\
=-\sum _{j=0} ^{d-1}s(j)+\sum _{j=2q_1+d-Q} ^{2q_1+2d-Q-1}s(j)
+(2q_1+d-Q)(s(c)-s(c+1)+1),
\label{AU}
\end{multline}
while, if $d> Q-q_1$, it is given by
(cf.\ \eqref{ANb})
\begin{multline}
\nu_2(\th_{q,q+2i})=\nu_2(\th_{q,q+2(cQ+d)})\\
-\sum _{j=0} ^{d-1}s(j)+\sum _{j=2q_1+d-Q} ^{2q_1+2d-Q-1}s(j)
-(2q_1+2d-2Q)+(Q-d)(s(c)-s(c+1)+1).
\label{AV}
\end{multline}
Now let $i>i_2$, that is, $d>Q-q_1$. Symmetry around the center
$i_2=(c+1)Q-q_1$ means to show that
$$\nu_2(\th_{q,q+2(cQ+2Q-q-d)})=\nu_2(\th_{q,q+2(cQ+d)}).$$
Using \eqref{AU}, the left-hand term is given by
\begin{align*}
\nu_2(&\th_{q,q+2((c+2)Q-2q_1-d)})\\
&=
-\sum _{j=0} ^{2Q-2q_1-d-1}s(j)
+\sum _{j=Q-d} ^{3Q-2q_1-2d-1}s(j)
+(Q-d)(s(c)-s(c+1)+1)\\
&=-\sum _{j=0} ^{Q-1}s(j)+\sum _{j=2Q-2q_1-d} ^{Q-1}s(j)
+\sum _{j=0} ^{2Q-1}s(j) -\sum _{j=0} ^{Q-d-1}s(j)
-\sum _{j=3Q-2q_1-2d} ^{2Q-1}s(j)\\
&\kern6cm
+(Q-d)(s(c)-s(c+1)+1)\\
&=\sum _{j=2Q-2q_1-d} ^{Q-1}s(j)
+\sum _{j=Q} ^{2Q-1}s(j)
-\sum _{j=0} ^{Q-d-1}s(j)
-\sum _{j=3Q-2q_1-2d} ^{2Q-1}s(j)\\
&\kern4cm
+(Q-d)(s(c)-s(c+1)+1)\\
&=\sum _{j=2Q-2q_1-d} ^{Q-1}s(j)
+\sum _{j=Q-d} ^{Q-1}s(j)+Q
-\sum _{j=3Q-2q_1-2d} ^{2Q-1}s(j)\\
&\kern4cm
+(Q-d)(s(c)-s(c+1)+1).
\end{align*}
Now we apply the reflection identity \eqref{AI} to all the sum of digit
functions. Thus, we obtain
\begin{align*}
\nu_2(&\th_{q,q+2((c+2)Q-2q_1-d)})
=(2q_1+2d-Q)\log_2Q-\sum _{j=0} ^{2q_1+d-Q-1}s(j)
-\sum _{j=0} ^{d-1}s(j)\\
&\kern1cm
-(2q_1+2d-Q)\log_2(2Q)+\sum _{j=0} ^{2q_1+2d-Q-1}s(j)\\
&\kern1cm
+Q+(Q-d)(s(c)-s(c+1)+1)\\
&=
-\sum _{j=0} ^{d-1}s(j)
+\sum _{j=2q_1+d-Q} ^{2q_1+2d-Q-1}s(j)
-(2q_1+2d-2Q)+(Q-d)(s(c)-s(c+1)+1)\\
&=\nu_2(\th_{q,q+2(cQ+d)}),
\end{align*}
by \eqref{AV}, as desired.
That the values at the extreme points of the interval $i=(c+1)Q-q$ and
$i=(c+1)Q$ are $0$, was already shown in Theorem~\ref{thm:2}, \textsc{Case~1}.

Next we determine the 2-adic value of $\th_{q,q+2i}$ at the
center
$i_2=(c+1)Q-q/2$ of the interval. By \eqref{AU}, we have
\begin{align*} \nu_2(\th_{q,q+2i_2})&=
\nu_2(\th_{q,q+2((c+1)Q-q_1)})\\
&=
-\sum _{j=0} ^{Q-q_1-1}s(j)
+\sum _{j=q_1} ^{Q-1}s(j)
+q_1(s(c)-s(c+1)+1)\\
&=
-\sum _{j=0} ^{q_1-1}s(j)
+\sum _{j=Q-q_1} ^{Q-1}s(j)
+q_1(s(c)-s(c+1)+1)\\
&=
q_1\log_2Q
-2\sum _{j=0} ^{q_1-1}s(j)
+q_1(s(c)-s(c+1)+1).
\end{align*}

Inequality \eqref{AT} was already implicitly proved in the proof of
Theorem~\ref{thm:2}, \textsc{Case~1d}. Namely, if $i=cQ+d$,
with $d\ge Q-q_1$, then the conditions of \textsc{Case~1d} are satisfied, and
there it was shown (see the paragraph after \eqref{ANb}) that
$$\nu_2(\th_{q,q+2i})=\nu_2(\th_{q,q+2(cQ+d)})\ge Q-d=(c+1)Q-i,$$
as desired.

To prove inequality \eqref{AS}, we compute the difference of the 2-adic
valuations of $\th_{q,q+2i_2}$ and $\th_{q,q+2i}$ for
$i\in[i_2,(c+1)Q]$: let again $i=cQ+d$,
$Q-q_1\le d\le Q$. Then, using \eqref{AV} again, we have
\begin{align*}
\nu_2&(\th_{q,q+2i_2})-\nu_2(\th_{q,q+2i})=
\nu_2(\th_{q,q+2((c+1)Q-q_1)})-\nu_2(\th_{q,q+2(cQ+d)})\\
&=-\sum _{j=0} ^{Q-q_1-1}s(j)
+\sum _{j=q_1} ^{Q-1}s(j)
+q_1(s(c)-s(c+1)+1)
+\sum _{j=0} ^{d-1}s(j)\\
&\kern1cm
-\sum _{j=2q_1+d-Q} ^{2q_1+2d-Q-1}s(j)
+(2q_1+2d-2Q)-(Q-d)(s(c)-s(c+1)+1)\\
&=\sum _{j=Q-q_1} ^{d-1}s(j)
+\sum _{j=q_1} ^{2q_1+d-Q-1}s(j)
-\sum _{j=Q} ^{2q_1+2d-Q-1}s(j)\\
&\kern4cm
+(q_1+d-Q)(s(c)-s(c+1)+3)\\
&=\sum _{j=Q-q_1} ^{d-1}s(j)
+\sum _{j=q_1} ^{2q_1+d-Q-1}s(j)
-\sum _{j=0} ^{2q_1+2d-2Q-1}s(j)\\
&\kern4cm
+(q_1+d-Q)(s(c)-s(c+1)+1)\\
&=\sum _{j=Q-q_1} ^{d-1}s(j)
+\sum _{j=q_1} ^{2q_1+d-Q-1}s(j)\\
&\kern4cm
-2\sum _{j=0} ^{q_1+d-Q-1}s(j)
+(q_1+d-Q)(s(c)-s(c+1)).
\end{align*}
Since $Q-q_1\ge q_1+d-Q$ and $q_1\ge q_1+d-Q$,
we may apply
the inequality (\ref{basins1}) twice to obtain
$$
\nu_2(\th_{q,q+2i_2})-\nu_2(\th_{q,q+2i})
\ge(q_1+d-Q)(s(c)-s(c+1)+2).
$$
As we used already quite often, $s(c)-s(c+1)+1\ge0$. Therefore,
$$\nu_2(\th_{q,q+2i_2})-\nu_2(\th_{q,q+2i})
\ge q_1+d-Q=i-i_2,$$
as desired.
\end{proof}

\begin{corol} \label{cor:9}
For a fixed even $q$, the values of the sequence
$(\nu_2(\th_{q,q+2i}))_{i\ge0}$ are never $1$ for
$i\in [(c+1)Q-q,(c+1)Q]$, except if $q$ is a power of $2$ and $c$ is
even. In the
latter case, $\nu_2(\th_{q,q+2i})$ is equal to $1$ for $i=(c+1)Q-q-1$
and for $i=(c+1)Q-1$.

\end{corol}

\begin{proof} By the symmetry of the values around the center of the
interval, and by the inequality \eqref{AT}, the only possible
candidates are $i=(c+1)Q-q+1$ and $i=(c+1)Q-1$. Let us concentrate
on $i=(c+1)Q-1$, which, as before, we write using the parameter $d$
as  $i=(c+1)Q-1=cQ+d$.
The formula \eqref{ANb} in
the proof of Theorem~\ref{thm:2}, \textsc{Case~1d}, (which we used to prove
\eqref{AT}), yields that $\nu_2(\th_{q,q+2i})$ is equal to
\begin{align*}
-\sum _{j=0} ^{Q-2}s(j)+\sum _{j=q-1} ^{q+Q-3}s(j)
&-(q-2)+(s(c)-s(c+1)+1)\\
&=-\sum _{j=0} ^{Q-1}s(j)+s(Q-1)+\sum _{j=q-1} ^{q+Q-2}s(j)-s(q+Q-2)\\
&\kern4cm
-(q-2)+(s(c)-s(c+1)+1)\\
&=\cl{\log_2q}-s(q+Q-2)+1+(s(c)-s(c+1)+1).
\end{align*}
For this expression to be equal to $1$, we must have $s(q+Q-2)=\cl{\log_2q}$
and $s(c)-s(c+1)+1=0$. The former is the case if and only if $q=Q$, that
is, if $q$ is a power of $2$, and the latter is the case if and only
if $c$ is even.
\end{proof}

In an analogous manner, one can prove the following two theorems,
with accompanying corollaries,
covering the case where $q$ is odd.

\begin{theo} \label{thm:10}
Let $q$ be a fixed positive odd integer,
$Q=2^{\cl{\log_2 q}}$, and let $c$ be any nonnegative integer.
Then the values of the sequence
$(\nu_2(\th_{q,q+2i}))_{i\ge 0}$ for
$i\in [cQ,(c+1)Q-q]$ are symmetric around its center $cQ+(Q-q)/2$.
The values at the extreme points of the interval
$i=cQ$ and $i=(c+1)Q-q$ are $0$, and
the value at the central points $i_3=cQ+(Q-q-1)/2$ and
$cQ+(Q-q+1)/2$ is
$$\nu_2(\th_{q,q+2i_3})=
\frac {Q-q-1} {2}\log_2\frac {Q} {2}
-2\sum _{j=0} ^{\frac {Q-q-3} {2}}s(j)
-s\(\frac {Q-q-1} {2}\),
$$
independent of $c$.  Furthermore,
for $i\in [cQ,i_3]$ the inequalities
\begin{equation} \label{AW}
\nu_2(\th_{q,q+2i})\ge i-cQ
\end{equation}
and
\begin{equation} \label{AX}
\nu_2(\th_{q,q+2i})\le \nu_2(\th_{q,q+2i_3})-\(i_3-i\)
\end{equation}
hold.
\end{theo}

\begin{corol} \label{cor:10}
For a fixed odd $q$, the values of the sequence
$(\nu_2(\th_{q,q+2i}))_{i\ge0}$ are never $1$ for
$i\in [cQ,(c+1)Q-q]$, except if $q$ has the form $2^M+2^m-1$, for
some positive integers $m$ and $M$, $m<M$. In the latter case, $\nu_2(\th_{q,q+2i})$
is equal to $1$ for $i=cQ+1$ and for $i=(c+1)Q-q-1$.

\end{corol}

\begin{proof} The arguments from the proof of Corollary~\ref{cor:8}
apply also here. Thus, again, the only possible
candidates are $i=cQ+1$ and $i=(c+1)Q-q-1$. Furthermore, we must have
$$s\(q+1-\frac {Q} {2}\)=1.$$
This is only possible if $q+1-\frac {Q} {2}$ is a power of $2$,
which means that $q$ has the form given in the statement of the corollary.
\end{proof}

\begin{theo} \label{thm:11}
Let $q$ be a fixed positive odd integer,
$Q=2^{\cl{\log_2 q}}$, and let $c$ be any nonnegative integer.
Then the values of the sequence
$(\nu_2(\th_{q,q+2i}))_{i\ge 0}$ for
$i\in [(c+1)Q-q,(c+1)Q]$ are symmetric around its center $(c+1)Q-q/2$.
The values at the extreme points of the interval $i=(c+1)Q-q$ and
$i=(c+1)Q$ are $0$, and
the value at the central points $i_4=(c+1)Q-(q+1)/2$ and
$(c+1)Q-(q-1)/2$ is
$$\nu_2(\th_{q,q+2i_4})=
\frac {q-1} {2}\log_2Q
-2\sum _{j=0} ^{\frac {q-3} {2}}s(j)
-s\(\frac {q-1} {2}\)+\frac {q-1} {2}(s(c)-s(c+1)+1).
$$
Furthermore,
for $i\in [i_4,(c+1)Q]$ the inequalities
\begin{equation} \label{AY}
\nu_2(\th_{q,q+2i})\ge (c+1)Q-i
\end{equation}
and
\begin{equation} \label{AZ}
\nu_2(\th_{q,q+2i})\le \nu_2(\th_{q,q+2i_4})-\(i-i_4\)
\end{equation}
hold.
\end{theo}

\begin{corol} \label{cor:11}
For a fixed odd $q$, the values of the sequence
$(\nu_2(\th_{q,q+2i}))_{i\ge0}$ are never $1$ for
$i\in [(c+1)Q-q,(c+1)Q]$.

\end{corol}

\begin{proof}  The arguments from the proof of Corollary~\ref{cor:9}
apply also here. The conclusion was that we can have
$\nu_2(\th_{q,q+2i})=1$, for some $i$, only if $q$ is a power of
$2$. This is a contradiction to our assumption that $q$ is odd.
\end{proof}


\section{Skew symmetric matrices
and the parity of $\varepsilon_{2p,n}$}

Let $\F=\C,\R$ and denote by $\gl(n,\F)\subset \rM_n(\F)$
the group
of $n\times n$ invertible matrices.  Recall
that $\rA_n(\F)$ is the linear space of $n\times n$ skew symmetric matrices
$A$ of order $n$
over $\F$, i.e., $A\trans =-A$.  Clearly $\dim
\rA_n(\F)={\binom n 2}$.
Two matrices $A,B\in\rA_n(\F)$ are called
\emph{congruent} if $A=TBT\trans$ for some $T\in \gl(n,\F)$.
Let
$S_2:=\left(\begin{smallmatrix}\hphantom{-}0&1\\-1&0\end{smallmatrix}\right)$.
The following result is well-known in the real case, but its
complex version
does not seem to appear in standard modern books on linear algebra.

\begin{prop}\label{eqvcsmat}  Let $\F=\R,\C$ and $A
\in\rA_n(\F)$.  Then $A$ has
even rank,
$2p$ say, and $A$ is
congruent over $\F$ to a direct sum of $p$ copies of $S_2$ and the
$(n-2p)\times (n-2p)$ zero matrix.  In particular,
$B\in \rA_n(\F)$ is congruent to $A$ over $\F$ if and only if
$\rank A=\rank B$.
\end{prop}
\begin{proof}  We first prove the fact that any $A\in\rA_n(\F)$ is
congruent to the direct sum of copies of $S_2$ and $0$.
The result is trivial if $A=0$.
Let $n=2$ and $\rank A=2$.  Then $A=aS_2$ for some $0\ne a\in\F$.
For $\F=\C$ we have $A=(\sqrt{a} I_2)S_2 (\sqrt{a} I_2)\trans$.
For $\F=\R$ and $a>0$ the above formula holds.  For $a<0$ we have
$A=(\sqrt{-a} P)S_2 (\sqrt{-a} P)\trans$, where
$P:=\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$.

Assume by induction that any $A\in\rA_n(\F)$ is congruent
to the direct sum of
copies of $S_2$ and $0\in\rM_n(\F)$ for $n=m\ge 2$.  Let $n=m+1$
and $A\in \rA_{m+1}(\F)$.  Suppose first that $\det A=0$.  Let
$\0\ne\x\in\F^n$ and $A\x=0$.  Let $Q\in\gl(m+1,F)$ such that the
last column of $Q\trans$ is $\x$.  Then $QAQ\trans =A_1\oplus 0$,
and $A_1\in \rA_{m}(\F)$.  Use the induction hypothesis to deduce
that $A$ is conjugate to the direct sum of copies of $S_2$ and
$0$.
It remains to study the case where $m+1$ is even and $\det
A\ne 0$.  Let $A=(a_{ij})_{i,j=1}^{m+1}$ and
$A_1=(a_{ij})_{i,j=1}^m\in \rA_m(\F)$.  Since $A$ has rank $m+1$,
$A_1$ has at least rank $m-1$.  Since $m$ is odd $A_1$ has
exactly rank $m-1$.  So $A_1$ is conjugate to a direct sum of
$\frac{m-1}{2}$ copies of $S_2$ and one copy of $0$.
Using the corresponding
congruence on $A$, we may assume without loss of
generality that $A_1=
\left(\bigoplus_{i=1}^{(m-1)/2} S_2\right) \oplus 0 $.
For each
$i=1,\ldots,m-1$ subtract from
column $m+1$ of
$A$ the corresponding multiple of column $i$ to obtain
the zero
element
for the $(i,m+1)$-entry.  Repeat these elementary operations
with the rows of $A$ to eliminate the
$(m+1,i)$-entry for
$i=1,\ldots,m-1$.  The
resulting matrix is of the form
$B= \left(\bigoplus_{i=1}^{(m-1/2)} S_2\right)\oplus bS_2$.
Clearly $A$ is congruent to $B$.  Hence $b\ne
0$. Since $bS_2$ is congruent
to $S_2$, we deduce that $A$ is congruent to
a direct sum of
copies of $S_2$.

Since a direct sum of copies of $S_2$ and $0$ has an even rank
we deduce that any $A\in\rA_n(\F)$ has
even rank.
\end{proof}


The following result is known to the experts. We bring its
proof for completeness.
\begin{prop}\label{irw2pc}  Let $\F=\C,\R$, $n\ge 2$, $p\ge 1$ be integers and
assume that $p\le \lfloor \frac{n}{2}\rfloor$.
Let $\P \rW_{2p,n}(\F)\subset \P \rA_n(\F)$ be the projective variety of all
{\em(}nonzero{\em)} skew symmetric matrices of rank
at most $2p$. Then $\P
\rW_{2p,n}(\F)$
is an irreducible projective variety in
$\P \rA_n(\F)$ of codimension $\binom {n-2p} 2$.
The variety of its singular points is $\P \rW_{2(p-1),n}(\F)$.
\end{prop}

\begin{proof}  Let $\rW_{2p,n}^o(\F)\subset \rA_n(\F)$ be the quasi-variety of all
$n\times n$ skew symmetric matrices of rank $2p$.
Proposition~\ref{eqvcsmat} yields that $\gl(n,\F)$ acts transitively on
$\rW_{2p,n}^o(\F)$.  Hence $\rW_{2p,n}^o(\F)$ is a homogeneous
space and a manifold.  Since $\rW_{2(p-1),n}(\F)$ is a strict affine
subvariety of $\rW_{2p,n}(\F)$ it follows that $\rW_{2p,n}^o(\F)$
is a subset of smooth points and $\dim \rW_{2p,n}^o(\F)=\dim
\rW_{2p,n}(\F)$.  The neighborhood of each point is obtained by the corresponding
action of the neighborhood of $I_n\in \gl(n,\F)$.
% In particular, the tangent bundle of $A\in\rW_{2p,n}^o(\F)$ is the quotient of
% $\rA_n(\F)$ by the subspace $\{X\in \rA_n(\F):\;XA+AX\trans=0\}$.
Proposition~\ref{eqvcsmat} yields that the orbit of any $B\in \rW_{2(p-1),n}(\F)$
does not contain any matrix in $\rW_{2p,n}^o(\F)$.  Hence $\rW_{2(p-1),n}(\F)$ is the
variety of singular points in $\rW_{2p,n}(\F)$.

We now find the dimension and codimension of $\rW_{2p,n}(\F)$.  For $p=\lfloor
\frac{n}{2}\rfloor$, we have $\rW_{2p,n}(\F)=\rA_n(\F)$.  Hence $\dim \rW_{2\lfloor
\frac{n}{2}\rfloor,n}(\F)= {\binom n 2}$ and $\codim \rW_{2\lfloor
\frac{n}{2}\rfloor,n}(\F)= \binom {n -2\lfloor\frac{n}{2}\rfloor }
2=0$.  Let $1\le p < \lfloor\frac{n}{2}\rfloor$.  Let $A=(a_{ij})_{i,j=1}^n\in
\rW_{2p,n}^o(\F)$.  Then $A$ has $2p$ independent rows.  Assume
for simplicity
that the first $2p$ rows are
linearly independent.  Hence the first $2p$ columns of $A$ are
linearly independent.  Hence $A_1=(a_{ij})_{i,j=1}^{2p}\in
\rA_{2p}(\F)$ is nonsingular.  Therefore there exists a unique block
lower triangular matrix
$T=\left(\begin{smallmatrix} I_{2p} &0\\R& I_{n-2p}\end{smallmatrix}\right)$
such that $TAT\trans =A_1\oplus 0$.
Equivalently $A=T^{-1}(A_1\oplus 0)(T\trans)^{-1}$.
Hence $\dim \rW_{2p,n}=\binom {2p} 2 + 2p(n-2p)$ and
$\codim \rW_{2p,n}=\binom {n-2p}2$.
Thus, $\codim \P\rW_{2p,n}=\binom {n-2p} 2$.
\end{proof}

\begin{theo}\label{degskn}  Let $\F=\R,\C$ and $4 \le n, 1\le p <
\lfloor\frac{n}{2}\rfloor$.  Let $\P\rW_{2p,n}(\F)$ be the
irreducible variety of all {\em(}projectivized{\em)} nonzero skew symmetric
$n\times n$ matrices of rank
at most $2p$ in the projective space
$\P\rA_n(\F)$ of all nonzero $n\times n$ skew symmetric matrices
over $\F$.  Then the degree of $\P\rW_{2p,n}(\C)$ is odd if and
only if either $p$ or $n-p$ is divisible by $2^{\lceil \log_2 (n-2p)\rceil}$.
Furthermore, if either $p$ or $n-p$ is divisible by $2^{\lceil \log_2
(n-2p)\rceil}$, then any $\left(\binom {n-p} 2+1\right)$-dimensional subspace of $n\times n$
real skew symmetric matrices
contains a nonzero matrix of rank
at most $2p$. For these values
of $n$ and $p$ the dimensions of subspaces are best possible, i.e.,
\eqref{asrc} holds.
\end{theo}
\begin{proof}  Recall from (\ref{degvk}) that $\varepsilon_{2p,n}:= \deg \P \rW_{2p,n}
=\delta_{2p+1,n}/2^{n-2p-1}$.  The definition
(\ref{deftqn}) of $\th_{q,n}$ yields that
$\nu_2(\varepsilon_{2p,n})=\nu_2(\th_{n-2p-1,n})-(n-2p-1)$.
Proposition~\ref{prop:1} yields that $\nu_2(\varepsilon_{2p,n})=
\nu_2(\th_{n-2p,n})$.  Use Theorem~\ref{thm:2} to deduce that
$\varepsilon_{2p,n}$ is odd if and only if
either $p$ or $n-p$ is divisible by $2^{\lceil \log_2 (n-2p)\rceil}$.

Assume that  either $p$ or $n-p$ is divisible by $2^{\lceil \log_2 (n-2p)\rceil}$.
Then the discussion in \S1 implies that any $\left(\binom {n-p}
2+1\right)$-dimensional subspace of $n\times n$ real skew symmetric matrices
contains a nonzero matrix of rank
at most $2p$.  The sharpness
of these dimensions follows from the fact that a complex subspace
$L$ of $\rA_n(\C)$ of dimension $\binom {n-p} 2$ in general
position will not contain a nonzero $A\in \rA_n(\C)$ of rank
at most $2p$.
\end{proof}

\begin{corol}\label{subcsm}
Let $n\equiv 2\;({\rm mod\;} 4)$.  Then any two-dimensional real
subspace of $n\times n$ skew symmetric matrices contains a
nonzero singular matrix.
\end{corol}

\section{Rectangular matrices
and the parity of $\gamma_{k,m,n}$}

In this section we consider the parity problem for
$\gamma_{k,m,n}$, the latter being defined in (\ref{degkmn}).  It is more convenient
to introduce the following symmetric quantity.
For $n\in\N$, let $\h(n)$ be the \emph{hyperfactorial} $\prod_{k=0}^{n-1} k!$.
Let $a,b,c\in\N$.  Then a straightforward calculation shows:
\begin{align} \nonumber
B(a,b,c)&=\prod_{i=1}^a\prod_{j=1}^b\prod_{k=1}^c
\frac {(i+j+k-1)} {(i+j+k-2)}\\
\nonumber
&=\prod _{i=1} ^{a}\frac {(b+c+i-1)!\,(i-1)!} {(b+i-1)!\,(c+i-1)!}\\
 \label{eq:M2}
&=\frac {\h(a)\,\h(b)\,\h(c)\,\h(a+b+c)}
{\h(a+b)\,\h(b+c)\,\h(c+a)}.
\end{align}
$B(a,b,c)$ is a symmetric function on $\N^3$.
We remark that $B(a,b,c)$ is
the number of plane partitions which are contained in an
$a\times b\times c$ box (see e.g.\
\cite{Bressoud}).
From the definition (\ref{degkmn}) of $\gamma_{k,m,n}$ it is obvious that
\begin{equation}\label{gambeq}
\gamma_{k,m,n}= B(n-k,m-k,k), \quad 1\le k\le \min(m,n).
\end{equation}

%June 25, 2005

For simplicity of notation we let
$$S(a):=\sum_{i=0}^{a-1} s(i)\quad \text{and}\quad  \nu(a,b,c):=\nu_2(B(a,b,c))
\quad \text{for any } a,b,c\in\N.$$
Then,
by Proposition~\ref{2adn!}, we have $\nu_2(\h(a))=\frac{(a-1)a}{2}-S(a)$ and
\begin{equation}
\label{eq:nu2}
\nu(a,b,c)=-S(a+b+c)-S(a)-S(b)-S(c)+S(a+b)+S(b+c)+S(a+c).
\end{equation}

\begin{lemma}\label{basnuid}  Let $a,b,c\in\N$.  Then the
following identities hold:
\begin{align}
\nu(2a,2b,2c)&=2\nu(a,b,c), \label{basnuid1}\\
\nu(2a,2b+1,2c+1)&=\nu(a,b+1,c)+\nu(a,b,c+1),
\label{basnuid2}\\
\nu(2a+1,2b,2c)&=\nu(a,b,c)+\nu(a+1,b,c), \label{basnuid3}\\
\nu(2a+1,2b+1,2c+1)&=\nu(a,b+1,c+1)+\nu(a+1,b,c) \nonumber\\
&\kern2cm +s(b+c)-s(b+c+1)+2. \label{basnuid4}
\end{align}
In particular,
\begin{enumerate}
\item[\em(1)]
$B(2a,2b,2c)$ is odd if and only if $B(a,b,c)$ is odd.
\item[\em(2)]
$B(2a,2b+1,2c+1)$ is odd if and only if both $B(a,b+1,c)$
and $B(a,b,c+1)$ are odd.
\item[\em(3)]
$B(2a+1,2b,2c)$ is odd if and only if both $B(a,b,c)$
and $B(a+1,b,c)$ are odd.
\item[\em(4)]
$B(2a+1,2b+1,2c+1)$ is always even.
\end{enumerate}
\end{lemma}

\begin{proof} Equality (\ref{fundeq}) is equivalent to
\begin{equation}\label{fundeqS}
S(2p)=2S(p)+p,\;S(2p+1)=S(p+1)+S(p)+p, \quad \text{for any } p\in\N.
\end{equation}
Use (\ref{eq:nu2}) and the above equalities to deduce
(\ref{basnuid1})--(\ref{basnuid3}) straightforwardly.
In particular, these equalities yield the
corresponding claims about the oddness of $B(u,v,w)$.

To obtain  (\ref{basnuid4}), we use in addition to
(\ref{eq:nu2}) and the above equalities the obvious equality
$S(p+1)=S(p)+s(p)$ for $p=b+c,b+c+1$.  Since $s(p+1)\le s(p)+1$
it follows that  $B(2a+1,2b+1,2c+1)$ is always even.
\end{proof}

\begin{rem} \label{rem2}
(1) Lemma~\ref{basnuid} provides us with an algorithm to compute
the parity of $B(a,b,c)$ from the binary expansions of $a,b,c$
directly, without having to actually compute $B(a,b,c)$.
Namely, given $a,b,c$, one determines the parities of $a,b,c$.
If all of $a,b,c$ are odd, then Conclusion~(4) in Lemma~\ref{basnuid}
says that $B(a,b,c)$ is even. If all of $a,b,c$ are even, then one
uses Conclusion~(1) to reduce the problem to the problem of
determining the parity of $B(a/2,b/2,c/2)$. If exactly two of $a,b,c$ should be
odd, then one uses Conclusion~(2) for a similar reduction, and
if only one of $a,b,c$ is odd, then one uses Conclusion~(3).
One sees quickly that this yields an algorithm which can be
most conveniently run on the binary expansions of $a,b,c$.
See Proposition~\ref{oddconb} for an attempt to turn this algorithm
into a concrete characterization of those $a,b,c$ for which
$B(a,b,c)$ is odd.

(2) For the interested reader, we remark that the inspiration for
Lemma~\ref{basnuid} comes from results on plane partitions due to
Stembridge and Eisenk\"olbl. More precisely, Stembridge showed in
\cite{Ste} that a certain $(-1)$-enumeration (for our purposes it
suffices to say that this means a weighted enumeration in which
some plane partitions count as $1$, as in ordinary enumeration,
and others count as $-1$) of plane partitions contained in an
$a\times b\times c$-box is equal (up to sign) to the number of
self-complementary plane partitions contained in the same box.
Since, by definition of self-complementary plane partitions, there
cannot exist any if all of $a,b,c$ are odd, this result implies
immediately Conclusion~(4) in Lemma~\ref{basnuid}. Subsequently, Eisenk\"olbl
\cite{EisTAF} has embarked on the $(-1)$-enumeration of self-complementary
plane partitions. Her result is that a certain $(-1)$-enumeration
of self-complementary plane partitions contained in an
$a\times b\times c$-box is equal to
\begin{align*}
B\(\tfrac a2,\tfrac b2, \tfrac c2\)^2 \quad &\text{for $a,b,c$ even,}\\
B\(\tfrac a2,\tfrac{b+1}2,\tfrac{c-1}2\) B\(\tfrac
a2,\tfrac{b-1}2,\tfrac{c+1}2\) \quad &\text{for $a$ even and $b$, $c$
  odd,}\\
B\(\tfrac {a+1}2,\tfrac{b}2,\tfrac{c}2\) B\(\tfrac
{a-1}2,\tfrac{b}2,\tfrac{c}2\)
\quad &\text{for $a$ odd and $b$, $c$
  even.}
\end{align*}
The first of the three cases implies Conclusion~(1) in
Lemma~\ref{basnuid}, the second implies Conclusion~(2),
and the third implies Conclusion~(3). The relations
(\ref{basnuid1})--(\ref{basnuid4}) refine these conclusions on the
level of $2$-adic valuations.
\end{rem}

Let $b$ and $c$ be two nonnegative integers.
Furthermore, let
$$b=\sum_{i=0}^{\infty} b_i2^i, \; c=\sum_{i=0}^{\infty}c_i2^i, \quad
b_i,c_i\in \{0,1\},\;i=0,\ldots,$$
be the binary expansion of $b$ and $c$, respectively.
We say that the pair $(b,c)$ has a \emph{disjoint binary expansion}
if $b_ic_i=0$ for $i=0,\ldots$.
The following proposition is straightforward to establish.
\begin{prop} \label{no1com}  Let $b$ and $c$ be nonnegative integers.
Then $s(b+c)\le s(b)+s(c)$, and $s(b+c)=s(b)+s(c)$ if and only
the pair $(b,c)$ has a disjoint binary expansion.
\end{prop}

To find out
under which conditions on the parameters $a,b,c$
the number $B(a,b,c)$  is odd, it is enough to consider the
case $a\le \min(b,c)$.
\begin{theo}\label{a=123}  Let $b,c\in\N$.  Then
\begin{enumerate}
\item
$B(1,b,c)$ is odd
if and only if $(b,c)$ has a disjoint binary expansion.
In particular, for any
$q\ge 0$, $B(2^q,2^q b,2^q c)$ is odd
if and only if $(2^q b,2^q c)$ has a disjoint binary expansion.

\item Let $\min(b,c)\ge 2$.
\begin{itemize}
\item
If $b$ and $c$ are even then $B(2,b,c)$ is odd if and only if
the pair $(b,c)$ has a disjoint binary expansion.
\item
If $b$ is even and $c$ is odd
then $B(2,b,c)$ is odd if and only if the pairs $(b,c)$ and $(b,c+1)$
have disjoint binary expansions.
\item
If $b$ and $c$ are odd then
$B(2,b,c)$ is odd if and only if the pairs $(b,c+1)$ and $(b+1,c)$
have disjoint binary expansions.
\end{itemize}
\item
Let $\min(b,c)\ge 3$.
\begin{itemize}
\item
Assume that $b$ and $c$ are even.

If $b,c\equiv 0$ \textrm{\em mod} $4$ then $B(3,b,c)$ is
odd if and only if the pair $(b,c)$ has a disjoint binary expansion.

If $b,c+2\equiv 0$ \textrm{\em mod} $4$ then $B(3,b,c)$ is
odd if and only if the pairs $(b,c)$ and $(b,c+2)$ have disjoint binary expansions.

If $b,c\equiv 2$ \textrm{\em mod} $4$ then
$B(3,b,c)$ is
odd if and only if the pairs $(b,c)$, $(b,c+2)$ and $(b+2,c)$ have disjoint
binary expansions.
\item
Assume that $b$ is even and $c$ is odd.

If $b\equiv 0$ \textrm{\em mod} $4$ then $B(3,b,c)$ is
odd if and only if the pairs $(b,c)$ and $(b,c+1)$ have
disjoint binary expansions.

If $b\equiv 2$ \textrm{\em mod} $4$ then $B(3,b,c)$ is
odd if and only if the pairs $(b,c+1)$ and $(b+2,c)$
have disjoint binary expansions.
\end{itemize}
\end{enumerate}
\end{theo}
\begin{proof}
\begin{enumerate}
\item
The expression (\ref{eq:M2}) yields that $B(1,b,c)=\binom {b+c} b$.
Hence $\nu(1,b,c)=s(b)+s(c)-s(b+c)$.  Thus $\nu(1,b,c)=0$ if and
only if $(b,c)$ has a disjoint binary expansion.
The last
assertion follows from (\ref{basnuid1}) and from the observation
that $(b,c)$ has a disjoint binary expansion if and only if $(2^q b,2^q c)$
has a disjoint binary expansion.

\item
\begin{itemize}
\item
If $b$ and $c$ are even then
Conclusion~(1) in Lemma~\ref{basnuid} and item~1, which we just established, yield that
$B(2,b,c)$ is odd if and only the pair $(b,c)$ has
a disjoint binary expansion.
\item
Assume that $b=2b'$, $c=2c'+1$, $b',c'\in\N$.
Since $\nu(a,b,c)$ is a symmetric function in $a,b,c$,
(\ref{basnuid3}) yields that
$\nu(2,2b',2c'+1)=\nu(1,b',c')+\nu(1,b',c'+1)$.
Hence $B(2,b,c)$ is odd if and only if $(b',c')$ and $(b',c'+1)$
have disjoint binary expansions.  This is equivalent to the assumption
that the two even pairs $(b,c-1)$ and $(b,c+1)$ have disjoint binary
expansions.  Since $b$ is even and $c$ odd the assumption that
$(b,c-1)$ has disjoint binary
expansions is equivalent to the assumption
that $(b,c)$ has a disjoint binary
expansion.
\item
Assume that $b=2b'+1$, $c=2c'+1$, $b',c'\in\N$.  Then
(\ref{basnuid2}) yields
$\nu(2,b,c)=\nu(1,b'+1,c)+\nu(1,b',c'+1)$.
Hence $B(2,b,c)$ is odd if and only if $(b'+1,c')$ and $(b',c'+1)$
have disjoint binary expansions.  This is equivalent
to the assumption
that the two even pairs $(b+1,c-1)$ and $(b-1,c+1)$ have disjoint binary
expansions.  This is also equivalent to to the assumption
that $(b+1,c)$ and $(b,c+1)$ have disjoint binary
expansions.
\end{itemize}

\item
\begin{itemize}
\item
Assume that $b=2b'$, $c=2c'$, $a',b'\ge 2$.
Relation~(\ref{basnuid3}) implies that $B(3,b,c)=B(1,b',c')+B(2,b',c')$.
Hence $B(3,b,c)$ is odd if an only if $B(1,b',c')$ and $B(2,b',c')$
are odd.  Recall that $B(1,b',c')$ is odd if and only if
$(b',c')$ has a disjoint binary expansion.  This is equivalent
to the assumption that $(b,c)$ has a disjoint binary expansion.

Assume that $b'$ and $c'$ are even.  Then $B(2,a',b')$ is
odd if and only $(b',c')$ has a disjoint binary expansion.

Assume now that $b'$ is even and $c'$ are is odd.
Then $B(2,b',c')$ is odd if $(b',c')$ and $(b',c'+1)$
have disjoint binary expansions.  This is equivalent to
the assumption that $(b,c)$ and $(b,c+2)$ have
disjoint binary expansions.

Assume now that $b'$ and $c'$ is odd.  Then $B(2,b',c')$ is odd
if and only if the pairs $(b',c'+1)$ and $(b'+1,c')$ have
disjoint binary expansions.  This is equivalent to the
assumption that the pairs $(b,c+2)$ and $(b+2,c)$
have disjoint binary expansions.

\item
Assume that $b=2b'$, $c=2c'+1$, $b'\ge 2$, $c'\ge 1$.
Then $\nu(3,2b',2c'+1)=\nu(1,b',c'+1)+\nu(2,b',c')$.
Hence $B(3,b,c)$ is odd if and only if
$B(1,b',c'+1)$ and $B(2,b',c')$ are odd.
$B(1,b',c'+1)$ is odd if and only if $(b',c'+1)$
has a disjoint binary expansion.  This is equivalent
to the assumption that $(b,c+1)$ has a disjoint binary expansion.

Assume that $b'$ is even, i.e., $4\mid b$.
Suppose first that $c'$ is even, i.e., $4\mid (c-1)$.
Then $B(2,b',c')$ is odd if and only if $(b',c')$
has a disjoint binary expansion.  This is equivalent
to the assumption that $(b,c)$ has a disjoint binary expansion.

Assume second that $c'$ is odd, i.e., $4\mid (c+1)$.
Then $B(2,b',c')$ is odd if and only if $(b',c')$ and
$(b',c'+1)$ have disjoint binary expansions.
This is equivalent to the assumption that
$(b,c)$ and $(b,c+1)$ have disjoint binary expansions.

Assume now that $b'$ is odd, i.e., $4\mid (b+2)$.
Suppose first that $c'$ is even.  Then $B(2,b',c')$
is odd if and only if $(b',c')$ and $(b'+1,c')$
have disjoint binary expansions.
This is equivalent to the assumption that $(b,c)$ and $(b+2, c)$
have disjoint binary expansions.

Suppose second that $c'$ is odd.
Assume first that $c'=1$, i.e., $c=3$. Then $B(2,b',1)$ is odd if and only if
$(b',2)$ has a disjoint binary expansion.  This is equivalent
to the assumption that $(b,4)=(b,c+1)$ has a disjoint binary expansion.
Since $4\mid (b+2)$ it follows that $(b+2,3)$ has a disjoint binary expansion.

Assume second that $c'\ge 3$. Then $B(2,b',c')$ is odd if and only if $(b'+1,c')$
and $(b',c'+1)$ have disjoint binary expansions.  This is
equivalent to the assumption that $(b+2,c)$ and $(b,c+1)$ have
disjoint binary expansions.
\end{itemize}
\end{enumerate}
\end{proof}

The above theorem can be generalized schematically as follows:

\begin{prop}\label{oddconb}
Let $a,b,c\in \N$ and assume that $\min(b,c)\ge a$.  Let
$q:=\lceil \log_2 a\rceil$ and assume that $b\equiv b_r$, $c\equiv
c_r$ \textrm{\em mod} $2^q$ for some
$b_r,c_r\in [0,2^q-1]$. Then there exists a sequence of nonnegative
integers $d_i,e_i \in [0,2^q-1]$, $i=1,\ldots,\break N(a,b_r,c_r)$, depending only on
$a,b_r,c_r$, such
that $B(a,b,c)$ is odd if and only $(b+d_i,c+e_i)$ has a
disjoint binary expansion for $i=1,\ldots,N(a,b_r,c_r)$.
\end{prop}

\begin{proof}
We prove the proposition by induction on $q$.
For $q=0,1$ the proposition holds in view of Theorem~\ref{a=123}.
Assume that the proposition holds for any $q\le p-1$, where $p\ge
2$ and any $b,c$ such that $\min(b,c)\ge a$.  Assume that
$\lceil \log_2 a\rceil=p$.

\begin{itemize}
\item
Let $a=2a'$, $a'\in\N$.
Then $\lceil \log_2 a\rceil= \lceil \log_2 a'\rceil + 1$.
Assume first that $b$ and $c$ are even.
Then $\nu(a,b,c)=2\nu(\frac{a}{2},\frac{b}{2},\frac{c}{2})$ and
the proposition follows straightforwardly from the induction hypothesis.

Assume now that $b=2b'$, $c=2c'+1$, $a'\le b',c'\in\N$.
Then $\nu(2a',2b',2c'+1)=\nu(a',b',c')+\nu(a',b',c'+1)$.
Hence $B(a,b,c)$ is odd if and only $B(a',b',c')$ and
$B(a',b',c'+1)$ are odd.  Use the induction hypothesis and the
observation that $(u,v)$ has a disjoint binary expansion
if and only if $(2u,2v+1)$ has one for the case $B(a',b',c')$
to deduce the proposition.  (For the case $B(a',b',c'+1)$
note that if $d_i\le 2^{p-1}-1$
then $2d_i+1 \le 2^p-1$.)

Use the equality $B(a,b,c)=B(a,c,b)$ to deduce the proposition
in the case that $b$ is odd and $c$ is even.

Assume now that $b=2b'+1$, $c=2c'+1$, $a'\le b',c'\in\N$.
Then $\nu(2a',2b'+1,2c'+1)=\nu(a',b'+1,c')+\nu(a',b',c'+1)$.
Hence $B(a,b,c)$ is odd if and only $B(a',b'+1,c')$ and
$B(a',b',c'+1)$ are odd.  Use the induction hypothesis
and the above remarks to deduce the proposition in this case.

\item
Assume that $a=2a'+1$, $a'\in \N$.  Then $\lceil \log_2 a\rceil
=\lceil \log_2 a+1\rceil=
\lceil \log_2 a'+1\rceil + 1$.

Since $B(a,b,c)=B(a,c,b)$ is even if $a,b,c$ are odd,
it is enough to consider the case $b=2b'$, $a'<b'\in\N$.
Assume first that $c=2c'$, $a'<c'\in\N$.
Then $\nu(2a'+1,2b',2c')=\nu(a',b',c')+\nu(a'+1,b',c')$.
Hence $B(a,b,c)$ is odd if and only $B(a',b',c')$ and
$B(a'+1,b',c')$ are odd.
Use the induction hypothesis and the above arguments to deduce
the proposition.

Assume finally that $c=2c'+1$, $a'\le c'$.
Then $\nu(2a'+1,2b',2c'+1)=\nu(a',b',c'+1)+\nu(a'+1,b',c')$.
Hence $B(a,b,c)$ is odd if and only $B(a',b',c'+1)$ and
$B(a'+1,b',c')$ are odd.  The case where $B(a',b',c'+1)$ is odd
is done by induction on $a'$.  For $c'>a'$ the case
$B(a'+1,b'c')$ is odd is done by induction on $a'+1$.
The case $c'=a'$, which is equivalent to the case
where $B(a'+1,b',a')=B(a',b',a'+1)$ is odd is done by induction
on $a'$.  (As in the case of $B(3,2b',3)$ in Theorem~\ref{a=123}.)
\end{itemize}
\end{proof}

In principle the proof of Proposition~\ref{oddconb} can be used
to find the sequences $d_i,c_i$, $i=1,\ldots,N(a,b_r,c_r)$,
recursively.  However, the explicit construction of all such sequences seems
complicated even in the simple case where $a=2^q$, $q=2,\ldots$.
Note that Case~1 of Theorem~\ref{a=123} finds the sequence
for $a=2^q$ and $b_r=c_r=0$.  The cases $b_r=0$, $c_r=1$ and
$b_r=c_r=1$ have simple results.

\begin{prop}\label{a=2qbc=1}
Let $q\in\N$ and $2^q\le b,c\in\N$.  If $2^q\mid b$, $2^q\mid (c-1)$, then
$B(2^q,b,c)$ is odd if and only if $(b,c)$ and $(b,c+2^{q}-1)$
have disjoint binary expansions.
If $2^q\mid (b-1)$, $2^q\mid (c-1)$, then
$B(2^q,b,c)$ is odd if and only if $(b+1,c)$, $(b,c+1)$, $(b+2^{q}-1,c)$, and $(b,c+2^{q}-1)$
have disjoint binary expansions.
\end{prop}

The proof of this proposition is left to the reader.

The results of \S1 yield the following theorem.
\begin{theo}\label{dkmncr}  Let $k,m,n\in\N$, assume that
$k< \min(m,n)$, and let $\gamma_{k,m,n}$ be the positive integer given
by \eqref{degkmn}.
Let $\rM_{m,n}(\R)$ be the space of all
$m\times n$ real valued matrices.  Let $L\subset \rM_{m,n}(\R)$
be a subspace of dimension $(m-k)(n-k)+1$.  If $\gamma_{k,m,n}$
is odd then $L$ contains a nonzero matrix rank
at most $k$.
\end{theo}

\begin{corol}\label{cordkmn} For the following positive integers
$1\le k <n\le m$ any $((m-k)(n-k)+1)$-dimensional subspace of
$\rM_{m,n}(\R)$ contains a nonzero matrix rank
at most $k$:
\begin{enumerate}
\item
$k=n-1$ and $(m-n+1,n-1)$ has a disjoint binary expansion.
\item
 $2 \le k=n-2$.
 \begin{itemize}
 \item
 $n$ and $m$ are even and $(n-2,m-n+2)$ has a disjoint binary expansion.
 \item
 $n$ is even, $m$ is odd, $(n-2,m-n+2)$ and $(n-2,m-n+3)$
 have disjoint binary expansions.
 \item
 $n$ is odd, $m$ is even, $(n-2,m-n+3)$ and $( n-1,m-n+2)$
 have disjoint binary expansions.
 \end{itemize}

 \item
 $3\le k=n-3$.
 \begin{itemize}
 \item
 $4\mid (n-3)$, $4\mid m$, and $(n-3,m-n+3)$ has a disjoint binary
 expansion.
 \item
 $4\mid (n-3)$, $4\mid (m+2)$, $(n-3,m-n+3)$ and $(n-3,m-n+5)$
 have disjoint binary expansions.
 \item
 $4\mid (n-1)$, $4\mid (m+2)$, $(n-3,m-n+3)$ and $(n-1,m-n+3)$
 have disjoint binary expansions.
 \item
 $4\mid (n-1)$, $4\mid m$, $(n-1,m-n+3)$ and $(n-3, m-n+5)$
 have disjoint binary expansions.
 \item
 $4\mid (n-3)$, $m$ odd, $(n-3,m-n+3)$ and $(n-3, m-n+4)$
 have disjoint binary expansions.
 \item
 $4\mid (n-1)$, $m$ odd, $(n-3,m-n+4)$ and $(n-1,m-n+3)$
 have disjoint binary expansions.
 \item
 $4\mid (m-n+3)$, $n$ is even, $(n-3,m-n+3)$ and $(n-2, m-n+3)$
 have disjoint binary expansions.
 \item
 $4\mid (m-n+5)$, $n$ is even, $(n-2,m-n+3)$ and $(n-3, m-n+5)$
 have disjoint binary expansions.

 \end{itemize}

\item
Let $q\in\N$.
\begin{itemize}
\item
 $n=k+2^q$, $2^q\mid k$, $2^q\mid m$, and $(k,m-k)$
 has a disjoint binary expansion.
 \item
 $n=k+2^q$, $2^q\mid k$, $2^q\mid (m-1)$, $(k,m-k)$ and $(k,m-k+2^q-1)$
 have disjoint binary expansions.
 \item
 $2^{q+1}<n=k+2^q$, $2^q\mid (k-1)$, $2^q\mid (m-1)$, $(k,m-k)$ and
 $(k+2^q-1,m-k)$
 have disjoint binary expansions.
 \item
 $2^{q+1}<n=k+2^q$, $2^q\mid (k-1)$, $2^q\mid (m-2)$, and the pairs $(k+1,m-k)$,
 $(k, m-k+1)$, $(k+2^q-1,m-k)$, and $(k,m-k+2^q-1)$
 have disjoint binary expansions.
 \end{itemize}



 \end{enumerate}



\end{corol}




\begin{thebibliography}{99}
\bibitem{Ada} J.F. Adams, Vector fields on spheres, {\it Annals of Math.} {\bf
75} (1962), 603--632.
\bibitem{ACGH}
E. Arbarello, M. Cornalba, P.A. Griffiths and J. Harris, {\it
Geometry of Algebraic Curves}, {\bf I}, Grundlehren der math.\
Wissenschaften, vol.~267, Springer, 1985.
\bibitem{BF}
M.A. Berger and S. Friedland, The generalized Radon--Hurwitz
numbers, {\it Compositio Mathematica} {\bf 59} (1986), 113--146.
\bibitem{Bressoud}
D.M. Bressoud, \emph{Proofs and confirmations:
The story of the alternating sign matrix conjecture},
Cambridge University Press, Cambridge, 1999.
\bibitem{EisTAF}
T.    Eisenk\"olbl, $(-1)$-enumeration of self-complementary
plane partitions, {\it Electron.\ J. Combin.} {\bf 12}(1) (2005),
Article~\#R7, 22~pp.
\bibitem{FFL}
D. Falikman, S. Friedland and R. Loewy, On spaces of matrices
containing a nonzero matrix of bounded rank,
{\it Pacific J. Math.} {\bf 207} (2002), 157--176.
\bibitem{FL} S. Friedland and A. Libgober, Generalizations of
the odd degree theorem and applications,
\emph{Israel J. Math.} {\bf 136} (2003), 353--371.
\bibitem{FRS} S. Friedland, J.W. Robbin and J.H. Sylvester, On the crossing
rule, {\it Comm.\ on Pure and Appl.\ Math.} {\bf 37} (1984), 19--37.
\bibitem{Gan} F.R. Gantmacher, \emph{The Theory of Matrices} II,
Chelsea Pub.\ Com., 1964.
\bibitem{GhorKrat}
S.R. Ghorpade and C. Krattenthaler,
The Hilbert series of Pfaffian rings, in: \emph{Algebra, Arithmetic and
Geometry with Applications} (C.~Christensen, G.~Sundaram, A.~Sathaye
and C.~Bajaj, eds.) Springer--Verlag, New York, 2004, pp.~337--356.
\bibitem{GrKnPa}
R.L. Graham, D.E. Knuth and O. Patashnik,
\emph{Concrete Mathematics},
Addison Wesley, Reading, Massachusetts, 1989.
\bibitem{HT1}
J. Harris and L.W. Tu, On symmetric and skew-symmetric
determinantal varieties, {\it Topology} {\bf 23} (1984), 71--84.
\bibitem{HT2}
J. Harris and L.W. Tu,  Chern numbers of kernels and cokernel
bundles, {\it Invent.\ Math.} 75 (1984), 467--475.
\bibitem{Hur}
A. Hurwitz, \"{U}ber die Komposition der quadratischen Formen,
{\it Math.\ Ann.} {\bf 88} (1922), 1--25.
\bibitem{King}
R.C. King and N.G.I. El-Sharkaway, Standard {Y}oung tableaux
and weight multiplicities of the classical {L}ie groups,
{\it J.~Physics A} {\bf 16} (1983), 3153--3178.
\bibitem{Lax}
P.D. Lax, The multiplicity of eigenvalues, {\it Bull.\ Amer.\ Math.\
Soc.} {\bf 6} (1982), 213--215.
\bibitem{Legendre}
A.M. Legendre, \emph{Th\'eorie des Nombres}, Firmin Didot Fr\`eres,
Paris, 1830.
\bibitem{Rad}
J. Radon, Lineare Scharen orthogonaler Matrizen, {\it Abh.\ Math.\
Sem.\ Univ.\ Hamburg} {\bf 1} (1923), 1--14.
\bibitem{MS} J.W. Milnor and J.D. Stasheff, \emph{Characteristic
Classes}, Annals of Math.\ Studies 76, Princeton Univ.\ Press, 1974.
\bibitem{Ste} J.R. Stembridge, Some hidden relations involving the ten symmetry
classes of plane partitions, {\it J.~Combin.\ Theory Ser.~A} {\bf 68}
(1994), 372--409.
\bibitem{Sundaram}
S. Sundaram, Tableaux in the representation theory of the classical
 {L}ie groups, in: \emph{Invariant Theory and Tableaux} (Dennis Stanton,
 ed.), Springer--Verlag, 1990, pp.~191--225.

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