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\begin{center}
\vskip 1cm{\LARGE\bf 
On the Number of Antichains and \\
\vskip .1in
Antichain Covers of Labeled Sets}
\vskip 1cm
\large
Pawe\l{} Marcin Kozyra \\
Institute of Mathematics\\
Pedagogical University of Krakow\\
Podchor\c{a}\.zych 2 \\
30-084 Krakow \\
Poland\\ 
\href{mailto:pawelmkozyra@gmail.com}{\tt pawelmkozyra@gmail.com} 
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\begin{abstract}
We present a formula for the number of antichains consisting of $m$ subsets of a labeled $n$-element set, as well as a formula for the number of all $m$-antichain covers of a labeled $n$-element set.  We also give a simple formula for the number of antichain covers that are composed of sets of the same cardinality.
\end{abstract}


\section{Introduction and preliminaries}
In 1897 Dedekind \cite{Dedekind} formulated the following question: 
\begin{question}
What is the number of antichains in the power set $\mathcal{P}_n$ of an $n$-element set (ordered by inclusion)?
\end{question}
Dedekind's problem and the related problem of determining the number of all antichain covers of a labeled $n$-element set remained unsolved for a long time \cite{Rival}. These are very difficult---and still baffling---counting problems. Before we formulate these problems precisely, let us define some necessary and fundamental notions. 
\begin{definition}\cite[Def. 1.1.1, p. 2]{Schroder}
An \emph{ordered set} (or partially ordered set or poset) is an ordered pair $(P, \leq)$ of a set $P$ and a binary relation $\leq$ contained in $P \times P$, called the order (or the partial order) on $P$, such that:
\begin{enumerate}
	\item The relation $\leq$ is reflexive, i.e., $\forall_{p\in P} (p\leq p)$.
	\item The relation $\leq$ is antisymmetric, that is, 
	$$\forall_{p,q\in P} \Bigl(\bigl((p\leq q) \wedge (q\leq p)\bigr) \Longrightarrow p=q\Bigr).$$
	\item The relation $\leq$ is transitive, i.e., 
	$$\forall_{p,q,r\in P} \Bigl(\bigl((p\leq q) \wedge (q\leq r)\bigr) \Longrightarrow p\leq r\Bigr).$$
\end{enumerate}
\end{definition}
The power set $\mathcal{P}_n$ of an $n$-element set with the inclusion $\subseteq$ relation is an example of an order set. Another key notion is an antichain defined as follows:
\begin{definition}\cite[Def.\ 2.5.1, p.\ 36]{Schroder} 
An ordered set $P$ is called an \emph{antichain} iff 
$$\forall_{p,q\in P} (p\neq q) \Longrightarrow (\neg p\leq q \wedge \neg q\leq p).$$
\end{definition}
Now we can define the $n$-th Dedekind number and related numbers.
\begin{definition}\cite[Def.\ 2.6.2, p.\ 39]{Schroder}
Let $n\in\mathbb{N}$. We define
\begin{enumerate}
	\item The \emph{$n$-th Dedekind number} $D_n$ to be the number of antichains in $\mathcal{P}_n$.
	\item $T_n$ to be the number of \emph{antichain covers of a labeled $n$-element set}, i.e., the number of antichains in $\mathcal{P}(\{1, \ldots , n\})$ such that the union of these antichains is $\{1, \ldots , n\}$.
\end{enumerate}
\end{definition}
Since the empty set is an antichain, we obtain straightforwardly: $D_0 = T_0 = 1$, $D_1 = 2$, and $T_1 = 1$. 
Hitherto only nine initial values of numbers $D_n$ and $T_n$ have been determined. They are listed in the {\it On-Line Encyclopedia of Integer Sequences} \cite{Sloane} as \seqnum{A000372} and \seqnum{A006126}, respectively. 
Dedekind's problem, in turn, is equivalent to the problem of determining the number of elements in the free bounded distributive lattice on $n$ generators which is the same as the number of monotone Boolean functions in $n$ arguments. $T_n$ is defined primarily in the {\it On-Line Encyclopedia of Integer Sequences} as the number of hierarchical models on $n$ labeled factors or variables with linear terms forced. Kilibarda and Janovic \cite{Jovovic_Kilibarda3} obtained the antichain cover interpretation of \seqnum{A006126}. 

An approach to solving Dedekind's problem and related problems is to break the problem into smaller subproblems. 
There is a simple relation between $D_n$ and $T_n$.
\begin{proposition}\cite[Prop.\ 2.6.3, p.\ 40]{Schroder}
\begin{equation}\label{eq:Schr1}
D_n = \sum_{k=0}^n{n\choose k }T_k.
\end{equation}
\end{proposition}
Any antichain (cover) in $\mathcal{P}_n$ is a Sperner family. Therefore
by Sperner's theorem \cite{Sperner}, for every antichain (cover)
$C=\{A_1,\ldots,A_m\}$ in $\mathcal{P}_n$ we have $m\leq {n\choose \lfloor
n/2\rfloor }$, where $\lfloor x\rfloor:=\max\{k\in\mathbb{Z}\colon k\leq
x\}$ for any real number $x\in\mathbb{R}$. It follows from the above
considerations that if
\begin{itemize}
	\item $D_n^m$ is the number of all $m$-element antichains in $\mathcal{P}_n$ 
	\item $T_n^m$ is the number of all $m$-element antichain covers ($m$-antichain covers) in $\mathcal{P}_n$
\end{itemize}
then
\begin{equation}\label{eq:ACCm}
D_n = \sum_{m=0}^{{n\choose \lfloor n/2\rfloor }} D_n^m\textrm{\quad and \quad}
T_n = \sum_{m=0}^{{n\choose \lfloor n/2\rfloor }} T_n^m
\end{equation}
for all $n\in\mathbb{N}$. 
The number $T_n^2$ of all $2$-antichain covers of a labeled $n$-element set coincides with Stirling numbers of second kind $S(n,3)$ as one can see in Tables \ref{tab:1}, \ref{tab:3}, and \ref{tab:4} as well as in the {\it On-Line Encyclopedia of Integer Sequences} \cite{Sloane} --- the sequence \seqnum{A000392}. 
Then $T_n^2$ can be easily determined by the following result.
\begin{proposition}\cite[Prop.\ 2.6.6, p.\ 40]{Schroder}
\begin{equation}\label{eq:Schr2}
T_n^2 = \frac12\sum_{k=1}^{n-1}{n\choose k}\left(2^k - 1\right),\textrm{ for all $n\geq 2$.}
\end{equation}
\end{proposition}
Similarly,  numbers of 3-antichain and 4-antichain covers of a labeled $n$-set are listed in the {\it On-Line Encyclopedia of Integer Sequences} \cite{Sloane} as the sequences \seqnum{A056046} and \seqnum{A056047}, respectively. Jovovic and Kilibarda \cite{Jovovic_Kilibarda2} consider these sequences in detail. In particular, they provide simple formulas for these sequences. 

In contrast to the problem of determining $T_n$, Nelsen and Schmidt
\cite{Nelsen} relatively easily enumerated the chains in the power set
of $X$. Similarly, Macula \cite{Macula} easily established the number
of all (proper) covers of an $n$-element set as well as the number of
all (proper) covers of length $m$ of an $n$-element set, where $1\leq
m \leq 2^{n}-1$. Clarke \cite{Clarke} considered this problem more
generally. In his approach, for a given $n$-element set $X$ ($n$-set)
and a positive integer $k$, a $k$-cover of $X$ is a collection of $k$
(not necessarily distinct) subsets of $X$, whose union is $X$.  A
$k$-cover is minimal if none of its proper subsets covers $X$. There are
four cases to consider: the element of $X$ may be taken to be identical
($X$ is unlabeled) or distinguishable ($X$ is labeled); the order of
the sets comprising the $k$-cover of $X$ may (cover is ordered) or may
not be material (cover is disordered). Clarke quoted formulas derived
by Hearne and Wagner \cite{Hearne,Wagner} for the number of (minimal)
ordered $k$-covers of a labeled $n$-set, the number of ordered $k$-covers
of an unlabeled $n$-set, and the number of (minimal) disordered $k$-covers
of a labeled $n$-set. Moreover, Clarke \cite{Clarke} presented and proved
a formula for the number of disordered $k$-covers of an unlabeled $n$-set.

Kisielewicz \cite{Kisielewicz} proved that 
\begin{equation}\label{eq:kisiel}
D_n = \sum_{k=1}^{2^{2^n}}\prod_{0\leq i &lt; j \leq 2^n-1} \big(1-b^k_i\cdot b^k_j\cdot f(i,j)\big),
\end{equation}
where $b^n_i = \lfloor n/2^i \rfloor - 2\lfloor n/2^{i+1} \rfloor$ for any $i,n\geq 0$, $\lfloor x\rfloor:=\max\{k\in\mathbb{Z}\colon k\leq x\}$ for any real number $x\in\mathbb{R}$,  
$$f(i,n)=\begin{cases}
1, &amp; \text{if $i\in S(n)$;}\\
0, &amp; \text{otherwise;}\end{cases}
=\begin{cases}
1, &amp; \text{if $i=0$;}\\
\prod_{m=0}^{\log_2(i)} \big(1-b^i_m+b^i_m b^n_m\big), &amp; \text{for $i\geq 1$;}
\end{cases}$$ 
and sets $S(n)$ are defined recursively as follows: 
\begin{align*}
S(0)&amp;=\{0\}\text{ and for $n\geq 0 $:}\\
S(2n)&amp;=\{2s\colon s\in S(n)\}\\
S(2n+1)&amp;=S(2n)\cup\{2s+1\colon s\in S(n)\}.
\end{align*}
Church \cite{Church1} computed the first six elements of the sequence \seqnum{A000372}, Ward \cite{Ward} determined $D_6$, Church \cite{Church2} calculated $D_7$, and Wiedemann \cite{Wiedemann} computed $D_8$. In recent years, Bakoev, Fidytek et al. \cite{Bakoev,Fidytek} tried to improve the computational time. Since obtaining the exact values of Dedekind numbers and related numbers is a very challenging task, many authors tried to determine upper bounds on the Dedekind numbers. 
In 1969 Kleitman \cite{Kleitman} found an upper bound on the logarithm of the Dedekind number. His result was next improved by Kleitman and Markowsky \cite{Kleitman2} in 1975. In 1981, Korshunov \cite{Korshunov} found more sophisticated and accurate estimates. Kahn \cite{Kahn} simplified proofs of these estimates using an `entropy' approach.

Baumann and Strass \cite{Baumann} proved that the number of bipolar Boolean functions in $n$ arguments is equal to $$b(n)=\sum_{i=0}^n 2^i\cdot{n\choose i}\cdot T_i$$ 
and the number of all monotone Boolean functions in $n$ arguments is equal to $D_n$.
A similar result comes from \cite[Prop.\ 2.6.7, p.\ 41]{Schroder}. 
\begin{proposition}
Let $P$ be a finite ordered set. Then the number of antichains in $P$ is equal to the number of order-preserving maps from $P$ into the two-element chain.
\end{proposition}

De Causmaecker and De Wannemacker \cite{Causmaecker} generalized Dedekind's problem to analysis of the number of antichains in intervals $[\alpha,\beta]=\{\chi\in\mathcal{A}_N\colon\alpha\leq\chi\leq\beta\}$ in a completely distributive lattice $(\mathcal{A}_N,\wedge,\vee)$ for any $\alpha,\beta\in\mathcal{A}_N$, where $\mathcal{A}_N$ is the set of all antichains in $N\subseteq\mathbb{N}$. Moreover if $N=\{1,2,\ldots,n\}$ then $\mathcal{A}_N$ is denoted by $\mathcal{A}_n$. Antichains $\alpha,\beta\in\mathcal{A}_N$ are partly ordered as 
$$\alpha\leq\beta\Longleftrightarrow \forall_{A\in\alpha}\exists_{B\in\beta}\colon A\subseteq B.$$
The operators $\wedge$ and $\vee$ are called \textit{meet} and \textit{join}, respectively, and are defined in the following way:
$$\alpha\vee\beta=\max(\alpha\cup\beta),\quad \alpha\wedge\beta=\max(\{A\cap B\colon A\in\alpha,B\in\beta\})$$
The max-operator on an arbitrary set of sets produces an antichain containing only the non dominated sets.
For $X\subseteq N$, $X^{-}=\{X\setminus\{x\}\colon x\in X\}$ and for antichain $\alpha$ denote 
$\alpha^-=\bigvee_{X\in\alpha}X^-$ and $\alpha^+=\bigvee_{X\in 2^N,X^-\leq\alpha}\{X\}$.
For each interval $[\alpha,\beta]$ the underlying interval poset of $[\alpha,\beta]$ is defined as follows 
$$\mathcal{P}_{[\alpha,\beta]}=(\{X\subseteq N\colon \alpha\vee \{X\}\in ]\alpha,\beta]\},\subseteq).$$
Given an interval $[\alpha,\beta]$, with $\alpha&lt;\beta$ so that $\mathcal{P}_{[\alpha,\beta]}\neq\emptyset$,
let $l_0$ denote the size of the smallest sets in $\mathcal{P}_{[\alpha,\beta]}$ and $l_i=l_0+i$. For any $l\in\mathbb{N}$ let $\lambda_l=\{X\in\mathcal{P}_{[\alpha,\beta]}\colon |X|=l\}$. For $\chi\subseteq\lambda_l$ define $\chi^+=\{X\in\lambda_{l+1}\colon X^-\cap\lambda_l\subseteq\chi\}$  and for $l&gt;l_0$, 
$\chi^-=\{X^-\cap\lambda_{l-1}\colon X\in\chi\}$.
De Causmaecker and De Wannemacker \cite{Causmaecker} proved that with the above notation, for $\alpha\leq\beta\in\mathcal{A}_N$, the size of the interval $[\alpha,\beta]$ is given by 
\begin{align}
|[\alpha,\beta]|&amp;=\sum_{\chi_0\subseteq\lambda_{l_0}}\ \sum_{\chi_2\subseteq\chi_0^{++}}
\ \sum_{\chi_4\subseteq\chi_2^{++}}\cdots 2^{|\chi_0^+|-|\chi_2^-|+|\chi_2^+|-|\chi_4^-|\ldots}\label{eq:int1}\\
&amp;=\sum_{\chi_1\subseteq\lambda_{l_1}}\ \sum_{\chi_3\subseteq\chi_1^{++}}
\ \sum_{\chi_5\subseteq\chi_3^{++}}\cdots 2^{|\lambda_{l_0}|-|\chi_1^-|+|\chi_1^+|-|\chi_3^-|\ldots}\label{eq:int2}
\end{align}
Let $\mathcal{B}_n\subseteq\mathcal{A}_n$ denote a basic interval of dimension $n$, i.e., 
$$\mathcal{B}_n=[\{\{1\},\{2\},\ldots,\{n\}\},\{\{1,2,\ldots,n\}\}].$$ Then from the well-known decomposition for the Dedekind numbers 
$D_n = |\mathcal{A}_n|=\sum\limits_{k=0}^n|\mathcal{B}_k|$ and from \eqref{eq:Schr1} it follows that 
$T_n = |\mathcal{B}_n|$ for all natural numbers $n$. Thus both $T_n$ and $D_n$ can be efficiently computed by \eqref{eq:int1} or \eqref{eq:int2} up to $n=6$.

Kilibarda \cite{Kilibarda} presents another generalization of Dedekind's
problem. He defines an antichain as a hypergraph satisfying the following
property: for every pair of distinct edges,
neither one is a subset of the other.
Kilibarda \cite{Kilibarda} enumerates antichains given on an $n$-set
having some of the following properties: being labeled or unlabeled;
being ordered or unordered; being a cover or a proper cover; and finally,
being a $T_0$-, $T_1$- or $T_2$-hypergraph.

The following part of this article is organized as follows. 
The second section is devoted to the derivation of the new elegant result --- Theorem \ref{thm:2}, concerning numbers $D^m_n$ and $T^m_n$ for all $m,n\geq 2$ and $m\leq {n \choose \lfloor n/2\rfloor}$. This theorem is preceded by Example \ref{ex:1}, which allows easier comprehension of the intuition behind Theorem \ref{thm:2}. Proof of Theorem \ref{thm:2} is divided into four Lemmas \ref{f0}--\ref{f3} that explicitly and accurately explain the computational process to obtain the formulae \eqref{eq:thm2:8} and \eqref{eq:thm2:9} from this theorem.

The third section concerns the number of covers of a labeled $n$-element set $X$ with antichains consisting of $m$ equinumerous subsets of $X$. 

The last section contains Example \ref{ex:2} demonstrating the use of Theorem \ref{thm:2} for computing the number of all $2$ and $3$-antichain covers as well as the number of all $2$ and $3$ element antichains in the power set $\mathcal{P}_n$.
In this section, some numerical experiments are also shown. In particular, this section presents:
\begin{itemize}
	\item values of \seqnum{A016269} --- sequence of numbers of monotone Boolean functions of $n-2$ variables with $2$ mincuts,  that is, sequence of numbers $D_n^2$ for $n\in\{4,\ldots,15\}$;
	\item values of sequences \seqnum{A047707} and \seqnum{A051112} of numbers of monotone Boolean functions of $n$ variables with $3$ and $4$ mincuts, respectively,  that is, sequences of numbers $D_n^3$ and $D_n^4$ for $n\in\{4,\ldots,15\}$;
	\item values of sequence \seqnum{A000392} of Stirling numbers of second kind $S(n+1,3)$, i.e., sequence of numbers $T_n^2$ for $n\in\{4,\ldots,50\}$;
	\item values of sequence \seqnum{A056046} of numbers $T_n^3$ for $n\in\{4,\ldots,50\}$;
	\item values of sequence \seqnum{A056047} of numbers $T_n^4$ for $n\in\{4,\ldots,15\}$;
	\item the numbers of covers of $7$-set with antichains consisting of equinumerous sets.
\end{itemize}


\section{Antichains and antichain covers consisting of $m$ subsets of a labeled $n$-element set}
Let $\mathbb{Z}_n$ denote the set $\{0,1,\ldots,n-1\}$, and we let $\mathbb{Z}^+_n$ denote the set $\{1,2,\ldots,n\}$, for all $n\in\mathbb{N}$. Let us denote the cardinality of a set $X$ by $\#X$.
Let $\mathcal{A}^m_n$ denote the set of all $m$-element antichains in the power set $\mathcal{P}_n$, and we let $\mathcal{AC}^m_n$ denote the set of all $m$-element antichain covers in $\mathcal{P}_n$.

The following lemma will be used later in this article. In particular, the lemma allows us to determine $T_n$ from $D_0,\ldots,D_n$ using formula \eqref{eq:Schr1}. 
\begin{lemma}\label{lem:2}
Let $\mathbb{N}_0=\mathbb{N}\cup\{0\}$ and $(a_n)_{n\in\mathbb{N}_0},(b_n)_{n\in\mathbb{N}_0}\in\mathbb{R}^{\mathbb{N}_0}$ be sequences such that 
\begin{equation}\label{eq:lem:2.1}
b_n=\sum_{k=n_0}^n {n\choose k} a_k
\end{equation}
for some $n_0\in\mathbb{N}_0$ and for all $n\geq n_0$. Then 
\begin{equation}\label{eq:lem:2.2}
a_n=\sum_{k=0}^{n-n_0}{n\choose n-k}\cdot(-1)^k\cdot b_{n-k}
\end{equation}
for all natural numbers $n\geq n_0$.
\end{lemma}
This lemma follows from the following more general result:
let $X$ be a finite set, and let $f,g\colon 2^X \to \mathbb{N}_0$ be two functions such that 
$$\forall_{A\subseteq X}\colon g(A) = \sum_{B\subseteq A} f(B).$$
Then $$\forall_{A\subseteq X} f(A) = \sum_{B\subseteq A} (-1)^{\#A - \#B} g(B).$$
We obtain Lemma \ref{lem:2} by putting 
\begin{align*}
f(A) &amp;:= \begin{cases}
0, &amp; \text{if $\#A&lt;n_0$;}\\
a_n, &amp; \text{if $\#A = n\geq n_0$;}
\end{cases}\\
g(A) &amp;:= \begin{cases}
0, &amp; \text{if $\#A&lt;n_0$;}\\
b_n, &amp; \text{if $\#A = n\geq n_0$;}
\end{cases},
\end{align*}
but we can also prove this lemma by an inductive procedure.
\begin{proof}
Equation \eqref{eq:lem:2.2} is satisfied for $n=n_0$, since by \eqref{eq:lem:2.1}, $b_{n_0}=a_{n_0}$. Assume inductively that equation \eqref{eq:lem:2.2} holds for all natural numbers $m$ such that $n_0\leq m\leq n$ for some natural number $n\geq n_0$. It will be shown that \eqref{eq:lem:2.2} is true for $n+1$, i.e., 
\begin{equation}\label{eq:lem:2.3}
a_{n+1} = b_{n+1} - \sum_{k=0}^{n-n_0}{n+1\choose n-k}\cdot(-1)^k\cdot b_{n-k}
\end{equation}
By \eqref{eq:lem:2.1}, $b_{n+1}=\sum_{k=n_0}^n {n+1\choose k} a_k + a_{n+1}$,
and hence by the induction hypothesis, we get 
$$a_{n+1} = b_{n+1} - \sum_{k=n_0}^{n} {n+1\choose k} \ \sum_{i=0}^{k-n_0}{k\choose k-i}\cdot(-1)^i\cdot b_{k-i}.$$
Hence after putting $s:=k-i$ we obtain
$$a_{n+1}= b_{n+1} - \sum_{k=n_0}^n {n+1\choose k }\ \sum_{s=n_0}^k {k\choose s} (-1)^{k-s} b_s.$$
After changing the order of summation, we have
$$a_{n+1}= b_{n+1} - \sum_{s=n_0}^n \ \sum_{k=s}^n {n+1\choose k}{k\choose s}(-1)^{k-s}b_s.$$
Thus, after putting $t:=n-s$, we obtain
\begin{equation}
a_{n+1}= b_{n+1} - \sum_{t=0}^{n-n_0}\ \sum_{k=n-t}^n {n+1\choose k}{k\choose n-t}(-1)^{k+t-n}b_{n-t}\label{eq:lem:2.4}
\end{equation}
We shall show that
\begin{equation}\label{eq:lem:2.5}
\sum_{k=n-t}^n {n+1\choose k}{k\choose n-t}(-1)^{k-n}={n+1\choose n-t} \textrm{, for all $t\in\{0,\ldots,n-n_0\}$.}
\end{equation}
Note that 
\begin{align*}
&amp;\sum_{k=n-t}^n {n+1\choose k}{k\choose n-t}(-1)^{k-n}\\
&amp;=\sum_{k=n-t}^n \frac{(n+1)!}{k!(n+1-k)!}\cdot \frac{k!}{(n-t)!(k+t-n)!}(-1)^{k-n}\\
&amp;=\sum_{k=n-t}^n \frac{(n+1)!}{(n-t)!(t+1)!}\cdot \frac{(t+1)!}{(n+1-k)!(k+t-n)!}(-1)^{k-n} \\
&amp;=\sum_{l=0}^t {n+1\choose n-t}{t+1\choose l+1}(-1)^{l}
\quad \text{(by setting $l:=n-k$)} \\
&amp;= {n+1\choose n-t},
\end{align*}
since $0=(1-1)^{t+1}=\sum_{l=0}^{t+1}{t+1\choose l}(-1)^l.$
By \eqref{eq:lem:2.4} and \eqref{eq:lem:2.5}, we obtain \eqref{eq:lem:2.3}. So by induction, \eqref{eq:lem:2.2} holds for all $n\geq n_0$.
\end{proof}

For small numbers $n$, formulae \eqref{eq:Schr1} and \eqref{eq:kisiel}, as well as Lemma \ref{lem:2} can be used to compute $T_n$:
\begin{corollary}
We have
$$T_n =  \sum_{k=0}^n {n\choose n-k}(-1)^k 
\ \sum_{l=1}^{2^{2^{n-k}}}\ \prod_{0\leq i &lt; j \leq 2^{n-k}-1} \big(1-b^l_i\cdot b^l_j\cdot f(i,j)\big),$$
where $b^n_i = \lfloor n/2^i \rfloor - 2\lfloor n/2^{i+1} \rfloor$ for any $i,n\geq 0$ and
$$f(i,n)=\begin{cases}
1, &amp; \text{if $i=0$;}\\
\prod_{m=0}^{\log_2(i)} \big(1-b^i_m+b^i_m b^n_m\big), &amp; \text{for $i\geq 1$.}
\end{cases}$$ 
\end{corollary}

The following theorem allows computing the number of all $m$-element antichains and antichain covers in the power set $\mathcal{P}_n$. Before formulating this theorem the simple example below is presented. It allows us easier comprehension of the intuition behind Theorem \ref{thm:2}.
\begin{example}\label{ex:1}
We now demonstrate the method for computing $T^m_n$, as well as $D^m_n$ for $m\in\{2,3\}$ and for all natural numbers $n\geq 3$. Theorem \ref{thm:2} is a generalization of this method for all natural numbers $n\geq 2$ and $2\leq m\leq {n\choose \lfloor n/2\rfloor}$.

Firstly, we shall show that for all sets $\emptyset \neq A^1, A^2 \subsetneq\mathbb{Z}^+_n$  if 
\begin{equation}\label{eq:ex0}
\#A^1 \leq \#A^2,
\end{equation}
then $\{A^1, A^2\}\in \mathcal{A}^2_n$, if and only if $A^2$ can be decomposed into two subsets  $A^2_1$ and $A^2_2$ of $\mathbb{Z}^+_n$ satisfying the following conditions:
\begin{align}
A^2 &amp;= A^2_1\cup A^2_2\label{eq:ex01}\\
A^2_1 &amp;\subsetneq A^1\label{eq:ex02}\\
A^2_2 &amp;\subseteq \mathbb{Z}^+_n\setminus A^1\label{eq:ex03}
\end{align}
Assume that $\{A^1, A^2\}\in \mathcal{A}^2_n$ and put $A^2_1:=A^2\cap A^1$ and $A^2_2:=A^2\cap (\mathbb{Z}^+_n\setminus A^1)$. Then \eqref{eq:ex01} and \eqref{eq:ex03} are straightforwardly satisfied. Moreover, $A^2_1\subseteq A^1$. If $A^2_1$ were equal to $A^1$, then $A^1$ would be included in $A^2$ and it would contradict the assumption $\{A^1, A^2\}\in \mathcal{A}^2_n$. Thus \eqref{eq:ex02} holds. For the proof of the converse implication assume that $A^1, A^2$ satisfy 
conditions \eqref{eq:ex01}--\eqref{eq:ex03}. It follows from \eqref{eq:ex02}, that there exists $x\in A^1\setminus A^2_1$. Then by \eqref{eq:ex03}, such $x\notin A^2_2$ and by \eqref{eq:ex01}, $x\notin A^2$. So $A^1\nsubseteq A^2$. Similarly, $A^2\nsubseteq A^1$, since otherwise by \eqref{eq:ex01} and \eqref{eq:ex03}, $A^2_2$ would have to be equal to $\emptyset$, but then $A^2 = A^2_1$ and by \eqref{eq:ex02}, $\#A^2 &lt; \#A^1$ --- a contradiction to our primary assumption \eqref{eq:ex0}.

Analogously, for all sets $\emptyset \neq A^1, A^2 \subsetneq\mathbb{Z}^+_n$ if  \eqref{eq:ex0} is satisfied, then $\{A^1, A^2\}\in \mathcal{AC}^2_n$ if and only if $A^2$ can be decomposed into two subsets  $A^2_1$ and $A^2_2$ of $\mathbb{Z}^+_n$ satisfying conditions \eqref{eq:ex01}--\eqref{eq:ex02}. However, instead of condition \eqref{eq:ex03} $A^2_2$ satisfies the condition:
\begin{equation}\label{eq:ex03'}
A^2_2 = \mathbb{Z}^+_n\setminus A^1.
\end{equation}

If $\emptyset \neq A^1, A^2 \subsetneq\mathbb{Z}^+_n$ and $A^1$, $A^2$, $A^2_1$, $A^2_2$ satisfy conditions \eqref{eq:ex0} as well as \eqref{eq:ex01}--\eqref{eq:ex03} and we denote $\#A^1$ by $a^1$, $\#A^2$ by $a^2$, and $\#A^2_i$ by $a^2_i$ for $i\in\{1,2\}$, then we obtain the following bounds:
\begin{equation}\label{eq:ex1}
\left\{\begin{array}{lllllll}
1 &amp; \leq &amp; a^1 &amp; \leq &amp; a^2 &amp; \leq &amp; n-1\\
&amp;&amp; 0 &amp; \leq &amp; a^2_1 &amp; \leq &amp; a^1 -1\\
&amp;&amp; 0 &amp; \leq &amp; a^2_2 &amp; \leq &amp; n - a^1 \textrm{ (or $a^2_2 = n - a^1$ for $\mathcal{AC}^2_n$)}
\end{array}\right.\end{equation}
Then $a^2_1\geq a^1 - a^2_2 \geq a^1-(n-a^1) = 2a^1 - n$ and $a^2_2 \geq a^1 - a^2_1\geq a^1 - (a^1-1) = 1$.

If $a^1 &lt; a^2$ then the number of all families of sets $\{A^1, A^2\}\in \mathcal{A}^2_n$, such that $A^1$, $A^2$, $A^2_1$, $A^2_2$ satisfy conditions \eqref{eq:ex01}--\eqref{eq:ex03} as well as $a^1 = \#A^1$, $a^2 = \#A^2$, $a^2_1 = \#A^2_1$, and $a^2_2 = \#A^2_2$, equals ${n \choose a^1}{a^1 \choose a^2_1}{n - a^1 \choose a^2_2}$. Otherwise, i.e., if $a^1 = a^2$ then the number of all such families of sets $\{A^1, A^2\}\in \mathcal{A}^2_n$, equals ${n \choose a^1}{a^1 \choose a^2_1}{n - a^1 \choose a^2_2}/2$, since $\{A^1, A^2\} = \{A^2, A^1\}$. However, if $\{A^1, A^2\}\in \mathcal{AC}^2_n$, then $a^2_2 = n - a^1$. Hence we obtain the following formulas:
\begin{align}
D^2_n &amp;= \sum_{a^1=1}^{n-1}\ \sum_{a^2_1 = \max(0, 2a^1-n)}^{a^1-1} \ \sum_{a^2_2 = a^1 - a^2_1}^{n-a^1}
\frac{{n \choose a^1}{a^1 \choose a^2_1}{n - a^1 \choose a^2_2}}{r(a^1, a^2)}\label{eq:ex2a}\\
T^2_n &amp;= \sum_{a^1=1}^{n-1}\ \sum_{a^2_1 = \max(0, 2a^1-n)}^{a^1-1} 
\frac{{n \choose a^1}{a^1 \choose a^2_1}}{r(a^1, a^2)},\label{eq:ex2b}
\end{align}
where $r(a,b)=\begin{cases}
1, &amp;\text{if $a&lt;b$;}\\
2, &amp;\text{otherwise.}
\end{cases}$

Similarly, we shall show that for all sets $\emptyset \neq A^1, A^2, A^3 \subsetneq\mathbb{Z}^+_n$ if  
\begin{equation}\label{eq:ex20}
\#A^1 \leq \#A^2 \leq \#A^3,
\end{equation}
then $\{A^1, A^2, A^3\}\in \mathcal{A}^3_n$, if and only if $A^2$ can be decomposed into two sets satisfying conditions \eqref{eq:ex01}--\eqref{eq:ex03} and $A^3$ can be decomposed into four subsets  $A^3_1$, $A^3_2$, $A^3_3$, $A^3_4$ of $\mathbb{Z}^+_n$ satisfying the following conditions:
\begin{align}
A^3 &amp;= A^3_1\cup A^3_2\cup A^3_3\cup A^3_4 \label{eq:ex201}\\
A^3_1&amp;\subseteq A^2_1 \label{eq:ex202}\\
A^3_2&amp;\subseteq A^1\setminus A^2_1 \label{eq:ex203}\\
A^3_3&amp;\subseteq A^2_2 \label{eq:ex204}\\
A^3_4 &amp;\subseteq \mathbb{Z}^+_n\setminus A^1\setminus A^2_2 \label{eq:ex205}\\
A^3_1\cup A^3_2&amp;\subsetneq A^1 \label{eq:ex206}\\
A^3_1\cup A^3_3&amp;\subsetneq A^2 \label{eq:ex207}
\end{align}
Assume that $\{A^1, A^2, A^3\}\in \mathcal{A}^3_n$. Then $\{A^1, A^2\}\in \mathcal{A}^2_n$. From the previous reasoning it follows that $A^2$ can be decomposed into two sets $A^2_1$ and $A^2_2$ satisfying conditions \eqref{eq:ex01}--\eqref{eq:ex03}. Put $A^3_1: = A^3\cap A^2_1$, $A^3_2 := A^3\cap(A^1\setminus A^2_1)$, 
$A^3_3 := A^3 \cap A^2_2$ and $A^3_4 := A^3\cap(\mathbb{Z}^+_n\setminus A^1\setminus A^2_2)$. Then straightforwardly conditions \eqref{eq:ex202}--\eqref{eq:ex205} are satisfied. 
Moreover, $A^3_1\cup A^3_2 = A^3\cap A^1$ and $A^3_3\cup A^3_4 = A^3 \setminus A^1$. So condition \eqref{eq:ex201} also holds and $A^3_1\cup A^3_2 \subseteq A^1$. Suppose that $A^3_1\cup A^3_2 = A^1$, then $A^3\cap A^1 = A^1$. Hence $A^1\subseteq A^3$ --- a contradiction to the assumption $\{A^1, A^2, A^3\}\in \mathcal{A}^3_n$. Therefore \eqref{eq:ex206} is true. By \eqref{eq:ex01}, $A^3_1\cup A^3_3 = A^3\cap A^2$. Thus If $A^3_1\cup A^3_3 = A^2 $, then $A^2\subseteq A^3$ --- a contradiction to the assumption $\{A^1, A^2, A^3\}\in \mathcal{A}^3_n$. Therefore condition \eqref{eq:ex207} is also satisfied. 

Assume that $A^2$ can be decomposed into two sets satisfying conditions \eqref{eq:ex01}--\eqref{eq:ex03} and $A^3$ can be decomposed into four subsets satisfying conditions \eqref{eq:ex201}--\eqref{eq:ex207}. We shall prove that $\{A^1, A^2, A^3\}\in \mathcal{A}^3_n$. By the previous reasoning $\{A^1, A^2\}\in \mathcal{A}^2_n$. By the assumption $\#A^1 \leq \#A^2 \leq \#A^3$ it suffices to show that $A^1\nsubseteq A^3$ and $A^2\nsubseteq A^3$. 
By \eqref{eq:ex03},\eqref{eq:ex201}, and \eqref{eq:ex204}--\eqref{eq:ex206}, $A^3\cap A^1  = A^3_1\cup A^3_2 \neq A^1$. So $A^1 \nsubseteq A^3$. Similarly, by \eqref{eq:ex01}--\eqref{eq:ex03}, \eqref{eq:ex201}--\eqref{eq:ex205}, $A^3\cap A^2 = A^3_1\cup A^3_3$. Hence by \eqref{eq:ex207}, $A^3\cap A^2 \neq A^2$. So $A^2 \nsubseteq A^3$.

Analogously, for all sets $\emptyset \neq A^1, A^2, A^3 \subsetneq\mathbb{Z}^+_n$ if condition \eqref{eq:ex20} is satisfied, then $$\{A^1, A^2, A^3\}\in\mathcal{AC}^3_n$$
if and only if $A^2$ can be decomposed into two sets satisfying conditions \eqref{eq:ex01}--\eqref{eq:ex03} as well as $A^3$ can be decomposed into four subsets  $A^3_1, A^3_2, A^3_3, A^3_4$ of $\mathbb{Z}^+_n$ satisfying conditions \eqref{eq:ex201}--\eqref{eq:ex207}. However, instead of condition \eqref{eq:ex205} $A^3_4$ satisfies the condition:
\begin{equation}\label{eq:ex205'}
A^3_4 = \mathbb{Z}^+_n\setminus A^1\setminus A^2_2.
\end{equation}
If $A^1$, $A^2$, $A^3$, $A^2_1$, $A^2_2$, $A^3_1$, $A^3_2$, $A^3_3$, $A^3_4$ satisfy conditions \eqref{eq:ex20}, \eqref{eq:ex01}--\eqref{eq:ex03} and \eqref{eq:ex201}--\eqref{eq:ex207}, we adopt the notation from the previous paragraph and we denote $\#A^3_i$ by $a^3_i$ for $i\in\{1,2,3,4\}$, then we obtain the following bounds for $a^3$ and $a^3_i$:\begin{equation}\label{eq:ex21}
\left\{\begin{array}{lllll}
0 &amp; \leq &amp; a^3_1 &amp; \leq &amp; a^2_1\\
0 &amp; \leq &amp; a^3_2 &amp; \leq &amp; a^1 - a^2_1\\
0 &amp; \leq &amp; a^3_3 &amp; \leq &amp; a^2_2\\
0 &amp; \leq &amp; a^3_4 &amp; \leq &amp; n - a^1 - a^2_2 \textrm{ (or $a^3_4 = n - a^1 - a^2_2$ for $\mathcal{AC}^3_n$)}\\
 &amp; &amp; a^3_1 + a^3_2 &amp; \leq &amp; a^1 - 1\\
 &amp; &amp; a^3_1 + a^3_3 &amp; \leq &amp; a^2 - 1 \\
a^2 &amp; \leq &amp; a^3 &amp; \leq &amp; n-1
\end{array}\right.\end{equation}
If $a^1 &lt; a^2 &lt; a^3$ then the number of all families of sets $\{A^1, A^2, A^3\}\in \mathcal{A}^3_n$, such that $A^1$, $A^2$, $A^3$, $A^2_1$, $A^2_2$, $A^3_1$, $A^3_2$, $A^3_3$, $A^3_4$ satisfy conditions \eqref{eq:ex01}--\eqref{eq:ex03} and \eqref{eq:ex201}--\eqref{eq:ex207}, as well as $a^1 = \#A^1$, $a^2 = \#A^2$, $a^2_1 = \#A^2_1$, $a^2_2 = \#A^2_2$, $a^3_i = \#A^3_i$ for $i\in\{1,2,3,4\}$, equals 
$${n \choose a^1}{a^1 \choose a^2_1}{n - a^1 \choose a^2_2}
{a^2_1 \choose a^3_1}{a^1 - a^2_1 \choose a^3_2}{a^2_2 \choose a^3_3}{n - a^1 - a^2_2 \choose a^3_4}.$$
If $a^1 &lt; a^2 = a^3$ or $a^1 = a^2 &lt; a^3$ then the number of all such families of sets $\{A^1, A^2, A^3\}\in \mathcal{A}^3_n$, equals 
$${n \choose a^1}{a^1 \choose a^2_1}{n - a^1 \choose a^2_2}
{a^2_1 \choose a^3_1}{a^1 - a^2_1 \choose a^3_2}{a^2_2 \choose a^3_3}{n - a^1 - a^2_2 \choose a^3_4}/2.$$
If $a^1 = a^2 = a^3$ then the number of all such families of sets $\{A^1, A^2, A^3\}\in \mathcal{A}^3_n$, equals 
$${n \choose a^1}{a^1 \choose a^2_1}{n - a^1 \choose a^2_2}
{a^2_1 \choose a^3_1}{a^1 - a^2_1 \choose a^3_2}{a^2_2 \choose a^3_3}{n - a^1 - a^2_2 \choose a^3_4}/3!.$$
However, if $\{A^1, A^2, A^3\}\in \mathcal{AC}^3_n$, then $a^3_4 = n - a^1 - a^2_2$. Hence we obtain the following formulae:
\begin{align}
D^3_n &amp;= \sum_{a^1=1}^{n-1}\ \sum_{a^2_1 = \max(0, 2a^1-n)}^{a^1-1} \ \sum_{a^2_2 = a^1 - a^2_1}^{n-a^1}
\ \sum_{(a^3_1,a^3_2,a^3_3,a^3_4) \textrm{ satisfying conditions \eqref{eq:ex21}}}\nonumber\\
&amp;\frac{{n \choose a^1}{a^1 \choose a^2_1}{n - a^1 \choose a^2_2}
{a^2_1 \choose a^3_1}{a^1 - a^2_1 \choose a^3_2}{a^2_2 \choose a^3_3}{n - a^1 - a^2_2 \choose a^3_4}}{r(a^1, a^2, a^3)}\label{eq:ex22a}\\
T^3_n &amp;= \sum_{a^1=1}^{n-1}\ \sum_{a^2_1 = \max(0, 2a^1-n)}^{a^1-1} 
\ \sum_{(a^3_1,a^3_2,a^3_3,a^3_4) \textrm{ satisfying conditions \eqref{eq:ex21}}}\nonumber\\
&amp;\frac{{n \choose a^1}{a^1 \choose a^2_1}{n - a^1 \choose a^2_2}
{a^2_1 \choose a^3_1}{a^1 - a^2_1 \choose a^3_2}{a^2_2 \choose a^3_3}}{r(a^1, a^2, a^3)},\label{eq:ex22b}
\end{align}
where $r(a,b,c)=\begin{cases}
1, &amp;\text{if $a&lt;b&lt;c$;}\\
2, &amp;\text{if $a&lt;b=c$ or $a=b&lt;c$;}\\
3!, &amp;\text{if $a=b=c$.}
\end{cases}$.
\end{example}
\begin{theorem}\label{thm:2}
Let 
\begin{equation}\label{eq:thm2:1}
I^m_i:=\bigcup_{k=0}^{2^{i-1}-1}\left[2k\cdot 2^{m-1-i}+1;(2k+1)\cdot 2^{m-1-i}\right]\cap\mathbb{N}
\end{equation}
for $m\geq 2$ and $i\in\mathbb{Z}^+_{m-1}$. 
For a natural number $n$ and sequence of sequences of natural numbers $(\underline{\textbf{a}}^1,\underline{\textbf{a}}^2,\ldots)$ such that $\underline{\textbf{a}}^m=(a^m_1,\ldots,a^m_{2^{m-1}})\in\mathbb{N}_0^{2^{m-1}}$ 
for $m\in\mathbb{N}$ define:
\begin{align}
a^m &amp;=\sum_{k=1}^{2^{m-1}}a^m_k\label{eq:thm2:2}\\
b^1_1&amp;:=n\label{eq:thm2:3}\\ 
b^m_k&amp;:=\begin{cases}
a^{m-1}_{(k+1)/2}, &amp;\text{if $2\nmid k$}\\
b^{m-1}_{k/2} - a^{m-1}_{k/2}, &amp;\text{if $2\mid k$}
\end{cases},\textrm{for $m\geq 2$ and $k\in\mathbb{Z}^+_{2^{m-1}}$}\label{eq:thm2:4}\\
S^n_m(\underline{\textbf{a}}^1,\ldots,\underline{\textbf{a}}^{m-1})&amp;:=\left\{\begin{array}{ll}
\underline{\textbf{a}}^m\in\mathbb{N}_0^{2^{m-1}}\colon 0\leq a^m_k\leq b^m_k, &amp; \textrm{for $k\in\mathbb{Z}^+_{2^{m-1}}$}\&amp;\\
a^{m-1}\leq a^{m}\leq n-1 &amp;\&amp;\\
\sum_{k\in I^m_i} a^m_k \leq a^i-1, &amp;\textrm{for $i\in\mathbb{Z}^+_{m-1}$}
\end{array}\right\},\label{eq:thm2:5}\\
&amp;\textrm{for $m\geq 2$}\nonumber\\
\overline{S^n_m}(\underline{\textbf{a}}^1,\ldots,\underline{\textbf{a}}^{m-1})&amp;:=\left\{
\underline{\textbf{a}}^m\in S^n_m(\underline{\textbf{a}}^1,\ldots,\underline{\textbf{a}}^{m-1})\colon a^m_{2^{m-1}}=b^m_{2^{m-1}}\right\},\textrm{ for $m\geq 2$}\label{eq:thm2:6}
\end{align}
For a non-decreasing sequence of natural numbers $(a_1,\ldots,a_m)$, natural numbers $m_1,\ldots, m_k$, and $M_1,\ldots,M_k$ such that:
\begin{enumerate}
	\item $m_1+m_2+\cdots+m_k=m$
	\item $M_1=0$
	\item $M_i:=m_1+\cdots+m_{i-1}$ for $2\leq i \leq k$
	\item $\forall_{i\in\mathbb{Z}^+_k}\forall_{j\in\mathbb{Z}^+_{m_i}} a_{M_i+j}=a_{M_i+1}$
	\item $\forall_{i\in\{2,\ldots,k\}} a_{M_i}&lt;a_{M_i+1}$.
\end{enumerate}
define function 
\begin{equation}\label{eq:thm2:7}
r(a_1,\ldots,a_m):=\prod_{i=1}^k m_i!.
\end{equation}
With the above notation, for all natural numbers $m,n\geq 2$ if $m\leq {n\choose \lfloor n/2\rfloor}$ then:
\begin{align}
D^m_n&amp;=\sum_{a^1=1}^{n-1}\ \sum_{\underline{\textbf{a}}^2\in S^n_2(a^1)}\ldots
\ \sum_{\underline{\textbf{a}}^{m-1}\in S^n_{m-1}(\underline{\textbf{c}}^{m-2})}
\ \sum_{\underline{\textbf{a}}^m\in S^n_m(\underline{\textbf{c}}^{m-1})}
\prod_{i=1}^m\prod_{j=1}^{2^{i-1}}\frac{{b^i_j\choose a^i_j}}{r(a^1,\ldots,a^m)}\label{eq:thm2:8}\\
T^m_n&amp;=\sum_{a^1=1}^{n-1}\ \sum_{\underline{\textbf{a}}^2\in S^n_2(a^1)}\ldots
\ \sum_{\underline{\textbf{a}}^{m-1}\in S^n_{m-1}(\underline{\textbf{c}}^{m-2})}
\ \sum_{\underline{\textbf{a}}^m\in \overline{S^n_m}(\underline{\textbf{c}}^{m-1})}
\prod_{i=1}^m\prod_{j=1}^{2^{i-1}}\frac{{b^i_j\choose a^i_j}}{r(a^1,\ldots,a^m)}\label{eq:thm2:9}
\end{align}
where $\underline{\textbf{c}}^{k} = \left(\underline{\textbf{a}}^{1},\underline{\textbf{a}}^{2},\ldots,\underline{\textbf{a}}^{k}\right)$.
\end{theorem}
\begin{remark}\label{rem:1}
The set $I^m_i$ defined in \eqref{eq:thm2:1} for $m\geq 2$ and $i\in\mathbb{Z}^+_{m-1}$ is intuitively the set of indices of sets belonging to the partition $\{A^m_1,\ldots,A^m_{2^{m-1}}\}$ of set $A^m$, which are included in $A^i$. Properties of these partitions are described by \eqref{eq:ex01}--\eqref{eq:ex03} and \eqref{eq:ex201}--\eqref{eq:ex207} in Example \ref{ex:1} for $A^2$ and $A^3$, respectively, as well as in Lemma \ref{f3} occurring later in this article. So the condition $\sum_{k\in I^m_i} a^m_k \leq a^i - 1$ from \eqref{eq:thm2:5} can be translated in the context of these partitions as $\bigcup_{k\in I^m_i} A^m_k \subsetneq A^i$ and it prevents $A^i \subseteq A^m$ for all $i\in\mathbb{Z}^+_{m-1}$. Thus this condition guarantees that sets $A^1,\ldots, A^m$, consisting of these partitions form an antichain.
\end{remark}

The following proposition gives an explicit formula for expressions defined recursively by \eqref{eq:thm2:3} and \eqref{eq:thm2:4}.
\begin{proposition}\label{prop:1}
Let $m,n\in\mathbb{N}$, $a^0_1:=n$ and $b^m_k$ be defined by \eqref{eq:thm2:3} and \eqref{eq:thm2:4}. Then 
\begin{equation}\label{eq:prop1:1}
b^m_k=a^{m-l-1}_{(r+1)/2} - \sum\limits_{i=1}^l a^{m-i}_{2^{l-i}r}.
\end{equation}
for all $r\in\mathbb{N}$ and $l\in\mathbb{Z}_{m}$ such that $2\nmid r$, $k=2^l\cdot r$, and $k\leq 2^{m-1}$.
\end{proposition}
\begin{proof}
The proof is by induction on $m$.
For the base step of the induction, assume that $m=1$. 
Then $k=1=2^0$ and by \eqref{eq:thm2:3}, 
$b^1_1 = n = a^0_1 = a^{m-l-1}_{(r+1)/2}$ for $l=0$ and $r=1$.

Assume that \eqref{eq:prop1:1} holds for some $m\in\mathbb{N}$.
We shall show that \eqref{eq:prop1:1} holds for $m+1$. We shall consider two cases.
Firstly, let us assume that $2\nmid k$. Then by \eqref{eq:thm2:4}, 
$b^{m+1}_k = a^m_{(k+1)/2} = a^{m+1-l-1}_{(r+1)/2}$ for $l=0$ and $r=k$. Therefore in this case \eqref{eq:prop1:1} is true. 

Assume that $k=2^l\cdot r$, where $0&lt;l\leq m$ and $2\nmid r$. Then by \eqref{eq:thm2:4}, 
\begin{equation}\label{eq:prop1:2}
b^{m+1}_k = b^m_{2^{l-1}\cdot r} - a^m_{2^{l-1}\cdot r}.
\end{equation}
By the induction hypothesis, 
\begin{equation}\label{eq:prop1:3}
b^m_{2^{l-1}\cdot r} = a^{m-(l-1)-1}_{(r+1)/2} - \sum_{i=1}^{l-1} a^{m-i}_{2^{l-1-i}r}.
\end{equation}
Therefore by \eqref{eq:prop1:2} and \eqref{eq:prop1:3}, 
\begin{align*}
b^{m+1}_k &amp;= a^{m-(l-1)-1}_{(r+1)/2} - \sum_{i=1}^{l-1} a^{m-i}_{2^{l-1-i}r} -  a^m_{2^{l-1}\cdot r}\\
&amp;= a^{m+1-l-1}_{(r+1)/2} - \sum_{i=1}^{l} a^{m+1-i}_{2^{l-i}r}.
\end{align*}
So \eqref{eq:prop1:1} holds for $m+1$, and by an inductive procedure, \eqref{eq:prop1:1} is true for all $m\in\mathbb{N}$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:2}]
\begin{lemma}\label{f0}
Let $I^m_i$ be defined by \eqref{eq:thm2:1}. Then 
\begin{equation}\label{eq:p0}
\bigcup_{i\in\mathbb{Z}^+_{m-1}} I^m_i = \mathbb{Z}^+_{2^{m-1}}\setminus\left\{2^{m-1}\right\}
\end{equation}
for all natural numbers $m\geq 2$.
\end{lemma}
\begin{proof}
Fix any natural number $m\geq 2$ and $x\in \mathbb{Z}^+_{2^{m-1}}\setminus\left\{2^{m-1}\right\}$. 
Let 
\begin{equation}\label{eq:f00}
i:=m-1-\lfloor \log_2(2^{m-1}-x)\rfloor 
\end{equation}
and $k:=2^{i-1}-1$. Note that $i\in\mathbb{Z}^+_{m-1}$.
We shall show that $$k\cdot 2^{m-i} &lt; x \leq k\cdot 2^{m-i} + 2^{m-i-1}:$$
Note that 
\begin{equation}\label{eq:f01}
k\cdot 2^{m-i} = 2^{m-1}-2\cdot 2^{\lfloor \log_2(2^{m-1}-x)\rfloor}, 
\end{equation}
and 
\begin{align}
2^{m-1}-2\cdot 2^{\lfloor \log_2(2^{m-1}-x)\rfloor} &amp;&lt; 2^{m-1}-2\cdot 2^{\log_2(2^{m-1}-x)-1} = x\label{eq:f02}\\
2^{m-1}-2\cdot 2^{\lfloor \log_2(2^{m-1}-x)\rfloor} &amp;\geq 2^{m-1}-2\cdot 2^{\log_2(2^{m-1}-x)} = 2x-2^{m-1}\label{eq:f03}
\end{align}
Thus by \eqref{eq:f01} and \eqref{eq:f02} we have $k\cdot 2^{m-i} &lt; x$ and 
\begin{align*}
x &amp;\leq 2^{m-1} - 2^{\lfloor \log_2(2^{m-1}-x)\rfloor}, \textrm{by \eqref{eq:f03}}\\
&amp;= k\cdot 2^{m-i} + 2^{\lfloor \log_2(2^{m-1}-x)\rfloor}, \textrm{by \eqref{eq:f01}}\\
&amp;= k\cdot 2^{m-i} + 2^{m-i-1},  \textrm{by \eqref{eq:f00}}
\end{align*}
Therefore $x\in I^m_i$.
\end{proof}

Let $m,n\geq 2$ be natural numbers and 
\begin{align}
\mathfrak{A}^n_m&amp;:=\bigg\{\left\{\left(A^1,A^2,\ldots,A^m\right),\left\{\left(A^k_1,\ldots,A^k_{2^{k-1}},B^k_1,\ldots,B^k_{2^{k-1}}\right)\colon k\in\mathbb{Z}^+_m\right\}\right\}\colon \\
&amp;\forall_{k\in\mathbb{Z}^+_m}\left(\emptyset\neq A^k\subsetneq\mathbb{Z}^+_n\quad\wedge\quad\#A^{k-1}\leq \#A^k\quad\wedge\quad\forall_{i\in\mathbb{Z}^+_{2^{k-1}}}\colon \left(A^k_i\subseteq B^k_i\subseteq \mathbb{Z}^+_n \right)\right)\bigg\},\nonumber
\end{align}
where $A^0 := \emptyset$.
\begin{lemma}\label{f1}
Let $m,n\geq 2$ be natural numbers and $\mathcal{A}\in\mathfrak{A}^n_m$. If $\mathcal{A}$ satisfies condition
\begin{equation}\label{eq:c1}
B^k_i=\begin{cases}
	A^{k-1}_{(i+1)/2},&amp;\text{if $2\nmid i$;}\\
	B^{k-1}_{i/2}\setminus A^{k-1}_{i/2},&amp;\text{otherwise;}
	\end{cases}
\end{equation}
for all $k\in\{2,\ldots,m\}$ and $i\in\mathbb{Z}^+_{2^{k-1}}$, where $B^1_1=\mathbb{Z}^+_n$ and $A^1_1=A^1$.
Then for all $k\in\{2,\ldots,m\}$:
\begin{align}
\bigcup_{i\in\mathbb{Z}^+_{2^{k-1}}} B^k_i&amp;=\mathbb{Z}^+_n \label{eq:p1}\\
B^k_i\cap B^k_j &amp;= \emptyset\textrm{, for all distinct $i,j\in\mathbb{Z}^+_{2^{k-1}}$}\label{eq:p2}
\end{align}
\end{lemma}
\begin{proof}
The proof is by induction on $k$.
For $k=2$ by \eqref{eq:c1}, $B^2_1=A^1_1=A^1$ and $B^2_2=B^1_1\setminus A^1_1=\mathbb{Z}^+_n\setminus A^1$. Thus \eqref{eq:p1} and \eqref{eq:p2} hold. 

Assume that \eqref{eq:p1} and \eqref{eq:p2} are satisfied for some natural number $2\leq k&lt;m$. 
We shall show that \eqref{eq:p1} holds for $k+1$: 
By \eqref{eq:c1} and the induction hypothesis we have
\begin{align*}
\bigcup_{i\in\mathbb{Z}^+_{2^k}}B^{k+1}_i &amp;= \bigcup_{i\in\mathbb{Z}^+_{2^{k-1}}}
\left(B^{k+1}_{2i-1}\cup B^{k+1}_{2i}\right)\\
&amp;=\bigcup_{i\in\mathbb{Z}^+_{2^{k-1}}}\left(A^k_{i}\cup \left(B^k_{i}\setminus A^k_{i}\right)\right)=
\bigcup_{i\in\mathbb{Z}^+_{2^{k-1}}} B^k_{i} = \mathbb{Z}^+_n .
\end{align*}
We shall show that \eqref{eq:p2} holds for $k+1$: 
Assume that  $i,j\in\mathbb{Z}^+_{2^k}$ and $i\neq j$.
\begin{enumerate}
	\item If $2\nmid i$ and $2\nmid j$ then by \eqref{eq:c1} and finally by the induction hypothesis, 
	$$B^{k+1}_i\cap B^{k+1}_j = A^k_{(i+1)/2}\cap A^k_{(j+1)/2}\subseteq B^k_{(i+1)/2}\cap B^k_{(j+1)/2}=\emptyset .$$
	\item If $2\nmid i$ and $2\mid j$ then
	\begin{itemize}
		\item if $(i+1)/2=j/2$ then by \eqref{eq:c1}, we have
		$$B^{k+1}_i\cap B^{k+1}_j = A^k_{j/2}\cap (B^k_{j/2}\setminus A^k_{j/2}) = \emptyset;$$
		\item if $(i+1)/2\neq j/2$, then by \eqref{eq:c1} and the induction hypothesis we have
		$$B^{k+1}_i\cap B^{k+1}_j = A^k_{(i+1)/2}\cap (B^k_{j/2}\setminus A^k_{j/2})\subseteq  B^k_{(i+1)/2}\cap B^k_{j/2} = \emptyset . $$
	\end{itemize}
	\item If $2\mid i$ and $2\mid j$ then by \eqref{eq:c1} and the induction hypothesis we have
	$$B^{k+1}_i\cap B^{k+1}_j = (B^k_{i/2}\setminus A^k_{i/2})\cap(B^k_{j/2}\setminus A^k_{j/2}) \subseteq B^k_{i/2}\cap B^k_{j/2}=\emptyset .$$
\end{enumerate}
\end{proof}
\begin{lemma}\label{f2}
Let $m,n\geq 2$ be natural numbers and $\mathcal{A}\in\mathfrak{A}^n_m$. If $\mathcal{A}$ satisfies condition \eqref{eq:c1} and the following condition:
\begin{equation}\label{eq:c2}
A^k=\bigcup\limits_{i=1}^{2^{k-1}} A^k_i
\end{equation}
for all $k\in\{2,\ldots,m\}$. 
Then 
\begin{align}
\bigcup_{i\in I^k_j} B^k_i &amp;= A^j \label{eq:p3}
\textrm{, for all $j\in\mathbb{Z}^+_{k-1}$}\\
B^k_{2^{k-1}} &amp;= \mathbb{Z}^+_n\setminus \bigcup_{i=1}^{k-1}A^i\label{eq:p4}
\end{align}
for all natural numbers $2\leq k\leq m$.
\end{lemma}
\begin{proof}
The proof of \eqref{eq:p3} is by induction on $k$.
For $k=2$, by \eqref{eq:thm2:1}, $I^2_1=\{1\}$. Hence $\bigcup_{i\in I^2_1} B^2_i = B^2_1 = A^1$. Therefore \eqref{eq:p3} holds.

Assume that\eqref{eq:p3} is satisfied for some natural number $k$ such that $2\leq k&lt;m$. 
Firstly we shall prove that $\bigcup_{i\in I^{k+1}_k} B^{k+1}_i = A^k$: 
By \eqref{eq:thm2:1}, $I^{k+1}_k = \{2i+1\colon i\in\mathbb{Z}_{2^{k-1}}\}$. Thus, by \eqref{eq:c1} and \eqref{eq:c2}, 
$$\bigcup_{i\in I^{k+1}_k} B^{k+1}_i = \bigcup_{i\in \mathbb{Z}_{2^{k-1}}} B^{k+1}_{2i+1} = \bigcup_{i\in \mathbb{Z}^+_{2^{k-1}}}A^k_i = A^k$$
Fix any $j\in\{1,\ldots,k-1\}$. Then by \eqref{eq:thm2:1}, $I^{k+1}_j$ is a union of 	$2^{j-1}$ $2^{k-j}$-element sets, furthermore 
$\{(i+1)/2\colon i\in I^{k+1}_j\quad\&amp; \quad2\nmid i\} = I^k_j$. Therefore by \eqref{eq:c1} and the induction hypothesis,
\begin{align*}
\bigcup_{i\in I^{k+1}_j} B^{k+1}_i &amp;= \bigcup_{{i\in I^{k+1}_j \atop 2\nmid i}} \left(B^{k+1}_i\cup B^{k+1}_{i+1}\right)\\
&amp;=\bigcup_{{i\in I^{k+1}_j \atop 2\nmid i}}\left( A^k_{(i+1)/2}\cup\left(B^k_{(i+1)/2}\setminus A^k_{(i+1)/2}\right)\right)
= \bigcup_{i\in I^k_j} B^k_i = A^j.
\end{align*}

To prove \eqref{eq:p4}, firstly we utilize \eqref{eq:p1} and \eqref{eq:p2} from Lemma \ref{f1}, from which it follows that $$B^k_{2^{k-1}} = \mathbb{Z}^+_n\setminus \bigcup_{i=1}^{2^{k-1}-1} B^k_i.$$ 
By \eqref{eq:p0} from Lemma \ref{f0}, and next by \eqref{eq:p3} from Lemma \ref{f2} we have
\begin{align*}
B^k_{2^{k-1}} &amp;= \mathbb{Z}^+_n\setminus \bigcup_{i\in\bigcup_{j\in\mathbb{Z}^+_{k-1}}I^k_j} B^k_i\\
&amp;= \mathbb{Z}^+_n\setminus \bigcup_{j\in\mathbb{Z}^+_{k-1}}\bigcup_{i\in I^k_j} B^k_i\\
&amp;= \mathbb{Z}^+_n\setminus \bigcup_{j\in\mathbb{Z}^+_{k-1}} A^j
\end{align*}
\end{proof}
\begin{lemma}\label{f3}
Let $m,n\geq 2$ be natural numbers, $m\leq{n\choose \lfloor n/2\rfloor}$, $A^k\subseteq\mathbb{Z}^+_n$ for all $k\in\mathbb{Z}^+_m$, and 
$1\leq \#A^1\leq\ldots \leq\#A^m\leq n-1$, then
\begin{enumerate}
	\item $\mathcal{A}=\{A^1,\ldots,A^m\}\in \mathcal{A}^m_n$ if and only if for each $k\in\{2,\ldots,m\}$ there exist sets $A^k_1,\ldots,A^k_{2^{k-1}}$ and $B^k_1,\ldots,B^k_{2^{k-1}}$ such that conditions \eqref{eq:c1}, \eqref{eq:c2}, and 
\begin{align}
A^k_i&amp;\subseteq B^k_i\textrm{, for all $i\in\mathbb{Z}^+_{2^{k-1}}$}\label{eq:c3}\\
\bigcup\limits_{i\in I^k_j} A^k_i&amp;\subsetneq A^j \textrm{, for all $j\in\mathbb{Z}^+_{k-1}$}\label{eq:c4}
\end{align} 
are satisfied for all $k\in\{2,\ldots,m\}$.
\item $\mathcal{A}=\{A^1,\ldots,A^m\}\in \mathcal{AC}^m_n$ if and only if for each $k\in\{2,\ldots,m\}$ there exist sets $A^k_1,\ldots,A^k_{2^{k-1}}$ and $B^k_1,\ldots,B^k_{2^{k-1}}$ such that conditions \eqref{eq:c1}, \eqref{eq:c2}, \eqref{eq:c3}, \eqref{eq:c4} are satisfied for all $k\in\{2,\ldots,m\}$ and $A^m_{2^{m-1}}=B^m_{2^{m-1}}$.
\end{enumerate}
\end{lemma}
\begin{proof}
The proof is by induction on $m$.
Firstly we shall prove this fact for $m=2$. 
Assume that $\{A^1,A^2\}\in \mathcal{A}^2_n$. Define $A^1_1:=A^1$, $B^1_1:=\mathbb{Z}^+_n$, $B^2_1:=A^1$, $B^2_2:=\mathbb{Z}^+_n\setminus A^1$, $A^2_1:=A^1\cap A^2$, and $A^2_2:=A^2\setminus A^1$. Then conditions \eqref{eq:c1}, \eqref{eq:c2}, and \eqref{eq:c3} are straightforwardly satisfied. Moreover $I^2_1=\{1\}$. Hence $\bigcup_{i\in I^2_1} A^2_i = A^2_1\subsetneq A^1$, since $A^1\nsubseteq A^2$. Thus condition \eqref{eq:c4} is also true. If additionally $\{A^1,A^2\}\in \mathcal{AC}^2_n$, then 
$\mathbb{Z}^+_n\setminus A^1\subseteq A^2$. Hence $A^2_2=B^2_2$.

Now assume that conditions \eqref{eq:c1}, \eqref{eq:c2}, \eqref{eq:c3}, \eqref{eq:c4} are satisfied. It follows from \eqref{eq:c4} that $A^2_1\subsetneq A^1$, and by \eqref{eq:c3} and \eqref{eq:c1}, $A^2_2\subseteq\mathbb{Z}^+_n\setminus A^1$. Therefore by \eqref{eq:c2}, $A^1\nsubseteq A^2$. Moreover, $\#A^1\leq \#A^2$. Hence $A^2\nsubseteq A^1$, since otherwise $A^2=A^1$ which contradicts $A^1\nsubseteq A^2$. So $\{A^1,A^2\}\in \mathcal{A}^2_n$. If $A^2_2=B^2_2$, then by \eqref{eq:c2} and \eqref{eq:c1}, $A^1\cup A^2 = A^1\cup A^2_2 = A^1\cup B^2_2 = \mathbb{Z}^+_n$. Thus $\{A^1,A^2\}\in \mathcal{AC}^2_n$ and Lemma \ref{f3} is true for $m=2$. 

Assume that Lemma \ref{f3} is true for some natural number $m$ such that $2\leq m&lt;{n\choose \lfloor n/2\rfloor}$. We shall show that it holds for $m+1$. 
Assume $\{A^1,\ldots,A^{m+1}\}\in \mathcal{A}^{m+1}_n$. Then $\{A^1,\ldots,A^{m}\}\in \mathcal{A}^{m}_n$ and by the induction hypothesis, for each $k\in\{2,\ldots,m\}$ there exist sets $A^k_1,\ldots,A^k_{2^{k-1}}$ and $B^k_1,\ldots,B^k_{2^{k-1}}$ such that conditions \eqref{eq:c1}, \eqref{eq:c2}, \eqref{eq:c3}, \eqref{eq:c4} are satisfied for all $k\in\{2\ldots,m\}$. 
Let $A^{m+1}_i:=A^{m+1}\cap B^{m+1}_i$, where $B^{m+1}_i$ defined by \eqref{eq:c1} exist by the induction hypothesis, for all $i\in\mathbb{Z}^+_{2^m}$. Then conditions \eqref{eq:c1} and \eqref{eq:c3} are also straightforwardly satisfied for $k=m+1$. By \eqref{eq:p1} from Lemma \ref{f1}, 
$$\bigcup_{i=1}^{2^m}A^{m+1}_i = A^{m+1}\cap \bigcup_{i=1}^{2^m}B^{m+1}_i = A^{m+1};$$
thus \eqref{eq:c2} is also satisfied for $m+1$. Therefore by \eqref{eq:p3} from Lemma \ref{f2},  we have
$\bigcup_{i\in I^{m+1}_j} B^{m+1}_i = A^j$ for all $j\in\mathbb{Z}^+_m$. So 
$$\bigcup_{i\in I^{m+1}_j} A^{m+1}_i = A^{m+1}\cap \bigcup_{i\in I^{m+1}_j} B^{m+1}_i = A^{m+1}\cap A^j\subsetneq A^j,$$
since otherwise $A^j\subseteq A^{m+1}$ which would contradict with assumption $\{A^1,\ldots,A^{m+1}\}\in \mathcal{A}^{m+1}_n$. It follows from the above reasoning that \eqref{eq:c4} is satisfied for all $k\in\{2,\ldots,m+1\}$.

Assume now that $\{A^1,\ldots,A^{m+1}\}\in \mathcal{AC}^{m+1}_n$. Then by the previous part of the proof, conditions \eqref{eq:c1}, \eqref{eq:c2}, \eqref{eq:c3}, \eqref{eq:c4} are satisfied for all $k\in\{2,\ldots,m+1\}$. We shall prove that $A^{m+1}_{2^m}=B^{m+1}_{2^m}$. By \eqref{eq:c3}, $A^{m+1}_{2^m}\subseteq B^{m+1}_{2^m}$. Suppose that $A^{m+1}_{2^m}\subsetneq B^{m+1}_{2^m}$. Then by 
\eqref{eq:p4}, 
$A^{m+1}_{2^m}\subsetneq \mathbb{Z}^+_n\setminus \bigcup_{i=1}^m A^i$, 
 and hence 
\begin{equation}\label{eq:f3.1}
A^{m+1}_{2^m}\cup\bigcup_{i=1}^m A^i\subsetneq \mathbb{Z}^+_n.
\end{equation} 
On the other hand, by \eqref{eq:c4}, $\bigcup_{j\in\mathbb{Z}^+_m}\bigcup_{i\in I^{m+1}_j} A^{m+1}_i\subseteq \bigcup_{j\in\mathbb{Z}^+_m} A^j$. So by Lemma \ref{f0} and \eqref{eq:c2}, $A^{m+1}\setminus A^{m+1}_{2^m} \subseteq \bigcup_{j\in\mathbb{Z}^+_m} A^j$. Therefore 
\begin{equation}\label{eq:f3.2}
\bigcup_{j\in\mathbb{Z}^+_{m+1}} A^j = \bigcup_{j\in\mathbb{Z}^+_m} A^j \cup A^{m+1}_{2^m}.
\end{equation} 
We see that \eqref{eq:f3.1} and \eqref{eq:f3.2} contradict assumption $\{A^1,\ldots,A^{m+1}\}\in \mathcal{AC}^{m+1}_n$.

Now assume that conditions \eqref{eq:c1}, \eqref{eq:c2}, \eqref{eq:c3}, \eqref{eq:c4} are satisfied for all $k\in\{2,\ldots,m+1\}$. We shall show that $\{A^1,\ldots,A^{m+1}\}\in \mathcal{A}^{m+1}_n$. By assumption $\#A^i\leq \#A^{i+1}$ for all $i\in\mathbb{Z}^+_m$ and the induction hypothesis, it suffices to prove that $A^j\nsubseteq A^{m+1}$ for all $j\in\mathbb{Z}^+_m$. Fix any $j\in\mathbb{Z}^+_m$. Then:
by \eqref{eq:p3}, $A^{m+1}\cap A^j =  A^{m+1}\cap\bigcup_{i\in I^{m+1}_j}B^{m+1}_i$; 
by \eqref{eq:c2}, $A^{m+1}\cap\bigcup_{i\in I^{m+1}_j}B^{m+1}_i =\bigcup_{i\in\mathbb{Z}^+_{2^m}}A^{m+1}_i \cap \bigcup_{i\in I^{m+1}_j}B^{m+1}_i$; 
by Lemma \ref{f0}, \eqref{eq:p1}, \eqref{eq:p2}, and \eqref{eq:c3}, $\bigcup_{i\in\mathbb{Z}^+_{2^m}}A^{m+1}_i \cap \bigcup_{i\in I^{m+1}_j}B^{m+1}_i =\bigcup_{i\in I^{m+1}_j}A^{m+1}_i$; 
hence by \eqref{eq:c4}, $A^{m+1} \cap A^j\subsetneq A^j$.
Therefore $A^j\nsubseteq A^{m+1}$. 

If additionally $A^{m+1}_{2^m}=B^{m+1}_{2^m}$, then by \eqref{eq:p4}, we have
$A^{m+1}_{2^m}\cup\bigcup_{i\in\mathbb{Z}^+_m} A^i = \mathbb{Z}^+_n$. So by \eqref{eq:c2}, $\bigcup_{i\in\mathbb{Z}^+_{m+1}} A^i = \mathbb{Z}^+_n$. Thus $\{A^1,\ldots,A^{m+1}\}\in \mathcal{AC}^{m+1}_n$.
\end{proof}

Using Lemma \ref{f3} we can determine the number of all antichains as well as the number of all antichain covers of an $n$-element set $X$ which consist of $m$ subsets of $X$. Let 
$$\mathfrak{C}^n_m:=\left\{(A^1,\ldots,A^m) \colon
\exists_{\mathcal{A}\in\mathfrak{A}^n_m} \colon (A^1,\ldots,A^m)\in\mathcal{A} \ \&amp;\ \mathcal{A}\textrm{ satisfies conditions \eqref{eq:c1}, \eqref{eq:c2}, \eqref{eq:c4}}\right\}$$ 
and 
$$\mathfrak{D}^n_m:=\left\{(A^1,\ldots,A^m)\in\mathfrak{C}^n_m\colon A^m_{2^{m-1}}=B^m_{2^{m-1}}\right\}$$ 
By Lemma \ref{f3}, if $\#A^1\leq\ldots\leq \#A^m$, then $\{A^1,\ldots,A^m\}\in \mathcal{A}^{m}_n$ iff $(A^1,\ldots,A^m)\in \mathfrak{C}^n_m$ and $\{A^1,\ldots,A^m\}\in \mathcal{AC}^{m}_n$ iff $(A^1,\ldots,A^m)\in \mathfrak{D}^n_m$.

If $(A^1,\ldots,A^m)\in \mathfrak{C}^n_m$, then $A^1$ can be chosen in ${b^1_1\choose a^1}$ ways, where $b^1_1=n$, $a^1=\#A^1$, and $a^1$ can differ from $1$ to $n-1$. Generally, it follows from Lemma \ref{f3}, that for fixed sets $A^1,\ldots,A^k$ of cardinality $a^1,\ldots,a^k$, respectively, set $A^{k+1}_i$ can be 
chosen in ${b^{k+1}_i\choose a^{k+1}_i}$ ways, for all $i\in\mathbb{Z}^+_{2^k}$, where $b^{k+1}_i$ are defined by \eqref{eq:thm2:4}, 
$\left(a^{k+1}_1,\ldots,a^{k+1}_{2^k}\right)\in S^n_m\left(\underline{a}^1,\ldots,\underline{a}^k\right)$, and $S^n_m$ is defined by \eqref{eq:thm2:5}. 
Moreover if $(A^1,\ldots,A^m)\in \mathfrak{D}^n_m$, then by Lemma \ref{f3}, $A^m_{2^{m-1}}=B^m_{2^{m-1}}$. So it can be chosen in one way. 
Furthermore, note that for any set $\{A^1,\ldots,A^m\}\in \mathcal{A}^{m}_n$ if $a_i=\#A^i$ for all $i\in\mathbb{Z}^+_m$ and 
$$\underbrace{a_1=\cdots =a_{m_1}}_{m_1}&lt;\underbrace{a_{m_1+1}=\cdots =a_{m_1+m_2}}_{m_2}&lt;\cdots&lt;
\underbrace{a_{M_k+1}=\cdots =a_{M_k+m_k}}_{m_k}$$
where $M_i:=m_1+\cdots+m_{i-1}$ for all $1&lt;i\leq k$, then the number of all $(B^1,\ldots,B^m)\in\mathfrak{C}^n_m$ such that 
$\{B^1,\ldots,B^m\}=\{A^1,\ldots,A^m\}$ is equal to $\prod_{i=1}^k m_i!$. Therefore \eqref{eq:thm2:8} and \eqref{eq:thm2:9} are true.
\end{proof}

\section{Covers of labeled sets with antichains consisting of equinumerous sets}
Let $X$ be an $n$-element set and numbers $k,m\in\mathbb{N}$ be such that $\frac{n}{k}\leq m\leq {n\choose k}$. 
We let $\mathcal{AC}^{m,k}(X)$ denote the set of all $m$-antichain covers $\{A_1,\ldots,A_m\}$ in the power set $\mathcal{P}(X)$ such that $\#A_i = k$ for all $i\in\mathbb{Z}^+_m$. If $X=\mathbb{Z}^+_n$, then $\mathcal{AC}^{m,k}(X)$ is denoted by $\mathcal{AC}^{m,k}_n$ and $T^{m,k}_n = \#\mathcal{AC}^{m,k}_n$.

Note that the assumption $\frac{n}{k}\leq m\leq {n\choose k}$ is a necessary condition for the correctness of this notion, since for any $n$-element set $X$ and $\{A_1,\ldots,A_m\}\in \mathcal{AC}^{m,k}(X)$: 
$$n=\# X = \#\left(\bigcup_{i=1}^m A_i\right)\leq \sum_{i=1}^m \#A_i=m\cdot k$$ 
and $A_1,\ldots,A_m$ are distinct $k$-element subsets of $X$ whose total number equals ${n\choose k}$. 
\begin{lemma}
\label{lem:1}
Let $X$ be an $n$-element set, $m,k\in\mathbb{N}$ be such that $m\leq {n\choose k}$,
$$C^m_k(X):=\big\{\{A_1,\ldots,A_m\}\colon A_i\subseteq X\textrm{ and } \# A_i=k\textrm{ for all }i\in\{1,\ldots,m\}\big\},$$ $n_0:=\min\left\{n'\in\mathbb{N}\colon{n'\choose k}\geq m\right\}$. Then:
\begin{align}
C^m_k(X)&amp;=\bigcup_{Y\subseteq X\colon n_0\leq\# Y\leq \min(n,m\cdot k)} \mathcal{AC}^{m,k}(Y)\label{eq:lem:1.1}\\
\# C^m_k(X) &amp;= {{n\choose k}\choose m}\label{eq:lem:1.2}\\
\sum_{l=n_0}^{\min(n,m\cdot k)} {n\choose l}\cdot T^{m,k}_l &amp;= {{n\choose k}\choose m}.\label{eq:lem:1.3}
\end{align}
\end{lemma}
\begin{proof}
\leavevmode
\begin{enumerate}
	\item Fix any $\{A_1,\ldots,A_m\}\in C^m_k(X)$. Let $Y:=\bigcup_{i=1}^m A_i$. 
	Then $\{A_1,\ldots,A_m\}\in \mathcal{AC}^{m,k}(Y)$, $Y\subseteq X$; moreover, we have, $n':=\# Y\leq \min(n,m\cdot k)$ and ${n' \choose k}\geq m$. Thus $n'\geq n_0$. The converse inclusion is obvious.
	\item The number of all $k$-element subsets of an $n$-element set equals ${n\choose k}$. Thus $\# C^m_k(X)$ as the number of all $m$-element subsets of the family of all $k$-element subsets of an $n$-element set equals ${{n\choose k}\choose m}$.
	\item It follows straightforwardly from \eqref{eq:lem:1.1} and \eqref{eq:lem:1.2}, since 
	$$\#\mathcal{AC}^{m,k}(Y)=\#\mathcal{AC}^{m,k}(Y')=T^{m,k}_l$$ for all $Y,Y'\subseteq X$ such that $n_0\leq l:=\# Y=\# Y'\leq \min(n,m\cdot k)$ and the number of all $l$-element subsets of $X$ equals ${n\choose l}$.
\end{enumerate}
\end{proof}
\begin{theorem}
\label{thm:1}
Assume that $m,k\in\mathbb{N}$ and $n_0:=\min\left\{n'\in\mathbb{N}\colon{n'\choose k}\geq m\right\}$. Then 
\begin{equation}
\label{eq:thm:1.1}
T^{m,k}_n = \sum_{t=0}^{n-n_0} {n\choose n-t}\cdot{{n-t\choose k}\choose m}\cdot (-1)^t
\end{equation}
for all $n\in\{n_0,\ldots,m\cdot k\}$
\end{theorem}
\begin{proof}
Fix $m,k\in\mathbb{N}$ and put $n_0:=\min\left\{n'\in\mathbb{N}\colon{n'\choose k}\geq m\right\}$. 
Let $a_n:=T^{m,k}_n$ and $b_n:={{n\choose k}\choose m}$. Then by \eqref{eq:lem:1.3} from Lemma \ref{lem:1}, 
$b_n:=\sum_{k=n_0}^n {n\choose k} a_k$. Thus assumptions of Lemma \ref{lem:2} are satisfied. Therefore by the same lemma, \eqref{eq:thm:1.1} holds.
\end{proof}

\section{Numerical results}
\begin{example}\label{ex:2}
Let us return to Example \ref{ex:1} and compute $D^m_n$ for all $n\geq 3$ and $m\in\{2,3\}$ using Theorem \ref{thm:2} and Proposition \ref{prop:1}. By formulas \eqref{eq:thm2:8} and \eqref{eq:thm2:9}, 
\begin{align*}
D^2_n &amp;= \sum_{a^1=1}^{n-1}\ \sum_{\underline{\textbf{a}}^2\in S^n_2(a^1)}
\frac{{b^1_1\choose a^1_1} {b^2_1\choose a^2_1} {b^2_2\choose a^2_2}}{r(a^1,a^2)}\\
T^2_n &amp;= \sum_{a^1=1}^{n-1}\ \sum_{\underline{\textbf{a}}^2\in \overline{S^n_2}(a^1)}
\frac{{b^1_1\choose a^1_1} {b^2_1\choose a^2_1} {b^2_2\choose a^2_2}}{r(a^1,a^2)}\\
D^3_n &amp;= \sum_{a^1=1}^{n-1}\ \sum_{\underline{\textbf{a}}^2\in S^n_2(a^1)}
\ \sum_{\underline{\textbf{a}}^3\in S^n_3(a^1, \underline{\textbf{a}}^2)}
\frac{{b^1_1\choose a^1_1} {b^2_1\choose a^2_1} {b^2_2\choose a^2_2} 
{b^3_1 \choose a^3_1} {b^3_2 \choose a^3_2} {b^3_3 \choose a^3_3} {b^3_4 \choose a^3_4}}
{r(a^1,a^2,a^3)}\\
T^3_n &amp;= \sum_{a^1=1}^{n-1}\ \sum_{\underline{\textbf{a}}^2\in S^n_2(a^1)}
\ \sum_{\underline{\textbf{a}}^3\in \overline{S^n_3}(a^1, \underline{\textbf{a}}^2)}
\frac{{b^1_1\choose a^1_1} {b^2_1\choose a^2_1} {b^2_2\choose a^2_2} 
{b^3_1 \choose a^3_1} {b^3_2 \choose a^3_2} {b^3_3 \choose a^3_3} {b^3_4 \choose a^3_4}}
{r(a^1,a^2,a^3)}
\end{align*}
By \eqref{eq:prop1:1}, $b^1_1 = a^0_1 = n$, $b^2_1 = a^1_1 = a^1$, $b^2_2 = a^0_1 - a^1_1 = n - a^1$, 
$b^3_1 = a^2_1$, $b^3_2 = a^1_1 - a^2_1 = a^1 - a^2_1$, $b^3_3 = a^2_2$, and $b^3_4 = a^0_1 - a^2_2 - a^1_1 = n - a^1 - a^2_2$. 
Note that by \eqref{eq:thm2:1}, $I^2_1 = \{1\}$, $I^3_1 = \{1, 2\}$, and $I^3_2 = \{1, 3\}$
So by \eqref{eq:thm2:2} and \eqref{eq:thm2:5}, 
\begin{align*}
S^n_2(a^1) &amp;= \{(a^2_1, a^2_2)\colon 0 \leq a^2_1 \leq a^1 - 1 \quad\&amp;\quad 0 \leq a^2_2 \leq n - a^1 \quad\&amp;\quad a^1 \leq a^2 \leq n-1\} \\
\overline{S^n_2}(a^1) &amp;= \{(a^2_1, a^2_2)\colon 0 \leq a^2_1 \leq a^1 - 1 \quad\&amp;\quad a^1 \leq a^2 \leq n-1 \quad\&amp;\quad a^2_2 = n - a^1\}  \\
S^n_3(a^1, \underline{\textbf{a}}^2) &amp;= \{(a^3_1, a^3_2, a^3_3, a^3_4)\colon 0 \leq a^3_1 \leq a^2_1 \quad\&amp;\quad 0 \leq a^3_2 \leq a^1 - a^2_1 \quad\&amp;\quad 0 \leq a^3_3 \leq a^2_2 \quad\&amp;\quad \\
&amp;  0 \leq a^3_4 \leq n - a^1 - a^2_2 \quad\&amp;\quad a^2 \leq a^3 \leq n-1 \quad\&amp;\quad\\
&amp; a^3_1 + a^3_2 \leq a^1 - 1 \quad\&amp;\quad a^3_1 + a^3_3 \leq a^2 - 1\}\\
\overline{S^n_3}(a^1, \underline{\textbf{a}}^2) &amp;= \{(a^3_1, a^3_2, a^3_3, a^3_4)\colon 0 \leq a^3_1 \leq a^2_1 \quad\&amp;\quad 0 \leq a^3_2 \leq a^1 - a^2_1 \quad\&amp;\quad 0 \leq a^3_3 \leq a^2_2 \quad\&amp;\quad \\
&amp; 0 \leq a^3_4 = n - a^1 - a^2_2 \quad\&amp;\quad a^2 \leq a^3 \leq n-1 \quad\&amp;\quad \\
&amp; a^3_1 + a^3_2 \leq a^1 - 1 \quad\&amp;\quad a^3_1 + a^3_3 \leq a^2 - 1\}\\
\end{align*}
It follows from the system of inequalities in $S^n_2(a^1)$, that:
$\max(0, 2a^1 - n)\leq a^2_1 \leq a^1 - 1$ and 
$$1 \leq a^1 - a^2_1 \leq a^2_2 \leq n-a^1 \leq n - 1 - a^2_1.$$ 
From the system of inequalities and equality from $\overline{S^n_2}(a^1)$ we obtain the same bound for $a^2_1$ as above. Hence we obtain the same formulae \eqref{eq:ex2a} and \eqref{eq:ex2b} as in Example \ref{ex:1}, as follows:
\begin{align*}
D^2_n &amp;= \sum_{a^1=1}^{n-1}\ \sum_{a^2_1 = \max(0, 2a^1 - n)}^{a^1 - 1}\sum_{a^2_2 = a^1 - a^2_1}^{n-a^1}
\frac{{n\choose a^1_1} {a^1\choose a^2_1} {n - a^1 \choose a^2_2}}{r(a^1,a^2)}\\
T^2_n &amp;= \sum_{a^1=1}^{n-1}\ \sum_{a^2_1 = \max(0, 2a^1 - n)}^{a^1 - 1}
\frac{{n\choose a^1_1} {a^1\choose a^2_1}}{r(a^1,a^2)} .
\end{align*}
Note that the systems of inequalities in $S^n_3(a^1, \underline{\textbf{a}}^2)$ and $\overline{S^n_3}(a^1, \underline{\textbf{a}}^2)$ are the same as in \eqref{eq:ex21} from Example \ref{ex:1}. Solving these systems we obtain the following formulas:
\begin{align*}
D^3_n &amp;= \sum_{a^1=1}^{n-1}\ \sum_{a^2_1 = \max(0, 2a^1 - n)}^{a^1 - 1}
\ \sum_{a^2_2 = a^1 - a^2_1}^{\min(n-a^1, n-2)}\sum_{a^3_1 = \max(0,a^2+a^2_1-n)}^{\min(a^2_1,n-a^2_2-2)}\\
&amp;\sum_{a^3_2 = \max(0,a^1+a^2_2-n+1,a^1+a^2-n-a^3_1)}^{\min(a^1-a^2_1,a^1-1-a^3_1)}
\ \sum_{a^3_3 = \max(0,a^2+a^2_2-n,a^1+a^2+a^2_2-n-a^3_1-a^3_2)}^{\min(a^2_2,a^2-1-a^3_1)}\\
&amp;\sum_{a^3_4 = \max(0,a^2-a^3_1-a^3_2-a^3_3)}^{\min(n-a^1-a^2_2,n-1-a^3_1-a^3_2-a^3_3)}
\frac{{n \choose a^1_1} {a^1 \choose a^2_1} {n-a^1 \choose a^2_2} 
{a^2_1 \choose a^3_1} {a^1 - a^2_1 \choose a^3_2} {a^2_2 \choose a^3_3} {n - a^1 - a^2_2 \choose a^3_4}}
{r(a^1,a^2,a^3)}\\
T^3_n &amp;= \sum_{a^1=1}^{n-1}\ \sum_{a^2_1 = \max(0, 2a^1 - n)}^{a^1 - 1}
\ \sum_{a^2_2 = a^1 - a^2_1}^{\min(n-a^1, n-2)}\sum_{a^3_1 = \max(0,a^2+a^2_1-n)}^{\min(a^2_1,n-a^2_2-2)}
\ \sum_{a^3_2 = \max(0,a^1+a^2_2-n+1,a^1+a^2-n-a^3_1)}^{\min(a^1-a^2_1,a^1-1-a^3_1)}\\
&amp;\sum_{a^3_3 = \max(0,a^2+a^2_2-n,a^1+a^2+a^2_2-n-a^3_1-a^3_2)}^{\min(a^2_2,a^2-1-a^3_1)}
\frac{{n \choose a^1_1} {a^1 \choose a^2_1} {n-a^1 \choose a^2_2} 
{a^2_1 \choose a^3_1} {a^1 - a^2_1 \choose a^3_2} {a^2_2 \choose a^3_3}}
{r(a^1,a^2,a^3)}.
\end{align*}
\end{example}

Table \ref{tab:1} gives the number of all summands in the sums \eqref{eq:thm2:8} denoted by $N_{A}^m(n)$, the number of all summands in the sums \eqref{eq:thm2:9} denoted by $N_{AC}^m(n)$, as well as values of $D^m_n$ and $T^m_n$ for $n\in\{4,5\}$ and $m\in\left\{2,\ldots,{n\choose \lfloor n/2 \rfloor}-1\right\}$. Trivial cases when $m\in\left\{0,1,{n \choose \lfloor n/2 \rfloor}\right\}$ are omitted in the table. They are considered in the following remark. 
\begin{remark}
For all natural numbers $n\in\mathbb{N}_0$:
\begin{enumerate}
	\item $D^0_n = T^0_n = 1$;
	\item $D^1_n = 2^n$ and $T^1_n = 1$;
	\item if $m={n \choose \lfloor n/2 \rfloor}$, then $D^{m}_n = T^{m}_n=
	\begin{cases}
	1,&amp;\text{if $2\mid n$;}\\
	2,&amp;\text{if $2\nmid n$;}\end{cases}$
	\item if $m={n \choose \lfloor n/2 \rfloor}-1$, then $T^{m}_n=
	\begin{cases}
	{n \choose \lfloor n/2 \rfloor},&amp;\text{if $2\mid n$;}\\
	2{n \choose \lfloor n/2 \rfloor},&amp;\text{if $2\nmid n$;}\end{cases}$
	\item if ${n-1\choose \lfloor (n-1)/2 \rfloor} &lt; m\leq {n \choose \lfloor n/2 \rfloor}$, then $T^m_n = D^m_n$;
\end{enumerate}
\end{remark}

\begin{table}[H]
\centering
\begin{tabular}{cccccc}
\hline\hline
$n$ &amp; $m$ &amp; $N_{A}^m(n)$ &amp; $N_{AC}^m(n)$ &amp; $D^m_n$ &amp; $T^m_n$\\
\hline
4 &amp; 2 &amp; 7 &amp; 4 &amp; 55 &amp; 25\\
  &amp; 3 &amp; 11 &amp; 9 &amp; 64 &amp; 56\\
  &amp; 4 &amp; 19 &amp; 19 &amp; 25 &amp; 25\\
  &amp; 5 &amp; 30 &amp; 30 &amp; 6 &amp; 6\\
\hline
5 &amp; 2 &amp; 13 &amp; 6 &amp; 285 &amp; 90\\
  &amp; 3 &amp; 40 &amp; 29 &amp; 1090 &amp; 790\\
  &amp; 4 &amp; 164 &amp; 145 &amp; 2020 &amp; 1895\\
  &amp; 5 &amp; 760 &amp; 730 &amp; 2146 &amp; 2116\\
  &amp; 6 &amp; 3180 &amp; 3150 &amp; 1380 &amp; 1375\\
  &amp; 7 &amp; 11148 &amp; 11148 &amp; 490 &amp; 490\\
  &amp; 8 &amp; 31104 &amp; 31104 &amp; 115 &amp; 115\\
	&amp; 9 &amp; 60480 &amp; 60480 &amp;  20 &amp;  20\\
\hline\hline
\end{tabular}
\caption{The number of all summands in the sums \eqref{eq:thm2:8}
--- $N_{A}^m(n)$, \eqref{eq:thm2:9} --- $N_{AC}^m(n)$, $D^m_n$, and
$T^m_n$ for $n\in\{4,5\}$ and $m\in\left\{2,\ldots,{n\choose \lfloor
n/2 \rfloor}-1\right\}$.}
\label{tab:1}
\end{table}
Tables \ref{tab:2} and \ref{tab:3} depict the number of all summands in the sums \eqref{eq:thm2:8} and \eqref{eq:thm2:9} --- $N_{A}^m(n)$ and $N_{AC}^m(n)$, respectively, as well as values of $D^m_n$ and $T^m_n$, respectively, for $n\in\{6,\ldots,15\}$ and $m\in\{2,3,4\}$.

\begin{table}[H]
\centering
\begin{tabular}{cllllll}
\hline\hline
$n$ &amp; $N_{A}^2(n)$ &amp; $D^2_n$ &amp; $N_{A}^3(n)$ &amp; $D^3_n$ &amp; $N_{A}^4(n)$ &amp; $D^4_n$\\
\hline
6 &amp; 22 &amp; 1351 &amp; 113 &amp; 14000 &amp; 913 &amp; 82115 \\
7 &amp; 34 &amp; 6069 &amp; 272 &amp; 153762 &amp; 3889 &amp; 2401910\\
8 &amp; 50 &amp; 26335 &amp; 585 &amp; 1533504 &amp; 13850 &amp; 58089465 \\
9 &amp; 70 &amp; 111645 &amp; 1154 &amp; 14356610 &amp; 43157 &amp; 1245331920\\
10 &amp; 95 &amp; 465751 &amp; 2129 &amp; 128722000 &amp; 121243 &amp; 24625121455 \\
11 &amp; 125 &amp; 1921029 &amp; 3718 &amp; 1119607522 &amp; 313162 &amp; 460316430970 \\
12 &amp; 161 &amp; 7859215 &amp; 6208 &amp; 9528462944 &amp; 754557 &amp; 8266174350005 \\
13 &amp; 203 &amp; 31964205 &amp; 9976 &amp; 79817940930 &amp; 1714126 &amp; 144171200793620 \\
14 &amp; 252 &amp; 129442951 &amp; 15520 &amp; 660876543600 &amp; 3702041 &amp; 2461016066613195\\
15 &amp; 308 &amp; 522538389 &amp; 23470 &amp; 5424917141282 &amp; 7650964 &amp; 41343340015862430 \\
\hline\hline
\end{tabular}
\caption{The number of all summands in the sums \eqref{eq:thm2:8} --- $N_{A}^m(n)$ and values of $D^m_n$ for $n\in\{6,\ldots,15\}$ and $m\in\{2,3,4\}$.}
\label{tab:2}
\end{table}
\begin{table}[H]
\centering
\begin{tabular}{cllllll}
\hline\hline
$n$ &amp; $N_{AC}^2(n)$ &amp; $T^2_n$ &amp; $N_{AC}^3(n)$ &amp; $T^3_n$ &amp; $N_{AC}^4(n)$ &amp; $T^4_n$\\
\hline
6 &amp; 9 &amp; 301 &amp; 73 &amp; 8380 &amp; 749 &amp; 70370 \\
7 &amp; 12 &amp; 966 &amp; 159 &amp; 76482 &amp; 2976 &amp; 1868650 \\
8 &amp; 16 &amp; 3025 &amp; 313 &amp; 638736 &amp; 9961 &amp; 41062035 \\
9 &amp; 20 &amp; 9330 &amp; 569 &amp; 5043950 &amp; 29307 &amp; 802349205 \\
10 &amp; 25 &amp; 28501 &amp; 975 &amp; 38390660 &amp; 78086 &amp; 14514339340 \\
11 &amp; 30 &amp; 86526 &amp; 1589 &amp; 285007162 &amp; 191919 &amp; 249104207000 \\
12 &amp; 36 &amp; 261625 &amp; 2490 &amp; 2079779416 &amp; 441395 &amp; 4120588431245 \\
13 &amp; 42 &amp; 788970 &amp; 3768 &amp; 14995363110 &amp; 959569 &amp; 66392465654515 \\
14 &amp; 49 &amp; 2375101 &amp; 5544 &amp; 107204473740 &amp; 1987915 &amp; 1049608974433110 \\
15 &amp; 56 &amp; 7141686 &amp; 7950 &amp; 761823557042 &amp; 3948923 &amp; 16365222591176550 \\ 
\hline\hline
\end{tabular}
\caption{The number of all summands in the sums \eqref{eq:thm2:9}
--- $N_{AC}^m(n)$ and values of $T^m_n$ for $n\in\{6,\ldots,15\}$ and
$m\in\{2,3,4\}$.}
\label{tab:3}
\end{table}

Table \ref{tab:4} presents values of $T^m_n$ for $n\in\{16,\ldots,50\}$ and $m\in\{2,3\}$.

Table \ref{tab:5}  presents values of $T^{m,k}_n$ for $n=7$, $m\in\{2,\ldots,35\}$, and all possible $k$, i.e., $k\geq\frac{7}{m}$ and ${7 \choose k}\geq m$.

Methods enabling the computation of $T^m_n$, $D^m_n$ for all
$n\in\mathbb{N}$ and $1\leq m \leq {n \choose \lfloor n/2\rfloor}$,
as well as a method allowing the determination of $T^{m,k}_n$, for all
$m,n,k\in\mathbb{N}$ such that $\frac{n}{k}\leq m \leq {n \choose k}$
were programmed in R. These methods were used to obtain results from
Tables \ref{tab:1}--\ref{tab:3}, \ref{tab:5}, and \ref{tab:6}. Methods
permitting the determination of $T^m_n$ for $m\in\{2,3,4\}$ and all
natural numbers $n\geq 4$ were programmed in Maple, and were used to
obtain the results in Table \ref{tab:4}. All methods written in R were
implemented on a PC with Intel Core i7-8750H 2.20-GHz CPU and 16 GB of
RAM. However, methods written in Maple were implemented on a PC with
Intel Core i5-3230M 2.60-GHz CPU and 6 GB of RAM.

Table \ref{tab:6} presents the running time of the method
$\Ind(n,m,\ind(n,m-1))$ determining the list $\ind(n,m)$ of all
possible sequences $\underline{a}^1,\ldots,\underline{a}^m$ for
$m\in\{2,3,4\}$ and $n\in\{12,13,14,15\}$ such that $\underline{a}^k\in
S^n_k(\underline{a}^1,\ldots,\underline{a}^{k-1})$, where
$S^n_k(\underline{a}^1,\ldots,\underline{a}^{k-1})$ are defined by
\eqref{eq:thm2:5}, as well as the method $\AC(n,m,\ind(n,m))$ computing
$D^m_n$ and $T^m_n$ according to \eqref{eq:thm2:8} and  \eqref{eq:thm2:9}
for $m\in\{2,3,4\}$, $n\in\{12,13,14,15\}$,  and the list of
sequences $\ind(n,m)$. The list $\indc(n,m)$ of all possible sequences
$\underline{a}^1,\ldots,\underline{a}^m$ suitable for \eqref{eq:thm2:9}
can be easily and quickly obtained from the list $\ind(n,m)$ of all
possible sequences $\underline{a}^1,\ldots,\underline{a}^m$ suitable
for \eqref{eq:thm2:8}. Therefore the running time of the procedure
$\convToCl$, converting one list into another, is omitted.

It can be seen in Table \ref{tab:6} that methods $\Ind$ and $\AC$,
based on Theorem \ref{thm:2}, allow us to effectively compute $D^m_n$
and $T^m_n$ for small $m\in\{2,3\}$ and relatively large $n$.

\begin{table}[H]
\centering
\begin{tabular}{cll}
\hline\hline
$n$ &amp; $T^2_n$ &amp; $T^3_n$\\
\hline
16 &amp;21457825 &amp;5390550296096 \\
17 &amp;64439010 &amp;38026057186270 \\
18 &amp;193448101 &amp;267656481977620 \\
19 &amp;580606446 &amp;1881017836414122 \\
20 &amp;1742343625 &amp;13204444871932776 \\
21 &amp;5228079450 &amp;92618543463601430 \\
22 &amp;15686335501 &amp;649270263511862300 \\
23 &amp;47063200806 &amp;4549607376865786402 \\
24 &amp;141197991025 &amp;31870882201493713456 \\
25 &amp;423610750290 &amp;223214539710301456590 \\
26 &amp;1270865805301 &amp;1563094445025734127780 \\
27 &amp;3812664524766 &amp;10944627831536630201882 \\
28 &amp;11438127792025 &amp;76627241923504206742136 \\
29 &amp;34314651811530 &amp;536464983051964328959750 \\
30 &amp;34314651811530 &amp;3755626545565209968614060 \\
31 &amp;308834550658326 &amp;26291245004929410021308562 \\
32 &amp;926505799458625 &amp;184048014423958633597002816 \\
33 &amp;2779521693343170 &amp;1288382611752502969073278910 \\
34 &amp;8338573669964101 &amp;9018910891539127690443801140 \\
35 &amp;25015738189761486 &amp;63133539508515093365692674442 \\
36 &amp;75047248929022825 &amp;441940593783822075229938159496 \\
37 &amp;225141815506545210 &amp;3093613246149811622777248142070 \\
38 &amp;675425583958589101 &amp;21655438185531666666398473609020 \\
39 &amp;2026277026753674246 &amp;151588794667806792127575720587522 \\
40 &amp;6078831630016836625 &amp;1061125199746752166787452564356176 \\
41 &amp;18236495989562137650 &amp;7427894584494498851907618171437230 \\
42 &amp;54709490167709668501 &amp;51995353026424486474819809382197700 \\
43 &amp;164128474901175516606 &amp;363967925874293633608399150016855802 \\
44 &amp;492385433499619572025 &amp;2547777754624695453699489227563016856 \\
45 &amp;1477156318091044760490 &amp;17834455650128183338493646918462812390 \\
46 &amp;4431468989457506370301 &amp;124841246390602318150989020114676767180 \\
47 &amp;13294407038741263288566 &amp;873889008936455232899697038492373607282 \\
48 &amp;39883221256961278221025 &amp;6117224483581236975434924744272427645536 \\
49 &amp;119649664052358811373730 &amp;42820578490258331781970654161998901275550 \\
50 &amp;358948992720026387542501 &amp;299744084957994372050034044992597868357460 \\
\hline\hline
\end{tabular}
\caption{Values of $T^m_n$ for $n\in\{16,\ldots,50\}$ and $m\in\{2,3\}$.}
\label{tab:4}
\end{table}


\begin{table}[H]
\centering
\begin{tabular}{lll|lll|lll|lll}
\hline\hline
$m$ &amp; $k$ &amp; $T^{m,k}_7$ &amp; $m$ &amp; $k$ &amp; $T^{m,k}_7$ &amp; $m$ &amp; $k$ &amp; $T^{m,k}_7$ &amp; $m$ &amp; $k$ &amp; $T^{m,k}_7$\\
\hline
2 &amp; 4 &amp; 70 &amp; 8 &amp; 2 &amp; 159390 &amp; 15 &amp; 2 &amp; 54257 &amp; 22 &amp; 3 &amp; 1476337800 \\ 
&amp; 5 &amp; 105 &amp;  &amp; 3 &amp; 22654975 &amp;  &amp; 3 &amp; 3247834632 &amp;  &amp; 4 &amp; 1476337800 \\ 
 &amp; 6 &amp; 21 &amp;  &amp; 4 &amp; 23490775 &amp;  &amp; 4 &amp; 3247943153 &amp; 23 &amp; 3 &amp; 834451800 \\ 
3 &amp; 3 &amp; 945 &amp;  &amp; 5 &amp; 203490 &amp;  &amp; 5 &amp; 54264 &amp;  &amp; 4 &amp; 834451800 \\ 
 &amp; 4 &amp; 3570 &amp; 9 &amp; 2 &amp; 259105 &amp; 16 &amp; 2 &amp; 20349 &amp; 24 &amp; 3 &amp; 417225900 \\ 
 &amp; 5 &amp; 1190 &amp;  &amp; 3 &amp; 69431950 &amp;  &amp; 3 &amp; 4059895035 &amp;  &amp; 4 &amp; 417225900 \\ 
 &amp; 6 &amp; 35 &amp;  &amp; 4 &amp; 70572425 &amp;  &amp; 4 &amp; 4059928950 &amp; 25 &amp; 3 &amp; 183579396 \\ 
4 &amp; 2 &amp; 315 &amp;  &amp; 5 &amp; 293930 &amp;  &amp; 5 &amp; 20349 &amp;  &amp; 4 &amp; 183579396 \\ 
 &amp; 3 &amp; 22820 &amp; 10 &amp; 2 &amp; 331716 &amp; 17 &amp; 2 &amp; 5985 &amp; 26 &amp; 3 &amp; 70607460 \\ 
 &amp; 4 &amp; 42910 &amp;  &amp; 3 &amp; 182286125 &amp;  &amp; 3 &amp; 4537559670 &amp;  &amp; 4 &amp; 70607460 \\ 
 &amp; 5 &amp; 5880 &amp;  &amp; 4 &amp; 183558375 &amp;  &amp; 4 &amp; 4537567650 &amp; 27 &amp; 3 &amp; 23535820 \\ 
 &amp; 6 &amp; 35 &amp;  &amp; 5 &amp; 352716 &amp;  &amp; 5 &amp; 5985 &amp;  &amp; 4 &amp; 23535820 \\ 
5 &amp; 2 &amp; 4410 &amp; 11 &amp; 2 &amp; 343161 &amp; 18 &amp; 2 &amp; 1330 &amp; 28 &amp; 3 &amp; 6724520 \\ 
 &amp; 3 &amp; 221396 &amp;  &amp; 3 &amp; 416050180 &amp;  &amp; 3 &amp; 4537566320 &amp;  &amp; 4 &amp; 6724520 \\ 
 &amp; 4 &amp; 303632 &amp;  &amp; 4 &amp; 417216345 &amp;  &amp; 4 &amp; 4537567650 &amp; 29 &amp; 3 &amp; 1623160 \\ 
 &amp; 5 &amp; 20307 &amp;  &amp; 5 &amp; 352716 &amp;  &amp; 5 &amp; 1330 &amp;  &amp; 4 &amp; 1623160 \\ 
 &amp; 6 &amp; 21 &amp; 12 &amp; 2 &amp; 290745 &amp; 19 &amp; 2 &amp; 210 &amp; 30 &amp; 3 &amp; 324632 \\ 
6 &amp; 2 &amp; 23604 &amp;  &amp; 3 &amp; 833570010 &amp;  &amp; 3 &amp; 4059928810 &amp;  &amp; 4 &amp; 324632 \\ 
 &amp; 3 &amp; 1356250 &amp;  &amp; 4 &amp; 834448615 &amp;  &amp; 4 &amp; 4059928950 &amp; 31 &amp; 3 &amp; 52360 \\ 
 &amp; 4 &amp; 1588125 &amp;  &amp; 5 &amp; 293930 &amp;  &amp; 5 &amp; 210 &amp;  &amp; 4 &amp; 52360 \\ 
 &amp; 5 &amp; 54257 &amp; 13 &amp; 2 &amp; 202755 &amp; 20 &amp; 2 &amp; 21 &amp; 32 &amp; 3 &amp; 6545 \\ 
 &amp; 6 &amp; 7 &amp;  &amp; 3 &amp; 1475795160 &amp;  &amp; 3 &amp; 3247943153 &amp;  &amp; 4 &amp; 6545 \\ 
7 &amp; 1 &amp; 1 &amp;  &amp; 4 &amp; 1476337065 &amp;  &amp; 4 &amp; 3247943160 &amp; 33 &amp; 3 &amp; 595 \\ 
 &amp; 2 &amp; 73755 &amp;  &amp; 5 &amp; 203490 &amp;  &amp; 5 &amp; 21 &amp;  &amp; 4 &amp; 595 \\ 
 &amp; 3 &amp; 6184400 &amp; 14 &amp; 2 &amp; 116175 &amp; 21 &amp; 2 &amp; 1 &amp; 34 &amp; 3 &amp; 35 \\ 
 &amp; 4 &amp; 6679475 &amp;  &amp; 3 &amp; 2319688080 &amp;  &amp; 3 &amp; 2319959400 &amp;  &amp; 4 &amp; 35 \\ 
 &amp; 5 &amp; 116280 &amp;  &amp; 4 &amp; 2319959295 &amp;  &amp; 4 &amp; 2319959400 &amp; 35 &amp; 3 &amp; 1 \\ 
 &amp; 6 &amp; 1 &amp;  &amp; 5 &amp; 116280 &amp;  &amp; 5 &amp; 1 &amp;  &amp; 4 &amp; 1 \\
\hline\hline
\end{tabular}
\caption{Values of $T^{m,k}_n$ for $n=7$, $m\in\{2,\ldots,35\}$, and all possible $k$.}\label{tab:5}
\end{table}

\section{Acknowledgment}
The author gratefully acknowledges Charlotte St\c{e}pien for checking the text as well as the anonymous referee for valuable comments and suggestions.


\begin{table}[H]
\centering
\begin{tabular}{cllll}
\hline\hline
 &amp; &amp; \multicolumn{3}{c}{Computational time (s)}\\
\cline{3-5}
$n$ &amp; $m$ &amp; $\Ind(n,m,\ind(n,m-1))$ &amp; $\AC(n,m,\ind(n,m))$ &amp; $\AC(n,m,\indc(n,m))$\\
\hline
12 &amp; 2 &amp;0.01 &amp;0.01 &amp;0 \\
   &amp; 3 &amp;0.19 &amp;0.05 &amp;0.01 \\
	 &amp; 4 &amp;205.98 &amp;12.89 &amp;7.76 \\
13 &amp; 2 &amp;0 &amp;0 &amp;0 \\
   &amp; 3 &amp;0.24 &amp;0.11 &amp;0.03 \\
	 &amp; 4 &amp;536.09 &amp;29.55 &amp;17.03 \\
14 &amp; 2 &amp;0 &amp;0 &amp;0 \\
   &amp; 3 &amp;0.32 &amp;0.18 &amp;0.06 \\
	 &amp; 4 &amp;1668.4 &amp;65.19 &amp;34.6 \\
15 &amp; 2 &amp;0 &amp;0 &amp;0 \\
   &amp; 3 &amp;0.47 &amp;0.25 &amp;0.079 \\
	 &amp; 4 &amp;5217.23 &amp;136.77 &amp;73.77 \\
\hline\hline
\end{tabular}
\caption{The running time of methods $\Ind(n,m)$, $\AC(n,m,A)$ for $m\in\{2,3,4\}$, $n\in\{12,13,14,15\}$, and $A\in\{\ind(n,m),\indc(n,m)\}$.}\label{tab:6}
\end{table}

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\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 05A15; Secondary 90B25.

\noindent \emph{Keywords: }
ordered set, antichain, cover, Dedekind number.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000372},
\seqnum{A000392},
\seqnum{A006126},
\seqnum{A016269},
\seqnum{A047707},
\seqnum{A051112},
\seqnum{A056046}, and
\seqnum{A056047}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received September 1 2020;
revised versions received  February 17 2021; April 16 2021.
Published in {\it Journal of Integer Sequences}, April 17 2021.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}

                                                                                

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