text:
\section{Introduction}
\label{sec:introduction}

A \emph{Steiner system of order $n$}, $S(t,k,n)$, is a pair $(Q,B)$, where $Q$ is an $n$-set
(whose elements are called \emph{points}) and $B$ is a collection
of $k$-subsets (called \emph{blocks}) of $Q$, such that each $t$-subset of $Q$ is contained
in exactly one block of $B$.

A \emph{large set} of Steiner systems $S(t,k,n)$, on an $n$-set $Q$, is a partition of all $k$-subsets
of $Q$ into Steiner systems $S(t,k,n)$. If we restrict ourself to Steiner systems $S(t-1,t,n)$, then exactly
two families of large sets are solved completely. A Steiner system $S(1,2,n)$ exists if and only if $n$ is even and its
large set is known as a one-factorization of the complete graph $K_n$. The existence of such one-factorizations
is a folklore and a survey can be found in~\cite{Wal97}. A Steiner system $S(2,3,n)$ is known as Steiner triple
system and the corresponding large set is known to exist for every admissible $n \equiv 1$ or $3 ~(\text{mod}~6)$, where $n > 7$.
It was first proved by Lu~\cite{Lu83,Lu84}, who left six open cases which were solved by Teirlinck~\cite{Tei91}.
An alternative shorter proof was given later by Ji~\cite{Ji05}.

A Steiner system $S(3,4,n)$ is also called a Steiner quadruple system of order~$n$
and it is denoted by SQS($n$). These systems which are
of special interest exist if and only if $n \equiv 2$ or $4~(\text{mod}~6)$~\cite{Han60}.
No construction of nontrivial large set of SQS($n$) is known, although it is a common belief that they exist
even for small $n$. Using probabilistic arguments, their existence for large enough $n$, $n \equiv 2$ or $4~(\text{mod}~6)$, was proved recently~\cite{Kee18}.
In practice, it is of interest to construct many disjoint systems.
Let $D(n)$ be the number of pairwise disjoint SQS($n$)s (PDQs of order $n$ in short, the
order is omitted when it is understood).
The size of an SQS($n$) is $\binom{n}{3} \Huge / 4$, and hence $D(n) \leq n-3$
and a large set contains $n-3$ PDQs of order $n$, i.e. $D(n)=n-3$, if a large set exists.

The first lower bound on $D(n)$ was given by Lindner~\cite{Lin77} who proved that
if $n \equiv 2$ or $4~(\text{mod}~6)$, then $D(2n) \geq n$. He further obtained
another bound~\cite{Lin85}, $D(4n) \geq 3n$ if $n \equiv 2$ or $4~(\text{mod}~6)$, where $n \geq 8$.
In a sequence of papers~\cite{KrMe74,Phe91,PhRo80} it was proved that $D(2n) \geq n$ for
$n= 5^a 13^b 17^c$, where $a,b,c \geq 0$ and $a+b+c>0$ and for $n=11$.
Etzion and Hartman~\cite{EtHa91} has proved that $D(2^k n) \geq (2^k -1)n$, $k \geq 2$, if
there exists a set of $3n$ PDQs of order $4n$ with a certain structure. The result can be applied
on $n=11$, $n \equiv 2$ or $4~(\text{mod}~6)$, and $n= 5^a 13^b 17^c$, where $a,b,c \geq 0$ and $a+b+c>0$.
In particular it is proved in~\cite{EtHa91} that $D( 5 \cdot 2^m ) \geq 5 \cdot 2^m -5$, which falls short
by just one system from a large set. This claim follows from the fact that $5 \cdot 2^m -4$ PDQs imply that the remaining
quadruples which are not contained in these PDQs form another SQS($5 \cdot 2^m )$.
The construction in~\cite{EtHa91} is recursive, based
on the existence of 2-chromatic PDQs. A recursive equation for $D(4n)$,
$n \equiv 1$ or $5~(\text{mod}~6)$, was given in~\cite{Etz93}, where it is proved that
$$
D(4n) \geq 2n + \min \{ D(2n),n-2 \}~.
$$
Our goal in this paper is to improve this recursive formula to the following theorem which
is the main result of the paper.

\begin{theorem}
\label{thm:main}
$~$
If $n \geq 7$ and $n \equiv 1$ or $5~(\text{mod}~6)$, then
$D(4n) \geq 2n + \min \{ D(2n),2n-7 \}$.
\end{theorem}
The rest of this paper is organized as follows. In Section~\ref{sec:review} we present the
two known recursive doubling construction for SQSs and for a set of PDQs.
The proof of Theorem~\ref{thm:main} is given in Section~\ref{sec:recursion} with its related construction.
We start with any SQS($2n$), a certain Latin square of order $2n$, and apply the first doubling
construction (DLS Construction) to obtain $2n$ PDQs of order $4n$. We analyse the quadruples which are contained in the
constructed $2n$ PDQs and continue with the second doubling construction (DB Construction) to form PDQs of order $4n$ which contain
quadruples which are not contained in the first $2n$ PDQs.
The construction for $n \equiv 5~(\text{mod}~6)$ is slightly simpler to explain than the one
for $n \equiv 1~(\text{mod}~6)$ and also
when $n$ is a prime some of the construction is slightly simpler. We make some distinction between
these cases in the proofs.
Examples of each step are presented throughout the description of the
related recursive construction. In Section~\ref{sec:conclusion} a conclusion, comparison with previous results,
and several problems for future research, are presented.
\vspace{-0.2cm}

\section{Two Recursive Constructions}
\label{sec:review}

In this section we will discuss two main recursive doubling constructions for
SQSs or more precisely, for a set of PDQs.
The first one is a doubling construction due to Lindner~\cite{Lin77}. It will
be called DLS (for Doubling Lindner Systems) Construction. We will present a slightly different variant from the one
given in~\cite{Lin77}. This variant was already presented in~\cite{Etz96}.
The second construction is a folklore doubling construction
which will be called DB (for doubling) Construction.
These two constructions will be the main ingredients
for the other constructions presented in this paper.

For the two constructions we need to define several well-known combinatorial designs
which are essential in the construction used to prove Theorem~\ref{thm:main} in Section~\ref{sec:recursion}.

A $v \times v$ \emph{Latin square} is a $v \times v$ array in which each row and each column
is a permutation of a $v$-set $Q$. A $k \times v$ array is called a \emph{Latin rectangle}
if each row is a permutation of $Q$ and each column contains $k$ distinct elements.
It is well known that such a rectangle can be completed to an $v \times v$ Latin square,
for example by applying the well-known Hall's marriage theorem~\cite{Hal35}. A Latin square has no $2 \times 2$ subsquares if any
$2 \times 2$ subsquare restricted to two rows and two columns does not form a Latin square~\cite{KLR75}.

A \emph{one-factorization} $F=\{F_0,F_1,\ldots,F_{v-2} \}$ of the complete graph $K_v$, where $v$ is an even positive integer,
is partition of the edges of $K_v$ into perfect matchings. Each $F_i$, $0 \leq i \leq v-2$, is a perfect matching of $K_v$,
which is also called a \emph{one-factor}. As mentioned before a one-factor is a Steiner system $S(1,2,v)$
and a one-factorization is a related large set.
Clearly, a one-factor contains $\frac{v}{2}$ pairs of vertices whose
union is the set of $v$ vertices in $K_v$.
\vspace{-0.2cm}

\subsection{DLS Construction}
\label{sec:DLS}

Let $(\Z_v,B)$ be an SQS($v$) and let $A$ be a $v \times v$ Latin square of order $v$, on the point set $\Z_v$, with
no $2 \times 2$ subsquares.
Denote by~$\alpha_i$ the permutation on $\Z_v$ defined by $\alpha_i(j)=y$ if and only if $A(i,j)=y$.
For each $i$, $0 \leq i \leq v-1$, we define a set of quadruples $B_i$ on $\Z_v \times \Z_2$ as follows:

\begin{enumerate}
\item For each quadruple $\{ x_1,x_2,x_3,x_4 \} \in B$, the following 8 quadruples are contained in~$B_i$.
$$
\{ (x_1,0),(x_2,0),(x_3,0),(\alpha_i(x_4),1) \},~~ \{ (x_1,1),(x_2,1),(x_3,1),(\alpha_i(x_4),0) \}
$$
$$
\{ (x_1,0),(x_2,0),(\alpha_i(x_3),1),(x_4,0) \},~~ \{ (x_1,1),(x_2,1),(\alpha_i(x_3),0),(x_4,1) \}
$$
$$
\{ (x_1,0),(\alpha_i(x_2),1),(x_3,0),(x_4,0) \},~~ \{ (x_1,1),(\alpha_i(x_2),0),(x_3,1),(x_4,1) \}
$$
$$
\{ (\alpha_i(x_1),1),(x_2,0),(x_3,0),(x_4,0) \},~~ \{ (\alpha_i(x_1),0),(x_2,1),(x_3,1),(x_4,1) \}
$$

\item For each pair $\{ x_1,x_2 \} \subset \Z_v$, the quadruple
$\{ (x_1,0),(x_2,0),(\alpha_i(x_1),1),(\alpha_i(x_2),1) \}$ is contained in~$B_i$.
\end{enumerate}

This DLS Construction is a variant~\cite{Etz96} of the Lindner Construction~\cite{Lin77}. It yields $v$ PDQs of order $2v$.
\vspace{-0.2cm}

\subsection{DB Construction}
\label{sec:DB}

Let $(\Z_v,B)$ be an SQS($v$), let $F=\{F_0,F_1,\ldots,F_{v-2} \}$ and
$F'=\{F'_0,F'_1,\ldots,F'_{n-2} \}$ be two one-factorizations of $K_v$ on the vertex set $\Z_v$, and
let $\alpha$ be any permutation on the set $\{ 0,1,\ldots,v-2 \}$. We define a collection of quadruples $B'$
on $\Z_v \times \Z_2$ as follows.

\begin{enumerate}
\item For each quadruple $\{ x_1,x_2,x_3,x_4 \} \in B$, the following two quadruples are
contained in~$B'$.
$$
\{ (x_1,0),(x_2,0),(x_3,0),(x_4,0) \},~~ \{ (x_1,1),(x_2,1),(x_3,1),(x_4,1) \}~.
$$

\item For each $i \in \{ 0,1,\ldots,n-2\}$ and $\{ x_1,x_2 \} \in F_i$, $\{ y_1,y_2 \} \in F'_j$,
where $j=\alpha (i)$, the following quadruple is contained in $B'$.
$$
\{ (x_1,0),(x_2,0),(y_1,1),(y_2,1) \}~.
$$
\end{enumerate}
The set of all such quadruples produced by $F_i$ and $F'_j$ are said to be the direct product of $F_i$ and~$F'_j$.
For the construction in Section~\ref{sec:recursion}, it is assumed that instead of one SQS($v$) $B$, we have
a set of $k$ PDQs of order $v$ and instead of $\alpha$ we
use $k$ permutations on $\{ 0,1,\ldots,v-2 \}$ defined by the rows of a $k \times (v-1)$ Latin rectangle.
In this case, the DB construction yields a set of $k$ PDQs of order $2v$.
\vspace{-0.2cm}

\subsection{Configurations}

Both constructions, the DLS Construction and the DB Construction, are based
on a partition of the point set $Q$ into two equal sets, say $P_1$ and $P_2$.
Given such a partition and a quadruple
$X = \{ x_1,x_2,x_3,x_4 \}$ of $Q$, we say that $X$ is a $4$-subset from \emph{configuration} $(i,j)$, where $i+j=4$, if
$|X \cap P_1|=i$ and $|X \cap P_2|=j$. There are clearly five possible configurations,
$(4,0)$, $(3,1)$, $(2,2)$, $(1,3)$, and $(0,4)$. The following trivial lemma can be easily verified.

\begin{lemma}
\label{lem:conf1_3}
In the DLS construction, based on the partition of $\Z_{v}\times\{0,1\}$ into $\Z_v \times \{0\}$ and $\Z_n \times \{1\}$,
each quadruple from configuration $(3,1)$ and each quadruple from configuration $(1,3)$
is contained exactly once in the $v$ PDQs of order $2v$.
\end{lemma}
\vspace{-0.2cm}

\section{An Improved Recursive Construction}
\label{sec:recursion}

In this section a recursive doubling construction for PDQs of order $4n$,
where $n$ is an odd integer not divisible by 3, is presented.
The construction proceeds in two parts, where we employ the two doubling constructions with $v=2n$, $n\equiv 1$ or 5 (mod 6). In the first part
the DLS Construction is applied with an appropriate $(2n) \times (2n)$ Latin square.
A set of $2n$ PDQs of order $4n$ is obtained in this part.
By Lemma~\ref{lem:conf1_3}, each quadruple from configurations $(3,1)$ and each quadruple
from configuration $(1,3)$ is contained in these $2n$ PDQs.
We analyse the quadruples from configuration $(2,2)$ contained in these $2n$ PDQs.
In the second part, the DB Construction is applied with a set of $k$ PDQs of order $2n$ and
a $k \times (2n-1)$ Latin rectangle which yields quadruples from configuration $(2,2)$
which are not obtained in the first part of the construction. The final outcome
from the two parts of the construction are $2n+k$ PDQs of order $4n$ which will imply Theorem~\ref{thm:main}.

\subsection{PDQs from the DLS Construction}
\label{sec:part1}

The first step in the first part of the construction is to define a $(2n) \times (2n)$ Latin square
to be used in the DLS Construction. This Latin square was defined in~\cite{KLR75} and used
in~\cite{Etz93} for constructions of PDQs of order $4n$.
Let $A_n$, $B_n$, be the two $n \times n$ Latin squares defined as follows.

$$
A_n (i,j) \equiv i+j-2~(\text{mod}~n), ~~~~ 1 \leq i,j \leq n,
$$
$$
B_n (i,j) \equiv i-j~(\text{mod}~n), ~~~~ 1 \leq i,j \leq n,
$$
where the values in $A_n$ and $B_n$ are reduced to the range $\{0,1,\ldots,n-1\}$.

Let $C_n$ and $D_n$ be the following two $n \times n$ Latin squares, where
for each $0 \leq i,j \leq n-1$, we have

$$
C_n(i,j) = A_n(i,j)~\text{reduced~modulo}~n~\text{to~the~range}~ \{n,n+1,\ldots,2n-1\},
$$
$$
D_n(i,j) = A_n(i,j)-1~\text{reduced~modulo}~n~\text{to~the~range}~\{n,n+1,\ldots,2n-1\}.
$$

The $(2n) \times (2n)$ Latin square $M_n$ which will be used in the DLS Construction is
given by
{\large
$$
M_n = \left[
\begin{array}{cc}
B_n & C_n \\
D_n & B_n
\end{array}
\right]
$$
}
It was proved in~\cite{KLR75} that $M_n$ has no $2 \times 2$ subsquares.

\begin{example}
\label{ex:M7}
Throughout this section, we will concentrate on a few examples, for $n=7$, $n=11$, an also $n=19$.

For $n=7$, the $14 \times 14$ Latin square $M_7$ is given by
$$
\left[
\begin{array}{ccccccccccccccc}
0 & 6 & 5 & 4 & 3 & 2 & 1 &  & 7 & 8 & 9 & 10 & 11 & 12 & 13\\
1 & 0 & 6 & 5 & 4 & 3 & 2 &  & 8 & 9 & 10 & 11 & 12 & 13 & 7\\
2 & 1 & 0 & 6 & 5 & 4 & 3 &  & 9 & 10 & 11 & 12 & 13 & 7 & 8\\
3 & 2 & 1 & 0 & 6 & 5 & 4 &  & 10 & 11 & 12 & 13 & 7 & 8 & 9\\
4 & 3 & 2 & 1 & 0 & 6 & 5 &  & 11 & 12 & 13 & 7 & 8 & 9 & 10\\
5 & 4 & 3 & 2 & 1 & 0 & 6 &  & 12 & 13 & 7 & 8 & 9 & 10 & 11\\
6 & 5 & 4 & 3 & 2 & 1 & 0 &  & 13 & 7 & 8 & 9 & 10 & 11 & 12\\ \\
13 & 7 & 8 & 9 & 10 & 11 & 12 &  & 0 & 6 & 5 & 4 & 3 & 2 & 1\\
7 & 8 & 9 & 10 & 11 & 12 & 13 &  & 1 & 0 & 6 & 5 & 4 & 3 & 2\\
8 & 9 & 10 & 11 & 12 & 13 & 7 &  & 2 & 1 & 0 & 6 & 5 & 4 & 3\\
9 & 10 & 11 & 12 & 13 & 7 & 8 &  & 3 & 2 & 1 & 0 & 6 & 5 & 4\\
10 & 11 & 12 & 13 & 7 & 8 & 9 &  & 4 & 3 & 2 & 1 & 0 & 6 & 5\\
11 & 12 & 13 & 7 & 8 & 9 & 10 &  & 5 & 4 & 3 & 2 & 1 & 0 & 6\\
12 & 13 & 7 & 8 & 9 & 10 & 11 &  & 6 & 5 & 4 & 3 & 2 & 1 & 0
\end{array}
\right]
$$

\end{example}

\subsection{PDQs from the DB Construction}
\label{sec:part2}

Now, we apply the DB Construction and obtain PDQs which are also disjoint
from the ones obtained in the first part via the DLS Construction. To this end,
we have first to analyse the quadruples from configuration $(2,2)$ which
are contained in the $2n$ PDQs obtained in the DLS Construction.
The reason is that by Lemma~\ref{lem:conf1_3} we can ignore quadruples from configuration
$(3,1)$ and configuration $(1,3)$ and clearly no quadruple from configurations
$(4,0)$ and $(0,4)$ is obtained in the DLS construction.

\subsubsection{Quadruples from Configuration $(2,2)$ in the DLS Construction}

The quadruples from configuration $(2,2)$ which are contained in the $2n$ PDQs of order $4n$
from the DLS Construction can be defined by observing the properties of
the Latin square $M_n$. First, note that four sets of pairs are defined.

$$
\cA_i \triangleq \{ \{ x,y \} ~:~ x,y \in \Z_n, ~ y - x \equiv i~(\text{mod}~n) \},~~ 1 \leq i \leq \frac{n-1}{2},
$$
$$
\cB_i \triangleq \{ \{ x,y \} ~:~ x,y \in \{n,n+1,\ldots,2n-1\}, ~ y - x \equiv i~(\text{mod}~n) \},~~ 1 \leq i \leq \frac{n-1}{2},
$$
$$
\cC_i \triangleq \{ \{x,y\}  ~:~  0 \leq x \leq n-1,~ n \leq y \leq 2n-1,~ x+y \equiv i~(\text{mod}~n)  \},~~0 \leq i \leq n-1,
$$
$$
\cD_i \triangleq \{ \{x,y\}  ~:~  0 \leq x \leq n-1,~ n \leq y \leq 2n-1,~ y-x \equiv i~(\text{mod}~n)  \},~~0 \leq i \leq n-1.
$$

\begin{example}
For $n=7$, the sets $\cA_i$'s, $\cB_i$'s, $\cC_i$'s, and $\cD_i$'s, are as follows.
$$
\cA_1 = \{ \{0,1\},\{1,2\},\{2,3\},\{3,4\},\{4,5\},\{5,6\},\{6,0\} \}
$$
$$
\cA_2 = \{ \{0,2\},\{1,3\},\{2,4\},\{3,5\},\{4,6\},\{5,0\},\{6,1\} \}
$$
$$
\cA_3 = \{ \{0,3\},\{1,4\},\{2,5\},\{3,6\},\{4,0\},\{5,1\},\{6,2\} \}
$$
\vspace{0.1cm}
$$
\cB_1 = \{ \{7,8\},\{8,9\},\{9,10\},\{10,11\},\{11,12\},\{12,13\},\{13,7\} \}
$$
$$
\cB_2 = \{ \{7,9\},\{8,10\},\{9,11\},\{10,12\},\{11,13\},\{12,7\},\{13,8\} \}
$$
$$
\cB_3 = \{ \{7,10\},\{8,11\},\{9,12\},\{10,13\},\{11,7\},\{12,8\},\{13,9\} \}
$$
\vspace{0.1cm}
$$
\cC_0 = \{ \{0,7\},\{1,13\},\{2,12\},\{3,11\},\{4,10\},\{5,9\},\{6,8\} \}
$$
$$
\cC_1 = \{ \{0,8\},\{1,7\},\{2,13\},\{3,12\},\{4,11\},\{5,10\},\{6,9\} \}
$$
$$
\cC_2 = \{ \{0,9\},\{1,8\},\{2,7\},\{3,13\},\{4,12\},\{5,11\},\{6,10\} \}
$$
$$
\cC_3 = \{ \{0,10\},\{1,9\},\{2,8\},\{3,7\},\{4,13\},\{5,12\},\{6,11\} \}
$$
$$
\cC_4 = \{ \{0,11\},\{1,10\},\{2,9\},\{3,8\},\{4,7\},\{5,13\},\{6,12\} \}
$$
$$
\cC_5 = \{ \{0,12\},\{1,11\},\{2,10\},\{3,9\},\{4,8\},\{5,7\},\{6,13\} \}
$$
$$
\cC_6 = \{ \{0,13\},\{1,12\},\{2,11\},\{3,10\},\{4,9\},\{5,8\},\{6,7\} \}
$$
\vspace{0.1cm}
$$
\cD_0 = \{ \{0,7\},\{1,8\},\{2,9\},\{3,10\},\{4,11\},\{5,12\},\{6,13\} \}
$$
$$
\cD_1 = \{ \{0,8\},\{1,9\},\{2,10\},\{3,11\},\{4,12\},\{5,13\},\{6,7\} \}
$$
$$
\cD_2 = \{ \{0,9\},\{1,10\},\{2,11\},\{3,12\},\{4,13\},\{5,7\},\{6,8\} \}
$$
$$
\cD_3 = \{ \{0,10\},\{1,11\},\{2,12\},\{3,13\},\{4,7\},\{5,8\},\{6,9\} \}
$$
$$
\cD_4 = \{ \{0,11\},\{1,12\},\{2,13\},\{3,7\},\{4,8\},\{5,9\},\{6,10\} \}
$$
$$
\cD_5 = \{ \{0,12\},\{1,13\},\{2,7\},\{3,8\},\{4,9\},\{5,10\},\{6,11\} \}
$$
$$
\cD_6 = \{ \{0,13\},\{1,7\},\{2,8\},\{3,9\},\{4,10\},\{5,11\},\{6,12\} \}
$$
\end{example}

The following two lemmas are immediate observations from the definitions of the
$\cA_i$'s, $\cB_i$'s, $\cC_i$'s, $\cD_i$'s, and $M_n$.

\begin{lemma}
\label{lem:pairsDLS}
By using the Latin square $M_n$, the following sets of quadruples are contained, in the first $2n$ PDQs
obtained via the DLS Construction.
$$
\{ (v,0),(x,0),(y,1),(z,1) \},~~\{v,x \} \in \cA_i \cup \cB_i,~~\{y,z \} \in \cA_i \cup \cB_i,~ 1 \leq i \leq \frac{n-1}{2},
$$
$$
\{ (v,0),(x,0),(y,1),(z,1) \},~~\{v,x \} \in \cC_i,~~\{y,z \} \in \cD_{i-1} \cup \cD_i,~ 0 \leq i \leq n-1,
$$
where subscripts are taken~ modulo $n$.
Each one of these quadruples is contained in exactly one of these $2n$ PDQs.
No other quadruple from configuration $(2,2)$ is contained in these $2n$ PDQs.
\end{lemma}

Note, that Lemma~\ref{lem:pairsDLS} implies that the quadruples from configuration $(2,2)$ formed by the
first $2n$ PDQs are of the form $\{ (v,0),(x,0),(y,1),(z,1) \}$.
The pair $\{ v,x \}$ is from one of the $2n-1$ sets $\cA_i$,
$\cB_i$, $1 \leq i \leq \frac{n-1}{2}$, or $\cC_j$, $0 \leq j \leq n-1$.
The pair $\{ y,z \}$ is from one of the $2n-1$ sets $\cA_i$,
$\cB_i$, $1 \leq i \leq \frac{n-1}{2}$, or $\cD_j$, $0 \leq j \leq n-1$.
These facts will be used in the sequel. Now, we state the second simple lemma which can be easily verified.

\begin{lemma}
\label{lem:one-factors}
$~$
\begin{enumerate}
\item Each one of the $\cC_i$'s and the $\cD_i$'s is a one-factor of $K_{2n}$.

\item $\bigcup_{i=1}^{(n-1)/2} (\cA_i \cup \cB_i) \cup \bigcup_{i=0}^{n-1} \cC_i = \{ \{x,y\} ~:~ 0 \leq x<y \leq 2n-1 \}$.

\item $\bigcup_{i=1}^{(n-1)/2} (\cA_i \cup \cB_i) \cup \bigcup_{i=0}^{n-1} \cD_i = \{ \{x,y\} ~:~ 0 \leq x<y \leq 2n-1 \}$.

\item $\sum_{i=1}^{(n-1)/2} (|\cA_i| +|\cB_i|)+ \sum_{i=0}^{n-1} |\cC_i| = n(2n-1)$.

\item $\sum_{i=1}^{(n-1)/2} (|\cA_i| +|\cB_i|)+ \sum_{i=0}^{n-1} |\cD_i| = n(2n-1)$.
\end{enumerate}
\end{lemma}

\subsubsection{One-Factorizations for the DB Construction}

The second part of the construction is to use the DB Construction to form more PDQs or order $4n$ with quadruples which were not
used in the first $2n$ PDQs. Such a set of new PDQs of order $4n$ depends on the value of $D(2n)$.
These PDQs of order $2n$ will form the quadruples from configuration $(4,0)$
and from configuration $(0,4)$ in the recursive construction. We distinguish now between the case
$n \equiv 5~(\text{mod}~6)$ and the case
$n \equiv 1~(\text{mod}~6)$.

\begin{lemma}
\label{lem:primesR}
$~$
Let $n\equiv 1$ or $5~(\text{mod}~6)$ be a prime. Then  there exists a one-factorization $\cR$ of $K_{2n}$
on the vertex set $\{0,1,\ldots, 2n-1\}$ satisfying the followings:
\begin{enumerate}
\item  if $n\equiv 5~(\text{mod}~6)$, then ${\cal R}=\{{\cal R}_{i,j}:1\le i\le {2n-1\over 3}, ~ 1 \leq j \leq 3 \},$ where
\begin{align}\label{redistr}{\cal R}_{i,1}\cup {\cal R}_{i,2}\cup {\cal R}_{i,3}={\cal A}_i\cup {\cal B}_i\cup{\cal C}_{i-1},1\le i\le {n-1\over 2},\end{align}
  and $${\cR}_{i,1}={\cC}_{3i-n-2}, {\cal R}_{i,2}= {\cC}_{3i-n-1}, {\cal R}_{i,3}={\cal C}_{3i-n},{n+1\over 2}\le i\le {2n-1\over 3}.$$

\item  if $n\equiv 1~(\text{mod}~6)$, then ${\cal R}=\{{\cal R}_{i,j}:1\le i\le {n-1\over 2}, ~ 1 \leq j \leq 3\}\cup\{{\cal C}_i:{n-1\over2}\le i\le n-1\}$, where Eq. {\rm (\ref{redistr})} also holds.
\end{enumerate}
\end{lemma}
\begin{proof}
By Lemma~\ref{lem:one-factors}, each ${\cC}_i$ is a one-factor of $K_{2n}$ and $\bigcup_{i=1}^{n-1\over 2}({\cal A}_i\cup{\cal B}_i)\cup(\bigcup_{i=0}^{n-1}{\cal C}_i)$ covers all pairs of $\{0,1,\ldots, 2n-1\}$. So, to prove the conclusion, we only need to show that there exist one-factors ${\cal R}_{i,j}$ with $1\le i\le {2n-1\over 3}$ and $j=1,2,3$ such that Eq. (\ref{redistr}) holds.

For a given $1\le i\le {n-1\over 2}$, form the sequences $s_j =\{j\cdot i, (j + 1) \cdot i\}, 0\le j \le n - 1$,
where the elements are taken modulo $n$, reduced to the range $\{0, 1, \ldots , n -1\}$.
Form the second sequence $q_j =\{ j \cdot i, (j +1) \cdot i\}, 0\le j \le n - 1$, where elements are
taken modulo $n$, reduced to the range $\{n, n + 1, \ldots , 2n -1\}$. It is obvious that ${\cA}_i=\{s_j:0\le j\le n-1\}$
and ${\cB}_i=\{q_j:0\le j\le n-1\}$. Since $n$ is prime, the $n$ pairs of ${\cA}_i$ form an $n$-cycle (a cycle of length $n$),
so are those of ${\cB}_i$. Note that
$s_0=\{0,i\}\in {\cA}_i,q_{m}=\{n+i-1,2n-1\}\in {\cB}_i$ where $n-1\equiv m\cdot i$ (mod $n$) ($0\le m\le n-1$).
Since $n$ is odd, applying the $n$-cycle of ${\cA}_i$ we can partition
${\cA}_i\setminus\{s_0\}$ into two subsets ${\cA}_{i,1}$ and  ${\cA}_{i,2}$ with equal size ${n-1\over 2}$ and
$$
\bigcup_{P\in{\cA}_{i,1}}P=\{0,1,\ldots,n-1\}\setminus\{0\},\bigcup_{P\in{\cA}_{i,2}}P=\{0,1,\ldots,n-1\}\setminus\{i\}.
$$
Similarly, we can partition ${\cal B}_i\setminus\{q_m\}$ into two subsets
${\cB}_{i,1}$ and  ${\cal B}_{i,2}$ with equal size ${n-1\over 2}$ and
$$
\bigcup_{P\in{\cal B}_{i,1}}P=\{n,n+1,\ldots,2n-1\}\setminus\{n+i-1\},\bigcup_{P\in{\cB}_{i,2}}P=\{n,n+1,\ldots,2n-1\}\setminus\{2n-1\}.
$$
Obviously we have
$\{0,n+i-1\},\{i,2n-1\}\in{\cC}_{i-1}$. So define
\begin{align*}
&{\cal R}_{i,1}={\cal A}_{i,1}\cup{\cal B}_{i,1}\cup\{\{0,n+i-1\}\},\\
&{\cal R}_{i,2}={\cal A}_{i,2}\cup{\cal B}_{i,2}\cup\{\{i,2n-1\}\},\\
&{\cal R}_{i,3}=({\cal C}_{i-1}\cup\{s_0,q_m\})\setminus\{\{0,n+i-1\},\{i,2n-1\}\}.
\end{align*}
It is easy to verify that each ${\cal R}_{i,j}$ is a one-factor ($1\le i\le {2n-1\over 3}$ and $1 \leq j \leq 3$) and Eq. (\ref{redistr}) holds. This completes the proof.
\end{proof}
A related result will be given in the following lemma for a one-factorization $\cT$ with almost an identical proof.
\begin{lemma}
\label{lem:primesT}
Let $n\equiv 1$ or $5~(\text{mod}~6)$ be a prime. Then there exists a one-factorization $\cT$ of $K_{2n}$
on the vertex set $\{0,1,\ldots, 2n-1\}$ satisfying the followings:
\begin{enumerate}
\item if $n\equiv 5$ {\rm (mod 6)}, then ${\cal T}=\{{\cal T}_{i,j}:1\le i\le {2n-1\over 3}~ 1 \leq j \leq 3\},$ where
\begin{align}\label{redistr2}{\cal T}_{i,1}\cup {\cal T}_{i,2}\cup {\cal T}_{i,3}={\cal A}_i\cup {\cal B}_i\cup{\cal D}_{i-1},1\le i\le {n-1\over 2},\end{align}
  and $${\cal T}_{i,1}={\cal D}_{3i-n-2}, {\cal T}_{i,2}= {\cal D}_{3i-n-1}, {\cal T}_{i,3}={\cal D}_{3i-n},{n+1\over 2}\le i\le {2n-1\over 3}.$$

\item   if $n\equiv 1$ {\rm (mod 6)}, then ${\cal T}=\{{\cal T}_{i,j}:1\le i\le {n-1\over 2~ 1 \leq j \leq 3 }\}\cup\{{\cal D}_i:{n-1\over2}\le i\le n-1\},$ where Eq. {\rm (\ref{redistr2})} also holds.
\end{enumerate}
\end{lemma}
\begin{proof}
The proof is very similar to that of Lemma~\ref{lem:primesR}. The difference is that~$\cD_i$
is used instead of~${\cC}_i$. To prove that, for $1\le i\le {n-1\over 2}$, ${\cal A}_i\cup {\cB}_i\cup{\cD}_{i-1}$
can be partitioned into three one-factors ${\cT}_{i,1}, {\cT}_{i,2}$, and~${\cT}_{i,3}$, we note
that $\{0,i\}\in{\cA}_i,\{n+i-1,n+2i-1\}\in{\cB}_i$ and $\{0,n+i-1\},\{i,n+2i-1\}\in{\cD}_{i-1}$. The rest of
the proof is exactly parallel to that of Lemma~\ref{lem:primesR}.
\end{proof}

\begin{example}
For $n=11$, the set $\cR$ and part of the set $\cT$ are given by the following sets.
$$
\cR_{1,1} = \{ \{1,2\},\{3,4\},\{5,6\},\{7,8\},\{9,10\},\{12,13\},\{14,15\},\{16,17\},\{18,19\},\{20,21\},\{0,11\}  \}
$$
$$
\cR_{1,2} = \{ \{2,3\},\{4,5\},\{6,7\},\{8,9\},\{10,0\},\{11,12\},\{13,14\},\{15,16\},\{17,18\},\{19,20\},\{1,21\}  \}
$$
$$
\cR_{1,3} = \{ \{2,20\},\{3,19\},\{4,18\},\{5,17\},\{6,16\},\{7,15\},\{8,14\},\{9,13\},\{10,12\},\{0,1\},\{11,21\}  \}
$$
$$
\cR_{2,1} = \{ \{2,4\},\{3,5\},\{6,8\},\{7,9\},\{1,10\},\{11,13\},\{14,16\},\{15,17\},\{18,20\},\{19,21\},\{0,12\}  \}
$$
$$
\cR_{2,2} = \{ \{1,3\},\{4,6\},\{5,7\},\{8,10\},\{0,9\},\{12,14\},\{13,15\},\{16,18\},\{17,19\},\{11,20\},\{2,21\}  \}
$$
$$
\cR_{2,3} = \{ \{1,11\},\{3,20\},\{4,19\},\{5,18\},\{6,17\},\{7,16\},\{8,15\},\{9,14\},\{10,13\},\{0,2\},\{12,21\}  \}
$$
$$
\cR_{3,1} = \{ \{3,6\},\{4,7\},\{5,8\},\{1,9\},\{2,10\},\{11,14\},\{12,15\},\{16,19\},\{17,20\},\{18,21\},\{0,13\}  \}
$$
$$
\cR_{3,2} = \{ \{1,4\},\{2,5\},\{6,9\},\{7,10\},\{8,0\},\{13,16\},\{14,17\},\{15,18\},\{11,19\},\{12,20\},\{3,21\}  \}
$$
$$
\cR_{3,3} = \{ \{1,12\},\{2,11\},\{4,20\},\{5,19\},\{6,18\},\{7,17\},\{8,16\},\{9,15\},\{10,14\},\{0,3\},\{13,21\}  \}
$$
$$
\cR_{4,1} = \{ \{1,5\},\{3,7\},\{4,8\},\{6,10\},\{2,9\},\{12,16\},\{15,19\},\{17,21\},\{11,18\},\{13,20\},\{0,14\}  \}
$$
$$
\cR_{4,2} = \{ \{2,6\},\{5,9\},\{0,7\},\{1,8\},\{3,10\},\{11,15\},\{13,17\},\{14,18\},\{16,20\},\{12,19\},\{4,21\}  \}
$$
$$
\cR_{4,3} = \{ \{1,13\},\{2,12\},\{3,11\},\{5,20\},\{6,19\},\{7,18\},\{8,17\},\{9,16\},\{10,15\},\{0,4\},\{14,21\}  \}
$$
$$
\cR_{5,1} = \{ \{1,6\},\{2,7\},\{3,8\},\{4,9\},\{5,10\},\{16,21\},\{11,17\},\{12,18\},\{13,19\},\{14,20\},\{0,15\}  \}
$$
$$
\cR_{5,2} = \{ \{0,6\},\{1,7\},\{2,8\},\{3,9\},\{4,10\},\{11,16\},\{12,17\},\{13,18\},\{14,19\},\{15,20\},\{5,21\}  \}
$$
$$
\cR_{5,3} = \{ \{1,14\},\{2,13\},\{3,12\},\{4,11\},\{6,20\},\{7,19\},\{8,18\},\{9,17\},\{10,16\},\{0,5\},\{15,21\}  \}
$$
$$
\cR_{6,1} = \cC_5,~  \cR_{6,2} = \cC_6,~ \cR_{6,3} = \cC_7,
$$
$$
\cR_{7,1} = \cC_8,~  \cR_{7,2} = \cC_9,~ \cR_{7,3} = \cC_{10},
$$
\vspace{0.1cm}
$$
\cT_{1,1} = \{ \{1,2\},\{3,4\},\{5,6\},\{7,8\},\{9,10\},\{12,13\},\{14,15\},\{16,17\},\{18,19\},\{20,21\},\{0,11\}  \}
$$
$$
\cT_{1,2} = \{ \{2,3\},\{4,5\},\{6,7\},\{8,9\},\{0,10\},\{11,21\},\{13,14\},\{15,16\},\{17,18\},\{19,20\},\{1,12\}  \}
$$
$$
\cT_{1,3} = \{ \{2,13\},\{3,14\},\{4,15\},\{5,16\},\{6,17\},\{7,18\},\{8,19\},\{9,20\},\{10,21\},\{0,1\},\{11,12\}  \}
$$
$$
\cT_{2,1} = \{ \{2,4\},\{3,5\},\{6,8\},\{7,9\},\{1,10\},\{11,13\},\{14,16\},\{15,17\},\{18,20\},\{19,21\},\{0,12\}  \}
$$
$$
\cT_{2,2} = \{ \{1,3\},\{4,6\},\{5,7\},\{8,10\},\{0,9\},\{13,15\},\{16,18\},\{17,19\},\{11,20\},\{12,21\},\{2,14\}  \}
$$
$$
\cT_{2,3} = \{ \{1,13\},\{3,15\},\{4,16\},\{5,17\},\{6,18\},\{7,19\},\{8,20\},\{9,21\},\{10,11\},\{0,2\},\{12,14\}  \}
$$
\end{example}

\vspace{0.2cm}

It is not difficult to accommodate the proofs of Lemmas~\ref{lem:primesR} and~\ref{lem:primesT}
to all positive integers $n \equiv 1$ or $5~(\text{mod}~6)$, where $n \geq 7$.
\begin{lemma}
\label{lem:generalRT}
Lemmas~\ref{lem:primesR} and~\ref{lem:primesT} hold for any positive integer $n \equiv 1$ or $5~(\text{mod}~6)$, where $n \geq 7$.
\end{lemma}
\begin{proof}
When $n \equiv 1$ or $5~(\text{mod}~6)$,
the proofs for Lemmas~\ref{lem:primesR} and~\ref{lem:primesT} are similar, so we only show the existence of
a one-factorization ${\cT}$ for Lemma~\ref{lem:primesT}. The proof is actually similar to the case of $n$ being prime.
The difference  is that  ${\cA}_i$ and ${\cB}_i$
may consist of  more than one cycle if $n$ is not a prime.
Nevertheless, we can use the same procedure. More precisely, we are done as follows.

Assume that we want to partition ${\cA}_i\cup {\cB}_i\cup{\cD}_{i-1},1\le i\le {n-1\over 2}$, into three one-factors.
The length of the cycles  generated from $\cA_i$ is $\ell= {n\over \gcd(i,n)}$ and the number
of cycles is $\rho=\gcd(i,n)$. We denote the $j$-th cycle, $0\le j\le \rho-1$,
by ${\cal A}_{i}^{(j)}$, which consists of the following $\ell$ edges (elements are reduced to the range $\{0,1,\ldots,n-1\}$):
$$\{j, i + j\}, \{i + j, 2i + j\}, \{2i + j, 3i + j\}, \ldots, \{n- i + j, j\}.$$
For $\cB_i$ we have the same $\rho$ cycles of length $\ell$, say ${\cal B}_{i}^{(j)},0\le j\le \rho-1$,
where the elements are reduced to the range $\{n,n+1,\ldots,2n-1\}$. Note that $\ell$ is odd
and $\{j,i+j\}\in{\cal A}_i^{(j)},\{\beta_{i,j},b_{i,j}\}\in \cB_i^{(j')}$,
where $\beta_{i,j} \equiv i+j-1$ (mod $n$), $b_{i,j}\equiv 2i+j-1$ (mod $n$) with $n\le \beta_{i,j},b_{i,j}\le 2n-1$,
and $j'\equiv j-1$ (mod $\rho$) with $0\le j'\le \rho-1$. We can partition ${\cal A}_i^{(j)}\setminus\{\{j,i+j\}\}$
into two subsets  ${\cal A}_{i,1}^{(j)}$, and ${\cal A}_{i,2}^{(j)}$ with equal
size ${\ell-1\over 2}$ and
$$
\bigcup_{P\in \cA_{i,1}^{(j)}}P=(\bigcup_{P\in \cA_{i}^{(j)}}P)\setminus\{j\},\bigcup_{P\in \cA_{i,2}^{(j)}}P=(\bigcup_{P\in \cA_{i}^{(j)}}P)\setminus\{i+j\}.
$$
Similarly, we can partition  $\cB_i^{(j')}\setminus\{\{\beta_{i,j},b_{i,j}\}\}$ into two subsets
$\cB_{i,1}^{(j')}$ and $\cB_{i,2}^{(j')}$ with equal size ${\ell-1\over 2}$ and
$$
\bigcup_{P\in \cB_{i,1}^{(j')}}P=(\bigcup_{P\in \cB_{i}^{(j')}}P)\setminus\{\beta_{i,j}\},\bigcup_{P\in \cB_{i,2}^{(j')}}P=(\bigcup_{P\in \cB_{i}^{(j')}}P)\setminus\{b_{i,j}\}.
$$
Obviously we  have $\{j,\beta_{i,j}\},\{i+j,b_{i,j}\}\in{\cal D}_{i-1}$. So define
\begin{align*}
&\cT_{i,1}=\bigcup_{j=0}^{\rho-1}(\cA_{i,1}^{(j)}\cup \cB_{i,1}^{(j')}\cup\{\{j,\beta_{i,j}\}\}),\\
&\cT_{i,2}=\bigcup_{j=0}^{\rho-1}(\cA_{i,2}^{(j)}\cup{\cal B}_{i,2}^{(j')}\cup\{\{i+j,b_{i,j}\}\}),\\
&\cT_{i,3}=({\cal D}_{i-1}\cup\{\{j,i+j\},\{\beta_{i,j},b_{i,j}\}:0\le j\le \rho-1\})\\
&\qquad\qquad \ \ \ \ \setminus\{\{j,\beta_{i,j}\},\{i+j,b_{i,j}\}:0\le j\le \rho-1\}.
\end{align*}
It is easy to verify that each $\cT_{i,k}$ is a one-factor ($1\le i\le {n-1\over 2}$ and $k=1,2,3$) and Eq. (\ref{redistr2}) holds.
\end{proof}

\subsubsection{A Latin Square for Direct Products}
\label{sec:LatinS}

This subsection is devoted only to the case when $n \equiv 5~(\text{mod}~6)$.
Let $\cR_i \triangleq \{ \cR_{i,1},\cR_{i,2},\cR_{i,3} \}$ and
$\cT_i \triangleq \{ \cT_{i,1},\cT_{i,2},\cT_{i,3} \}$, $1 \leq i \leq \frac{2n-1}{3}$.
The same definition for $1 \leq i \leq \frac{n-1}{2}$ is given for $n \equiv 1~(\text{mod}~6)$.
We construct a Latin square which makes a distinction between
the the quadruples which are contained in
the DLS Construction and those which will be constructed with the DB Construction.

Let $\nu = \frac{2n-1}{3} $ and let $\cM$ be
a $\nu \times \nu$ Latin square defined as follows.
The elements of $\cM$ are taken from the set $\{\cT_1,\cT_2,\ldots,\cT_\nu \}$.
The $i$-th column of $\cM$ is indexed by $\cR_i$, $1 \leq i \leq \nu$.
The first two rows of $\cM$ represent direct products of $\cR_i$'s by $\cT_j$'s
for which some of the related quadruples are contained in the first $2n$ PDQs of order $4n$,
and all the quadruples of the first $2n$ PDQs are contained in these direct products, which will be explained in the next subsection.
For this purpose let $\cM(1,\cR_i) =\cT_i$ for $1 \leq i \leq \nu$,
$\cM(2,\cR_i) = \cT_{i-1}$ for $2 \leq i \leq \nu$, and $\cM(2,\cR_1)=\cT_\nu$.
The last $\nu -2$ rows of $\cM$ can be chosen arbitrarily to form a Latin square.
One simple choice is to take these rows as cyclic shifts of the first row.

\begin{example}
\label{ex:LS_DB}
We continue the example for $n=11$.
The first two rows of $\cM$, represent the quadruples from configuration $(2,2)$
which are contained in the first 22 PDQs of order 44 obtained by the DLS Construction.
These rows are given by
$$
\begin{array}{lccccccc}
& \cR_1 & \cR_2 & \cR_3 & \cR_4 & \cR_5 & \cR_6 & \cR_7 \\ \hline
1 & \cT_1 & \cT_2 & \cT_3 & \cT_4 & \cT_5 & \cT_6 & \cT_7 \\
2 & \cT_7 & \cT_1 & \cT_2 & \cT_3 & \cT_4 & \cT_5 & \cT_6
\end{array}
$$
These two rows can be completed to a $7 \times 7$ Latin square as follows
$$
\begin{array}{lccccccc}
& \cR_1 & \cR_2 & \cR_3 & \cR_4 & \cR_5 & \cR_6 & \cR_7 \\ \hline
1 & \cT_1 & \cT_2 & \cT_3 & \cT_4 & \cT_5 & \cT_6 & \cT_7 \\
2 & \cT_7 & \cT_1 & \cT_2 & \cT_3 & \cT_4 & \cT_5 & \cT_6 \\ \hline
3 & \cT_6 & \cT_7 & \cT_1 & \cT_2 & \cT_3 & \cT_4 & \cT_5 \\
4 & \cT_5 & \cT_6 & \cT_7 & \cT_1 & \cT_2 & \cT_3 & \cT_4 \\
5 & \cT_4 & \cT_5 & \cT_6 & \cT_7 & \cT_1 & \cT_2 & \cT_3 \\
6 & \cT_3 & \cT_4 & \cT_5 & \cT_6 & \cT_7 & \cT_1 & \cT_2 \\
7 & \cT_2 & \cT_3 & \cT_4 & \cT_5 & \cT_6 & \cT_7 & \cT_1
\end{array}.
$$

\end{example}

\subsubsection{Proof of Theorem~\ref{thm:main}}

We distinguish again between the case
$n \equiv 5~(\text{mod}~6)$ and the case $n \equiv 1~(\text{mod}~6)$.

\noindent
{\bf Case 1:} $n \equiv 5~(\text{mod}~6)$.

We continue now to conclude the second part of the construction and obtain more
PDQs of order $4n$. We construct a $(2n-1) \times (2n-1)$ Latin square $\cH_{2n-1}$ from the Latin square $\cM$
as follows. Each row $i$ of $\cM$ is replaced by rows $(i,1)$, $(i,2)$, and $(i,3)$, in $\cH_{2n-1}$.
Each column $\cR_j$ of $\cM$ is replaced by columns $\cR_{j,1}$,
$\cR_{j,2}$, and $\cR_{j,3}$, in $\cH_{2n-1}$.
Any entry of $\cM(i,\cR_j)$ which is equal to $\cT_\ell$ is replaced by the $3 \times 3$ Latin square in the related
three rows and three columns of $\cH_{2n-1}$:

$$
\begin{array}{lcccc}
& \vline & \cR_{j,1} & \cR_{j,2} & \cR_{j,3}  \\ \hline
(i,1) & \vline & \cT_{\ell,1} & \cT_{\ell,2} & \cT_{\ell,3}  \\
(i,2) & \vline & \cT_{\ell,2} & \cT_{\ell,3} & \cT_{\ell,1}  \\
(i,3) & \vline & \cT_{\ell,3} & \cT_{\ell,1} & \cT_{\ell,2}
\end{array}.
$$

\noindent
Then each row of ${\cal H}_{2n-1}$ forms the one-factorization ${\cal T}$.
By Lemma~\ref{lem:pairsDLS}, all the quadruples from configuration $(2,2)$ in the first $2n$ PDQs from the DLS Construction are
contained in the direct products related to the first six rows of $\cH_{2n-1}$.
That is, the direct products of ${\cal R}_{j,k}$ and $\cH_{2n-1}((i,k'),\cR_{j,k})$
for all $1\leq j\le {2n-1\over 3}$, $i=1,2$, and $1\le k,k'\le 3$.
All the direct products related to the last $2n-7$ rows of $\cH_{2n-1}$
are pairwise disjoint and hence by using them in the DB Construction, combined with $k$
PDQs of order $2n$ in the DB Construction, where $k = \min \{ D(2n),2n-7\}$
we obtain with the first $2n$ PDQs,
a total of $2n +\min \{ D(2n),2n-7\}$ PDQs of order $4n$. Thus, we have proved
Theorem~\ref{thm:main} for $n \equiv 5~(\text{mod}~6)$.

\begin{example}
\label{ex:DB11}
Continue with the example for $n=11$. We first form the first six rows of ${\cal H}_{21}$
from the Latin square ${\cal M}$ in Example~\ref{ex:LS_DB}. If column $\cR_{i_1,j_1}$, where $1 \leq i_1 \leq 6$,
has entry $\cT_{i_2,j_2}$ then the DLS
construction has used some quadruples $\{ (v,0),(x,1),(y,2),(z,3) \}$, where $\{ v,x \} \in \cR_{i_1,j_1}$
and $\{ y,z \} \in \cT_{i_2,j_2}$.
Clearly, there are also many of the quadruples related to these six rows were not obtained in the first part of the construction.
Finally, note that instead of some $\cR_{i,j}$'s and $\cT_{i,j}$'s we use $\cC_{\ell}$'s and $\cD_{\ell}$'s.

{\fontsize{4.5}{12} \selectfont
$$
\begin{array}{ccccccccccccccccccccc}
\cR_{1,1} & \cR_{1,2} & \cR_{1,3} & \cR_{2,1} & \cR_{2,2} & \cR_{2,3} & \cR_{3,1} & \cR_{3,2} & \cR_{3,3} &
\cR_{4,1} & \cR_{4,2} & \cR_{4,3} & \cR_{5,1} & \cR_{5,2} & \cR_{5,3} & \cC_5 & \cC_6 & \cC_7 & \cC_8 & \cC_9 & \cC_{10}\\ \hline
\cT_{1,1} & \cT_{1,2} & \cT_{1,3} & \cT_{2,1} & \cT_{2,2} & \cT_{2,3} & \cT_{3,1} & \cT_{3,2} & \cT_{3,3} &
\cT_{4,1} & \cT_{4,2} & \cT_{4,3} & \cT_{5,1} & \cT_{5,2} & \cT_{5,3} & \cD_5 & \cD_6 & \cD_7 & \cD_8 & \cD_9 & \cD_{10} \\
\cT_{1,2} & \cT_{1,3} & \cT_{1,1} & \cT_{2,2} & \cT_{2,3} & \cT_{2,1} & \cT_{3,2} & \cT_{3,3} & \cT_{3,1} &
\cT_{4,2} & \cT_{4,3} & \cT_{4,1} & \cT_{5,2} & \cT_{5,3} & \cT_{5,1} & \cD_6 & \cD_7 & \cD_5 & \cD_9 & \cD_{10} & \cD_8 \\
\cT_{1,3} & \cT_{1,1} & \cT_{1,2} & \cT_{2,3} & \cT_{2,1} & \cT_{2,2} & \cT_{3,3} & \cT_{3,1} & \cT_{3,2} &
\cT_{4,3} & \cT_{4,1} & \cT_{4,2} & \cT_{5,3} & \cT_{5,1} & \cT_{5,2} & \cD_7 & \cD_5 & \cD_6 & \cD_{10} & \cD_8 & \cD_9 \\
\cD_8 & \cD_9 & \cD_{10} & \cT_{1,1} & \cT_{1,2} & \cT_{1,3} & \cT_{2,1} & \cT_{2,2} & \cT_{2,3} &
\cT_{3,1} & \cT_{3,2} & \cT_{3,3} & \cT_{4,1} & \cT_{4,2} & \cT_{4,3} & \cT_{5,1} & \cT_{5,2} & \cT_{5,3} & \cD_5 & \cD_6 & \cD_7 \\
\cD_9 & \cD_{10} & \cD_8 & \cT_{1,2} & \cT_{1,3} & \cT_{1,1} & \cT_{2,2} & \cT_{2,3} & \cT_{2,1} &
\cT_{3,2} & \cT_{3,3} & \cT_{3,1} & \cT_{4,2} & \cT_{4,3} & \cT_{4,1} & \cT_{5,2} & \cT_{5,3} & \cT_{5,1} & \cD_6 & \cD_7 & \cD_5 \\
\cD_{10} & \cD_8 & \cD_9 & \cT_{1,3} & \cT_{1,1} & \cT_{1,2} & \cT_{2,3} & \cT_{2,1} & \cT_{2,2} &
\cT_{3,3} & \cT_{3,1} & \cT_{3,2} & \cT_{4,3} & \cT_{4,1} & \cT_{4,2} & \cT_{5,3} & \cT_{5,1} & \cT_{5,2} & \cD_7 & \cD_5 & \cD_6 \\
\end{array}
$$}

A completed $21 \times 21$ Latin square $\cH_{21}$ whose last 15 rows form the $15 \times 21$ Latin rectangle
for the DB Construction is given next. In the first 6 rows we remove those entries for which
no quadruple of the related direct product is contained in the PDQs obtained
by the DLS Construction. Note, that each column contains different elements and each one of the
last 15 rows contains each one-factor of $\cT$ exactly once.

$$
\begin{array}{cccccccccccccc}
& \cR_{1,1} & \cR_{1,2} & \cR_{1,3} & \cdots & \cR_{5,1} & \cR_{5,2} & \cR_{5,3} & \cC_5 & \cC_6 & \cC_7 & \cC_8 & \cC_9 & \cC_{10}\\ \hline
1 & \cT_{1,1} & \cT_{1,2} & \cT_{1,3} & \cdots & \cT_{5,1} & \cT_{5,2} & \cT_{5,3} & \cD_5 & \cD_6 & \cD_7 & \cD_8 & \cD_9 & \cD_{10} \\
2 & \cT_{1,2} & \cT_{1,3} & \cT_{1,1} & \cdots & \cT_{5,2} & \cT_{5,3} & \cT_{5,1} &  &  &  &  &  &  \\
3 & \cT_{1,3} & \cT_{1,1} & \cT_{1,2} & \cdots & \cT_{5,3} & \cT_{5,1} & \cT_{5,2} &  & \cD_5 & \cD_6 & & \cD_8 & \cD_9 \\
4 & &  & \cD_{10} & \cdots & \cT_{4,1} & \cT_{4,2} & \cT_{4,3} & \cT_{5,1} &  &  &  &  &  \\
5 & & \cD_{10} &  & \cdots & \cT_{4,2} & \cT_{4,3} & \cT_{4,1} & \cT_{5,2} &  &  & &  &  \\
6 & \cD_{10} & & & \cdots & \cT_{4,3} & \cT_{4,1} & \cT_{4,2} & \cT_{5,3} &  &  & \cD_7 &  &  \\ \hline
7 & \cD_5 & \cD_6 & \cD_7 & \cdots & \cT_{3,1} & \cT_{3,2} & \cT_{3,3} & \cT_{4,1} & \cT_{4,2} & \cT_{4,3} & \cT_{5,1} & \cT_{5,2} & \cT_{5,3} \\
8 & \cD_6 & \cD_7 & \cD_5 & \cdots & \cT_{3,2} & \cT_{3,3} & \cT_{3,1} & \cT_{4,2} & \cT_{4,3} & \cT_{4,1} & \cT_{5,2} & \cT_{5,3} & \cT_{5,1} \\
9 & \cD_7 & \cD_5 & \cD_6 & \cdots & \cT_{3,3} & \cT_{3,1} & \cT_{3,2} & \cT_{4,3} & \cT_{4,1} & \cT_{4,2} & \cT_{5,3} & \cT_{5,1} & \cT_{5,2} \\
10 & \cT_{5,1} & \cT_{5,2} & \cT_{5,3} & \cdots & \cT_{2,1} & \cT_{2,2} & \cT_{2,3} & \cT_{3,1} & \cT_{3,2} & \cT_{3,3} & \cT_{4,1} & \cT_{4,2} & \cT_{4,3} \\
11 & \cT_{5,2} & \cT_{5,3} & \cT_{5,1} & \cdots & \cT_{2,2} & \cT_{2,3} & \cT_{2,1} & \cT_{3,2} & \cT_{3,3} & \cT_{3,1} & \cT_{4,2} & \cT_{4,3} & \cT_{4,1} \\
12 & \cT_{5,3} & \cT_{5,1} & \cT_{5,2} & \cdots & \cT_{2,3} & \cT_{2,1} & \cT_{2,2} & \cT_{3,3} & \cT_{3,1} & \cT_{3,2} & \cT_{4,3} & \cT_{4,1} & \cT_{4,2} \\
13 & \cT_{4,1} & \cT_{4,2} & \cT_{4,3} & \cdots & \cT_{1,1} & \cT_{1,2} & \cT_{1,3} & \cT_{2,1} & \cT_{2,2} & \cT_{2,3} & \cT_{3,1} & \cT_{3,2} & \cT_{3,3} \\
14 & \cT_{4,2} & \cT_{4,3} & \cT_{4,1} & \cdots & \cT_{1,2} & \cT_{1,3} & \cT_{1,1} & \cT_{2,2} & \cT_{2,3} & \cT_{2,1} & \cT_{3,2} & \cT_{3,3} & \cT_{3,1} \\
15 & \cT_{4,3} & \cT_{4,1} & \cT_{4,2} & \cdots & \cT_{1,3} & \cT_{1,1} & \cT_{1,2} & \cT_{2,3} & \cT_{2,1} & \cT_{2,2} & \cT_{3,3} & \cT_{3,1} & \cT_{3,2} \\
16 & \cT_{3,1} & \cT_{3,2} & \cT_{3,3} & \cdots & \cD_8 & \cD_9 & \cD_{10} & \cT_{1,1} & \cT_{1,2} & \cT_{1,3} & \cT_{2,1} & \cT_{2,2} & \cT_{2,3} \\
17 & \cT_{3,2} & \cT_{3,3} & \cT_{3,1} & \cdots & \cD_9 & \cD_{10} & \cD_8 & \cT_{1,2} & \cT_{1,3} & \cT_{1,1} & \cT_{2,2} & \cT_{2,3} & \cT_{2,1} \\
18 & \cT_{3,3} & \cT_{3,1} & \cT_{3,2} & \cdots & \cD_{10} & \cD_8 & \cD_9 & \cT_{1,3} & \cT_{1,1} & \cT_{1,2} & \cT_{2,3} & \cT_{2,1} & \cT_{2,2}  \\
19 & \cT_{2,1} & \cT_{2,2} & \cT_{2,3} & \cdots & \cD_5 & \cD_6 & \cD_7 & \cD_8 & \cD_9 & \cD_{10} & \cT_{1,1} & \cT_{1,2} & \cT_{1,3} \\
20 & \cT_{2,2} & \cT_{2,3} & \cT_{2,1} & \cdots & \cD_6 & \cD_7 & \cD_5 & \cD_9 & \cD_{10} & \cD_8 & \cT_{1,2} & \cT_{1,3} & \cT_{1,1}  \\
21 & \cT_{2,3} & \cT_{2,1} & \cT_{2,2} & \cdots & \cD_7 & \cD_5 & \cD_6 & \cD_{10} & \cD_8 & \cD_9 & \cT_{1,3} & \cT_{1,1} & \cT_{1,2}
\end{array}
$$

\end{example}

\noindent
{\bf Case 2:} $n \equiv 1~(\text{mod}~6)$.

In this case the direct products are not obtained in a straightforward way like
in Case 1 for $n \equiv 5~(\text{mod}~6)$. But, the general method is very similar.
We start with an example for $n=7$.

\begin{example}
\label{lem:DB7}
Continue with the example for $n=7$.
The last seven rows of the following array  form one of the possible direct products to
produce seven sets of quadruples from configuration $(2,2)$ for the DB Construction.
Taking the direct products of the column labels and the corresponding entries in the first six rows
contains all the quadruples obtained from the DLS Construction (and also some other quadruples).
$$
\begin{array}{ccccccccccccc}
\cR_{1,1} & \cR_{1,2} & \cR_{1,3} & \cR_{2,1} & \cR_{2,2} & \cR_{2,3} & \cR_{3,1} & \cR_{3,2} & \cR_{3,3} & \cC_3 & \cC_4 & \cC_5 & \cC_6 \\ \hline
\cT_{1,1} & \cT_{1,2} & \cT_{1,3} & \cT_{2,1} & \cT_{2,2} & \cT_{2,3} & \cT_{3,1} & \cT_{3,2} & \cT_{3,3} & \cD_3 & \cD_4 & \cD_5 & \cD_6 \\
\cT_{1,2} & \cT_{1,3} & \cT_{1,1} & \cT_{2,2} & \cT_{2,3} & \cT_{2,1} & \cT_{3,2} & \cT_{3,3} & \cT_{3,1} &  & \cD_3 & \cD_4 & \cD_5 \\
\cT_{1,3} & \cT_{1,1} & \cT_{1,2} & \cT_{2,3} & \cT_{2,1} & \cT_{2,2} & \cT_{3,3} & \cT_{3,1} & \cT_{3,2} &  &  & &  \\
\cD_6 &  &  & \cT_{1,1} & \cT_{1,2} & \cT_{1,3} & \cT_{2,1} & \cT_{2,2} & \cT_{2,3} & \cT_{3,1} &  & &  \\
 & \cD_6 &  & \cT_{1,2} & \cT_{1,3} & \cT_{1,1} & \cT_{2,2} & \cT_{2,3} & \cT_{2,1} & \cT_{3,2} &  & &  \\
 & & \cD_6 & \cT_{1,3} & \cT_{1,1} & \cT_{1,2} & \cT_{2,3} & \cT_{2,1} & \cT_{2,2} & \cT_{3,3} &  & &  \\ \hline
\cT_{3,1} & \cT_{3,2} & \cT_{3,3} & \cD_3 & \cD_4 & \cD_5 & \cD_6 & \cT_{1,2} & \cT_{1,3} & \cT_{2,3} & \cT_{2,2} & \cT_{2,1} & \cT_{1,1} \\
\cT_{3,2} & \cT_{3,3} & \cT_{3,1} & \cD_4 & \cD_5 & \cD_3 & \cT_{1,2} & \cD_6 & \cT_{1,1} & \cT_{2,2} & \cT_{1,3} & \cT_{2,3} & \cT_{2,1} \\
\cT_{3,3} & \cT_{3,1} & \cT_{3,2} & \cD_5 & \cD_3 & \cD_4 & \cT_{1,3} & \cT_{1,1} & \cD_6 & \cT_{2,1} & \cT_{2,3} & \cT_{1,2} & \cT_{2,2} \\
\cT_{2,1} & \cT_{2,2} & \cT_{2,3} & \cD_6 & \cT_{3,2} & \cT_{3,3} & \cD_3 & \cD_4 & \cD_5 & \cT_{1,1} & \cT_{1,2} & \cT_{1,3}& \cT_{3,1} \\
\cT_{2,2} & \cT_{2,3} & \cT_{2,1} & \cT_{3,2} & \cD_6 & \cT_{3,1} & \cD_4 & \cD_5 & \cD_3 & \cT_{1,2} & \cT_{1,1}& \cT_{3,3} & \cT_{1,3} \\
\cT_{2,3} & \cT_{2,1} & \cT_{2,2} & \cT_{3,3} & \cT_{3,1} & \cD_6 & \cD_5 & \cD_3 & \cD_4 & \cT_{1,3} & \cT_{3,2}& \cT_{1,1}& \cT_{1,2} \\
\cD_3 & \cD_4 & \cD_5 & \cT_{3,1} & \cT_{3,3} & \cT_{3,2} & \cT_{1,1} & \cT_{1,3} & \cT_{1,2} & \cD_6 & \cT_{2,1}& \cT_{2,2}& \cT_{2,3}
\end{array}
$$
\end{example}

If $n > 7$ consider a $(2n-1) \times (2n-1)$ array $\cH_{2n-1}$ ($\cH$ in short) whose first $\frac{3n-3}{2}$ columns are indexed by
$\cR_{i,j}$, $1 \leq i \leq \frac{n-1}{2}$, $1 \leq j \leq 3$,
and whose last $\frac{n+1}{2}$ columns are indexed by $\cC_i$, $\frac{n-1}{2} \leq i \leq n-1$.
The entries of the array are the elements $\cT_{i,j}$, $1 \leq i \leq \frac{n-1}{2}$, $1 \leq j \leq 3$,
and $\cD_i$, $\frac{n-1}{2} \leq i \leq n-1$. The first $\frac{3n-3}{2}$ columns of the first six rows of $\cH$ are defined
by abused of definition similar for $\cH_{2n-1}$,
in the case of $n\equiv 5$ (mod 6). That is, we choose the first two rows and the first $\frac{n-1}{2}$ columns
of the Latin square $\cM$ and substitute $\cR_i$ by $\cR_{i,1},\cR_{i,2},\cR_{i,3}$,
and replace every entry $\cT_\ell$ with a Latin square with elements $\cT_{\ell,1},\cT_{\ell,2},\cT_{l,3}$,
but note that the first three columns of the resultant rows
4, 5, and 6, form a $3 \times 3$ Latin square with the
elements $\cD_{n-3}$, $\cD_{n-2}$, $\cD_{n-1}$.
The other entries of the first six rows of $\cH$ are defined as follows.
\begin{itemize}
\item $\cH(1,\cC_i)= \cD_i$, $\frac{n-1}{2} \leq i \leq n-1$.

\item $\cH(2,\cC_i)= \cD_{i-1}$, $\frac{n+1}{2} \leq i \leq n-1$, and $\cH(2,\cC_{(n-1)/2}))=\cD_{n-1}$.

\item $\cH(3,\cC_i)= \cD_{i-2}$, $\frac{n+3}{2} \leq i \leq n-1$, $\cH(3,\cC_{(n-1)/2}))=\cD_{n-2}$, and $\cH(3,\cC_{(n+1)/2}))=\cD_{n-1}$.

\item Columns $\cC_{(n-1)/2}$, $\cC_{(n+1)/2}$, $\cC_{(n+3)/2}$, in rows 4, 5, and 6, form a $3 \times 3$ Latin square with the
elements $\cT_{(n-1)/2,j}$, $1 \leq j \leq 3$.

\item $\cH(4,\cC_i)= \cD_{i-3}$, $\frac{n+5}{2} \leq i \leq n-1$.

\item $\cH(5,\cC_i)= \cD_{i-4}$, $\frac{n+7}{2} \leq i \leq n-1$, and $\cH(5,\cC_{(n+5)/2}))=\cD_{n-4}$.

\item $\cH(6,\cC_i)= \cD_{i-5}$, $\frac{n+9}{2} \leq i \leq n-1$, $\cH(6,\cC_{(n+5)/2}))=\cD_{n-5}$, and $\cH(6,\cC_{(n+7)/2}))=\cD_{n-4}$.
\end{itemize}

One can easily verifies that these six rows form a $6 \times (2n-1)$ Latin rectangle.
It can be completed to the $(2n-1) \times (2n-1)$ Latin square $\cH$,
whose last $2n-7$ rows define the disjoint quadruples from configuration $(2,2)$ for the DB Construction.
Thus, we have proved Theorem~\ref{thm:main} for $n \equiv 1~(\text{mod}~6)$.

\begin{example}
\label{ex:LS19}
The six rows of the Latin rectangle for $n=19$ is given in the following array.
\begin{footnotesize}
$$
\begin{array}{cccccccccccccccccc}
\cR_{1,1} & \cR_{1,2} & \cR_{1,3} & \cR_{2,1} & \cR_{2,2} & \cR_{2,3} & \cdots & \cR_{9,1} & \cR_{9,2} & \cR_{9,3} & \cC_9 & \cC_{10} & \cC_{11} & \cC_{12} & \cC_{13} & \cdots & \cC_{17} & \cC_{18} \\ \hline
\cT_{1,1} & \cT_{1,2} & \cT_{1,3} & \cT_{2,1} & \cT_{2,2} & \cT_{2,3} & \cdots & \cT_{9,1} & \cT_{9,2} & \cT_{9,3} & \cD_9 & \cD_{10} & \cD_{11} & \cD_{12} & \cD_{13} & \cdots & \cD_{17} & \cD_{18} \\
\cT_{1,2} & \cT_{1,3} & \cT_{1,1} & \cT_{2,2} & \cT_{2,3} & \cT_{2,1} & \cdots & \cT_{9,2} & \cT_{9,3} & \cT_{9,1} & \cD_{18} & \cD_9 & \cD_{10} & \cD_{11} & \cD_{12} & \cdots & \cD_{16} & \cD_{17} \\
\cT_{1,3} & \cT_{1,1} & \cT_{1,2} & \cT_{2,3} & \cT_{2,1} & \cT_{2,2} & \cdots & \cT_{9,3} & \cT_{9,1} & \cT_{9,2} & \cD_{17} & \cD_{18} & \cD_9 & \cD_{10} & \cD_{11} & \cdots & \cD_{15} & \cD_{16} \\
\cD_{16} & \cD_{17} & \cD_{18} & \cT_{1,1} & \cT_{1,2} & \cT_{1,3} & \cdots & \cT_{8,1} & \cT_{8,2} & \cT_{8,3} & \cT_{9,1} & \cT_{9,2} & \cT_{9,3} & \cD_9 & \cD_{10} & \cdots & \cD_{14} & \cD_{15} \\
\cD_{17} & \cD_{18} & \cD_{16} & \cT_{1,2} & \cT_{1,3} & \cT_{1,1} & \cdots & \cT_{8,2} & \cT_{8,3} & \cT_{8,1} & \cT_{9,2} & \cT_{9,3} & \cT_{9,1} & \cD_{15} & \cD_9 & \cdots & \cD_{13} & \cD_{14} \\
\cD_{18} & \cD_{16} & \cD_{17} & \cT_{1,3} & \cT_{1,1} & \cT_{1,2} & \cdots & \cT_{8,3} & \cT_{8,1} & \cT_{8,2} & \cT_{9,3} & \cT_{9,1} & \cT_{9,2} & \cD_{14} & \cD_{15} & \cdots & \cD_{12} & \cD_{13} \\
\end{array}
$$
\end{footnotesize}
\end{example}

Finally, note that we have solved the case where $n \equiv 5~(\text{mod}~6)$ in exactly the same way as the
case $n \equiv 1~(\text{mod}~6)$. The advantage of the presented solution for $n \equiv 5~(\text{mod}~6)$
is that it provides an explicit construction of
the Latin square $\cH_{2n-1}$ from a Latin square $\cM$ of order $\frac{2n-1}{3}$.

\section{Conclusion, Discussion, and Future Research}
\label{sec:conclusion}

We have presented a recursive construction for PDQs
of order $4n$ from a set of PDQs
or order $2n$, where $n$ is congruent to 1 or 5 modulo 6. Having almost a large set
of PDQs order $2n$ the construction would yield an almost large set of PDQs order $4n$.
More precisely, if $n>5$ is an odd integer not divisible by 3, then
$$
D(4n) \geq 2n + \min \{ D(2n),2n-7 \}~.
$$

The first question that one might ask is whether the recursive formula can be improved to
$D(4n) = 2n +  \min \{ D(2n),2n-3 \}$ by partitioning the unused quadruples of configuration $(2,2)$
into four sets, each one of size $n^2(2n-1)$,
in which no triple is contained in more than one quadruple. Unfortunately, the answer
is negative. To see that one can easily verify that all the quadruples
of the form
$$
\{ (v,0),(x,0),(y,1),(z,1) \},~~\{v,x\} \subset \cA_1 \cup \cB_1 \cup \cA_2 \cup \cB_2, ~~\{y,z\} \subset \cD_0
$$
are not used in our construction,
which implies that $\cA_1 \cup \cB_1 \cup \cA_2 \cup \cB_2$ should be partitioned into four one-factors of $\Z_{2n}$
which is obviously impossible.

Next, we would like to know how the main construction is compared to previous results.
In~\cite{Etz93} we have the recursive formula
$$
D(4n) \geq 2n + \min \{ D(2n),n-2 \}~\text{for}~n \equiv 1~\text{or}~5~(\text{mod}~6)
$$
which is clearly inferior to the new result. An interesting comparison is with the lower bound on
$D(n)$ given in~\cite{EtHa91}. For $n \equiv 1$ or $5~(\text{mod}~6)$, it is proved that
$D(4n) \geq 3r$ if there exists a set of $r$ mutually 2-chromatic PDQs of order $2n$.
When $r=n$ we have that the set of $3n$ PDQs is maximal,
in the sense that the set cannot be extended to contain more PDQs. Our new result is better.
If there exists a set of $r$ PDQs of order $2n$, Theorem~\ref{thm:main} implies
that $D(4n) \geq 2n+r$ and if $r = n$ then the two results coincide, but while
the set of PDQs in~\cite{EtHa91} is maximal, our set might be extended. In fact,
if $D(2n) > n$, then Theorem~\ref{thm:main} implies $D(4n) > 3n$, while the
construction in~\cite{EtHa91} cannot be applied since in this case the PDQs of order
$2n$ cannot be mutually 2-chromatic.

There are many open questions which are remained to be answered in the future.
We list some of them in order of difficulty, from the most difficult one to the
easiest one to our opinion.
\begin{enumerate}
\item Construct an infinite family of large sets of PDQs of order $n$.
We conjecture that $n = 2^m$, $m \geq 4$, should be the easiest case, if there is an easy one.

\item Construct one nontrivial large set of SQSs of order $n$. We conjecture that $n=16$
should be the easiest target.

\item Prove that $D(2n) \geq n + D(n)$ for $n \equiv 2$ or $4~(\text{mod}~6)$.

\item Prove that $D(2n) \geq n + D(n)$ for an infinite sequence of values of $n$.

\item Prove that $D(2n) \geq n$ for all $n \equiv 1$ or $2~(\text{mod}~3)$ and $n > 4$.
Note that this holds for $n \equiv 2$ or $4~(\text{mod}~6)$.

\item Improve Theorem~\ref{thm:main} for all $n \equiv 2$ or $4~(\text{mod}~6)$.
\end{enumerate}

redpajama_set_name: RedPajamaArXiv


text:
A visual arts non-profit that connects creativity and community through a dynamic ecosystem of arts education, exhibitions, and studios.
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Indira Urrutia: Introspection
SoMA Grand, 1160 Mission St, San Francisco, California 94103
Exhibition Dates:
Aug 3, 2017 to Oct 2, 2017
Indira Urrutia is a Chilean born immigrant who moved to New Haven, Connecticut in 1986 with her family. Urrutia received her B.S. in Studio Art with a focus on photography from Southern Connecticut State University in 1994. Since then, she has followed her passion for photography and explored other formats such as mixed media, installation art, experimental video, and most recently, metal weaving.
Urrutia has spent the past year exploring the technique of metal weaving. Knitting and weaving have been part of Urrutia's life since childhood. The techniques of weaving have become embedded in Urrutia's DNA over the hundreds of years that the practice has been part of her family history. As such, Indira uses the process of hand knitting to reconnect with the female family members who have inspired her throughout her life. Just as she has done with photography in the past, Urrutia uses weaving to recollect the memories, tales, and history of her life.
Introspection belongs to a series of six pieces, each of which begins from the center. This wall piece was created out of thinking about the human nature of repeating things, regardless of the outcome. This body of work also reflects the influence of Ruth Asawa's work.
© 2020 Root Division. All rights reserved.
Address: 1131 Mission Street, San Francisco, CA 94103. Please buzz for entry.
Office Hours: Monday - Friday, 11am to 6pm / Gallery Hours: Wednesday - Saturday, 2pm to 6pm
415.863.7668 / info@rootdivision.org
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redpajama_set_name: RedPajamaCommonCrawl


We are currently recruiting for an experienced estate agent looking to advance their career with this fantastic opportunity in Peterborough, Cambridgeshire for a Branch Manager. Base salary is around £28,000 with on target earnings of £50,000. Your working week will include occasional Saturdays. You will have experience either working as a Branch Manager within estate agency and be keen to progress your career to the next level.
This vacancy is for a Branch Manager based in Peterborough, Cambridgeshire with a base salary on offer of around £28,000 with on target earnings of £50,000. A car allowance or company car will be offered.


Quinny Zapp Flex Plus Stroller (all colours) with Lux Carrycot and Changing Bag and Pebble Plus Car Seat. A seamless blend of flexibility and comfort, the Zapp Flex Plus lets you stroll around with ease, thanks to its four large wheels and closed pushbar for one-hand steering. Yet it packs all the flexibility of the Zapp Flex: compact folding, a two-way reclining seat, and compatibility with all three Quinny from-birth solutions. So whatever you do, with the Zapp Flex Plus you're more than ready.
TWO-WAY FLEX SEAT - Reclines fully flat both ways.
FOUR LARGE WHEELS - Always a smooth ride.
CLOSED PUSHBAR - Easy one-hand pushing.
EXTENDABLE SUN CANOPY - Shield your child from the sun, wind and rain (UV 50+).
WATER-RESISTANT FABRICS - Simply wipe away any spills.
HIGHLY COMPACT FOLD - Easy 3D folding mechanism.
EASY-ACCESS SHOPPING BASKET - Bring everything you want.
TRANSPORT LOCK - Click down and lift up.
RAIN COVER INCLUDED - Walk your way in any weather.
SWIVEL WHEELS - Highly manoeuvrable.
MAXI-COSI® CAR SEAT ADAPTERS INCLUDED - Easily click your Maxi-Cosi® Car Seat in or out.
Use from birth until 75 cm (approx. 1 year).
Compatible with Maxi-Cosi 2wayFix Base for the safest and complete i-Size compliant solution.
Part of the 2wayFamily collection; offering up to 4 years of rearward facing travel and safety.
New high tech impact-absorbing side protection system allows for extra comfort and greatly improves safety performance in a side impact collision.
Comfortable baby-hugg inlay offers a better fit and protection for newborn babies.
Memory buttons allow for one hand release from a Maxi-Cosi or Quinny pushchair.
Clear belt routing guides for securing with a 3-point seat belt.
One-pull harness system for adjusting the harness length.
Simultaneous safety harness and headrest adjustment.
Removable and washable cover without removing the harness.


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text:
Carena Liptak is a writer and editor in Nashville, Tenn. She is a fan of strong coffee and large-eared dogs. When not writing, she can be found chasin' that neon rainbow and livin' that honky-tonk dream.
Chris Janson's 'All I Need Is You' Is a Grown-Up Love Story [Listen]
The singer sets his priorities straight with this romantic new track, which is the lead single from his next album.
Wynonna Judd Doesn't Think The Judds Final Tour Will Extend Again: 'There Are No Plans'
Instead, the singer says, she's planning to focus on something new.
Luke Combs Reveals His New Album's Title and Cover Artwork
The new album has a special connection to his last release, 'Growin' Up.'
Florida Georgia Line's Hiatus Came as a Surprise to Duo's Tyler Hubbard
When his bandmate Brian Kelley first broached the idea of a break from the band, Hubbard admits it was a bit of a shock.
"I figured everybody would take it as sarcasm," Bryan explained.
Get to know the jaw-dropping real-life landscape featured in the hit TV show.
Naomi Judd's Death Brought Wynonna + Ashley Judd Together After Years of Estrangement
One good thing has come from a terrible situation.
Garth Brooks Calls to Make Scalping Illegal Amid Ticketmaster Investigation
The superstar is weighing in on the controversy.
Luke Combs, Brandi Carlile to Perform at the 2023 Grammy Awards
The Grammys just got a little more country.
Little Big Town Roll Out 2023 Friends of Mine Tour Dates
The fun kicks off in April.


What did individuals do with their time 20 years in the past? I know driving has change into infuriating as individuals actively chat on their cell telephones as opposed to concentrating on the street. I do. We didn't have loads of channels back then and the programming was in all probability quite a bit better than it was right now as solely the cream of the crop made it to the airwaves.\n\nJury instructions and indigent (lacking meals, clothing, and other requirements of life because of poverty; needy; poor; impoverished, destitute) defendants are needed with the court docket's permission for the presenting of such proof. When something new and "better" is revealed, the "older" expertise becomes extra inexpensive, permitting it for use in instructional settings, even when colleges are on a tight budget. \n\nOne of the important points of computer systems right now was invented in 1969. By way of numerous mediums, individuals can get help and seek information freely. I do know numerous different people who find themselves paying a heckova lot more for premium channels. The Web provides another dimension to studying , which may make it more interactive, comprehensible, and inspiring for young people.\n\nCalifornia Department of Training, Far West Laboratory for Instructional Analysis and Growth, California Schooling and Information Processing Association, "California Pupil Info System: A Examine of the Feasibility of Implementing a Statewide Course of for Electronically Sharing Student Data," Sacramento, CA.: California Division of Education, October 1992.


text:
Anthony Hoyt "Tony" Reich
June 18, 1961 - December 26, 2021
Anthony Hoyt "Tony" Reich, age 60, left to join his Heavenly Father on December 26, 2021, at his home in Saint Johns, Florida. Tony was born in Jennings, Louisiana on June 18, 1961, to Robert and Dorothy Reich. He was one of five children. Tony was a very devoted follower of Jesus Christ, a man whose life genuinely reflected godly character, love, and compassion. Tony was an ordained minister and engaged in various ministries during his lifetime. He was a graduate of Florida State University with a bachelor's degree in History. Tony also earned a master's degree in Theology from Lee University in Cleveland, Tennessee. Tony was an exceptional son, brother, and favorite uncle to many. He walked in humility and with a constant awareness of God. For Tony, material possessions held little significance. He kept his life simple. Tony walked by faith and not by sight. He leaned on God and trusted His eternal Father to guide him along this challenging path we call life. Tony is survived by his mother, Dorothy Reich of Saint Johns, FL; sister, Cindy (Chip) Doll of Saint Johns, FL; brothers, Michael (Jayne) Reich of Franklin, NC, Keith Reich of San Antonio, TX, and Scott Reich of Nashville, NC. He is also survived by nieces and nephews, Lana (Cam) Navis, Sashell (Michael) Dragich, Rachel (Nick) Bowyer, Bianca Doll, Lane Reich, Chase (Jessica) Doll, David Reich, Roby (Deanna) Doll, and Alex Jones; great-nieces and great-nephews, Kingston, Coen, Joshua, Christian, Lincoln, Austin, Lucas, Levi, Amelia, Scarlett, Kai, Lawson, and baby Ezra (on the way). Tony was preceded in death by his father, Robert Reich. The family will be holding a Celebration of Life on January 1, 2022, at 10:00 AM at the Switzerland Community Church. The church is located at 2179 State Rd 13, Saint Johns, FL 32259. Flowers may be sent to 405 Chicopee Court in St. Johns, Florida where Tony resided. Or in lieu of flowers, love gifts may be given to the Switzerland Community Church at the above listed address. St Johns Family Funeral Home & Crematory is in charge of arrangements.
Anthony Hoyt "Tony" Reich, age 60, left to join his Heavenly Father on December 26, 2021, at his home in Saint Johns, Florida. Tony was born in Jennings, Louisiana on June 18, 1961, to Robert and Dorothy Reich. He was one of five... View Obituary & Service Information
The family of Anthony Hoyt "Tony" Reich created this Life Tributes page to make it easy to share your memories.
Anthony Hoyt "Tony" Reich, age 60, left to join his Heavenly...
Send flowers to the Reich family.


Even now, thinking back on my experiences at Tokyo Disney Sea, I still can't get over how amazing the park is. There is such a richness in detail both from afar and up close, such grandeur in scale and scope, such hyper-realistic immersion... it is almost overwhelming. Tokyo Disney Sea is a photographer's dream, because pointing the camera anywhere yields a shot of something fantastic. In the American Waterfront, this especially holds true around the area of the S.S. Columbia and the Tower of Terror. The Tower of Terror is what happens when the Japanese get a hold of a rather simple facade and decide they want something more magnificent. So exit the California Spanish style and enter an intricate Moorish Revival style. And the Columbia? Nothing significant... it's just a restaurant and lounge... inside of a large scaled steamliner... that they built because they could.
Tokyo Disney Sea is insane... insanely detailed and beautiful... and I can't wait to go back.


Top Israeli defense official fears Iran election win for reformist leader.
A victory by Muhammad Khatami in the upcoming Iranian presidential elections would likely derail international efforts to stop Iran's race toward nuclear power, a top Israeli defense official involved in those efforts has told The Jerusalem Post. Earlier this week, Khatami, who held the presidency from 1997 to 2005, was reported to be seriously considering running in the June elections. Khatami is a popular leader in the reformist camp and has been a vocal critic of incumbent President Mahmoud Ahmadinejad. Despite Khatami's popularity and Ahmadinejad's economic policy failures, the battle for the office is far from over. The reformist camp in Iran is divided among a number of factions. One of them is led by former parliament speaker Mehdi Karoubi, chairman of the Etemad-e-Meli (National Confidence) party, who has also declared his bid for the presidency. But the Israeli official warned that if Khatami won, he would likely succeed in reducing the level of international pressure on Iran to stop its enrichment of uranium and pursuit of nuclear capability. "People tend to forget that Khatami as president also promoted the nuclear program," the official said. "If he wins, he would succeed in laundering the program in the eyes of the international community. In comparison to Ahmadinejad, he appears more moderate."


text:
Q: Filtering mongodb data I have the following model:
Base class:
public abstract class Identifiable{
    private ObjectId id;
    private string name;

    protected Identifiable(){
        id = ObjectId.GenerateNewId();
    }

    [BsonId]
    public ObjectId Id{
        get { return id; }
        set { id = value; }
    }

    [BsonRequired]
    public string Name{
        get { return name; }
        set { name = value; }
    }
}

The name is unique.
A channel class
 public class Channel : Identifiable{
    private DateTime creationDate;

    private string url;
    private DailyPrograming dailyPrograming;

    public DailyPrograming DailyPrograming{
        get { return dailyPrograming; }
        set { dailyPrograming = value; }
    }

    public DateTime CreationDate{
        get { return creationDate; }
        set { creationDate = value; }
    }

    public string Url{
        get { return url; }
        set { url = value; }
    }
}

Daily programs. The name property is the date stored as ddMMyyyy:
public class DailyPrograming : Identifiable{
    public DailyPrograming(){
        DailyPrograms = new List<Program>(30);
    }

    public IList<Program> DailyPrograms { get; set; }
}

The programs:
public class Program : Identifiable{
    private DateTime programDate;
    private string category;
    private string description;

    public DateTime ProgramDate{
        get { return programDate; }
        set { programDate = value; }
    }

    public string Category{
        get { return category; }
        set { category = value; }
    }

    public string Description{
        get { return description; }
        set { description = value; }
    }
}

Now, I want to filter the program of certain channel for specific date using:
public DailyPrograming GetProgramsForDate(string channelId, string prgDate){
        ObjectId id = new ObjectId(channelId);
        IMongoQuery query = Query.And(Query<Channel>.EQ(c => c.Id, id),
            Query<DailyPrograming>.EQ(dp => dp.Name, prgDate));
        var result = Database.GetCollection<DailyPrograming>(CollectionName).Find(query).FirstOrDefault();
        return result;
    }

But it never returns the existing data. How to retrieve the programings of a channel for a date?
     -

A: According to your sample I used id = "54c00c65c215161c7ce2a77c" and prgDate = "2212015"
then I changed the query to this:
var collection = database.GetCollection<Channel>("test6");
var id = new ObjectId("54c00c65c215161c7ce2a77c");
var query = Query.And(Query<Channel>.EQ(c => c.Id, id), Query<Channel>.EQ(c => c.DailyPrograming.Name, "2212015"));
var result = collection.Find(query).FirstOrDefault();

this query works fine
Some point:

Your collection type is Chanel not DailyPrograming
  When your collection is  Chanel you have to use Query<Channel> and query  nested DailyPrograming via Query<Channel>.EQ(c => c.DailyPrograming.Name, "2212015")

A: var builder = Builders<BsonDocument>.Filter;
var filt = builder.Eq("Price", "9.20")
         & builder.Eq("ProductName", "WH-208");
var list = await collection.Find(filt).ToListAsync();

We can use & instead of $and. See this post, for another example.

redpajama_set_name: RedPajamaStackExchange


For many, Valentine's Day conjures up images of love, romance, flowers and chocolate.
But for detective agencies across the United States, the romantic holiday is a boon for business as it is the ideal time for a spouse to catch a cheating mate.
"Valentine's Day is the biggest day of the year for private investigators," says Tony Delorenzo, of Private Detectives of America, a New Jersey-based company.
Delorenzo and several other sleuths contacted say that in the run-up to the February 14 holiday, they have been overwhelmed with appeals by men and women seeking to find out whether their partner is unfaithful.
He says Valentine's Day is an opportune time to catch a cheating partner as it is the one day lovers will battle all odds to be with each other.
Telltale signs of a cheating spouse are the surprise "business trip" or late office meeting, unexplained gifts and the depressed friend who needs a shoulder to cry on.
"This is the one day when infidelity and extramarital affairs reach their peak," says Ruth Houston, author of Is He Cheating on You: 829 Telltale Signs.
She says men or women who suspect their partners need to be extra alert on Valentine's Day and look for credit-card receipts, new jewellery or lingerie, and unusual appointments so as to nail their partners.
Jeanine Davis (41) says she nabbed her husband with another woman on Valentine's Day last year after hiring a private eye to track him down that evening.
"They followed him and found him having dinner with her at a restaurant, and then they took pictures of them kissing in the parking lot and he spent the night at her place," says Davis, who is in the process of getting a divorce.
Private Detectives of America's Delorenzo says he is working on about 50 infidelity cases this year. He charges between $500 and $1Â 000 to confirm to a client whether a spouse is cheating and $650 for five hours of surveillance.
One spouse who may be in for a Valentine's Day surprise this year is a Michigan man whose wife spotted an e-mail confirming his hotel reservation for Tuesday, when he was supposed to be undergoing a special training session at work.