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\theoremstyle{plain}
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\begin{center}
\vskip 1cm{\Large\bf 
On Dedekind Numbers and Two Sequences of Knuth
}
\vskip 1cm
\large
  J. Berman \\
  Department of Mathematics, Statistics, and Computer Science\\
  University of Illinois at Chicago \\
  Chicago, IL 60607\\
  USA\\
  \href{mailto:jberman@uic.edu}{\tt jberman@uic.edu}\\
  \ \\
  P. K\"ohler\footnote{Author's present address: 
  \href{mailto:peter@koehler-weilburg.de}{\tt peter@koehler-weilburg.de}.}
  \\
  Mathematisches Institut\\
  Justus-Liebig-Universit\"at Giessen\\
  Arndtstr.\ 2\\
  35392 Giessen \\
  Germany\\
  \href{mailto:peter.koehler@math.uni-giessen.de}{\tt peter.koehler@math.uni-giessen.de}\\
\end{center}

\vskip .2 in

\begin{abstract}
  We consider the sequence whose $n^{\rm th}$ term is the number $F(n)$
  of anti-chains in the partially ordered set whose elements are
  $0,1, \ldots, n-1$ and the  order relation is coordinate-wise
  on the binary representation of each integer.  
  This sequence is a sort of ``background" sequence to its more
  prominent subsequence of Dedekind numbers, that is, the sequence
  whose terms are $F(2^k)$.
  We also consider the sequence of first differences with terms $F(n)-F(n-1)$. 
  We discuss, state, and prove some (recursive) relations between the
  terms of these three sequences.
\end{abstract}

\section{Introduction}
\label{intro}
Let $P_n$ denote the partially ordered set (poset) whose elements are
$0,1, \ldots, n-1$ and the order relation is coordinate-wise with respect
to the binary representation of the integers in $P_n$.
Thus, if $n=2^k$, then $P_n$ is a Boolean lattice with top element
$2^k-1$ and atoms $2^i$ for $0 \le i  \le 2^{k-1}$.
 
For each $n$ let $F(n)$ be the number of anti-chains in $P_n$.
So $F(2^k)$ is the number of anti-chains in a Boolean lattice with $k$ atoms,
that is, $F(2^k)$ is a Dedekind number. The exact values of the Dedekind
numbers are known only for $k \le 8$. The computation for the case $k=8$
is described in Wiedemann \cite{wie}.
The sequence of Dedekind numbers is \seqnum{A000372} in the
{\it On-Line Encyclopedia of Integer Sequences} \cite{OEIS}.
The many entries in the comments and references
for this sequence there attest to the wide interest in it. 

The integer sequence with terms $F(n)$ is a background sequence for the
sequence of Dedekind numbers.
It is \seqnum{A132581} and was authored by D. E. Knuth.
The function $\Delta(n)=F(n)-F(n-1)$ gives the terms for the
first differences of \seqnum{A132581}; we tacitly extend the definition
by $\Delta(0) = 0$.

The sequence of $\Delta$ values is \seqnum{A132582} and is also
authored by Knuth. Another way to define this sequence would
be to say that $\Delta(n)$ is the number of anti-chains containing $n-1$
in the partially ordered set $P_n$.  

In this paper we investigate the poset $P_n$  and the two
functions $F$ and $\Delta$.
Our main interest is in finding expressions for $F(n)$ and for
$\Delta(n)$ in terms of $F$ or $\Delta$ applied to integers less than $n$.

In Section \ref{basics} we present the basics on posets and
lattices that we will use.  
Section \ref{f-formulas} contains lemmas that express the number
of anti-chains in an
arbitrary finite poset as the sum of the number of anti-chains in
certain subposets. 
We use these lemmas to find two formulas for Dedekind numbers $F(2^k)$
for arbitrary $k$  that involve the function $F$ applied to arguments less
than $2^k$. 
In Section \ref{delta-formulas} we present
formulas for $\Delta(n)$ that only involve the $F$ and $\Delta$
functions applied to arguments less than $n$.  
The final section of the paper is a summary that contains a list of the
formulas proved in the paper.
As an application, we show how these formulas may be used to determine
$\Delta(n)$  in the interval $16 \le n \le 32$.

\section{Basics}
\label{basics}
A {\sl semi-ideal} of a poset $P$ is a subset $I \subseteq P$,
which is closed downwards, i.e., it satisfies the condition
$p \in I, q \leq p \implies q \in I$.
A {\sl semi-filter} is a subset $F$, which is closed upwards, i.e., it
satisfies the condition
$p \in F, q \geq p \implies q \in F$.
An {\sl anti-chain} of $P$ is a set of mutually incomparable elements.
For a poset $P$ and an element $p \in P$ let $\sub(p)$ denote the semi-ideal
$\{q \in P \mid q \leq p\}$ and
$\super(p)$ denote the semi-filter $\{q \in P \mid p \leq q\}$ and let
$\cone(p)$ denote the union $\sub(p) \cup \super(p)$ of these. To facilitate
notation we will use the term $\cocone(p)$ as an abbreviation
of $P \setminus \cone(p)$, as well as $\cosub(p)$ for $P \setminus \sub(p)$
and $\cosuper(p)$ for $P \setminus \super(p)$. If the context of the poset
$P$ is necessary we will express this via a prefix $P.$ as in $P.\cosub(p)$
and $P.\cosuper(p)$. This comes in handy in expressions
like $P.\cosub(p).\cosub(q)$, which of course is equal to
$P \setminus (\sub(p) \cup \sub(q))$. We will also use $\cosub(p,q)$ as a
shorthand for this set, as well
as $\cosuper(p,q)$ for the poset $P \setminus (\super(p) \cup \super(q))$.
For other notation and terminology on partially ordered sets and Boolean
lattices we follow the monographs of Balbes and Dwinger \cite{baldwi}
and Birkhoff \cite{bir}.

It is well known that the set of semi-ideals and the set of semi-filters
of a poset $P$ form distributive lattices ${\cal I}(P)$ and ${\cal F}(P)$
with set intersection and set union. Both these lattices are isomorphic to
the lattice of ${\cal A}(P)$ of anti-chains of $P$, where the meet- and
join-operations are slightly more complicated. Let $\alpha(P)$ denote
the number of elements of ${\cal A}(P)$.

We recall from Section \ref{intro} the definition of the
poset $P_n$ as the set $\{0,\ldots,n-1\}$ with {\em binary} ordering,
i.e., coordinate-wise with respect
to the binary representation of the elements.
In particular, we note that 
$P_0$ is the empty poset. Since the usual ordering of $\mathbb{N}$ 
is different from the binary ordering we express the latter
via $\le_b$ and $\ge_b$ where necessary. We note for later reference
that the natural ordering is an {\em extension} of the binary ordering
in the sense that $p \le_b q$  implies $p \le q$, as well as
$p \ge_b q$ implies $p \ge q$ for all $ p, q \in \mathbb{N}$. To put
it differently, we have 
$P_n.\sub(p) \subseteq [0, p]$ and 
$P_n.\super(p) \subseteq [p, n-1]$, where we use the handy
{\em interval notation}
$[p,q]$ with respect to the natural ordering, i.e.,
$[p,q] = \{x \mid p \le x \le q\}$. 

Figure \ref{p} shows the poset $P_{16}$.
\begin{figure}[htb]
\begin{center}
\unitlength0.4cm
\thinlines
\vskip .3in
\begin{picture}(38,20)
\scriptsize
\put(18,4){\circle*{0.5}} \put(18.5,4) {$0$}
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\drawline(18,4)(15,8)
\drawline(18,4)(21,8)
\drawline(18,4)(29,7)
\put(10,8){\circle*{0.5}} \put(10.5,8) {$1$}
\drawline(10,8)(7,12)
\drawline(10,8)(13,12)
\drawline(10,8)(21,11)
\put(15,8){\circle*{0.5}} \put(15.5,8) {$2$}
\drawline(15,8)(7,12)
\drawline(15,8)(18,12)
\drawline(15,8)(26,11)
\put(7,12){\circle*{0.5}} \put(7.5,12) {$3$}
\drawline(7,12)(10,16)
\drawline(7,12)(18,15)
\put(21,8){\circle*{0.5}} \put(21.5,8) {$4$}
\drawline(21,8)(13,12)
\drawline(21,8)(18,12)
\drawline(21,8)(32,11)
\put(13,12){\circle*{0.5}} \put(13.5,12) {$5$}
\drawline(13,12)(10,16)
\drawline(13,12)(24,15)
\put(18,12){\circle*{0.5}} \put(18.5,12) {$6$}
\drawline(18,12)(10,16)
\drawline(18,12)(29,15)
\put(10,16){\circle*{0.5}} \put(10.5,16) {$7$}
\drawline(10,16)(21,19)
\put(29,7){\circle*{0.5}} \put(29.5,7) {$8$}
\drawline(29,7)(21,11)
\drawline(29,7)(26,11)
\drawline(29,7)(32,11)
\put(21,11){\circle*{0.5}} \put(21.5,11) {$9$}
\drawline(21,11)(18,15)
\drawline(21,11)(24,15)
\put(26,11){\circle*{0.5}} \put(26.5,11) {$10$}
\drawline(26,11)(18,15)
\drawline(26,11)(29,15)
\put(18,15){\circle*{0.5}} \put(18.5,15) {$11$}
\drawline(18,15)(21,19)
\put(32,11){\circle*{0.5}} \put(32.5,11) {$12$}
\drawline(32,11)(24,15)
\drawline(32,11)(29,15)
\put(24,15){\circle*{0.5}} \put(24.5,15) {$13$}
\drawline(24,15)(21,19)
\put(29,15){\circle*{0.5}} \put(29.5,15) {$14$}
\drawline(29,15)(21,19)
\put(21,19){\circle*{0.5}} \put(21.5,19) {$15$}
\end{picture}
%\vskip -.5in
\caption{\label{p} The Boolean poset $B_4 = P_{16}$ with 16 elements.}
\end{center}
\end{figure}

If $n$ is of the form $n = 2^k$, then $P_n$ is
a Boolean poset, which we also denote by $B_k$. The atoms of $B_k$ are
the elements $2^0,\ 2^1, \ldots, 2^{k-2},\ 2^{k-1}$
and the dual atoms are the elements
$2^k-2,\ 2^k-3, \ldots, 2^k-1-2^{k-1}$.
For this Boolean poset we have a slightly closer connection
between the binary and the linear ordering. That is,
\begin{align}
  \label{a1}
P_n.\super(p) &amp;= [p,n-1]\  \mbox{if and only if}\ p = 2^k - 2^m\ \mbox{for some}\ m \le k, \\
  \label{a2}
P_n.\sub(p) &amp;= [0, p]\  \mbox{if and only if}\ p = 2^m-1\ \mbox{for some}\ m \le k.
\end{align}
Both are immediate consequences of the sum formula 
$2^{m-1} - 1 = \sum \limits_{i=0}^{i=m-1} 2^i$.

Another way to see this is via the binary representation of the
elements of $B_k$ as a 0-1--sequence. So for $k=4$ these elements are
\begin{align*}
  2^k-2^0 &amp;= 15 =  1111,\\
  2^k-2^1 &amp;= 14 =  1110,\\
  2^k-2^2 &amp;= 12 =  1100,\\
  2^k-2^3 &amp;= 8 =  1000,\\
  2^k-2^4 &amp;= 0 =  0000.
\end{align*}
In general, these elements are those, whose binary representation is a
sequence with `leading ones'.

Every automorphism of $B_k$ is determined by its action on the atoms, and
therefore the automorpism group of $B_k$ is isomorphic to the symmetric
group on $k$ elements. Of particular interest in the following is the
{\em atom inflection} $\beta$, which maps the atom $2^0$ onto the atom
$2^{k-1}$,
the atom $2^1$ onto the atom $2^{k-2}$, and so on.
Figure \ref{beta} shows the order automorphism $\beta$ of $B_4$.
(Note that in order to avoid excessive arrows
we restrict ourselves to indicate the action by an appropriate labelling.)
\begin{figure}[htb]
\unitlength0.4cm
\thinlines
\begin{picture}(38,20)
\scriptsize
\put(18,4){\circle*{0.5}} \put(18.5,4) {$0=\beta(0)$}
\drawline(18,4)(10,8)
\drawline(18,4)(15,8)
\drawline(18,4)(21,8)
\drawline(18,4)(29,7)
\put(10,8){\circle*{0.5}} \put(10.5,8) {$1=\beta(8)$}
\drawline(10,8)(7,12)
\drawline(10,8)(13,12)
\drawline(10,8)(21,11)
\put(15,8){\circle*{0.5}} \put(15.5,8) {$2=\beta(4)$}
\drawline(15,8)(7,12)
\drawline(15,8)(18,12)
\drawline(15,8)(26,11)
\put(7,12){\circle*{0.5}} \put(7.5,12) {$3=\beta(12)$}
\drawline(7,12)(10,16)
\drawline(7,12)(18,15)
\put(21,8){\circle*{0.5}} \put(21.5,8) {$4=\beta(2)$}
\drawline(21,8)(13,12)
\drawline(21,8)(18,12)
\drawline(21,8)(32,11)
\put(13,12){\circle*{0.5}} \put(13.5,12) {$5=\beta(10)$}
\drawline(13,12)(10,16)
\drawline(13,12)(24,15)
\put(18,12){\circle*{0.5}} \put(18.5,12) {$6=\beta(6)$}
\drawline(18,12)(10,16)
\drawline(18,12)(29,15)
\put(10,16){\circle*{0.5}} \put(10.5,16) {$7=\beta(14)$}
\drawline(10,16)(21,19)
\put(29,7){\circle*{0.5}} \put(29.5,7) {$8=\beta(1)$}
\drawline(29,7)(21,11)
\drawline(29,7)(26,11)
\drawline(29,7)(32,11)
\put(21,11){\circle*{0.5}} \put(21.5,11) {$9=\beta(9)$}
\drawline(21,11)(18,15)
\drawline(21,11)(24,15)
\put(26,11){\circle*{0.5}} \put(26.5,11) {$10=\beta(5)$}
\drawline(26,11)(18,15)
\drawline(26,11)(29,15)
\put(18,15){\circle*{0.5}} \put(18.5,15) {$11=\beta(13)$}
\drawline(18,15)(21,19)
\put(32,11){\circle*{0.5}} \put(32.5,11) {$12=\beta(3)$}
\drawline(32,11)(24,15)
\drawline(32,11)(29,15)
\put(24,15){\circle*{0.5}} \put(24.5,15) {$13=\beta(11)$}
\drawline(24,15)(21,19)
\put(29,15){\circle*{0.5}} \put(29.5,15) {$14=\beta(7)$}
\drawline(29,15)(21,19)
\put(21,19){\circle*{0.5}} \put(21.5,19) {$15=\beta(15)$}
\end{picture}
\vskip-.2in
\caption{\label{beta} The action of $\beta$ on $B_4 = P_{16}$.}
\end{figure}
We also denote the isomorphism that
exchanges the atoms $2^i$ and $2^j$ and keeps all the other atoms fixed, by
$\rho_{i,j}$. 

Figure \ref{rho} shows $\rho_{0,1}$ on $B_4$.
\begin{figure}[htb]
\unitlength0.4cm
\thinlines
\vskip .5in
\begin{picture}(38,16)
\scriptsize
\put(18,4){\circle*{0.5}} \put(18.5,4) {$0=\rho_{0,1}(0)$}
\drawline(18,4)(10,8)
\drawline(18,4)(15,8)
\drawline(18,4)(21,8)
\drawline(18,4)(29,7)
\put(10,8){\circle*{0.5}} \put(10.5,8) {$1=\rho_{0,1}(2)$}
\drawline(10,8)(7,12)
\drawline(10,8)(13,12)
\drawline(10,8)(21,11)
\put(15,8){\circle*{0.5}} \put(15.5,8) {$2=\rho_{0,1}(1)$}
\drawline(15,8)(7,12)
\drawline(15,8)(18,12)
\drawline(15,8)(26,11)
\put(7,12){\circle*{0.5}} \put(7.5,12) {$3=\rho_{0,1}(3)$}
\drawline(7,12)(10,16)
\drawline(7,12)(18,15)
\put(21,8){\circle*{0.5}} \put(21.5,8) {$4=\rho_{0,1}(4)$}
\drawline(21,8)(13,12)
\drawline(21,8)(18,12)
\drawline(21,8)(32,11)
\put(13,12){\circle*{0.5}} \put(13.5,12) {$5=\rho_{0,1}(6)$}
\drawline(13,12)(10,16)
\drawline(13,12)(24,15)
\put(18,12){\circle*{0.5}} \put(18.5,12) {$6=\rho_{0,1}(5)$}
\drawline(18,12)(10,16)
\drawline(18,12)(29,15)
\put(10,16){\circle*{0.5}} \put(10.5,16) {$7=\rho_{0,1}(7)$}
\drawline(10,16)(21,19)
\put(29,7){\circle*{0.5}} \put(29.5,7) {$8=\rho_{0,1}(8)$}
\drawline(29,7)(21,11)
\drawline(29,7)(26,11)
\drawline(29,7)(32,11)
\put(21,11){\circle*{0.5}} \put(21.5,11) {$9=\rho_{0,1}(10)$}
\drawline(21,11)(18,15)
\drawline(21,11)(24,15)
\put(26,11){\circle*{0.5}} \put(26.5,11) {$10=\rho_{0,1}(9)$}
\drawline(26,11)(18,15)
\drawline(26,11)(29,15)
\put(18,15){\circle*{0.5}} \put(18.5,15) {$11=\rho_{0,1}(11)$}
\drawline(18,15)(21,19)
\put(32,11){\circle*{0.5}} \put(32.5,11) {$12=\rho_{0,1}(12)$}
\drawline(32,11)(24,15)
\drawline(32,11)(29,15)
\put(24,15){\circle*{0.5}} \put(24.5,15) {$13=\rho_{0,1}(14)$}
\drawline(24,15)(21,19)
\put(29,15){\circle*{0.5}} \put(29.5,15) {$14=\rho_{0,1}(13)$}
\drawline(29,15)(21,19)
\put(21,19){\circle*{0.5}} \put(21.5,19) {$15=\rho_{0,1}(15)$}
\end{picture}
\vskip -.3in
\caption{\label{rho} The action of $\rho$ on $B_4 = P_{16}$.}
\end{figure}

The {\em complementation} $\gamma$ defined by
$\gamma(x) = 2^k-1-x$ is an order anti-automorphism.
Figure \ref{gamma} shows $\gamma$ on $B_4$.
\begin{figure}[htb]
\unitlength0.4cm
\thinlines
\begin{picture}(38,18)
\scriptsize
\put(18,4){\circle*{0.5}} \put(18.5,4) {$0=\gamma(15)$}
\drawline(18,4)(10,8)
\drawline(18,4)(15,8)
\drawline(18,4)(21,8)
\drawline(18,4)(29,7)
\put(10,8){\circle*{0.5}} \put(10.5,8) {$1=\gamma(14)$}
\drawline(10,8)(7,12)
\drawline(10,8)(13,12)
\drawline(10,8)(21,11)
\put(15,8){\circle*{0.5}} \put(15.5,8) {$2=\gamma(13)$}
\drawline(15,8)(7,12)
\drawline(15,8)(18,12)
\drawline(15,8)(26,11)
\put(7,12){\circle*{0.5}} \put(7.5,12) {$3=\gamma(12)$}
\drawline(7,12)(10,16)
\drawline(7,12)(18,15)
\put(21,8){\circle*{0.5}} \put(21.5,8) {$4=\gamma(11)$}
\drawline(21,8)(13,12)
\drawline(21,8)(18,12)
\drawline(21,8)(32,11)
\put(13,12){\circle*{0.5}} \put(13.5,12) {$5=\gamma(10)$}
\drawline(13,12)(10,16)
\drawline(13,12)(24,15)
\put(18,12){\circle*{0.5}} \put(18.5,12) {$6=\gamma(9)$}
\drawline(18,12)(10,16)
\drawline(18,12)(29,15)
\put(10,16){\circle*{0.5}} \put(10.5,16) {$7=\gamma(8)$}
\drawline(10,16)(21,19)
\put(29,7){\circle*{0.5}} \put(29.5,7) {$8=\gamma(7)$}
\drawline(29,7)(21,11)
\drawline(29,7)(26,11)
\drawline(29,7)(32,11)
\put(21,11){\circle*{0.5}} \put(21.5,11) {$9=\gamma(6)$}
\drawline(21,11)(18,15)
\drawline(21,11)(24,15)
\put(26,11){\circle*{0.5}} \put(26.5,11) {$10=\gamma(5)$}
\drawline(26,11)(18,15)
\drawline(26,11)(29,15)
\put(18,15){\circle*{0.5}} \put(18.5,15) {$11=\gamma(4)$}
\drawline(18,15)(21,19)
\put(32,11){\circle*{0.5}} \put(32.5,11) {$12=\gamma(3)$}
\drawline(32,11)(24,15)
\drawline(32,11)(29,15)
\put(24,15){\circle*{0.5}} \put(24.5,15) {$13=\gamma(2)$}
\drawline(24,15)(21,19)
\put(29,15){\circle*{0.5}} \put(29.5,15) {$14=\gamma(1)$}
\drawline(29,15)(21,19)
\put(21,19){\circle*{0.5}} \put(21.5,19) {$15=\gamma(0)$}
\end{picture}
\vskip -.4in
\caption{\label{gamma} The action of $\gamma$ on $B_4 = P_{16}$.}
\end{figure}

The composition $\delta = \beta \gamma$ is an order anti-automorphism
that acts like a {\em top-down inflection}.
Figure \ref{delta} shows $\delta$ on $B_4$.
\begin{figure}[htb]
\unitlength0.4cm
\thinlines
\begin{picture}(38,20)
\scriptsize
\put(18,4){\circle*{0.5}} \put(18.5,4) {$0=\delta(15)$}
\drawline(18,4)(10,8)
\drawline(18,4)(15,8)
\drawline(18,4)(21,8)
\drawline(18,4)(29,7)
\put(10,8){\circle*{0.5}} \put(10.5,8) {$1=\delta(7)$}
\drawline(10,8)(7,12)
\drawline(10,8)(13,12)
\drawline(10,8)(21,11)
\put(15,8){\circle*{0.5}} \put(15.5,8) {$2=\delta(11)$}
\drawline(15,8)(7,12)
\drawline(15,8)(18,12)
\drawline(15,8)(26,11)
\put(7,12){\circle*{0.5}} \put(7.5,12) {$3=\delta(3)$}
\drawline(7,12)(10,16)
\drawline(7,12)(18,15)
\put(21,8){\circle*{0.5}} \put(21.5,8) {$4=\delta(13)$}
\drawline(21,8)(13,12)
\drawline(21,8)(18,12)
\drawline(21,8)(32,11)
\put(13,12){\circle*{0.5}} \put(13.5,12) {$5=\delta(5)$}
\drawline(13,12)(10,16)
\drawline(13,12)(24,15)
\put(18,12){\circle*{0.5}} \put(18.5,12) {$6=\delta(9)$}
\drawline(18,12)(10,16)
\drawline(18,12)(29,15)
\put(10,16){\circle*{0.5}} \put(10.5,16) {$7=\delta(1)$}
\drawline(10,16)(21,19)
\put(29,7){\circle*{0.5}} \put(29.5,7) {$8=\delta(14)$}
\drawline(29,7)(21,11)
\drawline(29,7)(26,11)
\drawline(29,7)(32,11)
\put(21,11){\circle*{0.5}} \put(21.5,11) {$9=\delta(6)$}
\drawline(21,11)(18,15)
\drawline(21,11)(24,15)
\put(26,11){\circle*{0.5}} \put(26.5,11) {$10=\delta(10)$}
\drawline(26,11)(18,15)
\drawline(26,11)(29,15)
\put(18,15){\circle*{0.5}} \put(18.5,15) {$11=\delta(2)$}
\drawline(18,15)(21,19)
\put(32,11){\circle*{0.5}} \put(32.5,11) {$12=\delta(12)$}
\drawline(32,11)(24,15)
\drawline(32,11)(29,15)
\put(24,15){\circle*{0.5}} \put(24.5,15) {$13=\delta(4)$}
\drawline(24,15)(21,19)
\put(29,15){\circle*{0.5}} \put(29.5,15) {$14=\delta(8)$}
\drawline(29,15)(21,19)
\put(21,19){\circle*{0.5}} \put(21.5,19) {$15=\delta(0)$}
\end{picture}
\vskip -.5in
\caption{\label{delta} The action of $\delta$ on $B_4 = P_{16}$.}
\end{figure}

And since $\gamma$
commutes with all the order isomorphisms, the group of all
order automorphisms and order anti-automorphisms of $B_k$ is is
isomorphic to the (direct) product of the symmetric group and an
inflection group of order 2.

Let $D_n$ denote the lattice ${\cal I}(P_n$).
Then for $P_8$ as shown in Figure \ref{b3},
the lattice $D_8$ consists of the 20 semi-ideals
  $$ \emptyset,
  \{0\},
  \{0,1\},\{0,2\},\{0,4\},
  $$
  $$
  \{0,1,2\},\{0,1,4\},\{0,2,4\},
  \{0,1,2,3\},\{0,1,2,4\},\{0,1,4,5\},\{0,2,4,6\},
  $$
  $$
  \{0,1,2,3,4\},\{0,1,2,4,5\},\{0,1,2,4,6\},
  \{0,1,2,3,4,5\},\{0,1,2,4,6\},\{0,1,2,5,6\},
  $$
  $$
  \{0,1,2,3,4,5,6\}
  \{0,1,2,3,4,5,6,7\}
  $$
and has the lattice diagram of Figure \ref{fd3}.
\begin{figure}[htb]
\begin{center}
\unitlength0.4cm
\thinlines
\begin{picture}(18,11)
\scriptsize

\put(11,10){\circle*{0.5}}     \put(12,10) {$7$}

\put(14,8){\circle*{0.5}}    \put(15,8){$6$}
\put(14,8){\line(-3, 2){3}}

\put(11,8){\circle*{0.5}}    \put(12,8){$5$}
\put(11,8){\line(0, 1){2}}

\put(8,8){\circle*{0.5}}    \put(9,8){$3$}
\put(8,8){\line(3, 2){3}}

\put(14,6){\circle*{0.5}}     \put(15,6) {$4$}
\put(14,6){\line(-3, 2){3}}
\put(14,6){\line(0, 1){2}}

\put(11,6){\circle*{0.5}}     \put(12,6) {$2$}
\put(11,6){\line(-3, 2){3}}
\put(11,6){\line(3, 2){3}}

\put(8,6){\circle*{0.5}}     \put(9,6) {$1$}
\put(8,6){\line(0, 1){2}}
\put(8,6){\line(3, 2){3}}

\put(11,4){\circle*{0.5}}     \put(12,4) {$0$}
\put(11,4){\line(-3, 2){3}}
\put(11,4){\line(0, 1){2}}
\put(11,4){\line(3, 2){3}}

\end{picture}
\vskip -.5in
\caption{\label{b3} The Boolean poset $B_3 = P_8$ with 8 elements.}
\end{center}
\end{figure}
\begin{figure}[htb]
\begin{center}
\unitlength0.4cm
\thinlines
\begin{picture}(18,19)
\scriptsize

\put(11,18){\circle*{0.5}}

\put(11,16){\circle*{0.5}}
\put(11,16){\line(0, 1){2}}

\put(14,14){\circle*{0.5}}
\put(14,14){\line(-3, 2){3}}

\put(11,14){\circle*{0.5}}
\put(11,14){\line(0, 1){2}}

\put(8,14){\circle*{0.5}}
\put(8,14){\line(3, 2){3}}

\put(14,12){\circle*{0.5}}
\put(14,12){\line(0, 1){2}}
\put(14,12){\line(-3, 2){3}}

\put(11,12){\circle*{0.5}}
\put(11,12){\line(-3, 2){3}}
\put(11,12){\line(3, 2){3}}

\put(8,12){\circle*{0.5}}
\put(8,12){\line(0, 1){2}}
\put(8,12){\line(3, 2){3}}

\put(17,10){\circle*{0.5}}
\put(17,10){\line(-3, 2){3}}

\put(14,10){\circle*{0.5}}
\put(14,10){\line(-3, 2){3}}

\put(11,10){\circle*{0.5}}
\put(11,10){\line(-3, 2){3}}
\put(11,10){\line(0, 1){2}}
\put(11,10){\line(3, 2){3}}

\put(5,10){\circle*{0.5}}
\put(5,10){\line(3, 2){3}}

\put(14,8){\circle*{0.5}}
\put(14,8){\line(-3, 2){3}}
\put(14,8){\line(3, 2){3}}

\put(11,8){\circle*{0.5}}
\put(11,8){\line(0, 1){2}}
\put(11,8){\line(3, 2){3}}

\put(8,8){\circle*{0.5}}
\put(8,8){\line(-3, 2){3}}
\put(8,8){\line(3, 2){3}}

\put(14,6){\circle*{0.5}}
\put(14,6){\line(-3, 2){3}}
\put(14,6){\line(0, 1){2}}

\put(11,6){\circle*{0.5}}
\put(11,6){\line(-3, 2){3}}
\put(11,6){\line(3, 2){3}}

\put(8,6){\circle*{0.5}}
\put(8,6){\line(0, 1){2}}
\put(8,6){\line(3, 2){3}}

\put(11,4){\circle*{0.5}}
\put(11,4){\line(-3, 2){3}}
\put(11,4){\line(0, 1){2}}
\put(11,4){\line(3, 2){3}}

\put(11,2){\circle*{0.5}}
\put(11,2){\line(0, 1){2}}

\end{picture}
\vskip -.1in
\caption{\label{fd3} The lattice $D_8$.}
\end{center}
\end{figure}

If we let $D_{01}$
denote the variety of all distributive lattices with $0$ and $1$,
and $\FD_{01}(n)$ denote
the free algebra of $D_{01}$ on $n$ generators,
then it is well known that 
$D_{2^n} \cong \FD_{01}(n)$ for all $n \in \mathbb{N}$.

The values of \seqnum{A132581} and its
difference sequence \seqnum{A132582}
have been computed for the first 212 terms (J. M. Aranda).
In order to illustrate the results of the following sections, we list the first 32 terms
in Table \ref{fndeltan}.

\begin{table}[htb]
\begin{center}
\begin{small}
\begin{tabular}{|l|l|l||l|l|l|}
  \hline
  \bf{$n$} &amp; \bf{$F(n)$} &amp; \bf{$\Delta(n)$} &amp; \bf{$n$} &amp; \bf{$F(n)$} &amp; \bf{$\Delta(n)$} \\ \hline
1 &amp; 2 &amp;       1 &amp; 17 &amp; 335  &amp; 167                   \\ \hline
2 &amp; 3 &amp;       1 &amp; 18 &amp; 483  &amp; 148                   \\   \hline
3 &amp; 5 &amp;       2 &amp; 19 &amp; 765  &amp; 282                   \\   \hline
4 &amp; 6 &amp;       1 &amp; 20 &amp; 849  &amp;  84                   \\   \hline
5 &amp; 11 &amp;      5 &amp; 21 &amp; 1466 &amp; 617                   \\   \hline
6 &amp; 14 &amp;      3 &amp; 22 &amp; 1681 &amp;215                    \\   \hline
7 &amp; 19 &amp;      5 &amp; 23 &amp; 1988 &amp;307                    \\   \hline
8 &amp; 20 &amp;      1 &amp; 24 &amp; 2008 &amp;20                     \\   \hline
9 &amp; 39 &amp;     19 &amp; 25 &amp; 3700 &amp;1692                   \\ \hline
10 &amp; 53 &amp;    14 &amp; 26 &amp; 4414 &amp;714                    \\ \hline
11 &amp; 78 &amp;    25 &amp; 27 &amp; 5489 &amp;1075                   \\ \hline
12 &amp; 84 &amp;     6 &amp; 28 &amp; 5573 &amp;84                     \\ \hline
13 &amp; 134 &amp;    50 &amp; 29 &amp; 7265 &amp;1692                  \\ \hline
14 &amp; 148  &amp;  14 &amp; 30 &amp; 7413 &amp;148                    \\ \hline
15 &amp; 167  &amp;  19 &amp; 31 &amp; 7580 &amp;167                    \\ \hline
16 &amp; 168  &amp;   1 &amp; 32 &amp; 7581 &amp;1                      \\ \hline
\end{tabular}
\end{small}
\end{center}
\caption{$F(n)$ and $\Delta(n)$.}
\label{fndeltan}
\end{table}

\section{Formulas for $F(n)$}
\label{f-formulas}
An old result allowing a (recursive) computation of the
number of anti-chains of a poset is taken from Berman and K\"ohler
\cite[Thm.~2.1]{berkoe}.
\begin{lemma}
\label{th0}
For every poset $P$ and every $p \in P$, we have
$\alpha(P) = \alpha(P\setminus \{p\}) + \alpha(\cocone(p))$.
\end{lemma}
Applying this to the poset $P_n$ and the element $n-1$ gives
\begin{equation}
\label{z1}
F(n) = F(n-1) + \alpha(\cocone(n-1)).
\end{equation}
    
Actually (\ref{z1}) is still the basis for the computation of the
known sequence terms. For later use we introduce the notion of
a {\em pivot}. For a poset $Q$ and an element $q\in Q$ we say
that the poset $Q$ {\it pivots} on the element $q$
to produce the two smaller sets  
$Q\setminus \{q\}$ and $Q.\cocone(q)$.

Using the arguments of Lemma \ref{th0} in a slightly more
sophisticated way one can obtain the following alternative from
a Campo and Ern\'e \cite[Cor.~3.2]{camer}.

\begin{lemma}
\label{th1}
For every poset $P$ and every $p \in P$, we have
$\alpha(P) = \alpha(\cosuper(p)) + \alpha(\cosub(p))$.
\end{lemma}

Applying this to the Boolean poset $B_n$ and an atom of the
form $2^k$ we get

\begin{formula}
\label{fo1}
For all $n, k \in \mathbb{N}$ with $k \leq n$, we have
$F(2^n) = F(2^n-2^k) + F(2^n - 2^{n-k})$.
\end{formula}

This gives the following interesting special cases
\begin{align*}
F(2^n) &amp;= F(2^n - 1) + 1, \\
F(2^n) &amp; F(2^n - 2) + F(2^{n-1}), \\
F(2^n) &amp;= F(2^n - 4) + F(3 \cdot 2^{n-2}).
\end{align*}
In particular, the case where $n$ is even and $k = \frac{n}{2}$ yields
$F(2^n) = 2 \cdot F(2^n - 2^k)$,
and this is another proof of the fact that $|\FD(n)|$ is even for even $n$.

Formula \ref{fo1} was stated without proof by J.\ M.\ Aranda in the Formula section of \seqnum{A132581}.

Before starting the proof of Formula \ref{fo1} we begin with some general remarks
about the elements $2^n-2^k$ of $B_n$.
As already mentioned in Section \ref{basics} (\ref{a1}),(\ref{a2}),
integers of the form $2^n -2^k$ have the unique and special property that
$[2^n-2^k, 2^n-1] = \super(2^n-2^k)$.
The elements $2^n-2^k$ form a maximal chain in $B_n$ that appears on the right side of the
diagram of $B_n$.
The complement of $2^n-2^k$ is $2^k-1$ and all of these complements appear as a maximal chain 
on the left side in the diagram of $B_n$.
And it is also easy to see that the anti-automorphism $\delta$
maps $2^n-2^k$ onto $2^n-2^{n-k}$.
As a result, the poset $\cone(2^n -2^k)$ is dually isomorphic to $\cone(2^n-2^{n-k})$.  
In particular, $\cone(2^n-2^k)$ is isomorphic to the poset $B_k$ placed above the poset 
$B_{n-k}$ with the bottom element of $B_k$ equal to the top element of $B_{n-k}$. This can
also be expressed as
\begin{equation}
  \label{x1}
  \cone(2^n-2^k) = [2^n-2^k, 2^n-1] \ \cup \ 
  \{i \cdot 2^k \mid i=0,1, \ldots, 2^{n-k}-1\}.
\end{equation}

In addition
\begin{align}
  \label{x2}
  &amp; B_n.\cosuper(2^n-2^k)  \mbox{\ is isomorphic to \ } P_{2^n-2^k}\\
  \label{x3}
  &amp; B_n.\cosub(2^n-2^k)   \mbox{\ is dually isomorphic to \ } P_{2^n-2^{n-k}}.
\end{align}
We also note that (\ref{x2}) and (\ref{x3}) hold for every element of
rank $n-k$, i.e., for every element above exactly $n-k$ atoms.

\begin{proof}
(of Formula \ref{fo1})
  Combining (\ref{x2}), (\ref{x3}), and Lemma \ref{th1} gives the result.
\end{proof}

Using the principle of inclusion and exclusion one can easily extend the
result of Lemma \ref{th1} to a summation of more terms.
\begin{lemma}
\label{th2}
For every poset $P$ and every $p, q \in P$, we have
$\alpha(P) = \alpha(\cosuper(p,q)) + \alpha(\cosub(p))
+ \alpha(\cosub(q)) - \alpha(\cosub(p,q))$.
\end{lemma}

\begin{lemma}
\label{th2b}
For every poset $P$ and every $p, q \in P$, we have
$\alpha(P) = \alpha(\cosub(p,q)) + \alpha(\cosuper(p))
+ \alpha(\cosuper(q)) - \alpha(\cosuper(p,q))$.
\end{lemma}

With the help of Lemma \ref{th2} and and two conveniently chosen elements of $B_n$ we get
\begin{formula}
\label{fo2}
For $k, n \in \mathbb{N}$ with $1 \leq k &lt; n$, we have
$F(2^n) = F(2^n-2^{n-k}-2^{n-k-1}) + 2 F(2^n-2^k) - F(2^n-2^k-2^{k-1})$.
\end{formula}

E.g., for $n=5, k=2$ we have
$F(32) = F(20)+2\cdot F(28)-F(26)$ and for $n=5, k=3$ we have $F(32) = F(26) + 2 \cdot F(24) - F(20)$.

\begin{proof}
  Let $P = B_n$. We consider the two elements $a_1 = 2^n-2^{n-k}$
  and $a_2 = 2^n-2^{n-k}-2^{n-k-1}$ and note that
  $a_2$ is the image of $a_1$ under the automorphism
  $\rho_{n-k,n-k-1}$, that exchanges the atoms $2^{n-k}$ and $2^{n-k-1}$
  and keeps the other atoms fixed. 

  By Lemma \ref{th2} we have
  $F(2^n) = \alpha(P.\cosuper(a_1,a_2)) + \alpha(P.\cosub(a_1))
  + \alpha(P.\cosub(a_2)) - \alpha(P.\cosub(a_1,a_2))$.
  We first observe that $P.\cosuper(a_1,a_2) = P_{a_2}$,
  which accounts for the first term in the formula.

  Moreover $\alpha(P.\cosub(a_1)) = F(2^n-2^k)$ by (\ref{x2}).
  And as $a_1$ is mapped onto $a_2$ by the automorphism
  $\rho_{n-k,n-k-1}$, we have the same result for
  $P.\cosub(a_2)$. This accounts for the second term in the formula.

  To account for the subtraction of the last term
  we use the anti-automorphism $\delta$ as
  in the derivation of (\ref{x1}). Let $b_i = \delta(a_1)$
  for $i=1,2$. Then
  $P.\cosub(a_1,a_2)$ is dually isomorphic to
  $P.\cosuper(b_1,b_2)$ and by the argument of the first
  part this shows that $\alpha(\cosub(a_1,a_2) = \alpha(P.\cosuper(b_1, b_2)) = F(b_2)$.
\end{proof}

In addition, it would be possible to formulate an analog
to Lemma \ref{th2} for three elements. There was,
however, no immediate application of that to gain another recursion formula
involving $F$-terms.

Other recursions are possible, but they are more conveniently formulated using
the terms of the $\Delta$-sequence---we will discuss that in the following
section.

\section{Formulas for $\Delta(n)$}
\label{delta-formulas}
Let $d(n)$ denote the poset $P_n.\cosub(n-1)$. As $n-1$ is a maximal element
of $P_n$ we have $P_n.\cosuper(n-1) = P_n \setminus \{n-1\}$ and 
$P_n.\cosub(n-1) = P_n.\cocone(n-1)$ and therefore 
$\Delta(n)=\alpha(d(n))$ by Lemma \ref{th0} and Lemma \ref{th1}.

We wish to find formulas for $\Delta(n)$ that involve the functions $\Delta$ and $F$
applied to arguments that are less than $n$.

\begin{formula}
\label{fo3}
For all $n,k \in \mathbb{N}$ with $0\le k\le n$, we have
$\Delta(2^n-2^k+1) = \Delta(2^n-2^{n-k}+1)$.
\end{formula}

E.g., for $n=5, k=2$ we have
$\Delta(29) = \Delta(25)$.

\begin{proof}
In order to prove the formula it suffices to show that the poset
$d(2^n-2^k+1)$ is dually isomorphic to $d(2^n+2^{n-k}+1)$.
By definition, we have 
$$d(2^n-2^k+1) = P_{2^n-2^k+1}.\cocone(2^n-2^k).$$
And by (\ref{x2}) we have $B_n= [0,2^n-2^k]\ \cup \ \super(2^n-2^k)$.
Therefore 
$$B_n.\cocone(2^n-2^k) = P_{2^n-2^k+1}.\cocone(2^n-2^k) =d(2^n-2^k+1).$$
And as $\delta(2^n-2^k) = 2^n-2^{n-k}$, we get
$$\delta(d(2^n-2^k+1)) = \delta(B_n.\cocone(2^n-2^k)) = B_n.\cocone(2^n-2^{n-k}) = d(2^n-2^{n-k}+1).$$
\end{proof}

The fact that $B_n.\cocone(2^n-2^k) = d(2^n-2^k+1)$ allows for a description of the poset
 $d(2^n-2^k+1)$ as follows.
The minimal elements of this poset are the $k$ atoms 
$2^0, 2^1, \ldots, 2^{k-1}$ of $B_n$.
The maximal elements of this poset are the $n-k$ co-atoms 
$$(2^n-1)-2^{n-1}, (2^n-1)-2^{n-2}, \ldots, 2^k$$
of $B_n$.
An element $b$ of $B_n$ is in $d(2^n-2^k+1)$ if and only if 
there exist elements $a$ and $c$ such that $a \le b \le c$
with $a$ one of the $k$ atoms in this list of minimal elements and $c$ one of the 
$n-k$ co-atoms in this list of maximal elements.
Despite this description,
we do not have a direct formula for the number of anti-chains in this poset.

We do have, however, a sum formula, which can be proven along the same lines
as Formula \ref{fo2}.

\begin{formula}
\label{fo11}
For all $k,n \in \mathbb{N}$ with $1 \le k \le n-1$, we have
$$\Delta(2^n-2^k+1) = \Delta(2^n-2^k-2^{k-1}+1) + \Delta(2^n-2^{n-k}-2^{n-k-1}+1).$$
\end{formula}

E.g., for $n=5, k=2$ we have
$\Delta(29) = \Delta(27)+ \Delta(21)$ and, combining this with Formula \ref{fo3}, 
$\Delta(27) = \Delta(25) - \Delta(21)$.

\begin{proof}
Let $a = 2^n - 2^k$ and $P = P_{a+1}$,
Then $d(a+1) = P.\cosub(a)$.
Let 
$$x = \rho_{k-1,k}(a) = 2^n - 2^k - 2^{k-1}.$$
Then $P.\super(x) =  \{i \mid x \le i &lt; a \}$, since $x$
is covered by the $k-1$ elements $x + 2^0, \ldots , x + 2^{k-2}$.
As a result, $P.\cosuper(x) = P \setminus \{i \mid x \le i &lt; a \}$
and that is easily seen to be isomorphic to $P_{x+1}$
via the isomorphism $\rho_{k-1,k}$.

We form $d_1 = d(a+1).\cosub(x)$ and $d_2 = d(a+1).\cosuper(x)$.
By Lemma \ref{th1} we have $\Delta(a+1) = \alpha(d_1) + \alpha(d_2)$.
So it remains to show that $\alpha(d_2) = \alpha(d(2^n-2^k-2^{k-1}+1))$
and $\alpha(d_1) = \alpha(d(2^n-2^{n-k}-2^{n-k-1}+1))$.

For $d_2$ this is rather immediate, since
$d_2 = P.\cosuper(x).\cosub(a)$ and that is, by the remark above,
isomorphic to $P_{x+1}.\cosub(x)$, which is $d(x+1)$.

For $d_1$ this is not as obvious,
but we can make use of the dual isomorphism between $d(a+1)$ and
$d(2^n + 2^{n-k} +1)$ of the proof of Formula \ref{fo3}. In fact,
$\delta(a) = 2^n - 2^{n-k}$
and $\delta(x) = 2^n - 2^{n-k} - 2^{n-k-1}$. This implies that
$d_1$ is dually isomorphic to 
$$d(2^n + 2^{n-k} +1).\cosuper(2^n - 2^{n-k} - 2^{n-k-1}).$$
And this finally is, by the same reasoning as for $d_2$ above, isomorphic
to $d(2^n-2^{n-k}-2^{n-k-1}+1)$.
\end{proof}

Applying Formula \ref{fo11} for specific combinations of $n, k$ gives
some interesting connections.
For even $n$, say $n=2q$, we get
$$\Delta(2^{2q}-2^q+1) = 2 \cdot \Delta(2^{2q}-2^q-2^{q-1}+1),$$
e.g., $\Delta(13) = 2 \cdot \Delta(11)$ or $\Delta(57) = 2 \cdot \Delta(53)$.

An additional curiosity here is that $\Delta(13)=50$ and 50 is the largest
$\Delta$ value among all $\Delta(i)$ for $i\le 2^4$ and $\Delta(57)=2453690$
is the largest $\Delta$ value among all $\Delta(i)$ for $1\le i \le 2^6$.
This suggests the general result for $n=2q$ that $\Delta(2^{2q}-2^q+1)$ 
is the largest value of $\Delta(i)$ for $i \le n$.

If we continue to speculate in this way for $n$ odd, say, $n=2q-1$ and
if we look for the value of $\Delta(i)$ that is largest among all $i \le 2^n$, then
\begin{itemize}
\item[]
  for $n=2\cdot 2-1=3$ the largest value is $5=\Delta(5)=\Delta(7)$,
\item[]
for $n=2\cdot 3-1=5$ the largest value is $1692=\Delta(25)=\Delta(29)$,
\item[]
for $n=2\cdot 4-1=7$ the largest value is $510955171111=\Delta(113)=\Delta(121)$.
\end{itemize}
This suggests that in general the largest $\Delta(i)$ value for $i \le n=2q-1$ is obtained by 
$\Delta(2^{2q-1} - 2^{q-1}+1)$ and by $\Delta(2^{2q-1} - 2^q+1)$.
Proving this may be quite difficult.

We also note that from Table \ref{fndeltan} we see that the
equalities
\begin{align*}
\Delta(2^5-3)&amp;=6\Delta(2^4+3) \quad \text{ and } \\
\Delta(2^4-3)&amp;=2 \cdot \Delta(2^3+3)
\end{align*} hold.
But $\Delta(2^6-3)=580655=5 \cdot 116131$,
and 116131 is prime and is not a $\Delta$ or an $F$ value.

Here are some other recursion formulas.

\begin{formula}
\label{fo4}
For all $k,n \in \mathbb{N}$ with $0 \le k \le n-1$, we have
$\Delta(2^n-2^k) =F(2^{n-1}-2^k)$.
\end{formula}

\begin{proof}
Let $P = P_{2^n}$ and $a = 2^n - 2^k$.
Then by (\ref{x2}) $P.\cosuper(a) = P_a$
and therefore $d(a) = P.\cosuper(a).\cosub(a-1)$.
We now observe that $a-1 = 2^n - 1 - 2^k$
is a dual atom of $P$ and, as noted right after (\ref{x3}),
this implies that the poset $Q$
with 
$$Q = P.\cosub(a-1)$$
is dually isomorphic to $P_{2^{n-1}}$.
If $\phi$ is this dual isomorphism then
$$\phi(Q.\cosuper(a)) = P_{2^{n-1}}.\cosuper(\phi(a))$$
and $\phi(a)$ has rank $k$ in $P_{2^{n-1}}$. As a
result, $\phi(Q.\cosuper(a))$ is isomorphic to
$P_{2^{n-1} - 2^k}$, which proves Formula \ref{fo4}.
\end{proof}

\begin{formula}
\label{fo5}
For all $n, k \in \mathbb{N}$ with $0 \le k \le n-1$, we have
$\Delta(2^n+2^k) =F(2^n-2^k)$.
\end{formula}
 
\begin{proof}
$\Delta(2^n+2^k)=\alpha(P_{2^n+2^k}.\cocone(2^n+2^k-1))$.
The element $2^n+2^k-1$ is the join in the lattice $B_{n+1}$ of the atom $2^n$ and $2^k-1$.
So $2^k-1$ is covered by $2^n+2^k-1$.
Thus, the semi-ideal $\sub(2^n+2^k-1)$ is the union of the interval 
$[0,2^k-1]$ and the interval $[2^n, 2^n+2^k-1]$.  
Therefore, the poset $P_{2^n+2^k}.\cocone(2^n+2^k-1)$
consists of $P_{2^n} \setminus P_{2^k}$, which is dually isomorphic to $P_{2^n-2^k}$.
\end{proof}

We next find formulas for $\Delta(2^n+i)$ for $i \le 7$ that only involve
the functions $\Delta$ and $F$ applied to arguments less than $2^n+i$.
By virtue of Formula \ref{fo5} we need only consider $i=3,5, 6, 7$.
The proofs will involve frequent use of the pivot notion introduced
right after Lemma \ref{th0}.

Let $Q$ denote $P_{2^n+i} .\cocone(2^n+i-1)$.
Then  $\Delta(2^n+i)=\alpha(Q)$ and  for any $p \in Q$ by Lemma \ref{th0}
we have
$\alpha(Q)$ is the sum $\alpha(Q\setminus\{p\}) +\alpha(Q.\cocone(p))$.
If both summands are known then we are done.  
If not then choose a poset for which the $\alpha$ value is not known and
apply Lemma \ref{th0}.
Continue in this way until all the $\alpha$ values are known
and add up these values.  
This will be a formula for $\Delta(2^n+i)$.

In the context in which we are working a poset $S$ has the value of
$\alpha(S)$ known if $S$ or its dual is isomorphic to a $P_k$ for
$k&lt; 2^n+i$ or to a poset whose $\alpha$ value was determined at an
earlier stage in the proof.  

An important fact to consider in choosing the element on which a poset is
to pivot is that for any $j \le i$ the poset $[0,2^n - 1] \cap \cocone(2^n+j)$
is dually isomorphic to $P_{2^n-j}$,
which is a poset with a known $\alpha$ value.
We use the heuristic of choosing pivot elements of the form $2^n+j$
wherever possible so as to obtain posets that are either entirely in
$[0,2^n-1]$ that are dually isomorphic to $P_r$ for $r \le 2^n$ or are
isomorphic to a $P_r$  for $r=2^n+j$ for which $\alpha(P_r)$ has already
been determined.  This method suffices in our proofs for $i=3,6,$ and 7.
For determining $\alpha(Q)$ for $Q \subseteq [0,2^n-1]$ that are not
isomorphic or dually isomorphic to a $P_k$ we look for a poset $Q'$ for
which $\alpha(Q')$ had been determined by the use of Lemma \ref{th0} earlier in
the proof.  This is the case for our proof of the formula for $\Delta(2^n+5)$.

\begin{formula}
\label{fo6}
For every $n \in \mathbb{N}$ with $n \ge 2$, we have
$\Delta(2^n+3)=F(2^n-2) + F(2^n-3)$.
\end{formula}

\begin{proof}  The poset $Q = P_{2^n+3}.\cocone(2^n+2)$ consists of
  $[0, 2^n+2] \setminus \{0, 2^1,2^n,2^n+2\}$.
  Pivoting it with $p = 2^n + 1$ yields
  $Q \setminus \{p\} = P_{2^n}\setminus\{0,1\}$ and
  $Q.\cocone(p) = P_{2^n}\setminus\{0,1,2\}$.
  And these posets are dually isomorphic to $P_{2^n-2}$ and $P_{2^n-3}$ as desired.
\end{proof}

\begin{formula}
  \label{fo9}
  For every $n \in \mathbb{N}$ with $n \ge 1$, we have
  $$\Delta(2^n+5) = 2 \cdot \Delta(2^n+3) + \Delta(2^n+4) -
  \Delta(2^{n-1}+3) - \Delta(2^{n-1}+4).$$
\end{formula}

E.g., $\Delta(21)=2 \cdot \Delta(19)+\Delta(20)-\Delta(11)-\Delta(12)=
2\cdot 282 + 84 - 25 - 6 = 617$.

\begin{proof}
Let $Q = d(2^n+5)$. Then by definition 
$$Q = P_{2^n+5}.\cocone(2^n+4) = P_{2^n+5}\setminus \{0,4,2^n,2^n+4\}.$$

First pivot $Q$ on the element $2^n+3$. Then we get the pivot
components $Q_1$ and $Q_2$ with $Q_1 = Q \setminus \{2^n+3\}$ and
$Q_2 = Q.\cocone(2^n+3) = [5,2^n-1]$, which is easily seen
to be dually isomorphic to $P_{2^n-5}$. This pivot of $Q$ gives
\begin{align}
  \label{b1}
  \alpha(Q) &amp;= \alpha(Q_1) + \alpha(Q_2) \\
  \label{b2}
  \alpha(Q_2) &amp;= F(2^n-5).
\end{align}
Another pivoting of $Q_1$
at $2^n+2$ leads to the components $Q_{11} = Q_1 \setminus \{2^n+2\}$ and
$Q_{12} = Q_1 .\cocone(2^n+2)$. Now $Q_{11}$ is easily seen to be
isomorphic to $d(2^n+3)$ via the isomorphism $\rho_{1,2}$ of $B_n$ that
interchanges the atoms $2^1$ and $2^2$.
This accounts for one of the occurrences of $\Delta(2^n+3)$
in the formula.
\begin{align}
  \label{b3x}
  \alpha(Q_1) &amp;= \alpha(Q_{11}) + \alpha(Q_{12}) \\
  \label{b4}
  \alpha(Q_{11}) &amp;= \alpha(d(2^n+3)))
\end{align}
To obtain the other parts, we show that they arise as pivot components
of known posets. In fact, if we pivot $Q_{11}$ at $2$, we get
$Q_{111} = Q_{11} \setminus \{2\}$ and $Q_{112} = Q_{11}.\cocone(2)$.
We see that $Q_{111} = Q_{12}$ and $Q_{112}$ is mapped onto
$d(2^{n-1}+3)$ by the automorphism $\eta$ of $P_{n+1}$
that fixes the atom $2^0$ and 
cycles the other atoms via $2^n \mapsto \cdots \mapsto 2^2 \mapsto 2^1 \mapsto 2^n$.
Figure \ref{eta} shows $\eta$ on $B_4$.
\begin{figure}[htb]
\unitlength0.4cm
\thinlines
\begin{picture}(38,20)
\scriptsize
\put(18,4){\circle*{0.5}} \put(18.5,4) {$0=\eta(0)$}
\drawline(18,4)(10,8)
\drawline(18,4)(15,8)
\drawline(18,4)(21,8)
\drawline(18,4)(29,7)
\put(10,8){\circle*{0.5}} \put(10.5,8) {$1=\eta(1)$}
\drawline(10,8)(7,12)
\drawline(10,8)(13,12)
\drawline(10,8)(21,11)
\put(15,8){\circle*{0.5}} \put(15.5,8) {$2=\eta(4)$}
\drawline(15,8)(7,12)
\drawline(15,8)(18,12)
\drawline(15,8)(26,11)
\put(7,12){\circle*{0.5}} \put(7.5,12) {$3=\eta(5)$}
\drawline(7,12)(10,16)
\drawline(7,12)(18,15)
\put(21,8){\circle*{0.5}} \put(21.5,8) {$4=\eta(8)$}
\drawline(21,8)(13,12)
\drawline(21,8)(18,12)
\drawline(21,8)(32,11)
\put(13,12){\circle*{0.5}} \put(13.5,12) {$5=\eta(9)$}
\drawline(13,12)(10,16)
\drawline(13,12)(24,15)
\put(18,12){\circle*{0.5}} \put(18.5,12) {$6=\eta(12)$}
\drawline(18,12)(10,16)
\drawline(18,12)(29,15)
\put(10,16){\circle*{0.5}} \put(10.5,16) {$7=\eta(13)$}
\drawline(10,16)(21,19)
\put(29,7){\circle*{0.5}} \put(29.5,7) {$8=\eta(2)$}
\drawline(29,7)(21,11)
\drawline(29,7)(26,11)
\drawline(29,7)(32,11)
\put(21,11){\circle*{0.5}} \put(21.5,11) {$9=\eta(3)$}
\drawline(21,11)(18,15)
\drawline(21,11)(24,15)
\put(26,11){\circle*{0.5}} \put(26.5,11) {$10=\eta(6)$}
\drawline(26,11)(18,15)
\drawline(26,11)(29,15)
\put(18,15){\circle*{0.5}} \put(18.5,15) {$11=\eta(7)$}
\drawline(18,15)(21,19)
\put(32,11){\circle*{0.5}} \put(32.5,11) {$12=\eta(10)$}
\drawline(32,11)(24,15)
\drawline(32,11)(29,15)
\put(24,15){\circle*{0.5}} \put(24.5,15) {$13=\eta(11)$}
\drawline(24,15)(21,19)
\put(29,15){\circle*{0.5}} \put(29.5,15) {$14=\eta(14)$}
\drawline(29,15)(21,19)
\put(21,19){\circle*{0.5}} \put(21.5,19) {$15=\eta(15)$}
\end{picture}
\vskip -.5in
\caption{\label{eta} The action of $\eta$ on $B_4 = P_{16}$.}
\end{figure}

This gives
\begin{equation}
  \label{b5}
  \alpha(Q_{11}) = \alpha(d(2^n+3)) = \alpha(Q_{12}) + \alpha(d(2^{n-1}+3)).
\end{equation}
Finally we consider
$d(2^n+4) = \{4,\ldots,2^n-1\}$ and pivot it on $4$. Obviously
$d(2^n+4) \setminus \{4\} = Q_2$. And
$d(2^n+4).\cocone(4)$ is mapped onto $d(2^{n-1}+4)$ by the automorphism
$\rho_{2,n-1}$ that exchanges the atoms $2^2$ and $2^{n-1}$. This gives
\begin{equation}
  \label{b6}
  \alpha(d(2^n+4)) = \alpha(d(2^{n-1}+4))+ \alpha(Q_2).
\end{equation}
Combining all these intermediate results we end up with
\begin{align}
  \label{b7}
  \alpha(Q_2) &amp;= \alpha(d(2^n+4)) - \alpha(d(2^{n-1}+4)) \mbox{\ by (\ref{b6})} \\
  \label{b8}
  \alpha(Q_{12}) &amp;= \alpha(d(2^n+3)) - \alpha(d(2^{n-1}+3)) \mbox{\ by (\ref{b5})} \\
  \label{b9}
  \alpha(Q_1) &amp;= 2\alpha(d(2^n+3)) - \alpha(d(2^{n-1}+3)) \mbox{\ by (\ref{b3x}), (\ref{b4}) and (\ref{b8})} \\
  \label{b10}
  \alpha(Q) &amp;= 2\alpha(d(2^n+3)) + \alpha(d(2^n+4)) \ - \\
  &amp; \ \alpha(d(2^{n-1}+4) - \alpha(d(2^{n-1}+3)) \mbox{\ by (\ref{b1}), (\ref{b7}) and (\ref{b9}),}\nonumber
\end{align}
and this is Formula \ref{fo9}.
\end{proof}

\begin{formula}
\label{fo7}
For every $n \in \mathbb{N}$ with $n \ge 3$, we have
  $$\Delta(2^n + 2^2 + 2^1)=\Delta(2^n+6)=F(2^n-4)+F(2^n-5)+F(2^n-6).$$
\end{formula}

E.g., $\Delta(22)=F(12)+F(11)+F(10)=53+78+84=215$.

\begin{proof}  By definition the poset $d(2^n+2^2+2^1)$ is $P_{2^n+2^2+2^1}.\cocone(2^n+2^2+2^0)$, which is
$P_{2^n+2^2}\setminus \{0,1,4,5,2^n, 2^n+1\}$.
Let $P$ denote $d(2^n+2^2+2^1)$.
First pivot $P$ on the element $2^n+2+1$.
We have $P.\cocone(2^n+2+1)=P_{2^n} \setminus \{0,1,2,3,4,5\}$,
which is dually isomorphic to $P_{2^n-6}$.

Let $Q$ denote  $P \setminus \{2^n+2+1\}$. 
We pivot $Q$ on the element $2^n+2$.
We have 
$$Q.\cocone(2^n+2) =  P_{2^n} \setminus \{0,1,2,4,5\},$$
which is dually isomorphic to $P_{2^n-5}$.
We also have $Q\setminus \{2^n+2\} = P_{2^n} \setminus \{0,1,4,5\}$, 
which is dually isomorphic to $P_{2^n-4}$. Thus, $\Delta(2^n+2^2+2^1)$ is the sum of $F(2^n-6),  F(2^n-5),$  and  $F(2^n-4)$.
\end{proof}

\begin{formula}
\label{fo8}
For every $n \in \mathbb{N}$ with $n \ge 3$, we have
$$\Delta(2^n+2^2+2^1+2^0)=\Delta(2^n+7)=F(2^n-7)+F(2^n-6)+\Delta(2^n+6).$$
\end{formula}

E.g., $\Delta(23)=F(9)+F(10)+\Delta(22)=39+53+215=307$.

Note that this formula for a $\Delta$ value involves both $F$ values and a 
$\Delta$ value.

\begin{proof}
Let $P$ be the poset 
\begin{align*}
P &amp;= d(2^n+2^2+2^1+2^0)=d(2^n+7)=P_{2^n+7} .\cocone(2^n+6) \\
&amp; =(P_{2^n} \setminus\{0,2,4,6\} )\cup \{2^n+1, 2^n+3, 2^n+5\}.
\end{align*}
We pivot $P$ on the element $2^n+5$.  

Then $P.\cocone(2^n+5)=(P_{2^n}\setminus \{0,1,2,4,5,6\})  \cup \{2^n +3\}$.
We pivot $P.\cocone(2^n+5)$ on the element  $2^n+3$ to get the posets
$P_{2^n}\setminus \{0,1,2,4,5,6\}$ and $P_{2^n} \setminus \{0,1,2,3,4,5,6\}$.
These are dually isomorphic to $P_{2^n-6}$ and $P_{2^n-7}$, respectively.

We also have $P_{2^n+5}=(P_{2^n} \setminus\{0,2,4,6\}) \ \cup \ \{2^n+1, 2^n+3\}$.
But then 
$$d(2^n+6)= P_{2^n+6}.\cocone(2^n+5)=(P_{2^n} \setminus \{0,1,4,5\}) \cup\{2^n+2,2^n+3\},$$
 which is isomorphic to the poset $P_{2^n+5}$ by the automorphism $\rho_{0,1}$ of the 
Boolean lattice $B_{2^{m+1}}$ that interchanges the atoms $2^0$ and $2^1$ and 
leaves all the other atoms fixed.
Hence the number of anti-chains in the poset $P_{2^n+5}$ is equal to 
$\Delta(2^n+6)$.  Thus $\Delta(2^n+7)=\Delta(2^n+6) +F(2^n-6)+F(2^n-7).$
\end{proof}

The proof of our final formula does not involve pivots but does make
use of the inclusion-exclusion argument of Lemma \ref{th2b}.

\begin{formula}
\label{fo10}
For every $n \in \mathbb{N}$ with $n \ge 1$, we have
$$\Delta(2^n+2^{n-1}+1)=F(2^n+2^{n-1}-1)-2 \cdot F(2^n-2).$$
\end{formula}

E.g., $\Delta(25) = F(23) - 2\cdot F(14)$.

\begin{proof}
Let $a = 2^n+2^{n-1}$, and let $Q$ be the poset $P_a$.
Then $Q = \{0,\ldots,a-1\}$ and it is easily seen
that $d(a+1) = Q \setminus \{0, 2^{n-1}, 2^n\} =
Q.\cosub(2^{n-1}, 2^n)$.
 
 From Lemma \ref{th2b} we infer that
\begin{align*}
\alpha(Q) =\ &amp; \alpha(Q.\cosub(2^{n-1}, 2^n)) + \alpha(Q.\cosuper(2^{n-1})\ + \\
&amp; \alpha(Q.\cosuper(2^n)) - \alpha(Q.\cosuper(2^{n-1},2^n)).
\end{align*}
Again it is easily seen that $Q.\cosuper(2^n)$ is 
equal to $P_{2^n}$ and via the isomorphism $\rho_{n-1,n}$
of $P_{n+1}$ we see that
$Q.\cosuper(2^{n-1})$ is isomorphic to $P_{2^n}$.

Similarly we infer that
$Q.\cosuper(2^{n-1}, 2^n)$ is equal to
$P_{2^{n-1}}$.

Combining all these observations we see that
\begin{equation}
  \label{c0}
 F(a) = \Delta(a+1) + 2\cdot F(2^n) - F(2^{n-1}).
\end{equation}
Now by the definition of $\Delta$, we have
\begin{equation}
  \label{c2}
 F(a) = F(a-1) + \Delta(a)
\end{equation}
and by Formula \ref{fo5}
\begin{equation}
  \label{c3}
 \Delta(a) = F(2^n - 2^{n-1}) = F(2^{n-1}).
\end{equation}
Substituting first (\ref{c2}) and then (\ref{c3}) into (\ref{c0}) we get
\begin{equation}
  \label{c5}
 \Delta(a+1) = F(a-1) + 2\cdot F(2^{n-1}) - 2\cdot F(2^n).
\end{equation}
By Formula \ref{fo1} we have
\begin{equation}
  \label{c6}
 F(2^n) = F(2^n-2^1) + F(2^n- 2^{n-1}) = F(2^n-2) + F(2^{n-1})
\end{equation}
Finally substituting (\ref{c6}) into (\ref{c5}) we arrive at 
\begin{equation}
  \label{c8}
 \Delta(a+1) = F(a-1) - 2\cdot F(2^n-2),
\end{equation}
which is Formula \ref{fo10}.
\end{proof}

\section{Summary}

These are the formulas obtained so far.

For $k, n \in \mathbb{N}$ with $0&lt;k&lt;n$
\begin{itemize}
\item[] Formula \ref{fo1}:
 $F(2^n) = F(2^n-2^k) + F(2^n-2^{n-k})$,
\item[] Formula \ref{fo2}:
  $F(2^n) = F(2^n-2^{n-k}-2^{n-k-1}) + 2 F(2^n-2^k) - F(2^n-2^k-2^{k-1})$,
\item[]  Formula \ref{fo3}:
$\Delta(2^n-2^k+1)=\Delta(2^n-2^{n-k}+1)$,
\item[]  Formula \ref{fo11}:
$\Delta(2^n-2^k+1) = \Delta(2^n-2^k-2^{k-1}+1) + \Delta(2^n-2^{n-k}-2^{n-k-1}+1)$,
\item[]  Formula \ref{fo4}:
$\Delta(2^n-2^k)=F(2^{n-1}-2^k)$,
\item[]  Formula \ref{fo5}:
$\Delta(2^n+2^k)=F(2^n-2^k)$,
\item[]  Formula \ref{fo6}:
$\Delta(2^n+2^1+2^0)=F(2^n-2)+F(2^n-3)$,
\item[] Formula \ref{fo9}:
  $\Delta(2^n+2^2+2^0) = 2 \cdot \Delta(2^n+3) + \Delta(2^n+4) -
  \Delta(2^{n-1}+3) - \Delta(2^{n-1}+4)$,
\item[]  Formula \ref{fo7}:
$\Delta(2^n + 2^2 + 2^1)=F(2^n-4)+F(2^n-5)+F(2^n-6)$,
\item[]  Formula \ref{fo8}:
  $\Delta(2^n+2^2+2^1+2^0)=F(2^n-7)+F(2^n-6)+\Delta(2^n+6)$,
\item[] Formula \ref{fo10}:
$\Delta(2^n+2^{n-1}+1)=F(2^n+2^{n-1}-1)-2 \cdot F(2^n-2)$.
\end{itemize}

An application of these formulas gives numerical values for the values
of $\Delta$ for $n = 16, \ldots, 32$---with the exception of $\Delta(26)$,
where we leave it as an open question whether the formula given there
is a special case of a more general one. Thus we have

\begin{itemize}
\item[] $\Delta(16)=1$ by Formula \ref{fo1},
\item[] $\Delta(17)=F(15)=167$  by Formula \ref{fo5},
\item[] $\Delta(18)=F(14)=148$  by Formula \ref{fo5},
\item[] $\Delta(19)=F(13)+F(14)=282$  by Formula \ref{fo6},
\item[] $\Delta(20)=F(12)=84$ by Formula \ref{fo5},
\item[] $\Delta(21)= 2 \cdot \Delta(19)+\Delta(20)-\Delta(11)-\Delta(12)= 617$  by Formula \ref{fo9},
\item[] $\Delta(22)=F(12)+F(11)+F(10)=215$ by Formula \ref{fo5},
\item[] $\Delta(23)=F(9)+F(10)+\Delta(22)=307$  by Formula \ref{fo8},
\item[] $\Delta(24)=F(8) = 20$  by Formula \ref{fo5},
\item[] $\Delta(25)= F(23)-2 \cdot F(14) = 1692$ by Formula \ref{fo10},
\item[] $\Delta(26) = F(18)+F(13)+F(11)+F(7)=714$,
\item[] $\Delta(27) = \Delta(25) - \Delta(21) = 1075$
by Formulas \ref{fo3} and \ref{fo11},
\item[] $\Delta(28)= F(12) = 84$ by Formula \ref{fo4},
\item[] $\Delta(29)=\Delta(25)=1692$ by Formula \ref{fo3},
\item[] $\Delta(30)=F(14)=148$  by Formula \ref{fo4},
\item[] $\Delta(31)=F(15)=167$  by Formula \ref{fo4},
\item[] $\Delta(32)=1$ by Formula \ref{fo1}.
\end{itemize}

\section{Acknowledgments}
We acknowledge and thank  the referee for the many suggestions
that improved our paper.

\begin{thebibliography}{9}
 \bibitem{baldwi} R.\ Balbes and P.\ Dwinger,
   \emph{Distributive Lattices},
   Univ.\ of Missouri Press, 1974.
 \bibitem{berkoe} J.\ Berman and P.\ K\"ohler,
   Cardinalities of finite distributive lattices,
   \emph{Mitt. Math.\ Sem. Giessen} {\bf 121} (1976), 103--124.
   Available at: \url{https://oeis.org/A006356/a006356.pdf}.
 \bibitem{bir} G.\ Birkhoff,
   \emph{Lattice Theory},
   Amer.\ Math.\ Soc.\ Coll.\ Publ., Vol.\ 25, 3rd edition, 1967.
 \bibitem{camer} F.\ a Campo and M.\ Ern\'e,
   Exponential functions of finite posets and the number of
   extensions with a fixed set of minimal points,
   \emph{J.\ Combin.\ Math.\ Combin.\ Comput.} {\bf 110} (2019), 125--156.
 \bibitem{erne} M.\ Ern\'e,
   \emph{Ordnungskombinatorik}, Lecture Notes, University of Hannover, 2005.
 \bibitem{OEIS}
  N.\ J.\ A. Sloane et al.,
  \emph{On-Line Encyclopedia of Integer Sequences}, available at 
  \url{https://oeis.org}.
 \bibitem{wie} D.\ Wiedemann,
   A computation of the eighth Dedekind number,
   \emph{Order} {\bf 8} (1991), 5--6.
\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2020 {\it Mathematics Subject Classification}: Primary 06A07; Secondary 06A11, 06B25.

\noindent \emph{Keywords: } Dedekind number, anti-chain.

\bigskip
\hrule
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\noindent (Concerned with sequences
\seqnum{A000372},
\seqnum{A006356},
\seqnum{A132581}, and
\seqnum{A132582}.)

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\noindent
Received July 13 2021;
revised versions received  December 23 2021; December 27 2021.
Published in {\it Journal of Integer Sequences}, December 27 2021.

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