Remember the Occupy Wall Street movement that swept across the U.S. in 2011? The idea behind this was to highlight earnings inequities between the highest paid one percent of Americans and the other 99 percent of American workers.
After a viral start, the movement faded, although there are still sporadic rallies around the country.
I'd like to discuss a different one percent philosophy, one sweeping the industrial world. I'm referring to the incredible cost savings that can be gained by eliminating one percent of waste across numerous industry verticals.
GE has been talking a lot lately about the "Industrial Internet," or its vision for how the rapidly emerging Internet of Things can benefit industrial companies in asset-heavy industries like healthcare. On its website, GE notes that the healthcare industry could save $63 billion over 15 years — or more than $4 billion annually — by reducing capital expenditures by just one percent.
Think about that. If we were lucky enough to triple that reduction in capital to three percent – still a highly attainable figure — the healthcare industry stands to save some $12 billion a year. RFID, which is being rapidly deployed across healthcare, will play a major role in helping to achieve those financial gains, which could be substantial, considering that the global health care industry accounted for 10 percent of global GDP in 2011.
It has been estimated that more than 10 percent of those health expenditures are wasted from inefficiencies, such as expired products or items that end up in the dumpster because they are not stored at the required temperatures during transit or at the healthcare facility. That means the global cost of health care inefficiency is at least $731 billion per year, according to data from PricewaterhouseCoopers.
According to the GE report, clinical and operations inefficiencies, which can be most directly impacted by the Industrial Internet, account for 59 percent of healthcare inefficiencies, representing $429 billion per year.
It is estimated that deployment of the Industrial Internet can help to drive these costs down roughly 25 percent, or about $100 billion per year in savings, according to Ibid.
The GE report also says that the Industrial Internet application opens the possibility of creating a "care traffic control system" for hospitals. Hospitals are comprised of thousands of pieces of critical equipment, much of which is mobile.
GE says that the key is knowing where it all resides, and having a system that can alert doctors, nurses and technicians to changes in status, and provide metrics to improve resource utilization and patient and business outcomes.
Although the GE report surprisingly doesn't mention RFID as the key enabler to solving the above problem, the RFID-based solutions being deployed today to track highly expensive assets like heart monitors and hospital beds play a major role in the beginning of the Industrial Internet in health care.
GE Healthcare estimates that these innovations can translate into a 15 to 30 percent reduction in hospital equipment costs and permit healthcare workers to gain an additional hour of productivity on each shift.
That's an incredible amount of cost savings, not to mention the ability for each healthcare worker to gain one hour of productivity per shift. Hospitals run 24-7, so this adds up to substantial soft cost savings over three shifts.
Here at Terso Solutions, we've seen the benefits gained from deploying RFID-enabled freezers and refrigerators to more than 100 customers around the globe. In addition, Promega, our own parent company, has saved close to $2 million since deploying RFID-enabled refrigerators and freezer solutions into its daily operations. Prior to using RFID storage solutions from Terso, Promega was losing approximately 15-18 percent of its on-site inventory to shrinkage, equal to $1.2 million per year.
And it's not just healthcare that can reap gains from the Industrial Internet movement. According to the GE report, the oil and gas industry has the potential to save $90 billion over the next 15 years by reducing its capital expenditures by just one percent through automation. And a one percent fuel savings delivered by more efficient jet engines, enabled in part by RFID-tagged engine parts, would save the aviation industry $30 billion.
So here's to the Power of One Percent! However, I'm advocating that one percent is too conservative of a goal. I hope in a few years time we're toasting to the Power of Five Percent and the billions saved each year. Through continued innovative gains in technology, I'm sure we can get there.


text:
\section{Introduction}
 A {\em partition} $\lambda$ of a positive integer $n$ is defined to
be a sequence of nonnegative  integers $(\lambda_1, \lambda_2, \ldots, \lambda_m)$ such that $\lambda_1+\lambda_2+\cdots+
\lambda_m=n$ and $\lambda_1\geq \lambda_2\cdots \geq \lambda_m$.
    The {\em Young diagram} of $\lambda$ is defined to be an up- and left- justified array of $n$ boxes with $\lambda_i$ boxes in the $i$-th row.  The {\em conjugate } of $\lambda$, denoted by $\lambda'$, is the partition whose Young diagram is the reflection along the main diagonal of $\lambda$'s diagram, and $\lambda$ is said to be {\em self-conjugate} if   $\lambda=\lambda'$. The {\em hook} of each box $B$ in $\lambda$ consists of the box $B$ itself and boxes directly to the right and directly below $B$. For example, Figure \ref{Young} illustrates
the Young diagram and the hook lengths of  a self-conjugate partition $\lambda=(5,3,3,1,1)$.

\begin{figure}[h!]
\begin{center}
 \begin{ytableau}
     9& 6 & 5 & 2&1\\
     6 & 3 & 2\\
    5 & 2 &1\\
     2\\
     1
 \end{ytableau}
 \end{center}
\caption{ The Young  diagram  of  a self-conjugate partition $\lambda=(5,3,3,1,1)$.}\label{Young}
\end{figure}

For a positive integer $s$, a partition is said to be an {\em $s$-core partition}, or simply an {\em $s$-core},  if it contains no box whose hook length is a multiple of $s$.  Furthermore,  a partition is said to be {\em $(t_1, t_2, \ldots, t_m)$-core} if it is simultaneously $t_1$-core, $t_2$-core, \ldots,  and $t_m$-core.  Anderson \cite{And}  showed that the number of $(s,t)$-core partitions is the
rational Catalan number  ${1\over s+t}{s+t\choose s}$
when $s$ and $t$ are coprime to each other.

  Simultaneous core partitions have attracted much attention since  Anderson's work  \cite{And}.  Various  results on the enumeration of such   partitions are archived in  \cite{Agg,Arm, Baek, Chen, Joh, Sta2, Str, Wang, Xiong, Yan, Zaleski,Zaleski2}.

  Ford, Mai and Sze \cite{Ford} characterized the set of hook lengths
of diagonal cells in self-conjugate  $(s,t)$-core partitions, and they showed that the number of
self-conjugate $(s,t)$-core partitions is given by ${\lfloor {s\over 2}\rfloor+\lfloor {t\over 2}\rfloor\choose \lfloor {s\over 2}\rfloor}$.
Concerning the enumeration of $(s,s+1, s+2)$-core partitions, Amdeberhan \cite{Amd}  conjectured that the number of $(s, s+1, s+2)$-core  partitions is equal to the number of Motzkin paths of length $s$. This conjecture was first  confirmed    by Yang, Zhong and Zhou \cite{Yang}.
Amdeberhan and Leven  \cite{Amd-lev} further proved that  the number of $(s,s+1, \ldots, s+k)$-core partitions is equal to the number of $(s, k)$-Dyck paths.
The  largest  size of such  partitions was derived by Xiong \cite{Xiong2}.

A  { ballot $(s,k)$-path} of height $n$  is a lattice path  from $(0,0)$ to $(s,n)$
 using up steps $U_k=({k\over 2},1)$, down steps $D_k=({k\over 2},-1) $ and horizontal steps $H_{\ell}=(\ell,0)$ for some positive integer $\ell$ with  $1\leq \ell <k$,  and never
  lying below the $x$-axis.  A  ballot $(s,k)$-path of height $0$ is said to be an {\em (s,k)-Dyck} path. Notice that an (s,2)-Dyck  path is just a Motzkin path of length $s$. We say that an
$(s, k)$-Dyck path is {\em symmetric} if its reflection about the line $x={s\over 2}$ is itself. For example, a  symmetric $(10,4)$-Dyck path $P=U_2H_2H_2H_2D_2$ is shown in Figure \ref{sym}.

\begin{figure}[h!]
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\caption{ A symmetric $(10,4)$-Dyck path.}\label{sym}
\end{figure}

In this paper, we are mainly concerned with  the enumeration of  self-conjugate  $(s,s+1,  \ldots, s+k)$-core partitions, confirming the following  conjecture  posed by Cho, Huh and Sohn \cite{Cho}.

\begin{conj}\label{con1}
 For given positive integers $s$ and $k$, the number of self-conjugate $(s, s +
1, \ldots, s + k)$-core partitions is equal to the number of symmetric $(s, k)$-Dyck paths.
\end{conj}

 The case when $k=2$ has been confirmed by Cho, Huh and Sohn \cite{Cho}. To be more precise, they showed that the number of self-conjugate $(s, s +
1, s+2)$-core partitions is given by
$$
\sum_{i\geq 0}{\lfloor {s\over 2}\rfloor \choose i}{i\choose \lfloor {i\over 2}\rfloor  },
$$
which counts the    number of symmetric Motzkin paths of length $s$.

\section{Proof of Conjecture \ref{con1}}

In this section,  we aim to confirm Conjecture \ref{con1}. We begin with some definitions and notations.

  Following  the poset terminology given
by Stanley \cite{Sta},  for two elements $x$ and $y$ in a poset $(P, \prec)$, we say that $y$ {\em covers} $x$ if $x\prec y$ and there exists no element $z\in P$ such that $x\prec z\prec y$.
An element $x$ in $P$ is said to be of {\em rank} $s$ if it  covers an element of rank $s-1$. Note that the elements of rank $0$   in $P$ are just the minimal elements.
The {\em Hasse diagram} of a finite poset $P$ is a graph whose vertices are the elements of $P$,  whose edges are the cover relations, and such that if $y$  covers $x$ then there is an edge connecting $x$ and $y$ and
 $y$ is placed above $x$.
     An {\em order ideal} of $P$ is a subset $I$ such that if any $y\in I$ and $x\prec y$ in $P$, then $x\in I$.

For a partition $\lambda$, let $MD(\lambda)$ denote the set of      main diagonal hook lengths. It is apparent that  $MD(\lambda)$ is a set of distinct odd integers when $\lambda$ is self-conjugate. In the following theorem,  Ford, Mai and Sze \cite{Ford} provided a characterization of the set $MD(\lambda)$ of any self-conjugate $t$-core partition $\lambda$.
\begin{thm} { \upshape   (  \cite{Ford}) }\label{thford}
Let $\lambda$ be a self-conjugate partition. Then $\lambda$ is $t$-core if and only if  $MD(\lambda)$ verifies the following properties.
\begin{itemize}
\item[{\upshape (\rmnum{1})} ]  $MD(\lambda)$ is a set of  distinct odd integers;
\item[{\upshape (\rmnum{2})} ] If $h\in MD(\lambda)$ with $h>2t$, then $h-2t\in MD(\lambda)$;
\item[{\upshape (\rmnum{3})} ]  If $h_1, h_2\in MD(\lambda)$, then $  h_1+h_2$ is not a multiple of $2t$.
\end{itemize}
\end{thm}

 \begin{definition}
 Define  $L(s,k)=\bigcup_{j\geq 0} A_j,$ and $R(s,k)=\bigcup_{j\geq 0} B_j,$
  where
 $$
 A_j=\{2i-1+2sj\mid jk+1\leq i\leq \lfloor{s\over 2}\rfloor \},
 $$

 and
 $$
 B_j=\{2i-1+2sj\mid  jk+ \lceil{s+k\over 2} \rceil+1 \leq   i\leq s\}.
 $$
 \end{definition}

 \begin{lem}\label{lem3}
Let $x$ be a  positive odd  integer with $x>2s$. Then  $x$ is contained in $L(s,k) $ (resp. $R(s,k)$) if and only if  $x-2t\in L(s,k) $ (resp. $x-2t\in R(s,k) $ ) for all $s\leq t\leq s+k$.
\end{lem}
\pf
 The necessity follows directly from the definition of $L(s,k) $ (resp. $R(s,k)$).
Now we proceed to prove the sufficiency part of the theorem. Let   $x=2i-1+2js$  with  $1\leq i\leq s $ and $j\geq 1$. Assume that  $x-2t\in L(s,k) $ for all $s\leq t\leq s+k$. We proceed to show that $x\in L(s,k)$.  We claim that $i>k$. Since $x-2s-2k\in L(s,k)$, we have either $i>k$ or $j\geq 2$. Notice that   $x-2s=2i-1+2(j-1)s$. Then   $x-2s\in L(s,k)$ implies that $\lfloor{s\over 2}\rfloor\geq i\geq k(j-1)+1$. If $j\geq 2$, then  we also have $i>k$.      Then $x-2s-2k=2(i-k)-1+2(j-1)s\in L(s,k) $  implies that $i-k \geq k(j-1)+ 1$, which is equivalent to  $i\geq kj+1$. On the other hand,    $x-2s=2i-1+2(j-1)s\in L(s,k)$ implies that $  i  \leq \lfloor{s\over 2}\rfloor$.  Hence we have $x\in L(s,k) $.

    Assume that $x-2t\in R(s,k) $ for all $s\leq t\leq s+k$. We proceed to verify that $x\in R(s,k)$ by considering the following  two cases.

     If $j\geq 2$, then $x-2s=2i-1+2(j-1)s\in R(s,k)$ implies that $i-k(j-1)=i-kj+k\geq \lceil{s+k\over 2}\rceil+1$, which leads to   $  i- k \geq  \lceil{s+k\over 2}\rceil+1$.     Then $x-2s-2k=2(i-k)-1+2(j-1)s\in R(s,k) $  implies that $i-kj=i-k-k(j-1)\geq \lceil{s+k\over 2}\rceil+1$. Hence, we have $x\in R(s,k) $.

       If $j=1$, then $x-2s-2k=2i-1-2k\in R(s,k)$ implies that $i-k\geq \lceil{s+k\over 2}\rceil+1 $, which is equivalent to $i\geq k+ \lceil{s+k\over 2}\rceil+1$.  Thus,  we   have $x\in R(s,k)$.   This completes the proof. \qed

 \begin{definition}
Let $$P(s,k)=L(s,k)   \cup  R(s,k) $$
 with the following poset structure:  if $x, y\in P(s,k) $, then $y$ covers $x$ if and only if    $y=x+2s+2t$ for some $0\leq t\leq k$.
 \end{definition}
  Lemma \ref{lem3} tells us that $P(s,k)$ can be viewed as the disjoint union of the posets $L(s,k)$ and $R(s,k)$.   In the Hasse diagram of $L(s, k)$ (resp. $R(s, k)$),
 each element $x$ of rank $\ell$    covers exactly $ k+1$ elements   $ x-2s, x-2s-2$, $\ldots, x-2s-2k $ of rank $\ell-1$ for all $\ell\geq 1$. For example,  the Hasse   diagrams of $P(20,3)$ and $ P(20,4)$ are illustrated in Figure \ref{P{20,3}} and \ref{P{20,4}}, respectively.

An order ideal $I$ of $P(s,k)$ is said to be  {\em nice } if  $I$ does not contain  two elements $h_1, h_2$ such that  $h_1+h_2\in  \{2s, 2s+2, \ldots, 2s+2k\}$.

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\caption{ The Hasse diagram of the poset  $P(20,3)$.}\label{P{20,3}}
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\caption{ The Hasse diagram of the poset  $P(20,4)$.}\label{P{20,4}}
\end{figure}

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\caption{ The Hasse diagram of the poset  $P'(20,3)$.}\label{P'{20,3}}
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\caption{ The Hasse diagram of the poset  $P'(20,4)$.}\label{P'{20,4}}
\end{figure}

In the following theorem, we establish a correspondence between self-conjugate  $(s, s+1, \ldots, s+k)$-core partitions and the  nice order ideals of the poset $P(s,k) $.
\begin{thm}\label{main1}
Let $\lambda$ be a self-conjugate partition. Then $\lambda$ is    $(s, s+1, \ldots, s+k)$-core if and only if $MD(\lambda)$ is a nice order ideal of the poset $P(s,k) $.
 \end{thm}
\pf Assume that $\lambda$ is a self-conjugate  $(s, s+1, \ldots, s+k)$-core partition, we proceed to show that $MD(\lambda)$ is a nice order ideal of  $P(s,k)$. By   {\upshape (\rmnum{2})} and {\upshape (\rmnum{3})} of Theorem \ref{thford}, it suffices to show that    $MD(\lambda)$ is a subset of $P(s,k)$. Suppose to the contrary,  there exists an element $x$ of  $MD(\lambda)$ such that $x$ is not  contained in $P(s,k)$. Choose $x$ to the smallest such element. Examining  the definition of the poset $P(s,k)$, we have $x>2s$.  We claim that $x>2s+2k$. If not, suppose that $x=2s+2t+1$ for some $0\leq t\leq k-1$.  Then by {\upshape (\rmnum{2})} of Theorem \ref{thford}, $MD(\lambda)$ contains the element $y=1$. Then there are  two elements  $x=2s+2t+1$ and $y=1$ in $MD(\lambda)$ such  that $x+y=2s+2t+2$, contradicting {\upshape (\rmnum{3})}.   By {\upshape (\rmnum{2})}  of Theorem \ref{thford} and  the selection  of $x$, we have $x-2t\in P(s,k) $ for all $s\leq t\leq s+k$. From Lemma \ref{lem3}, it follows that $x\in P(s,k) $, which contradicts the selection of $x$. Hence we have reached the conclusion  that $MD(\lambda)$ is a subset of  $P(s,k)$.

Conversely, given a nice order ideal $I$ of $P(s,k)$, we wish to show that $I=MD(\lambda)$ for some self-conjugate $(s, s+1, \ldots, s+k)$-core partition $\lambda$. It suffices to prove that $I$ verifies properties {\upshape (\rmnum{1})}- {\upshape (\rmnum{3})} of Theorem \ref{thford}. It is easy to check that $I$ verifies properties {\upshape (\rmnum{1})} and   {\upshape (\rmnum{2})}. We proceed to show that $I$ also has property {\upshape (\rmnum{3})}.  Let    $h_1, h_2\in I$.  For any $1\leq t\leq k $, we have  $h_1=2p-1+2(s+t)q$ and $h_2=2p'-1+2(s+t)q'$ where $1\leq p,p'\leq s+t$, and $q,q'\geq 0$. Let $x=2p-1$ and $y=2p'-1$.  Since $I$ is a nice order ideal of $P(s,k)$, the elements $x$ and $y$ are contained in $I$ and $x+y\neq 2(s+t)$. This implies that $h_1+h_2$ is not a multiple of $2(s+t)$ for all $0\leq t\leq k$. Thus we have concluded that $I$ verifies  {\upshape (\rmnum{3})}.  This completes the proof. \qed

 \begin{definition}
 Define  $L'(s,k)=\bigcup_{j\geq 0} A'_j,$ and $R'(s,k)=\bigcup_{j\geq 0} B'_j,$
  where
 $$
 A'_j=\{(i-{k\over 2}j,j)\mid  jk+1\leq i\leq \lfloor{s\over 2}\rfloor,\,\, i\in \mathcal{Z}\},
 $$

 and
 $$
 B'_j=\{(s+1-i+{k\over 2}(j+1),-j-1)\mid  jk+ \lceil{s+k\over 2} \rceil+1 \leq   i\leq s,\,\, i\in \mathcal{Z}\}.
 $$
 \end{definition}

\begin{definition}
Let $$P'(s,k)=L'(s,k)   \cup  R'(s,k) $$
 with the following poset structure:
 \begin{itemize}
  \item If $(a, b), (a', b')\in L'(s,k) $, then $(a, b)$ covers $(a', b')$ if and only if  $|a-a'|\leq {k\over 2}$ and $b=b'+1$;
  \item If $(a, b), (a', b')\in R'(s,k) $, then $(a, b)$ covers $(a', b')$ if and only if  $|a-a'|\leq {k\over 2}$ and $b=b'-1$.
    \end{itemize}
 \end{definition}
For example, the Hasse diagrams of $P'(20,3)$ and $P'(20,4)$ are shown in Figure \ref{P'{20,3}} and Figure \ref{P'{20,4}},   respectively.

 Now we present a map $\chi: P(s, k)\rightarrow P'(s, k)$ as follows.
\begin{itemize}
 \item  If $x=2i-1+2sj$ with $jk+1\leq i\leq \lfloor{s\over 2}\rfloor$ and $j\geq 0$, set  $\chi(x)=(i-{k\over 2}j, j)$;
 \item  If $x=2i-1+2sj$ with  $ jk+ \lceil{s+k\over 2} \rceil+1 \leq   i\leq s$ and $j\geq 0$,  set  $\chi(x)=(s+1-i+{k\over 2}(j+1),-j-1)$.
\end{itemize}
 It is easily seen that the map $\chi: P(s, k)\rightarrow P'(s, k)$ is an isomorphism of posets.
 Furthermore,  $\chi$  maps a nice order ideal of $P(s,k)$ to an order  ideal of $P'(s,k)$ containing    no elements $(a, 0)$ and $(a',-1)$ such that $|a-a'|\leq {k\over 2}$.

 \begin{thm}\label{chi}
 For  given positive integers $s$ and $k$,  the nice order ideals of $P(s, k)$ are in one-to-one correspondence with order  ideals of $P'(s,k)$ containing    no elements $(a, 0)$ and $(a',-1)$ such that $|a-a'|\leq {k\over 2}$
 \end{thm}
 Denote by $J(P'(s,k))$    the set of order  ideals of $P'(s,k)$  containing    no elements $(a, 0)$ and $(a',-1)$ such that $|a-a'|\leq {k\over 2}$.
\subsection{On self-conjugate $(s, s+1, \ldots, s+2k)$-core partitions}

It is routine to check that  the poset $P'(2m,2k)$ is identical with the poset $P'(2m+1, 2k)$.  Relying on Theorems \ref{main1} and \ref{chi}, we are led to the following result.

\begin{thm}\label{mainth2}
For positive integers $m$ and $k$,  self-conjugate $(2m, 2m+1, \ldots, 2m+k)$-core partitions are equinumerous with  self-conjugate $(2m+1, 2m+2, \ldots, 2m+k+1)$-core partitions.
\end{thm}

Combining  Theorems \ref{main1}  and  \ref{mainth2},   we will focus on    the enumeration of   order ideals $I\in J(P'(2m,2k))$  for any positive integers $m$ and $k$. We begin with some definitions and notations.

      If a ballot $(s,k)$-path   is allowed to go below the $x$-axis, then it is called a {\em free} ballot $(s,k)$-path.  Denote by $\mathcal{FB}_n(s,k)$ the set of all free  ballot $(s,k)$-paths of height $n$.
  Let  $\mathcal{SD}(s,k)$ denote the set of all symmetric $(s, k)$-Dyck paths.

Throughout this paper we identify a  (free)  ballot $(s,k)$-path with a word of $U_{k\over 2}$'s, $D_{k\over 2}$'s and $H_{\ell}$'s .    Let $p$ be a step  running from the point $(x_1, y_1)$ to the point $(x_2, y_2)$. Then we say that the point $(x_1,y_1)$ is its  {\em starting
   } point, and the  point $(x_2, y_2)$ is its {\em ending} point. The {\em starting} point and the {\em ending } point  of a path are defined analogously.

 Let $\mathcal{FB}'_n(m,k)$ be the set of free ballot $(m, k)$-paths of height  $n$ which start  with either a horizontal step or a down step.
Let $$\mathcal{Q}(m,k)=(\bigcup_{i=0}^{k-2}\mathcal{FB}'_0(m-i,2k))\cup \mathcal{FB}'_{-1}(m+1,2k).$$

 We describe a map $\varphi:J(P'(2m,2k))\rightarrow \mathcal{Q}(m,k)$ as follows.  Let $I\in J(P'(2m,2k)) $. First we color the  point $(a,b)$ with $a, b\in \mathcal{Z}$ and $0\leq a\leq m+1$ in the plane by black and white according to the following rules.
\begin{itemize}
\item If $b\geq 0$, then
    the point $(a,b)$  is colored by  black if and only   if $(a,b) \in I  $;
    \item If $b<0$, then the point $(a,b)$  is colored by  black if and only    if $(a,b)\notin I  $.
 \end{itemize}
 Now we recursively  define a sequence   $P_0,P_1, \ldots P_n$ of points where  $P_0=(0,0)$. Assume that $  P_{i-1}$ has  been already determined. Let   $P_{i-1}=(a_{i-1}, b_{i-1})$.   We proceed to demonstrate how to get $P_i$.  There are two cases.\\
 Case 1.  There exists a black point   $(a_{i-1}+\ell, b_{i-1})$  with  $1\leq \ell\leq  2k-1$.  If  the point $ (a_{i-1}+k, b_{i-1}+1)$ is black, then  set $P_i=(a_{i-1}+k, b_{i-1}+1)$. Otherwise, we choose $\ell$ to the smallest such integer and set $P_i=(a_{i-1}+\ell, b_{i-1})$.   \\
   Case 2. There does not exist  a black point   $(a_{i-1}+\ell, b_{i-1})$  with  $1\leq \ell\leq  2k-1$. If the point $ (a_{i-1}+k, b_{i-1}-1)$ is black, set $P_i=(a_{i-1}+k, b_{i-1}-1)$. Otherwise, set $P_{i}=P_{i-1}$ and  $n=i-1$.\\
Let $\varphi(I)$ be the resulting  path with lattice points $P_0, P_1, \ldots, P_n$.
For example, let $I=\{(1,0), (2,0), (3,0), (4,0), (5,0), (3,1), (8,-1), (10,-1)\}\in J(P'(20,4))$. By applying the map $\varphi$, we get a  path $\varphi(I)$ as shown in Figure \ref{phi}.

\begin{figure}[h!]
\begin{center}
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  \put(0,6){\circle{0.2}} \put(1,6){\circle{0.2}} \put(2,6){\circle{0.2}} \put(3,6){\circle{0.2}} \put(4,6){\circle{0.2}} \put(5,6){\circle{0.2}}
  \put(6,6){\circle{0.2}} \put(7,6){\circle{0.2}} \put(8,6){\circle{0.2}}  \put(9,6){\circle{0.2}} \put(10,6){\circle{0.2}} \put(11,6){\circle{0.2}}

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\put(-1.2,3.8){\small$(0,0)$}

\end{picture}
\end{center}
\caption{ The Corresponding path  $\varphi(I)$.}\label{phi}
\end{figure}

\begin{figure}[h!]
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   \put(24,4){\circle{0.2}}
\put(0,2){\line(3,0){3}} \put(3,2){\line(2,1){2}}  \put(5,3){\line(1,0){1}} \put(6,3){\line(2,1){2}} \put(8,4){\line(2,-1){2}} \put(10,3){\line(2,-1){2}}
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\end{center}
\caption{ The decomposition  of a path $ P=H_3U_2H_1U_2D_2D_2D_2H_1D_2H_2H_1U_2U_2$.}\label{alpha1}
\end{figure}

\begin{figure}[h!]
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\caption{ The decomposition of a path $ Q=H_2H_1U_2U_2U_2H_1U_2U_2U_2D_2H_1D_2H_3$.}\label{alpha2}
\end{figure}

\begin{figure}[h!]
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\put(11,2){\line(2,-1){2}}  \put(13,1){\line(1,0){1}}   \put(14,1){\line(2,-1){2}}  \put(16,0){\line(3,0){3}}  \put(19,0){\line(2,1){2}} \put(21,1){\line(1,0){1}} \put(22,1){\line(2,0){2}}

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\caption{ The decomposition of a path $P=D_2U_2H_1U_2D_2H_2D_2H_1D_2H_3U_2H_1H_2$.}\label{beta1}
\end{figure}

\begin{figure}[h!]
\begin{center}
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\caption{ The decomposition of a path $Q=U_2H_1U_2D_2H_2U_2U_2H_1U_2H_3D_2H_1H_2$.}\label{beta2}
\end{figure}

\begin{lem}\label{lemphi3}
For any order ideal $I\in J(P'(2m,2k))$,  the resulting path $\varphi(I)$ verifies the following properties.
\begin{itemize}
 \item[{\upshape (\rmnum{1})}] The points   lying     above the path $\varphi(I)$ are white;
\item[ {\upshape (\rmnum{2})}] The points   lying     below the path $\varphi(I)$ are black.
\end{itemize}
 \end{lem}
\pf Here we only verify  {\upshape (\rmnum{1})}.  By similar arguments,  one can prove {\upshape (\rmnum{2})}. We follow notations in the definition of $\varphi$.     Let $P^{(\ell)}$ be  the partial path of $\varphi(I)$  whose lattice points are $P_0, P_1, \ldots, P_{\ell}$.  We prove   by induction on $\ell$.   Clearly the assertion holds for $\ell=0$.  Assume that  the points   lying    above the path $P^{(\ell-1)} $ are white.  Let $P_{\ell-1}=(a,b)$.   Now we proceed to show that the assertion also holds for $P^{(\ell)}$.
 We have three cases.

 Case 1. $P_{\ell}=(a+k, b+1)$.     When $y>b+1$ and $a<x\leq a+k$,  the point $(x,y)$ must be white. If not, by the cover relation of $I$, the point $(a, y-1)$  is also black, which contradicts the induction hypothesis. It remains to show that the point $(x,b+1)$ is white for any $a<x<a+k$. If not,  Let $c$ be the smallest integer such that  the point $(c,b+1)$ is black and $a<c<a+k$.     By  the cover relation of $I$, it follows that  the point $(c-k, b)$ is black.  By the induction hypothesis, we must have $P_{\ell-2}=(a-k, b+1)$.  Then by the construction of the path, we must  have $P_{\ell-1}=(c,b+1)$, which  contradicts the selection of $P_{\ell-1}$.

 Case 2.  $P_{\ell}=(a+i, b)$ for some $1\leq i\leq 2k-1$.  We claim that the point $(x,b+1)$ is white for any $a<x\leq a+i$. If not,  Let $c$ be the smallest integer such that  the point $(c,b+1)$ is black and $a<c\leq a+i$.  According to the selection of $P_\ell$,  we have $c\neq a+k$.  By the cover relation of $I$, the point $(c-k, b)$ is black.  Hence,  by the induction hypothesis, we must have $P_{\ell-2}=(a-k, b+1)$.  Then by the construction of the path, we must  have $P_{\ell-1}=(c,b+1)$, which  contradicts the selection of $P_{\ell-1}$.   Hence,  by the cover relation of $I$, the point $(x,y)$ must be white for any $y>b+1$ and $a<x\leq a+i$.

 Case 3.  $P_{\ell}=(a+k, b-1)$.  Due to the selection of the lattice point $P_{\ell}$,  the point $(x,b)$ is white for any $a<x\leq  a+k$. Then by the cover relation of $I$,    the point $(x,y)$ must be white for any $y>b+1$ and $a<x\leq a+k$.  This completes the proof. \qed

   \begin{lem}\label{lemphi1}
 The map $\varphi$ is well defined, that is, for any $I\in J(P'(2m,2k))$  we have   $ \varphi(I)\in\mathcal{Q}(m,k)$.
\end{lem}
\pf Let $\varphi(I)$ be   path with lattice points $P_0, P_1, \ldots, P_n$.
   It is apparent that the resulting path is a lattice path  consisting of up steps $U_k$, down steps $D_k$ and horizontal steps $H_\ell$ for some $1\leq \ell\leq 2k-1$. Moreover, since $(k, 1)$ is not contained in $P'(2m,2k)$, $\varphi(I)$ must start with either a down step or a horizontal step. In order to show that $\varphi$ is well defined, it remains  to verify that we have either $P_{n}=(m-i, 0)$ for some $0\leq i\leq k-2$  or $P_n=(m+1, -1)$. Notice that $(k,-1)$ is not contained in  $P'(m,k)$. Thus the point $(k,-1)$ is always colored by black. This ensures that we have $n>0$.

 Suppose that $P_n=(a,b)$. According to the construction of the path,  the point $P_n$ is black.  If  $b> 0$,  then we have $(a,b)\in I$ and $(a+k, b-1)\in I$ by the cover relation of $I$.  It follows that  the point $(a+k, b-1)$ is black  and  the path can not terminate at the   point $(a,b)$.  So we have $b\leq 0$.

 We claim that $P_n\neq (m-i, 0)$ for any $i\geq k-1$. If not, we have $(m-i, 0)\in I$, which implies that   $(m-i+k, -1)\notin I$.  It follows  that the point $(m-i+k,-1)$ is black  and the path can not terminate at the   point $(a,b)$.

 Suppose that $b<0$ and $a\leq m+1-k$.     Since the point $(a, b)$ is the ending point of $\varphi(I)$, the point $(a+k, b-1) $ must be  white.  This implies that we have $(a+k,b-1)\in I$. Recall that  $ (a+k,b-1) $ covers $(a,b)$. Therefore,   we have $(a, b)\in I$ and  the point   $(a,b)$ is white,   a contradiction.

 Suppose that $b<0$ and $  m+2-k\leq a\leq m$.   Since $m+1-a\leq k-1$ and $\varphi(I)$ terminates at $(a,b)$, the point   $(m+1,b)$ must be  white.  By the cover relation of $I$, we have $(m+1,-1)\in I$,    a contradiction.

 We claim that $P_n\neq (m+1, b)$ for any $b<-1$. If not,  by {\upshape (\rmnum{1})} of Lemma \ref{lemphi3}, the point $(m+1, -1)$ is white,  yielding a contradiction with the fact that the point $(m+1, -1)$ is black.

    So far, we have reached the conclusion that   we have either $P_{n}=(m-i, 0)$ for some $0\leq i\leq k-2$  or $P_n=(m+1, -1)$. This implies that the map $\varphi$ is well defined, completing the  proof. \qed

\begin{lem}\label{lemphi2}
For any order ideal $I\in J(P'(2m,2k))$,  the resulting path $\varphi(I)$ verifies the following properties.
\begin{itemize}
\item[{\upshape (\rmnum{1})}] The points   on  the path $\varphi(I)$  are black if and only if they are ending points of steps;
\item[ {\upshape (\rmnum{2})}] For $b<0$, the point $(a,b)$       to the right of the ending point of  the path $\varphi(I)$ is black;
\item[ {\upshape (\rmnum{3})}] For $b\geq 0$, the point $(a,b)$        to the right of the ending point of  the path $\varphi(I)$ is white.
 \end{itemize}
\end{lem}

\pf  Point  {\upshape (\rmnum{1})} follows  directly from the  construction of  $\varphi(I)$.

Recall that the ending point of $\varphi(I)$ is either the point $(m-i, 0)$ for some $0\leq i\leq k-2$ or the point $(m+1, -1)$. Clearly, it suffices to prove  $\varphi(I)$ verifies properties    {\upshape (\rmnum{2})} and  {\upshape (\rmnum{3})} for the former case.  Since $\varphi(I)$ terminates at $(m-i,0)$, the point  $(a, 0)$ is white for any $m-i<a\leq m+1$. Hence by the cover relation of $I$, the point $(a,b)$ is white for any $m-i<a\leq m+1$ and $b\geq 0$.  Moreover,  the point  $(a, -1)$ is black for any $m-i<a\leq m+1$. This implies that  the point $(a,b)$ is black for any $m-i<a\leq m+1$ and $b\leq -1$. Thus, $\varphi(I)$ has  properties    {\upshape (\rmnum{2})} and  {\upshape (\rmnum{3})}. This completes the proof. \qed

\begin{thm}\label{mainth3}
The map $\varphi$ is a bijection between  $J(P'(2m,2k)) $ and $\mathcal{Q}(m,k)$.
\end{thm}
\pf   First we give  the inverse map $\varphi'$ from the set $\mathcal{Q}(m,k)$ to the set $J(P'(2m,2k))$. Let $P\in \mathcal{Q}(m,k)$. First we color the  point $(a,b)$ with $a, b\in \mathcal{Z}$ and $0\leq a\leq m+1$ in the plane by black and white according to the following rules.
\begin{itemize}
\item If the point $(a,b)$ lies below the path, we color it by black;
    \item  If the point $(a,b)$ lies above the path, we color it by white;
    \item   For a point $(a,b)$   on the path, we color it by black if and only if it is  an    ending point  of a step;
        \item For  the point (a,b)   to the right to the endpoint of   $P$, we color it by white when $b\geq 0$,  and color it by black when $b<0$.
  \end{itemize}
For $b<0$,  let $(a,b)\in \varphi'(P)$  if and only if the point $(a,b)$ is white. For $b\geq 0$, let $(a,b)\in \varphi'(P)$ if and only if the point $(a,b)$ is black.  It is easily seen  that $\varphi'(P) \in J(P'(2m,2k))$.
Lemmas \ref{lemphi3} and \ref{lemphi2} guarantee  that the maps $\varphi $ and $\varphi'$ are inverses of each other, and thus the map $\varphi$ is a bijection. This completes the proof. \qed

In the following, we aim to establish a bijection between $\mathcal{Q}(m,k)$ and $\mathcal{SD}(2m,2k)$.

Let $\mathcal{B}^e(s,k)$ and $\mathcal{B}^o(s,k)$ denote the set of ballot $(s,k)$-paths of even height and the set of ballot $(s,k)$-paths of odd height, respectively.  The {\em complement } of the path $P$, denoted by $   P^c$,  is  obtained from $P$ by interchanging the up steps and the down steps.   The {\em reverse complement } of the path $P$,
denoted by $ P^{^{rc}}$,  is obtained from $P$ by   encoding the steps of $P$ from right to left and  interchanging the up steps and the down steps.
For example, the complement  of the path  $P=H_1U_3D_3D_3U_3D_3H_2$ is given by  $H_1D_3U_3U_3D_3U_3H_2  $, while the reverse complement of the path $P$ is given by $H_2U_3D_3U_3U_3D_3H_1$.

\begin{thm} \label{alpha}
There is a bijection $\alpha$  between  the set $\mathcal{FB}_0(s,k)$ and the set $\mathcal{B}^{e}(s,k)$.
\end{thm}
\pf  Let $P\in \mathcal{FB}_0(s,k) $. Suppose that $a$ is  the lowest point of $P$. If there are more than one such lowest point, we choose $a$ to be the
leftmost one. Now we proceed to construct a path $\alpha(P)$.
    Clearly, $P$ can be uniquely decomposed as  $P=P^{(1)}P^{(2)}$,
    where $P^{(1)}$ is the subpath of $P$ which goes from the starting point of $P$ to the point $a$, and $P^{(2)}$ is the remaining subpath of $P$.
          Set $\alpha(P)=P^{(2)}   ( P^{(1)})^{rc}$.
 It is apparent that  the resulting path  is a ballot $(s,k)$-path of  even height.

 In order to show that the map $\alpha$ is a bijection, we describe a map $\alpha':\mathcal{B}^{e}(s,k) \rightarrow \mathcal{FB}_0(s,k)$. Let $Q $ be a ballot $(s,k)$-path of height $2\ell$ for some $\ell\geq 0$. Now we proceed to demonstrate how to get a path $\alpha'(Q)$.  Let $b$ the rightmost lattice point of $Q$ which is on the line $y=\ell$. Then $Q$ can be uniquely decomposed as $Q=Q^{(1)}Q^{(2)}$, where $Q^{(1)}$ is the subpath of $Q$ which goes from the starting point of $Q$ to the point $b$, and $Q^{(2)}$ is the remaining subpath of $Q$. Set $\alpha'(Q)= (Q^{(2)})^{rc} Q^{(1)}$. One can easily verify that the resulting path $\alpha'(Q)\in \mathcal{FB}_0(s,k)$. Moreover, the point $b$ becomes the leftmost  lowest point of the resulting path.  This ensures that the maps $\alpha$ and $\alpha'$ are inverses of each other. Thus the map $\alpha$ is a bijection, completing the proof. \qed

\begin{example}
The decomposition of a path $$P=H_3U_2H_1U_2D_2D_2D_2H_1D_2H_2H_1U_2U_2$$ is shown in Figure \ref{alpha1}. By applying the map $\alpha$, its corresponding path is illustrated in Figure \ref{alpha2}.
\end{example}

\begin{example}
The decomposition of a path $$Q=H_2H_1U_2U_2U_2H_1U_2U_2U_2D_2H_1D_2H_3$$ is shown in Figure \ref{alpha2}. By applying the map $\alpha'$, its corresponding path is illustrated in Figure \ref{alpha1}.
\end{example}

\begin{thm}\label{beta}
There is a bijection $\beta$ between  the set $\mathcal{FB}_{-1}(s,k)$ and the set $\mathcal{B}^{o}(s,k)$.
\end{thm}
 \pf  Let $P\in \mathcal{FB}_{-1}(s,k) $.
      Now we proceed to construct a path $\beta(P)$ as follows. Suppose that $a$ is  the rightmost lattice  point of $P$ which is on the $x$-axis.
    Then $P$ can be uniquely decomposed as  $P=P^{(1)}P^{(2)}$,
    where $P^{(1)}$ is the subpath of $P$ which goes from the starting point of $P$ to the point $a$, and $P^{(2)}$ is the remaining subpath of $P$.
          Set $\beta(P)= \alpha(P^{(1)}) (P^{(2)})^{c}$.
 It is routine to check that   the resulting path $\beta(P)$ is a ballot $(s,k)$-path of   odd height.

 In order to show that the map $\beta$ is a bijection, we describe a map $\beta':\mathcal{B}^{o}(s,k) \rightarrow \mathcal{FB}_{-1}(s,k)$. Let $Q $ be a ballot $(s,k)$-path  of odd height $2\ell+1$ for some nonnegative  integer $\ell$ . Now we proceed to demonstrate how to get a path $\beta'(Q)$.  Let $b$ the rightmost lattice point of $P$ which is on the line $y=2\ell$. Then $Q$ can be uniquely decomposed as $Q=Q^{(1)}Q^{(2)}$, where $Q^{(1)}$ is the subpath of $Q$ which goes from the starting point of $Q$ to the point $b$, and $Q^{(2)}$ is the remaining subpath of $Q$. Set $\beta'(Q)= \alpha^{-1}(Q^{(1)})(Q^{(2)})^c $. One can easily verify that the resulting path $\beta'(Q)\in \mathcal{FB}_{-1}(s,k)$. Moreover, the ending point of $\alpha^{-1}(Q^{(1)})$ becomes the  rightmost lattice point which lies on the $x$-axis.    This ensures that the maps $\beta$ and $\beta'$ are inverses of each other. Thus the map $\beta$ is a bijection, completing the proof. \qed
\begin{example}
The decomposition of a path $$P=D_2U_2H_1U_2D_2H_2D_2H_1D_2H_3U_2H_1H_2$$ is shown in Figure \ref{beta1}. By applying the map $\beta$, its corresponding path is illustrated in Figure \ref{beta2}.
\end{example}

\begin{example}
The decomposition of a path $$Q=U_2H_1U_2D_2H_2U_2U_2H_1U_2H_3D_2H_1H_2$$ is shown in Figure \ref{beta2}. By applying the map $\beta'$, its corresponding path is illustrated in Figure \ref{beta1}.
\end{example}

 With abuse of notation, we use  $H_0$ to denote the empty step. Now we are in the position to establish a bijection between the set $\mathcal{Q}(m,k)$ and the set  $\mathcal{SD}(2m,2k)$.
\begin{thm}\label{delta}
There is a bijection between the  set $\mathcal{Q}(m,k)$ and the set  $\mathcal{SD}(2m,2k)$.
\end{thm}
\pf We first describe a map $\delta:  \mathcal{Q}(m, k) \rightarrow \mathcal{SD}(2m,2k) $. Let $P$ be a path in $\mathcal{Q}(m, k)$,  and  let $Q$ be the path obtained form $P$ by removing its first step.  We proceed to  construct a path $\delta(P)$ as follows.
\begin{itemize}
  \item[{\upshape (\rmnum{1})}] If $P\in \mathcal{FB}'_0(m-i,2k) $ for some $0\leq i\leq k-2$ and $P$ starts with a horizontal step $H_{\ell}$ for some $1\leq \ell\leq  2k-1$,  set $\delta(P)=\alpha(Q)H_{\ell}H_{2i}H_{\ell}  (\alpha(Q))^{rc} $.
  \item[{\upshape (\rmnum{2})}] If $P\in \mathcal{FB}'_0(m-i,2k) $ for some $0\leq i\leq k-2$ and $P$ starts with a down step $D_k$,  set $\delta(P)=\beta( Q ^{c})D_kH_{2i}U_k  (\beta( Q^{c}))^{rc}$.
     \item[{\upshape (\rmnum{3})}] If $P\in \mathcal{FB}'_{-1}(m+1,2k) $  and $P$ starts with a horizontal step $H_{\ell}$ for some $1\leq \ell\leq  k$,  set $\delta(P)=\beta(Q)H_{2\ell-2}  (\beta(Q))^{rc}$.
       \item[{\upshape (\rmnum{4})}] If $P\in \mathcal{FB}'_{-1}(m+1,2k) $  and $P$ starts with a horizontal step $H_{\ell}$ for some $k+1\leq \ell\leq 2k-1$,  set $\delta(P)=\beta(Q)U_kH_{2( \ell-k-1)}D_k(\beta(Q))^{rc}$.
           \item[{\upshape (\rmnum{5})}] If $P\in \mathcal{FB}'_{-1}(m+1,2k) $  and $P$ starts with a  down step $D_k$,  set $\delta(P)=\alpha(Q) H_{2k-2} (\alpha(Q))^{rc}$.
\end{itemize}
It is easy to check that  $\delta(P)\in SD(2m,2k)$, and thus the map $\delta$ is well defined.

In order to show the map $\delta$ is a bijection, we describe a map $\delta':  \mathcal{SD}(2m,2k) \rightarrow  \mathcal{Q}(m, k)$. Let $P$ be a path in $ \mathcal{SD}(2m,2k)$, we proceed to construct a path $\delta'(P)$ as follows.

\begin{itemize}
  \item[{\upshape ( \rmnum{1 }$'$ )}] If  $P= Q H_{\ell}H_{2i}H_{\ell}Q^{rc}$  with  $1\leq \ell\leq 2k-1$ and $0\leq i\leq k-2$ and   $Q$ is of even height, set $\delta'(P)=H_{\ell}\alpha^{-1}(Q)$.
  \item[{\upshape ( \rmnum{2 }$'$ )}] If  $P= Q  D_kH_{2i}U_k Q^{rc}$  with   $0\leq i\leq k-2$ and  $Q$ is of odd height, set $\delta'(P)=D_k(\beta^{-1}(Q))^c$.
     \item[{\upshape ( \rmnum{3 }$'$ )}] If    $P= Q   H_{2i} Q^{rc}$ with    $0\leq i\leq k-1$  and  $Q$ is of odd height, set $\delta'(P)=H_{i+1}\beta^{-1}(Q)$.
       \item[{\upshape ( \rmnum{4 }$'$ )}] If $P= Q  U_k H_{2i} D_k Q^{rc}$  with    $0\leq i\leq k-2$  and  $Q$ is of odd height, set $\delta'(P)=H_{k+1+i}\beta^{-1}(Q)$.
           \item[{\upshape ( \rmnum{5 }$'$ )}] If $P= Q   H_{2k-2}  Q^{rc}$  and  $Q$ is of even  height, set $\delta'(P)=D_k\alpha^{-1}(Q)$.
\end{itemize}
It is not difficult to check that the resulting path $\delta'(P)\in \mathcal{Q}(m,k)$. This implies that the map $\delta$ is well defined.  One can easily verify that {\upshape ( \rmnum{1 }$'$ )}, {\upshape ( \rmnum{2 }$'$ )},  {\upshape ( \rmnum{3 }$'$ )}, {\upshape ( \rmnum{4 }$'$ )} and {\upshape ( \rmnum{5 }$'$ )} respectively reverse the
procedures of {\upshape ( \rmnum{1 }  )}, {\upshape ( \rmnum{2 }  )},  {\upshape ( \rmnum{3 }  )}, {\upshape ( \rmnum{4 }  )} and {\upshape ( \rmnum{5 }  )}. This implies that the maps  $\delta$ and $\delta'$  are inverses of
each other. Hence, the map $\delta$ is a bijection, completing the proof. \qed

\begin{lem}\label{lemeven}
There is a bijection between $\mathcal{SD}(2m, 2k)$ and $\mathcal{SD}(2m+1,2k)$.
\end{lem}
 \pf  First we discribe a map $\xi:\mathcal{SD}(2m, 2k) \rightarrow \mathcal{SD}(2m+1,2k)$. For any path $P\in \mathcal{SD}(2m, 2k) $, it can be uniquely decomposed as $P=QH_{2\ell}Q^{rc}$ for some $0\leq \ell<{k}$. Let   $\xi(P)=QH_{2\ell+1}Q^{rc}$.  It is easily seen that the map $\xi$ is reversible, and thus is a bijection. This completes the proof.  \qed

  Theorems \ref{main1},  \ref{chi}, \ref{mainth2},  \ref{mainth3} and \ref{delta}  together with Lemma \ref{lemeven}  lead to   the following result.
\begin{thm}\label{theven}
For given positive integers $s$ and $ k$, the number of self-conjugate $(s, s+
1, \ldots, s + 2k)$-core partitions is equal to the number of symmetric $(s, 2k)$-Dyck paths.
\end{thm}

\subsection{On self-conjugate $(s, s+1, \ldots, s+2k+1)$-core partitions}
Let $$\mathcal{Q'}(m,k)=(\bigcup_{i=0}^{k-1}\mathcal{FB}'_0(m-i,2k+1))\cup \mathcal{FB}'_{-1}(m+{1\over 2},2k+1).$$
In the plane,   a point $(x,y)$  is said to be {\em visible} if  either $x=a+{1\over 2}$ for some $a\in \mathcal{Z}$ and $y$ is an  odd integer, or $x=a$ for some $a\in \mathcal{Z}$ and $y$ is an  even integer.

 We describe a map $\psi:J(P'(2m,2k+1))\rightarrow \mathcal{Q'}(m,k)$ as follows.  Let $I\in J(P'(2m,2k+1)) $. For the visible  point $(a,b)$  in the plane  where $ b\in \mathcal{Z}$ and   $0\leq a\leq m+{1\over 2}$, we color it  by black and white according to the following rules.
\begin{itemize}
\item If $b\geq 0$ , then
    the point $(a,b)$  is colored by  black if and only   if $(a,b) \in I  $;
    \item If $b<0$, then the point $(a,b)$  is colored by  black if and only    if $(a,b)\notin I  $.
    \end{itemize}

 Now we recursively  define a sequence   $P_0,P_1, \ldots P_n$ of points where  $P_0=(0,0)$. Assume that $  P_{i-1}$ has  been already determined. Let   $P_{i-1}=(a_{i-1}, b_{i-1})$.   We proceed to  get $P_i$ by the following procedure.\\
 Case 1. There exists a black point $(a_{i-1}+\ell, b_{i-1})$  with    $1\leq \ell\leq  2k$.  If the point $ (a_{i-1}+k+{1\over 2}, b_{i-1}+1)$ is black, then  set $P_i=(a_{i-1}+k+{1\over 2}, b_{i-1}+1)$. Otherwise,  choose $\ell$ to the smallest such integer and set  $P_i=(a_{i-1}+\ell, b_{i-1})$.\\
 Case 2. There does not exist  a black point $(a_{i-1}+\ell, b_{i-1})$  with    $1\leq \ell\leq  2k$. If the point $ (a_{i-1}+k+{1\over 2}, b_{i-1}-1)$ is black, set $P_i=(a_{i-1}+k+{1\over 2}, b_{i-1}-1)$. Otherwise, set $P_{i}=P_{i-1}$ and  $n=i-1$.\\
Let $\psi(I)$ be the resulting  path with lattice points $P_0, P_1, \ldots, P_n$.
For example, let $I=\{(1,0), (6,-2), (7.5,-1), (6.5, -1), (5.5,-1), (4.5,-1)\}\in J(P'(20,3))$. By applying the map $\psi$, we get a  path $\psi(I)$ as shown in Figure \ref{psi}.

 \begin{figure}[h!]
\begin{center}
\begin{picture}(280,160)
\setlength{\unitlength}{8mm}

 \put(0,0){\circle*{0.2}} \put(1,0){\circle*{0.2}} \put(2,0){\circle*{0.2}} \put(3,0){\circle*{0.2}} \put(4,0){\circle*{0.2}} \put(5,0){\circle*{0.2}}
  \put(6,0){\circle*{0.2}} \put(7,0){\circle*{0.2}} \put(8,0){\circle*{0.2}}  \put(9,0){\circle*{0.2}} \put(10,0){\circle*{0.2}}

  \put(0.5,1){\circle*{0.2}} \put(1.5,1){\circle*{0.2}} \put(2.5,1){\circle*{0.2}} \put(3.5,1){\circle*{0.2}} \put(4.5,1){\circle*{0.2}} \put(5.5,1){\circle*{0.2}}
  \put(6.5,1){\circle*{0.2}} \put(7.5,1){\circle*{0.2}} \put(8.5,1){\circle*{0.2}}  \put(9.5,1){\circle*{0.2}} \put(10.5,1){\circle*{0.2}}

\put(0,2){\circle*{0.2}} \put(1,2){\circle*{0.2}} \put(2,2){\circle*{0.2}} \put(3,2){\circle*{0.2}} \put(4,2){\circle*{0.2}} \put(5,2){\circle*{0.2}}
  \put(6,2){\circle{0.2}} \put(7,2){\circle*{0.2}} \put(8,2){\circle*{0.2}}  \put(9,2){\circle*{0.2}} \put(10,2){\circle*{0.2}}

  \put(0.5,3){\circle*{0.2}} \put(1.5,3){\circle*{0.2}} \put(2.5,3){\circle*{0.2}} \put(3.5,3){\circle*{0.2}} \put(4.5,3){\circle{0.2}} \put(5.5,3){\circle{0.2}}
  \put(6.5,3){\circle{0.2}} \put(7.5,3){\circle{0.2}} \put(8.5,3){\circle*{0.2}}  \put(9.5,3){\circle*{0.2}} \put(10.5,3){\circle*{0.2}}

  \put(0,4){\circle{0.2}} \put(1,4){\circle*{0.2}} \put(2,4){\circle{0.2}} \put(3,4){\circle{0.2}} \put(4,4){\circle{0.2}} \put(5,4){\circle{0.2}}
  \put(6,4){\circle{0.2}} \put(7,4){\circle{0.2}} \put(8,4){\circle{0.2}}  \put(9,4){\circle{0.2}} \put(10,4){\circle{0.2}}

  \put(0.5,5){\circle{0.2}} \put(1.5,5){\circle{0.2}} \put(2.5,5){\circle{0.2}} \put(3.5,5){\circle{0.2}} \put(4.5,5){\circle{0.2}} \put(5.5,5){\circle{0.2}}
  \put(6.5,5){\circle{0.2}} \put(7.5,5){\circle{0.2}} \put(8.5,5){\circle{0.2}}  \put(9.5,5){\circle{0.2}} \put(10.5,5){\circle{0.2}}

  \put(0,6){\circle{0.2}} \put(1,6){\circle{0.2}} \put(2,6){\circle{0.2}} \put(3,6){\circle{0.2}} \put(4,6){\circle{0.2}} \put(5,6){\circle{0.2}}
  \put(6,6){\circle{0.2}} \put(7,6){\circle{0.2}} \put(8,6){\circle{0.2}}  \put(9,6){\circle{0.2}} \put(10,6){\circle{0.2}}

\put(0,4){\line(1,0){1}}\put(1,4){\line(3,-2){1.5}}\put(2.5,3){\line(1,0){1}}\put(3.5,3){\line(3,-2){1.5}}\put(5,2){\line(2,0){2}} \put(7,2){\line(3,2){1.5}}\put(8.5,3){\line(1,0){1}}\put(9.5,3){\line(1,0){1}}

\put(-1.2,3.8){\small$(0,0)$}

\end{picture}
\end{center}
\caption{ The corresponding path $\psi(I)$.}\label{psi}
\end{figure}

By the same reasoning  as in the proof of Lemmas \ref{lemphi3}, \ref{lemphi1}, \ref{lemphi2} and Theorem \ref{mainth3}, we can deduce the following results.

   \begin{lem}\label{lempsi1}
 The map $\psi$ is well defined, that is, for any $I\in J(P'(2m,2k+1))$  we have   $ \psi(I)\in\mathcal{Q}'(m,k)$.
\end{lem}

\begin{lem}\label{lempsi2}
For any   $I\in J(P'(2m,2k+1))$,  the resulting path $\psi(I)$ verifies the following properties.
\begin{itemize}
\item[{\upshape (\rmnum{1})}] The visible points   on  the path $\psi(I)$  are black if and only if they are ending points of steps ;
\item[ {\upshape (\rmnum{2})}] For $b<0$, the visible point  $(a,b)$       to the right of the ending point of  the path $\psi(I)$ is black ;
\item[ {\upshape (\rmnum{3})}] For $b\geq 0$, the visible point $(a,b)$        to the right of the ending point of  the path $\psi(I)$ is white;
 \item[{\upshape (\rmnum{4})}] The visible points   lying     above the path $\psi(I)$ are white;
\item[ {\upshape (\rmnum{5})}] The visible points   lying     below the path $\psi(I)$ are black;

\end{itemize}
\end{lem}

\begin{thm}\label{mainth4}
The map $\psi$ is a bijection between  $J(P'(2m,2k+1)) $ and $\mathcal{Q}'(m,k)$.
\end{thm}

In the following theorem, we establish a bijection between  the set $\mathcal{Q}'(m,k)$ and the set  $\mathcal{SD}(2m,2k+1)$.

\begin{thm}\label{gamma}
There is a bijection   between the  set $\mathcal{Q}'(m,k)$ and the set  $\mathcal{SD}(2m,2k+1)$.
\end{thm}
\pf We first describe a map $\gamma:  \mathcal{Q}'(m,k) \rightarrow \mathcal{SD}(2m,2k+1) $. Let $P$ be a path in $\mathcal{Q}'(m,k) $ and let $Q$ be the path obtained from $P$ by removing its first step. Now  we proceed to demonstrate the procedure to get a path $\gamma(P)$.
\begin{itemize}
  \item[{\upshape (\rmnum{1})}] If $P\in \mathcal{FB}'_0(m-i,2k+1))$ for some $0\leq i\leq k-1$ and $P$ starts with a horizontal step $H_{\ell}$ for some $1\leq \ell\leq  2k$,  set $\gamma(P)=\alpha(Q)H_{\ell}H_{2i}H_{\ell}(\alpha(Q))^{rc}$.
  \item[{\upshape (\rmnum{2})}] If $P\in \mathcal{FB}'_0(m-i,2k+1))$ for some $0\leq i\leq k-1$ and $P$ starts with a down step $D_{k+{1\over 2}}$,  set $\gamma(P)=\beta(Q^c)D_{k+{1\over 2}}H_{2i}U_{k+{1\over 2}}(\beta(Q^c))^{rc}$.
     \item[{\upshape (\rmnum{3})}] If $P\in \mathcal{FB}'_{-1}(m+{1\over 2},2k+1))$  and $P$ starts with a horizontal step $H_{\ell}$ for some $1\leq \ell\leq  k$,  set $\gamma(P)=\beta(Q)H_{2\ell-1} (\beta(Q))^{rc}$.
       \item[{\upshape (\rmnum{4})}] If $P\in \mathcal{FB}'_{-1}(m+{1\over 2},2k+1))$  and $P$ starts with a horizontal step $H_{\ell}$ for some $k+1\leq \ell\leq 2k$,  set $\gamma(P)=\beta(Q)U_{k+{1\over 2}}H_{2( \ell-k-1)}D_{k+{1\over 2}}(\beta(Q))^{rc}$.
           \item[{\upshape (\rmnum{5})}] If $P\in \mathcal{FB}'_{-1}(m+{1\over 2},2k+1))$  and $P$ starts with a  down step $D_{k+{1\over 2}}$,  set $\gamma(P)=\alpha(Q) H_{2k} (\alpha(Q))^{rc} $.
\end{itemize}
One can easily  check that  $\gamma(P)\in \mathcal{SD}(2m,2k+1)$, and thus the map $\gamma$ is well defined.

Next we describe a map $\gamma':  \mathcal{SD}(2m,2k+1) \rightarrow  \mathcal{Q}'(m,k) $. Let $P$ be a path in $ \mathcal{SD}(2m,2k+1)$, we proceed to construct a path $\gamma'(P)$ by the following procedure.

\begin{itemize}
  \item[{\upshape ( \rmnum{1 }$'$ )}] If  $P= Q H_{\ell}H_{2i}H_{\ell}Q^{rc}$ with  $1\leq \ell\leq 2k$ and $0\leq i\leq k-1$ and $Q$ is of even height, set $\gamma'(P)=H_{\ell}\alpha^{-1}(Q)$.
  \item[{\upshape ( \rmnum{2 }$'$ )}] If  $P= Q  D_{k+{1\over 2}}H_{2i}U_{k+{1\over 2}}Q^{rc}$ with    $0\leq i\leq k-1$  and $Q$ is of odd height, set $\gamma'(P)=D_{k+{1\over 2}}(\beta^{-1}(Q))^c$.
     \item[{\upshape ( \rmnum{3 }$'$ )}] If    $P= Q   H_{2i-1} Q^{rc}$ with   $1\leq i\leq k$  and  $Q$ is of odd height, set $\gamma'(P)=H_{i}\beta^{-1}(Q)$.
       \item[{\upshape ( \rmnum{4 }$'$ )}] If $P= Q  U_{k+{1\over 2}} H_{2i} D_{k+{1\over 2}} Q^{rc}$  with  $0\leq i\leq k-1$  and  $Q$ is of odd height, set $\gamma'(P)=H_{k+1+i}\beta^{-1}(Q)$.
           \item[{\upshape ( \rmnum{5 }$'$ )}] If $P= Q    H_{2k}  Q^{rc}$  and  $Q$ is of even  height, set $\gamma'(P)=D_{k+{1\over 2}}\alpha^{-1}(Q)$.
\end{itemize}
It is not difficult to check that the resulting path $\gamma'(P)\in \mathcal{Q}'(m,k) $. This implies that the map $\gamma$ is well defined. Furthermore, one can easily verify that {\upshape ( \rmnum{1 }$'$ )}, {\upshape ( \rmnum{2 }$'$ )},  {\upshape ( \rmnum{3 }$'$ )}, {\upshape ( \rmnum{4 }$'$ )} and {\upshape ( \rmnum{5 }$'$ )} respectively reverse the
procedures of {\upshape ( \rmnum{1 }  )}, {\upshape ( \rmnum{2 }  )},  {\upshape ( \rmnum{3 }  )}, {\upshape ( \rmnum{4 }  )} and {\upshape ( \rmnum{5 }  )}. This implies that the maps  $\gamma$ and $\gamma'$  are inverses of
each other. Hence, the map $\gamma$ is a bijection, completing the proof. \qed

Combining   Theorems \ref{main1},   \ref{chi}, \ref{mainth4} and \ref{gamma}, we are    led  to   the following result.
\begin{thm}\label{thodd}
For given positive integer  $m$ and  nonnegative integer $ k$, the number of self-conjugate $(2m, 2m+
1, \ldots, 2m + 2k+1 )$-core partitions is equal to the number of symmetric $(2m, 2k+1)$-Dyck paths.
\end{thm}

Let $ J^*(P'(2m+2, 2k+1))$ denote   the order ideals $ I \in  J(P'(2m+2, 2k+1))$ such that $I$ does not contain the element $(m+1, 0)$.   It is easily seen that \begin{equation}\label{eq1}
J^*(P'(2m+2, 2k+1))= J(P'(2m+1, 2k+1)).
\end{equation}
Let  $\mathcal{Q}^*(m,k)=\mathcal{Q}'(m,k)\setminus \mathcal{FB}'_0(m, 2k+1)$. The construction of the map $\psi$ together with Lemma \ref{lempsi2} leads to  the following result.

\begin{thm}\label{psi'}
The map $\psi$ induces a bijection between the set $J^*(P'(2m+2, 2k+1))$ and the set  $\mathcal{Q}^*(m+1,k)$.
\end{thm}

In the following theorem, we aim to establish a bijection between the  set $\mathcal{Q}^*(m+1,k)$ and the set  $\mathcal{SD}(2m+1,2k+1)$.
\begin{thm}\label{eta}
There is a bijection   between the  set $\mathcal{Q}^*(m+1,k)$ and the set  $\mathcal{SD}(2m+1,2k+1)$.
\end{thm}
\pf We first describe a map $\eta:  \mathcal{Q}^*(m+1,k) \rightarrow \mathcal{SD}(2m+1,2k+1) $. Let $P$ be a path in $\mathcal{Q}^*(m+1,k) $ and let $Q$ be the path obtained from $P$ by removing its first step. Now  we proceed to demonstrate the procedure to get a path $\eta(P)$.
\begin{itemize}
  \item[{\upshape (\rmnum{1})}] If $P\in \mathcal{FB}'_0(m+1-i,2k+1))$ for some $1\leq i\leq k-1$ and $P$ starts with a horizontal step $H_{\ell}$ for some $1\leq \ell\leq  2k$,  set $\eta(P)=\alpha(Q)H_{\ell}H_{2i-1}H_{\ell}(\alpha(Q))^{rc}$.
  \item[{\upshape (\rmnum{2})}] If $P\in \mathcal{FB}'_0(m+1-i,2k+1))$ for some $1\leq i\leq k-1$ and $P$ starts with a down step $D_{k+{1\over 2}}$,  set $\eta(P)=\beta(Q^c)D_{k+{1\over 2}}H_{2i-1}U_{k+{1\over 2}}(\beta(Q^c))^{rc}$.
     \item[{\upshape (\rmnum{3})}] If $P\in \mathcal{FB}'_{-1}(m+1+{1\over 2},2k+1))$  and $P$ starts with a horizontal step $H_{\ell}$ for some $1\leq \ell\leq  k+1$,  set $\eta(P)=\beta(Q)H_{2\ell-2} (\beta(Q))^{rc}$.
       \item[{\upshape (\rmnum{4})}] If $P\in \mathcal{FB}'_{-1}(m+1+{1\over 2},2k+1))$  and $P$ starts with a horizontal step $H_{\ell}$ for some $k+2\leq \ell\leq 2k$,  set $\eta(P)=\beta(Q)U_{k+{1\over 2}}H_{2( \ell-k-1)-1}D_{k+{1\over 2}}(\beta(Q))^{rc}$.
           \item[{\upshape (\rmnum{5})}] If $P\in \mathcal{FB}'_{-1}(m+1+{1\over 2},2k+1))$  and $P$ starts with a  down step $D_{k+{1\over 2}}$,  set $\eta(P)=\alpha(Q) H_{2k-1} (\alpha(Q))^{rc} $.
\end{itemize}
It is easy to verify that $\eta(P)\in \mathcal{SD}(2m+1,2k+1)$, and thus the map $\eta$ is well defined.

Next we describe a map $\eta':  \mathcal{SD}(2m+1,2k+1) \rightarrow  \mathcal{Q}^*(m+1,k) $. Let $P$ be a path in $ \mathcal{SD}(2m+1,2k+1)$, we proceed to construct a path $\eta'(P)$ by the following procedure.

\begin{itemize}
  \item[{\upshape ( \rmnum{1 }$'$ )}] If  $P= Q H_{\ell}H_{2i-1}H_{\ell}Q^{rc}$ with  $1\leq \ell\leq 2k$ and $1\leq i\leq k-1$ and $Q$ is of even height, set $\eta'(P)=H_{\ell} \alpha^{-1}(Q)$.
  \item[{\upshape ( \rmnum{2 }$'$ )}] If  $P= Q  D_{k+{1\over 2}}H_{2i-1}U_{k+{1\over 2}}Q^{rc}$ with    $1\leq i\leq k-1$  and $Q$ is of odd height, set $\eta'(P)=D_{k+{1\over 2}}(\beta^{-1}(Q))^c$.
     \item[{\upshape ( \rmnum{3 }$'$ )}] If    $P= Q   H_{2i} Q^{rc}$ with   $0\leq i\leq k$  and  $Q$ is of odd height, set $\eta'(P)=H_{i+1}\beta^{-1}(Q)$.
       \item[{\upshape ( \rmnum{4 }$'$ )}] If $P= Q  U_{k+{1\over 2}} H_{2i-1} D_{k+{1\over 2}} Q^{rc}$  with  $1\leq i\leq k-1$  and  $Q$ is of odd height, set $\eta'(P)=H_{k+1+i}\beta^{-1}(Q)$.
           \item[{\upshape ( \rmnum{5 }$'$ )}] If $P= Q    H_{2k-1}  Q^{rc}$  and  $Q$ is of even  height, set $\eta'(P)=D_{k+{1\over 2}}\alpha^{-1}(Q)$.
\end{itemize}
It is routine  to check that the resulting path $\eta'(P)\in \mathcal{Q}^*(m+1,k) $. Therefore, the map $\eta'$ is well defined. Furthermore, one can easily verify that {\upshape ( \rmnum{1 }$'$ )}, {\upshape ( \rmnum{2 }$'$ )},  {\upshape ( \rmnum{3 }$'$ )}, {\upshape ( \rmnum{4 }$'$ )} and {\upshape ( \rmnum{5 }$'$ )} respectively reverse the
procedures of {\upshape ( \rmnum{1 }  )}, {\upshape ( \rmnum{2 }  )},  {\upshape ( \rmnum{3 }  )}, {\upshape ( \rmnum{4 }  )} and {\upshape ( \rmnum{5 }  )}. This implies that the maps  $\eta$ and $\eta'$  are inverses of
each other. Hence, the map $\eta$ is a bijection, completing the proof. \qed

 Theorems \ref{main1},   \ref{chi}, \ref{psi'} and \ref{eta} together with Equation (\ref{eq1})  yield  the following result.
\begin{thm}\label{thodd1}
For given positive integer  $m$ and  nonnegative integer $ k$, the number of self-conjugate $(2m+1, 2m+
2, \ldots, 2m + 2k+2 )$-core partitions is equal to the number of symmetric $(2m+1, 2k+1)$-Dyck paths.
\end{thm}

 Combining Theorems \ref{theven},  \ref{thodd} and \ref{thodd1}, we complete the  proof of Conjecture \ref{con1}.

\noindent{\bf Acknowledgments.} This work was supported by  the National Natural Science Foundation of China (11671366).

redpajama_set_name: RedPajamaArXiv


Install passwords even where not required. The only exception is with unreliable programs that might fail and lock you out.
Upper and lower case letters and numbers should all be present.
The first one is easiest to crack because your use or nonuse of the shift key is steady and easier for a spy to remember. The second alternates, and that's more secure, but it is regular enough that an observer can guess it. The third is a random arrangement; I made this one by tossing a coin for each letter. For 10 letters in a password, random choice just between capitals and lower case can end 1,024 possible ways and that should frustrate most spies. If that's too few for your needs, a password with 20 letters (not counting numbers in your password) can give 1,048,576 possibilities. If you know a spy who can cope with that by watching you type, practice typing faster. Keep one finger just over the shift key so that pressing down is hardly noticed by anyone. This won't work against a video recording in slow playback or software sneaked into your computer to do keyboard logging or input logging, but it's a start.
I don't like special characters unless required. I might use a keyboard that doesn't have a special character I've used and then I might be locked out.
Randomness: Choose every character to have no relationship to any other, so a missing character can't be guessed. For example, you can probably guess what's missing from "b_cycle". The difficulty with randomness is that your brain imposes order and so your brain is lousy at picking characters randomly. A spreadsheet program likely has a randomness function (maybe called Rand) that you can use for the values 1–62, just enough for capital and lower-case letters and ten numerals. If you keep the spreadsheet permanently, put the randomness formula into so many cells, at least a thousand, so that no one can figure out what password you made or will make.
On the other hand, randomness does not forbid being memorable. It's just that you add memorability after you've picked your random sequence of characters. After you have your password that could be patented as gibberish, discover patterns. I've seen a President's initials, an abbreviation for a natural feature of Earth, a number that could be a money amount, a common English word that's very short, a symbol for a financial expression, and a host of other patterns. They don't make much sense except that they help me remember a password. These are mnemonics.
I'm not a fan of websites creating passwords for me, because I don't know whether they really use good methods or only say they do and whether they store data about users' password choices (don't rely on privacy policies for your protection, especially against network snooping). If you're not concerned about that, like because you're only looking for ideas, then take a look at https://passwordcreator.org/ (no hyphen).
If you can't pick characters randomly, a less-good alternative is to memorize a line of poetry, a song lyric, a sentence, or some other text you'll always remember exactly the same way and that other people won't associate with you or hear you say, then use the first letter of every word to make a brand-new word. Don't use that word if it might be in a dictionary in any language, because security crackers use dictionary attacks. If the word passes that test, mix up which letters are capitals and which are lower-case and insert a number.
Double one character, consecutively. This discourages shoulder-surfing. If someone watches what you type, they may not notice that you quickly tapped the same key twice. An alternative is to slide your finger over a key as if it's a password character but without actually pressing it, or to do that a bunch of times in one password. That should confuse most shoulder-surfers. Let them complain. They'll reveal themselves.
If you write your password down, which is reasonable since we may have many passwords that we rarely use, encrypt your passwords. Someone might find where you wrote them. Use an encryption systemn you can reliably remember over a long time, because if you forget it you'll be locked out. The exception requiring literal unencrypted storage is for a password that someone else shares with you, because that person could reverse-engineer (figure out) your encryption system and decrypt that password and therefore your other passwords.
Software that stores passwords is susceptible to attack, so I don't use that method. Maybe its strength is that it's easier to memorize a single password when you use that password all the time, but if anyone figures out that one password they'll have all of your passwords. And how good are you at withstanding a sledgehammer across your kneecap?
No system can be perfect, but more steps toward security are usually better than fewer.
The page's principal nonadvertising content was prepared ca. May 29, 2017 , and revised through December 4, 2017 .
Passwords will have better security with these suggestions.


You are here: Home / World news / Consumer Electronic Giant Videocon Industries set to acquire Dr Datsons Labs for INR 350 cr!
Consumer Electronic Giant Videocon Industries set to acquire Dr Datsons Labs for INR 350 cr! Reviewed by PHARMALEADERS on May 7, 2014 .
Videocon Industries is set to acquire Dr Datsons Labs Ltd (DTL) at an enterprise value of Rs 350 crore. Sources said negotiations in this regard were in advanced stages and a deal might be signed soon.
"We are in talks with a couple of merchant bankers for a possible sale. At the same time, we are also evaluating other options, including investments into the company, to meet our proposed capital expenditure plan for expansion projects. By next week, we will be finalising the name of the merchant banker for a possible stake sale," said DTL Vice-chairman Kannan Vishwanath.
Sources said Videocon had written to Bank of Baroda to extend Rs 35 crore of credit to acquire additional stake in DTL.
Videocon Industries Chairman Venugopal Dhoot declined to comment.
Promoters hold about 40 per cent stake in DTL. In the quarter ended March, foreign portfolio investors, including Davos International Fund, Leman Diversified Fund, Stream Value Fund and Kuvera Fund, acquired 9.35 per cent in DTL from various domestic individual and institutional investors.
During the March quarter, Videocon, through its subsidiaries, had increased its stake in DTL to 19 per cent. In the previous quarter, it held nine per cent stake.
A leading broker on BSE and the National Stock Exchange said the market was abuzz with speculation of Videocon acquiring DTL.
DTL, a research-focused pharmaceuticals company, manufactures active pharmaceutical ingredients (APIs) and finished dosage forms and provides contract research and manufacturing services. It focuses on five niche business areas — anti-malaria, multi-therapeutic lozenges, oncology products, HIV products and codeine-based cough syrups. The market leader in the anti-malarial API segment, it is present in 61 countries.
DTL's market capitalisation stands at Rs 121 crore. The company stock fell to Rs 38.4 at close of trade on BSE, against its previous close of Rs 45.7.


As an animal lover and financial supporter of no-kill animal shelters myself, I understand the importance and significance that these programs hold in saving the lives of homeless and 'unwanted' animals. It's an unfortunate problem that not many people address today.
There are currently over 70 million stray animals in the United States, of which only 6 to 8 million dogs and cats find animals shelters annually. The only reasons why these animals successfully make it to our country's shelters is because of their owners either giving them up, or animal control officers finding them in the wild. But, even then, many of the cats and dogs rescued are subjected to a similar fate by being given to facilities that promote euthanization to control population sizes.
Approximately 1.5 million shelter animals are euthanized every year, largely due to the fact that most shelters do not have the space permitting for the number of cats and dogs received on a regular basis, along with adoption rates not being able to keep up. However, depending on the type of shelter, many homeless or 'unwanted' animals are able to once again find their way back to a loving home. Private shelters, for example, often have contracts that allow them to work with their local governments for funding sources, additional help in finding homes for these animals, and enforcing stricter animal care laws.
The standard practices within these facilities also differ in terms of dealing with the increasing size of the animal population. For example, standard shelters typically euthanize cats or dogs who have not been claimed in a shockingly small 72-hour window, thus stressing the importance of no-kill shelters.
The only reason why a cat or dog may be euthanized within a no-kill shelter is because they are suffering from an irreparable disease or injury, which is then for the good of the animal. Doing so as a means of controlling the shelter numbers is never policy. Because of this, communities working together to save homeless and 'unwanted' animals are able to rescue up to 90% of those found, without worrying about euthanization.
No-kill shelters are also responsible for largely promoting adoption, spay and neuter programs, donations, and fellow programs benefitting an animal's well-being. They strive to work collaboratively with other organizations dedicated to placing animals in loving homes, and avoid senselessly putting down those without them simply due to the fact that their facilities are not big enough. Though some people may argue that allowing at-risk animals through their doors is unsafe, that is the point of no-kill shelters; to give each animal a fighting chance.
Stray animals may appear hostile when approached by any individual, but trained professionals working with no-kill shelters are able to rescue those without homes regardless of age or breed. Sadly, pit-bulls and "bully breeds" are often discriminated against depending on the community's laws. Nearly every no-kill shelter though, rejects this false belief that one breed is more dangerous than another, accepting all animals into their facilities.
Today, the number of animals euthanized has thankfully decreased in recent decades because of the no-kill shelter movement. This humane trend is beginning to take effect in many cities across the United States, leading to a higher number of rescued animals and more volunteers willing to devote their time to this cause. While establishing no-kill shelters in every community across America is unlikely, there are strategies we can put in place, along with the combined efforts of those passionate enough to make a change, that can lead to this ideal goal.


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