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\begin{center}
\vskip 1cm{\LARGE\bf Finite Topologies and Partitions
}
\vskip 1cm
\large
Moussa Benoumhani\footnote{Author's current address:
Department of Mathematics,
Al-Imam University,
Faculty of Sciences,
P. O. Box 90950, Riyadh 11623,
Saudi Arabia.}
and Messaoud Kolli\\
Faculty of Science\\
Department of Mathematics\\
King Khaled University \\
Abha \\
Saudi Arabia\\
\href{mailto:mbenoumhani@kku.edu.sa}{\tt benoumhani@yahoo.com }\\
\href{mailto:kmessaud@kku.edu.sa}{\tt kmessaud@kku.edu.sa}\\
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\vskip .2 in

\begin{abstract}
Let $E$ be a set with $n$ elements, and let $T(n,k)$ be the number of
all labeled topologies having $k$ open sets that can be defined on $E$.
In this paper, we compute these numbers for $k \leq 17$, and arbitrary
$n$, as well as $t_{N0}(n,k)$, the number of all unlabeled non-$T_0$
topologies on $E$ with $k$ open sets, for $3 \leq k \leq 8. $
\end{abstract}

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\section{Introduction}

Let $E$ be an $n$-element set, and let $T(n)$ be the total number of labeled topologies one can
define on $E$. Using the one-to-one correspondence between finite topologies and idempotent
0--1-matrices with $1$ in the diagonal (see Krishnamurty \cite{Krishnamurthy}), it is possible to
establish explicit formulae for $T(n)$ (see Ern\'e \cite{Erne1}), but they are too complex for a
numerical evaluation in reasonable time, even on a modern computer. Comtet \cite{Comptet}, Evans et
al. \cite{E.H.L}, Renteln \cite{Renteln} and others reduced the computation of the numbers $T(n)$
to that of the numbers $\gamma_n$ of labeled acyclic transitive graphs, in other words, of
partially ordered sets with $n$ points. These numbers are connected via the formula
$T(n)=\sum_{l=1}^{n} S_{n,l} \gamma_l$, where the $S_{n,l}$ are the Stirling numbers of the second
kind. That formula is evident in view of the one-to-one correspondence between quasiorders and
topologies, respectively, between partial orders and $T_0$-topologies, observed by Alexandrov
\cite{Alexan} already in the 1930's, and pursued later by many authors (see, e.g.,
\cite{Benoumhani, B.M1, Erne1, Erne2, Erne-Stege1, Erne-Stege2, Heitzig}). Consequently, $\gamma_n$
is simultaneously the number $T_0(n)$ of all $T_0$-topologies on $n$ points. Combining the above
formula with computer methods, Evans et al.\ obtained the numbers $T(n)$ for $n\leq 7$ in $1966$;
using suitable combinatorial identities, Ern\'e computed $T_0(n)$ and $T(n)$ for $n\leq 8$ in
$1970$ and for $n\leq 9$ in $1972$ (see \cite{Erne1}). Later the calculations were pushed forward,
via sophisticated computer techniques, by Das \cite{Das}, Stege \cite{Stege}, Heitzig and Reinhold
\cite{Heitzig}, and finally by Brinkmann and McKay \cite{B.M1, B.M2} until $n = 18.$

\bigskip

We shall use a refinement of the above formula involving the Stirling numbers, due to Ern\'e (see
\cite{Erne1, Erne2}, saying that for any property $p$ of topologies that is invariant under lattice
isomorphisms, the number $T^p(n)$ of all topologies on $n$ points with property $p$ is obtained
from the corresponding numbers $T^p_0(n)$ of $T_0$-topologies by the identity $T^p(n) =\sum_{l=1}^n
S_{n,l} T_0^p(l).$ For example, this applies to the numbers $T(n, k)$ of all topologies on $n$
points having exactly $k$ open sets, or even more specifically, to the numbers $T_a(n, k)$ of all
such topologies with exactly $a$ atoms, i.e., minimal (nonempty) open sets. Alternatively, $T(n,
k)$ is the number of all quasiordered sets with $k$ downsets or antichains, respectively. Sharp
\cite{Sharp2} and Stephen \cite{Stephen} showed that the maximal size of non-discrete topologies on
$E$ is $3\cdot 2^{n-2}$. Stanley \cite{Stanley} enumerated all labeled topologies on $E$ with $k
\geq 7 \cdot 2^{n-4}$ open sets, and Kolli \cite{Kolli} extended the range to $k \geq 6 \cdot
2^{n-4}$. Ern\'e and Stege \cite{Erne-Stege1, Erne-Stege2} determined $T_0(n, k)$ and $T(n, k)$ for
$n \leq 12$ and, using these numbers, $T_0(n)$ and $T(n)$ for $n \leq 14$. In their paper
\cite{Erne-Stege2}, using another approach via an elegant combinatorial identity, they obtained
formulae for $T_0(n, n+k+1)$ and $k \leq \min \{n, 12 \}$. Combining the results of the two papers
\cite{Erne-Stege1} and \cite{Erne-Stege2}, one could compute $T_0(n, k)$ and $T(n, k)$ for $n \leq
12$ and $k \leq 25$, but not for all larger values. Independently, Benoumhani \cite{Benoumhani}
computed by a direct method, for all $n$, the number of labeled topologies with $k \leq 12$ open
sets, and gave alternate proofs of some known facts in the field. The enumeration of unlabeled
($T_0$) topologies seems to be an even more difficult task. For an early approach, see Knopfmacher
\cite{Knopfmacher}, and for later more extensive calculations, \cite{B.M1, B.M2, Erne-Stege1,
Erne-Stege2, Heitzig}.

\bigskip

The results in this paper are presented as follows: in Section 2, we continue the calculation of
$T(n, k)$. The method used is one that was developed in \cite{Kolli}. The tedious calculations in
Section 2 illustrate how hard it is to determine, for instance, $T_0(10, n).$ In Section $3$, we
compute the number of unlabeled non-$T_0$ topologies with $k \leq 8$ open sets, thereby extending
the work of Stanley \cite{Stanley}. In the two main results, the presence of the sequences related to
partitions of sets or numbers, respectively, is evident: labeled topologies create set partitions
(induced by the equivalence relation identifying points with the same closure), and as the elements
of the blocks are distinguishable in that case, the Stirling numbers of the second kind are
involved. When counted up to homeomorphism, finite topologies are related, in a similar way, to
partitions of integers, because in that case, only the number of elements in the single blocks is
relevant. This explains the appearance of the numbers of partitions of the integer $n$ in the
corresponding formulae.

\section{Computation of $T(n,k), k \leq 17$}

The following (known) properties of  Stirling numbers of the second
kind will be needed in Example~\ref{ex24}. We put them in
\begin{lemma}
For all $n$, and $1\leq k \leq n-1$, the following relations hold
\begin{eqnarray*}
\sum_{l=1}^{n-1} \binom{n}{l} S_{l,k}&=&(k+1)S_{n,k+1}\\
 \sum_{l=1}^{n-1}\binom{n}{l}
S_{l,k} S_{n-l,m}&=&\binom{m+k}{k} S_{n,k+m}.
\end{eqnarray*}
\end{lemma}
For a proof of this result, and other properties concerning Stirling
numbers, see \cite {GKP}.\\

For the sake of brevity, we recall the
definition of an ideal in a topology $\mathcal{T}$ (this is the definition of an ideal in a poset).
\begin{definition}
Let $\mathcal{T}$ be a topology on $E$. The set $\mathcal{A} \subset \mathcal{T}$ is called an
ideal if it satisfies
$$
\forall A,\; B \in \mathcal{A}, \quad \forall O \in \mathcal{T},  \qquad A \cup B \in \mathcal{A},
\text{ and}\quad A \cap O \in \mathcal{A}.
$$
\end{definition}

From \cite{Kolli} recall the definition of a minimal open set.
\begin{definition}
 The element $A \in \mathcal{T}$ is called a minimal open set if for all $O$ in $\mathcal{T}$, we have
$$
A \cap O \in \{\emptyset, A\}.
$$
\end{definition}
Using the lattice theoretic terminology, a minimal open set is in
fact an element covering the least element that is, an atom.\\

We designate by $\mathcal{T}_a (n, k)$ the set of all labeled topologies, having $k$ open sets, and
$a$ minimal open sets, $\mathcal{T}(n, k)$ the set of all topologies on $E$ having $k$ open sets.
Let $T_a (n,k)=|\mathcal{T}_a (n, k)|$, and $T(n, k)=|\mathcal{T}(n, k)|$. It is easy to see that
$$
T(n,k)=\sum_{a, 2^{a} \leq k} T_a (n,k).
$$
For $a=1$, Kolli  \cite{Kolli} proved
\begin{equation}
\label{eq2.1} T_1(n,k)=\sum_{l=1}^{n-1}\binom{n}{l}T(l,k-1),~\ \ \ \
\ \ \ \forall ~n\geq 1.
\end{equation}
To compute  the other remaining numbers $T_a (n,k)$, we use the method developed in \cite{Kolli}:
if $\mathcal{T} \in \mathcal{T}_a (n, k)$, write it as
$$
\mathcal{T}=\{ \varnothing,\;  A_1,\; \ldots,\; A_a,\; \ldots,\; E \},
$$
where $A_1,\; \ldots, \; A_a$ are the minimal open sets of the topology $\mathcal{T}$. Let
$A=\bigcup_{i=1}^{a}A_i$, and define the mapping
$$
\Phi (\mathcal{T})=\mathcal{T}'=\{O-A,\; O \in \mathcal{T}\}.
$$
It is clear that $\mathcal{T}'$ is a topology on $E-A$,  having $r$ open sets, where
\begin{equation}
\label{eq2.2}
 k-2^a+1 \geq r \geq \left \lceil\frac{k}{2^{a-1}}\right\rceil-1.
\end{equation}
Conversely, if $\mathcal{T}'$ is in $\mathcal{T}(l,r)$, with $l < n$ and $r$ satisfies
(\ref{eq2.2}), we may restitute all the topologies $\mathcal{T}$ from $\mathcal{T}'$ by determining
all the possibilities of $r$. This means, we determine all the topologies $\mathcal{T}$ having $k$
open sets and $a$ minimal open sets, and such that $\Phi (\mathcal{T})=\mathcal{T}'$.

\begin{example}
To illustrate our method, we compute $T(n,10)$ in all details. Using
 formula (\ref{eq2.1}), we have
$$
T_1(n,10)=\sum_{l=1}^{n-1} \binom{n}{l}T(l,9).
$$
Now, let $\mathcal{T}=\{ \varnothing,\; A_1,\;A_2,\; A_3=A_1\cup A_2, \;A_4,\; \ldots,\; A_8, \;E \} \in
\mathcal{T}_2(n,10),$ and let $ \mathcal{T}' = \{O-A_3,\; O \in \mathcal{T} \}$. This topology has
$r$ open sets, and according to (\ref{eq2.2}), the only possibilities for $r$ are $7,\;6,\;5,\;4$. Let us
examine these cases one by one. For $r=7$, let $ \mathcal{T}'=\{\varnothing,\; B_1,\; \ldots, \;B_5,\;
E-A_3 \},$ and since a topology is stable by union, then
$$
\mathcal{T}=\{\varnothing,\; A_1,\; A_2, \; A_3=A_1\cup A_2, \; A_4=B_1\cup A_3,\; \ldots,\; A_8=B_5\cup A_3,\; E\}.
$$
The topology $\mathcal{T}$ is then completely determined by $\mathcal{T}'$, and the total number of
such topologies is
$$
\sum_{l=1}^{n-1} \binom{n}{l}S_{l,2}T(n-l,7).
$$
For $r=6$,  let $ \mathcal{T}'=\{\varnothing,\; B_1, \; B_2, \; B_3,\; B_4,\; ,\; E-A_3 \}$. The topology
$\mathcal{T}$ will be obtained as follows:
$$
\mathcal{T}=\{\varnothing,\; A_1,\; A_2,\; A_3=A_1\cup A_2,\; A_4=B_1\cup A_3,\;\ldots, \;A_7=B_4 \cup A_3,\; X,\;
E \}.
$$
 We have to determine the form of $X$. Note that if $\{ \varnothing, B_1
\}$ is an ideal of $\mathcal{T}'$, then we have two solutions: $X=B_1 \cup A_l, \, l=1,2$. So, the
wanted number is
$$
\sum_{l=1}^{n-1} \binom{n}{l}S_{l,2}
 \biggl( 2T_1(n-l,6)+4T_2(n-l,6) \biggr).
$$
For $r=5$, let $ \mathcal{T}'=\{\varnothing, \; B_1,\; B_2,\; B_3,\; E-A_3\}$. The topology $\mathcal{T}$
has the form
$$
\mathcal{T}=\{\varnothing,\; A_1,\; A_2,\; A_3=A_1\cup A_2,\; A_4=B_1\cup A_3,\; \ldots,\; A_6=B_3 \cup A_3,\; X,\;
Y,
 E \}.
$$
As above, if $\{ \varnothing,\; B_1,\; B_2 \}$ is an ideal of $\mathcal{T}'$, then we have two
solutions:
$$
\left( X=B_1 \cup A_l,\; Y=B_2 \cup A_l
\right) \, l=1,\;2,
$$
and if $\{ \varnothing,\; B_1\}$, $\{\varnothing, \; B_2 \}$ are two distinct ideals of $\mathcal{T}'$,
then also we have two solutions:
$$
\left( X=B_1 \cup A_1,\; Y=B_2 \cup A_2 \right),
$$
and
$$\left( X=B_1\cup A_2, \;Y=B_2 \cup A_1 \right).
$$
So, the wanted number of such topologies is
$$
\sum_{l=1}^{n-1} \binom{n}{l}S_{l,2} \biggl( 2T_1(n-l,5)+4T_2(n-l,5)
\biggr).
$$
For $r=4$, $ \mathcal{T}'=\{\varnothing,\; B_1,\;  B_2,\; E-A_3\}$, the topology $\mathcal{T}'$ is
either a chain or its proper open sets form a partition of the whole set (see Figure 1). We
restitute $\mathcal{T}$ as follows:
$$
\mathcal{T}=\{\varnothing,\; A_1, \;A_2, \; A_3=A_1\cup A_2,\; A_4=B_1\cup A_3,\;A_5=B_2 \cup
A_3,\; X,\;Y,\;Z,\;E \}.
$$
 If $\mathcal{T}'$ is a chain topology, then we have two solutions:
$$
(X=B_1 \cup A_1,\; Y=B_2\cup A_1,\; Z=E-A_2),
$$
and
$$
(X=B_1 \cup A_2,\; Y=B_2\cup A_2,\; Z=E-A_1).
$$
In the other case, there is
no solution, and then the total number of such topologies is
$$
2 \sum_{l=1}^{n-1} \binom{n}{l}S_{l,2} T_1(n-l,4).
$$
All these numbers added, we obtain $T_2(n,10).$ \\
Now,  we compute $T_3(n,10)$. Let
$$
\mathcal{T}=\{ \varnothing,\; A_1,\; A_2, \; A_3,\; \ldots, \; A_7=A_1\cup A_2\cup A_3,\;  A_8,\; E
\} \in \mathcal{T}_3(n,10),
$$
where $A_1,\; A_2,\; A_3$ are its three minimal open sets. Let  $\mathcal{T}' = \{O-A_7, O \in
\mathcal{T} \}.$ This topology has $r$ open sets with $r=3$ or $r=2$. For $r=3$, $
\mathcal{T}'=\{\varnothing,\; B_1, \; E-A_7 \},$ then
$$
\mathcal{T}=\{\varnothing, A_1,\; A_2, \; A_3,\;  \ldots, \; A_7=A_1\cup A_2\cup A_3,\;
A_8=B_1\cup A_7, \; E \}.
$$
The topology $\mathcal{T}$ is then totally determined by $\mathcal{T}'$.  The total number of
possibilities to reconstruct $\mathcal{T}$ from $\mathcal{T}'$ is
$$
\sum_{l=1}^{n-1} \binom{n}{l}S_{l,3}T(n-l,3).
$$
For $r=2$, $\mathcal{T}'=\{\varnothing,\; E-A_7 \}$. The topology  $\mathcal{T}$ has the form
$$
\mathcal{T}=\{\varnothing,\; A_1,\; A_2,\; A_3,\;  \ldots, A_7=A_1\cup A_2\cup A_3,\; X,\;  E \},
$$
and the three possibilities for the open set $X$ are:
$$
X=E-A_3,\; {\rm or}\; X=E-A_2, \; {\rm or} \;  X=E-A_1.
$$
The number of such topologies is $3\sum_{l=1}^{n-1}\binom{n}{l} S_{l,3}.$ Adding these numbers for
$r=2, \; 3$, yields $ T_3(n,10),$ and then $T(n,10).$
\label{ex24}
\end{example}

According to Ern\'e's result \cite{Erne1}:
$$ T_a (n,k)=\sum_{l=1}^{n-1} T_{a, 0}(l,k)
S_{n,l}= \sum_{l=1}^{n-1} C_a (l,k) l! S_{n,l}.$$ This means that the numbers $ T_a (n,k)$ are
linear combinations of the $(l! S_{n,l})_{l=1}^{k-1}.$ So, we only have to determine the
coefficients $C_a (l,k)$. This is done in the following tables for $  10 \leq k \leq 17$. Note that
we omitted the easy computations for $k \leq 9$.
\begin{center}
$C_a (l, 10).$\\
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline   & $l=4$ & $l=5$ & $l=6$ & $l=7$ & $l=8$ & $l=9$  \\
\hline
$a=1$ &  $0$ & $\displaystyle\frac{5}{6}$ & $5$ &  $\displaystyle\frac{11}{2}$ & $3$ & $1$ \\
\hline
$a=2$ & $\displaystyle\frac{1}{2}$ & $\displaystyle\frac{9}{2}$ & $\displaystyle\frac{33}{8}$ &  $2$ &  $\displaystyle\frac{1}{2}$ & $0$  \\
\hline $a=3$ & $\displaystyle\frac{1}{2}$ & $\displaystyle\frac{1}{6}$ & $0$ & $0$ & $0$ & $0$ \\
\hline
\end{tabular}
\end{center}


\begin{center}
$C_a (l, 11).$\\
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline  & $l=5$ & $l=6$ & $l=7$ & $l=8$ & $l=9$ & $l=10$  \\
\hline $a=1$ &  $1$ & $\displaystyle\frac{11}{2}$ & $\displaystyle\frac{73}{8}$ &
$\displaystyle\frac{15}{2}$
& $\displaystyle\frac{7}{2}$ & $1$ \\
\hline $a=2$ & $\displaystyle\frac{31}{12}$ & $\displaystyle\frac{15}{2}$ &
$\displaystyle\frac{43}{8}$ &  $\displaystyle\frac{9}{4}$&
 $\displaystyle\frac{1}{2}$ & $0$  \\
\hline
$a=3$ & $\displaystyle\frac{7}{12}$ & $\displaystyle\frac{1}{6}$ & $0$ & $0$ & $0$ & $0$  \\
\hline
\end{tabular}
\end{center}

\medskip

\begin{center}
$C_a (l, 12).$\\
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline  & $l=4$ & $l=5$ & $l=6$ & $l=7$ & $l=8$& $l=9$& $l=10$ &$l=11$ \\
\hline $a=1$ & $0$& $0$&  $\displaystyle\frac{25}{6}$ & $\displaystyle\frac{79}{6}$ &
$\displaystyle\frac{29}{2}$ & $\displaystyle\frac{39}{4}$
& $4 $& $1$ \\
\hline $a=2$ & $0$ & $\displaystyle\frac{7}{2}$ & $\displaystyle\frac{67}{6}$ &
$\displaystyle\frac{45}{4}$ &  $\displaystyle\frac{27}{4}$&
 $\displaystyle\frac{5}{2}$ &  $\displaystyle\frac{1}{2}$ & $0 $ \\
\hline
$a=3$ & $\displaystyle\frac{1}{2}$ & $1$ & $\displaystyle\frac{2}{3}$ & $\displaystyle\frac{1}{6}$& $0 $& $0$&$0$&$0$  \\
\hline
\end{tabular}
\end{center}
\begin{center}
$C_a (l, 13).$\\
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline & $l=5$ & $l=6$ & $l=7$ & $l=8$& $l=9$& $l=10$ & $l=11$ & $l=12$ \\
\hline $a=1$ & $\displaystyle\frac{1}{2}$& $\displaystyle\frac{9}{2}$& $16$ &
$\displaystyle\frac{295}{12}$ & $\displaystyle\frac{85}{4}$ &
$\displaystyle\frac{49}{4}$ & $\displaystyle\frac{9}{2}$ & $1$ \\
\hline $a=2$ & $\displaystyle\frac{1}{2}$& $\displaystyle\frac{125}{12}$&
$\displaystyle\frac{79}{4}$ & $\displaystyle\frac{253}{16}$ & $\displaystyle\frac{33}{4}$ &
$\displaystyle\frac{11}{4}$ & $\displaystyle\frac{1}{2}$ & $0 $\\
\hline $a=3 $& $1 $& $\displaystyle\frac{17}{12}$ & $\displaystyle\frac{3}{4}$ &
$\displaystyle\frac{1}{6}$ &
$0$&$0$ & $ 0$ & $0$  \\
\hline
\end{tabular}
\end{center}

\vfill\eject

\begin{center}
$C_a (l, 14).$\\
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline  & $ l=5$ & $l=6$ & $l=7$ & $l=8$& $l=9$& $l=10$ & $l=11$ & $l=12$ & $l=13$ \\
\hline $a=1$ & $0$ &  $2$  & $\displaystyle\frac{49}{3}$ & $\displaystyle\frac{73}{2}$ &
$\displaystyle\frac{649}{16}$ & $\displaystyle\frac{59}{2}$ & $15$ &
$5$ &  $1$ \\
\hline $a=2$ & $\displaystyle\frac{3}{2}$ & $13$ &  $\displaystyle\frac{365}{12}$ &
$\displaystyle\frac{94}{3}$ & $\displaystyle\frac{85}{4}$ & $\displaystyle\frac{79}{8}$ &
$3$ & $\displaystyle\frac{1}{2}$ & $0$ \\
\hline $a=3$ &  $\displaystyle\frac{7}{4}$& $3$ & $\displaystyle\frac{15}{8}$ &
$\displaystyle\frac{5}{6}$ & $\displaystyle\frac{1}{6}$&
$0$&$0$ & $ 0$ & $0$  \\
\hline
\end{tabular}
\end{center}


\begin{center}
$C_a (l, 15).$\\
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline  & $l=5$ & $l=6$ & $l=7$ & $l=8$& $l=9$& $l=10$ &$l=11$ & $l=12$ & $l=13$ & $l=14$ \\
\hline $a=1$ & $0$&  $\displaystyle\frac{13}{4}$ & $18$ &  $\displaystyle\frac{389}{8}$&
$\displaystyle\frac{206}{3}$& $\displaystyle\frac{2975}{48}$ & $\displaystyle\frac{315}{8}$ & $18$
&
$\displaystyle\frac{11}{2}$ & $ 1$ \\
\hline $a=2$ & $\displaystyle\frac{1}{2}$& $\displaystyle\frac{33}{4}$ &
$\displaystyle\frac{209}{6}$& $\displaystyle\frac{439}{8}$ & $\displaystyle\frac{557}{12} $&
$\displaystyle\frac{221}{8}$ &
$\displaystyle\frac{93}{8}$ & $\displaystyle\frac{13}{4}$ & $\displaystyle\frac{1}{2}$& $0$ \\
\hline $a=3$ &  $\displaystyle\frac{1}{2}$& $\displaystyle\frac{127}{36}$ &
$\displaystyle\frac{13}{3}$ & $\displaystyle\frac{19}{8}$ & $\displaystyle\frac{11}{12}$&
$\displaystyle\frac{1}{6}$&
$0$&$0$ & $ 0$ & $0$  \\
\hline
\end{tabular}
\end{center}

\begin{center}
$C_a (l, 16).$\\
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
 & $l=4$ & $l=5$ & $l=6$ & $l=7$ & $l=8$& $l=9$& $l=10$ &$l=11$ & $l=12$ & $l=13$ & $l=14$ & $l=15$ \\
\hline $a=1$ & $0$ & $0$& $1$ &  $\displaystyle\frac{541}{36}$& $\displaystyle\frac{343}{6}$ &
$\displaystyle\frac{847}{8}$& $116$ & $\displaystyle\frac{4309}{48}$ & $51$ &
$\displaystyle\frac{85}{4}$ & $6$ &  $1$ \\
\hline $a=2$ & $0$ & $0$ & $\displaystyle\frac{41}{4}$& $\displaystyle\frac{139}{3}$ &
$\displaystyle\frac{685}{8}$& $\displaystyle\frac{2155}{24}$ & $\displaystyle\frac{2097}{32}$& $35$
& $\displaystyle\frac{27}{2}$ &
$\displaystyle\frac{7}{2}$ & $\displaystyle\frac{1}{2}$& $0 $\\
\hline $a=3$ & $0 $&  $\displaystyle\frac{3}{2}$& $\displaystyle\frac{23}{4}$ &
$\displaystyle\frac{293}{36}$ & $\displaystyle\frac{143}{24} $& $\displaystyle\frac{35}{12}$& $1$&
$\displaystyle\frac{1}{6}$&
$0$&$0$ & $ 0$ & $0$  \\
\hline $a=4$ & $\displaystyle\frac{1}{24}$& $0$&$0$ & $ 0$ & $0$ & $0$&$0$ & $ 0$ &
$0$&$0$ & $ 0$ & $0$  \\
\hline
\end{tabular}
\end{center}

\vfill\eject

\begin{center}
$C_a (l, 17).$\\
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline & $l=5$ & $l=6$ & $l=7$ & $l=8$& $l=9$& $l=10$ &$l=11$ & $l=12$ & $l=13$ & $l=14$ & $l=15$& $l=16$ \\
\hline$ a=1$ & $\displaystyle\frac{1}{24}$ & $\displaystyle\frac{3}{2}$ & $17$ &
$\displaystyle\frac{139}{2}$& $\displaystyle\frac{595}{4}$ & $\displaystyle\frac{2383}{12}$ &
$\displaystyle\frac{5841}{32}$ &
$\displaystyle\frac{1999}{16}$ & $\displaystyle\frac{129}{2}$ & $\displaystyle\frac{99}{4} $& $\displaystyle\frac{13}{2}$ &  $1$ \\
\hline $a=2$ &$ 0$ & $\displaystyle\frac{11}{2} $& $\displaystyle\frac{1013
}{24}$ &
$\displaystyle\frac{677}{6}$& $\displaystyle\frac{2503}{16}$ & $\displaystyle\frac{413}{3}$ &
$\displaystyle\frac{8567}{96}$ &
$\displaystyle\frac{695}{16}$ & $\displaystyle\frac{31}{2}$& $\displaystyle\frac{15}{4}$ & $\displaystyle\frac{1}{2}$&$0$ \\
\hline $a=3$ & $0$ &  $\displaystyle\frac{9}{2}$& $\displaystyle\frac{91}{8}$ &
$\displaystyle\frac{149}{12}$ & $\displaystyle\frac{379}{48}$ & $\displaystyle\frac{7}{2}$&
$\displaystyle\frac{13}{12}$& $\displaystyle\frac{1}{6}$&
$0$&$0$ & $ 0$ & $0$  \\
\hline
 $a=4$ & $ \displaystyle\frac{1}{24}$& $0$ &
$0$ & $0 $& $0$& $0$& $0$&
$0$&$0$&$0$ & $ 0$ & $0$  \\
\hline
\end{tabular}
\end{center}


\section{ Unlabeled topologies for small $k$ }
In this section, we investigate the number of unlabeled topologies, with $k \leq 8$ open
sets, denoted by $t_{N0}(n,k)$. Note that a $T_0-$ topology must satisfy $k> n$, and since these topologies are calculated for $n\geq k$, they are all non-$T_0$.
We have the

\begin{theorem} For  $n\geq k$, the numbers $t_{N0}(n,k)$ satisfy the following identities:

 \begin{eqnarray*}
t_{N0}(n,2)&=& 1\\
t_{N0}(n,3)&=& n-1 \\
 t_{N0}(n,4)&=& \frac{2n^2-4n+3}{4}+\frac{(-1)^n}{4} \\
t_{N0}(n,5)&=& \frac{2n^3-6n^2+10n-9}{12}-\frac{(-1)^n}{4}\\
  t_{N0}(n,6)&=&
  \frac{2n^4-8n^3+40n^2-112n+87}{48}+(-1)^n\frac{3}{16}\\
t_{N0}(n,7)&=& \frac{2n^5-10n^4+100n^3-560n^2+1113n-675}{240}+(-1)^n \frac{(n-3)}{16}
  \\
  t_{N0}(n,8)&=& \frac{4n^6-24n^5+400n^4-2820n^3+8056n^2-10386n+5130}{2880}-(-1)^n \frac{(3n-7)}{32}
  +\left \{ \frac{n^2}{12} \right \}
\end{eqnarray*}
Note that the number of partitions of $n$ in to $2$ and $3$ summands, $P(n,2),\;P(n,3)$ are given by
 $P(n,2)=\left \lfloor \frac{n}{2} \right\rfloor, \;P(n,3)=\left \{ \frac{n^2}{12} \right \}$, where $\{x\}$ is the nearest integer to $x$.
\label{thm31}
\end{theorem}

The proof of the previous theorem is based on the following lemma:
 \begin{lemma}
 We have
 \begin{eqnarray*}
 \sum_{k=2}^{n-1}\left \lfloor \frac{k}{2} \right\rfloor&=& \left
\lfloor
  \frac{n}{2}\right \rfloor \left
\lfloor
  \frac{n-1}{2}\right \rfloor  = \left \lfloor
  \frac{n}{2}\right \rfloor \left ( \left \lfloor
  \frac{n}{2}\right \rfloor-\frac{(-1)^{n}+1}{2}\right)
\\
 \sum_{k=2}^{n-2}k \left \lfloor \frac{k}{2} \right\rfloor&=&
\frac{1}{12}\left \lfloor \frac{n-1}{2}\right \rfloor
 \left ( 4n^2-(15-2(-1)^{n})n+12+ \frac{3}{2}(1-(-1)^{n})\right ) \\
 \sum_{k=2}^{n-2} \left \lfloor \frac{k}{2}
\right\rfloor^2&=&\frac{(n-2)}{4}\left (
\frac{(n-1)(n-3)}{3}+\frac{(1+(-1)^n}{2}\right)\\
 \sum_{k=2}^{n-2}
\left \lfloor \frac{k}{2} \right\rfloor\left \lfloor -\frac{k-1}{2}
\right\rfloor&=&-\sum_{k=2}^{n-2} \left \lfloor \frac{k}{2} \right\rfloor^2= -\frac{(n-2)}{4}\left
(
\frac{(n-1)(n-3)}{3}+\frac{1+(-1)^n}{2}\right)\\
\sum_{k=3}^{n-1}\left \lfloor \frac{k}{2} \right\rfloor\left \lfloor \frac{k-1}{2}
\right\rfloor&=& \left \lfloor
  \frac{n}{2}\right \rfloor \left \lfloor
  \frac{n-2}{2}\right  \rfloor
  \left( \frac{2n-5}{6}+\frac{(1-(-1)^{n})}{3}\right )\\
\sum_{k=4}^{n-1} k\left \lfloor \frac{k}{2}\right \rfloor
    \left \lfloor \frac{k-2}{2} \right \rfloor&=&\frac{1}{4}\left \lfloor
  \frac{n}{2}\right \rfloor \left \lfloor
  \frac{n-3}{2}\right  \rfloor
  \left(n^2-2n+\frac{(1-(-1)^{n})}{2}\right )\\
  \sum_{k=4}^{n-1}(-1)^{k-1} \left \lfloor \frac{k}{2}\right \rfloor
    \left \lfloor \frac{k-2}{2} \right \rfloor&=&-\frac{(n-1)(n-3)}{8}(1-(-1)^n)
\end{eqnarray*}
\label{lem32}
\end{lemma}

\begin{proof} First, note that
 \begin{eqnarray*}
 \left \lfloor \frac{k}{2} \right\rfloor &= &\frac{2k-1+(-1)^k}{4}\\
\left \lfloor- \frac{k}{2} \right\rfloor&=&-\left \lfloor
\frac{k}{2} \right\rfloor -\frac{1-(-1)^k}{2}\\
\left \lfloor\frac{k-1}{2} \right\rfloor&=&\left \lfloor \frac{k}{2}
\right\rfloor-\frac{(-1)^{k}+1}{2}.
\end{eqnarray*}
  So, the  second formula of the lemma is
$$\sum_{k=2}^{n-2}k \left \lfloor \frac{k}{2} \right\rfloor=
\sum_{k=2}^{n-2}k \left ( \frac{2k-1+(-1)^k}{4} \right).$$ Using the known summations $
\displaystyle \sum_{k}^{l}k^{j}$ and $ \displaystyle \sum_{k}^{l}(-1)^{k}k^{j}$, we obtain
\begin{eqnarray*}
\sum_{k=2}^{n-2}k \left \lfloor \frac{k}{2} \right\rfloor&=&
\sum_{k=2}^{n-2}k \left ( \frac{2k-1+(-1)^k}{4} \right)\\
&=& \frac{n^3}{6}-\frac{7n^2}{8}+\frac{35n}{24}-\frac{13}{16}
-\frac{(-1)^{n-1}n}{8}+\frac{3(-1)^{n-1}}{16}.
\end{eqnarray*}
This may be written
$$
\sum_{k=2}^{n-2}k \left \lfloor \frac{k}{2} \right\rfloor=
\begin{cases}
\frac{(n-2)(4n^{2}-13n +12)}{24}, \text{if $n=2l$;} \\
\frac{(4n-5)(n-1)(n-3)}{24}, \text{if $n=2l+1$.}
\end{cases}
$$

In a more compact formula, the previous summation becomes
$$\sum_{k=2}^{n-2}k \left \lfloor \frac{k}{2} \right\rfloor=
\frac{1}{12}\left \lfloor \frac{n-1}{2}\right \rfloor
 \left ( 4n^2-(15-2(-1)^{n})n+12+ \frac{3}{2}(1-(-1)^{n})\right )$$
 For the fourth one, using the observations in the beginning of the proof,
 we obtain
 \begin{eqnarray*}
\sum_{k=2}^{n-2} \left \lfloor \frac{k}{2} \right\rfloor\left \lfloor -\frac{k-1}{2}
\right\rfloor&=&-\sum_{k=2}^{n-2} \left \lfloor \frac{k}{2} \right\rfloor^2.
\end{eqnarray*}
But \begin{eqnarray*} \sum_{k=2}^{n-2} \left \lfloor \frac{k}{2}
\right\rfloor^2&=&\sum_{k=2}^{n-2} \left ( \frac{2k-1+(-1)^k}{4} \right)^2 \\
 &=&\frac{(n-2)}{4}\left (
\frac{(n-1)(n-3)}{3}+\frac{(1+(-1)^n}{2}\right ).\end{eqnarray*} The
remaining cases are proved similarly.
\end{proof}

\bigskip

\noindent {\bf Proof of Theorem~\ref{thm31}.}\\
\begin{proof}
Before starting the proof, we point out the following easy but important facts:\\ 1) Two finite
topologies are homeomorphic if and only if, as a poset, they have the same Hasse diagram and all the
corresponding open sets have the same cardinality.\\
2) The number of unlabeled non-$T_0$ chain topologies having $k$ nontrivial open sets
(i.e.; open sets different from $\phi$ and $E$) is $\binom{n-1}{k}$. This is also, the
number of strictly increasing sequences having $k$ terms, taken from a set of $n$ integers.\\
For the sake of simplicity and shortness, we performed the calculations using the floor brackets.\\
The computations of $t_{N0}(n,k)$ will be the same as in \cite{Benoumhani}: We list all the
possible forms, and then count them. Calculation of $t_{N0}(n,3)$ is easily obtained: those are
chains with three elements, their number is $\binom{n-1}{1}$. For $t_{N0}(n,4)$, we have two kinds
of topologies: chains with 4 open sets and topologies having the form below
$$
\begin{minipage}{6truecm} \unitlength=.75truemm
\begin{picture}(40,30)(5,-5)
\put(39.25,-.10){$\bullet$} %\put(39,-4){$\emptyset$}
 \put(40,1){\line(-1,1){10}}
\put(29.25,10){$\bullet$} %\put(26,10){$A$}
 \put(30,11){\line(1,1){10}}
 \put(39.25,20){$\bullet$}
 %put(39,23){$E$)}
 \put(40,21){\line(1,-1){10}}
 \put(49.25,10){$\bullet$}
 %\put(52,10){$B$}
\put(40,1){\line(1,1){10}} \put(30,-6) {Figure 1}
\end{picture}

\end{minipage} $$

For the first one the number is $\binom{n-1}{2}$. In the second case, the
total number of different non-$T_0$ topologies is
$\left \lfloor
\frac{n}{2}\right
\rfloor,$  (this is also the number of partitions of $n$ into exactly 2 summands; $P(n,2)$).\\
Furthermore

 $$t_{N0}(n,4)=\binom{n-1}{2}+  \left \lfloor
  \frac{n}{2}\right \rfloor   =\binom{n-1}{2}+P(n,2)=\frac{2n^2-4n+3}{4}+\frac{(-1)^n}{4}.$$
  For $t_{N0}(n,5)$, there are 3 forms: Chains and topologies having one of the two last forms as indicated in the figure below
 (note that the two last diagrams have the same number).
$$ \unitlength=1truemm
\begin{picture}(40,40)(25,-13)
\put(-.75,-15){$\bullet$}  \put(0,-14){\line(0,1){10}}
  \put(-0.75,-5){$\bullet$} \put(2,-15){$\emptyset$}
 \put(0,-4){\line(0,1){10}}
\put(-0.75,5){$\bullet$} %\put(2,5){$B$}
 \put(0,6){\line(0,1){10}}
%\put(2,-5){$A$
 \put(-0.75,15){$\bullet$}
 %\put(2,15){$C$}
 \put(0,15){\line(0,1){10}}
 \put(-0.75,24.5){$\bullet$}
 \put(2,25){$E$}
 %%%
 \put(39.25,0){$\bullet$}
 \put(40,1){\line(-1,1){10}}
\put(29.25,10){$\bullet$} %\put(26,10){$A$}
 \put(30,11){\line(1,1){10}}
 \put(39.25,20){$\bullet$}
 \put(39,23){$E$}
 \put(40,21){\line(1,-1){10}}
 \put(49.25,10){$\bullet$}
 %\put(52,10){$B$}
\put(40,1){\line(1,1){10}} \put(40,1){\line(0,-1){10}} %\put(42,0){$C$}
 \put(39,-14){$\emptyset$}
\put(39.25,-10){$\bullet$}
%%%
%%%
 \put(79.25,-10){$\bullet$}
 \put(80,-9){\line(-1,1){10}}
\put(69.25,0){$\bullet$} %\put(66,0){$A$}
 \put(70,1){\line(1,1){10}}
 \put(79.25,10){$\bullet$}
 %\put(82,12){$C$}
 \put(80,11){\line(1,-1){10}}
 \put(89.25,0){$\bullet$}
 %\put(92,0){$B$}
\put(80,-9){\line(1,1){10}} \put(80,11){\line(0,1){10}}
 \put(79.5,-14){$\emptyset$}
 \put(79,23){$E$}
\put(79.25,20){$\bullet$} \put(30,-20) {Figure 2}
\end{picture}
$$
\\ \\
 For the chains, their number is $\binom{n-1}{3}$. For the
remaining cases the number is counted as follows:
   take $ 2\leq k \leq n-1$, and then partition into 2 parts, to obtain a topology
 of four open sets on a set of cardinality $k$. To obtain a topology on $E$, add the whole set $E$.
 The number of ways to do that is  $$\sum_{k=2}^{n-1}\left \lfloor \frac{k}{2}\right\rfloor .$$
 The previous number is counted twice because of the symmetry of the cases. By
 Lemma~\ref{lem32}, this gives
 $$ 2\sum_{k=2}^{n-1}\left \lfloor \frac{k}{2}\right \rfloor=2\left
\lfloor
  \frac{n}{2}\right\rfloor \left( 2\left \lfloor
  \frac{n}{2}\right \rfloor-(-1)^{n}-1\right)
  =2\left \lfloor
  \frac{n}{2}\right \rfloor \left \lfloor
  \frac{n-1}{2}\right \rfloor
  $$
   Thus, the total number in this case is

\begin {eqnarray*}
t_{N0}(n,5) &=& \binom{n-1}{3}+ 2
  \left \lfloor
  \frac{n}{2}\right \rfloor \left \lfloor
  \frac{n-1}{2}\right \rfloor \\
  &=& \binom{n-1}{3}+\frac{2n^2-4n+1}{4}-\frac{(-1)^n}{4}\\
  &=&\frac{2n^3-6n^2+10n-9}{12}-\frac{(-1)^n}{4}
  \end{eqnarray*}

 For $t_{N0}(n,6)$, the forms are illustrated below

\par\noindent
 \unitlength=1truemm
\begin{picture}(40,40)(-30,-10) \put(-.75,-25){$\bullet$}
 \put(0,-24){\line(0,1){10}}
\put(-.75,-15){$\bullet$}  \put(0,-14){\line(0,1){10}}
  \put(-0.75,-5){$\bullet$} \put(2,-25){$\emptyset$}
 \put(0,-4){\line(0,1){10}}
\put(-0.75,5){$\bullet$} %\put(2,-5){$B$}
 \put(0,6){\line(0,1){10}}
%\put(2,-15){$A$}
 \put(-0.75,15){$\bullet$}
% \put(2, 5){$C$}
 \put(0,15){\line(0,1){10}}
 \put(-0.75,24.5){$\bullet$}
 %\put(2,15){$D$}
 \put(2,25){$E$}
 %%%2

 \put(39.25,0){$\bullet$}
 \put(40,1){\line(-1,1){10}}
\put(29.25,10){$\bullet$} %\put(26,10){$C$}
 \put(30,11){\line(1,1){10}}
 \put(39.25,20){$\bullet$}
 \put(39,23){$E$}
 \put(40,21){\line(1,-1){10}}
 \put(49.25,10){$\bullet$}
 %\put(52,10){$B$}
\put(40,1){\line(1,1){10}} \put(40,1){\line(0,-1){10}}
 %\put(42,0){$A$}
 %\put(42,-10){$D$}
 \put(39,-24){$\emptyset$}
\put(40,-9){\line(0,-1){10}} \put(39.25,-10){$\bullet$}
 \put(39.25,-20){$\bullet$}
%%%3
%%%
 \put(79.25,-10){$\bullet$}
 \put(80,-9){\line(-1,1){10}}
\put(69.25,0){$\bullet$} %\put(66,0){$A$}
 \put(70,1){\line(1,1){10}}
 \put(79.25,10){$\bullet$}
 %\put(82,12){$C$}
 \put(80,11){\line(1,-1){10}}
 \put(89.25,0){$\bullet$}
 %\put(92,0){$B$}
\put(80,-9){\line(1,1){10}} \put(80,11){\line(0,1){10}}
 \put(80,-19){\line(0,1){10}}
 \put(79.5,-24){$\emptyset$}
 \put(79,23){$E$}
\put(79.25,20){$\bullet$}
 %\put(82,-11){$D$}
 \put(79.25,-20){$\bullet$}
\end{picture}

\par\vspace{1.8cm}
 \unitlength=1truemm
\begin{picture}(40,40)(60,-10)
\put(119.25,-25){$\bullet$}
 \put(120,-24){\line(-1,1){10}}
\put(109.25,-15){$\bullet$} \put(106,-15){$A$}
 \put(110,-14){\line(1,1){10}}
 \put(119.25,-5){$\bullet$}
% \put(122,-3){$C$}
 \put(120,-4){\line(1,-1){10}}
 \put(129.25,-15){$\bullet$}
 %\put(132,-15){$B$}
\put(120,-24){\line(1,1){10}}
 \put(120,-4){\line(0,1){10}}%
 \put(120,6){\line(0,1){10}}%
 %\put(122,5){$D$}
 \put(122,14){$E$}
\put(119.25,5){$\bullet$}
 \put(119,-29){$\emptyset$}
 \put(119.25,15){$\bullet$}
 %%%5
 \put(159.25,-25){$\bullet$}
 \put(160,-24){\line(-1,1){10}}
 \put(179.96,-4){\line(-1,1){10}}
\put(149.25,-15){$\bullet$}
 \put(146,-15){$A$}
 \put(150,-14){\line(1,1){20}}
 \put(159.25,-5){$\bullet$}
 \put(157,-3){$C$}
 \put(160,-4){\line(1,-1){10}}
 \put(169.25,-15){$\bullet$}
 \put(172,-15){$B$}
\put(160,-24){\line(1,1){20}}
 %\put(160,-4){\line(0,1){10}}%
 %\put(160,6){\line(0,1){10}}%
 \put(170,8){$E$}
 \put(182,-4){$D$}
\put(169.25,5){$\bullet$}
 \put(159,-29){$\emptyset$}
 \put(179.25,-5){$\bullet$}
 \put(135,-30){Figure 3}
\end{picture}
\par\vspace{2.5cm}
The number of chain topologies is $\binom{n-1}{4}$.  The number of topologies corresponding to the next three graphs
 is counted as follows. Take $ 3\leq k \leq n-1$,
 there are
$$
\frac{1}{2}\left\lfloor
  \frac{k}{2}\right\rfloor \left (2 \left \lfloor
  \frac{k}{2}\right\rfloor
  -(-1)^{k}-1  \right)=\left \lfloor
  \frac{k}{2}\right \rfloor \left \lfloor
  \frac{k-1}{2}\right \rfloor
  $$
 non-$T_0$ topologies with 5 open sets, on a $k$-element set, having the second (or the third) form in Figure 2. A topology
 of the wanted form is obtained  by adjoining the whole set $E$. This sum has to be taken three times.
So, the wanted number is

$$3\sum_{k=3}^{n-1} \left \lfloor
  \frac{k}{2}\right \rfloor \left \lfloor
  \frac{k-1}{2}\right \rfloor
  $$
using the fifth formula in Lemma~\ref{lem32}, we obtain
\begin{eqnarray*}
3\sum_{k=3}^{n-1} \left \lfloor
  \frac{k}{2}\right \rfloor \left \lfloor
  \frac{k-1}{2}\right \rfloor &=& 3 \left \lfloor
  \frac{n}{2}\right \rfloor \left \lfloor
  \frac{n-2}{2}\right  \rfloor
  \left( \frac{2n-5}{6}+\frac{(1-(-1)^{n})}{3}\right )\\
 &=&
 \frac{1}{2}\left \lfloor
  \frac{n}{2}\right \rfloor \left \lfloor
  \frac{n-2}{2}\right \rfloor\left ( 2n-2(-1)^{n}-3\right ).\end{eqnarray*}
  For the number of topologies corresponding to the last graph in Figure 3, let $ 2 \leq k=|C| \leq n-1$. Partition it into two parts,
  $k=l+j,\;|A|=l,\; |B|=j$. To each partition of $k$, it corresponds two different topologies: $\displaystyle \tau_1= \{ \phi,\;A_1,\; B_1,\; C,\; D_1,\; E \},\; |A_1|=l,\; |B_1|=j, \; |D_1|=n-k+j$
  and $\displaystyle  \tau_2=\{ \phi,\;A_2,\; B_2,\; C,\; D_2,\; E \},\; |A_2|=j,\; |B_2|=l,\; |D_2|=n-k+l $, unless $l=j$, in this case we have only one topology. So, the number of topologies corresponding to $k$ is
  $$
\alpha_k=
 \begin{cases} 2  \left \lfloor \frac{k}{2} \right \rfloor,
 \text{if $k=2l+1$}; \\
  2 \left \lfloor\frac{k}{2} \right \rfloor -1 ,
\text{if $k=2l$}.
\end{cases}
$$

  The wanted number is $\displaystyle \sum_{k=2}^{n-1}{\alpha_k}=1+2+3+ \cdots +n-2=\binom{n-1}{2}.$

 Finally, summing and replacing  $\displaystyle \left \lfloor
  \frac{l}{2}\right \rfloor $ by  $\displaystyle \frac{2l-1+(-1)^l}{4}$, and the binomial coefficients by their values
we obtain
\begin{eqnarray*}
t_{N0}(n,6)&= & \binom{n-1}{4}+ \binom{n-1}{2}+\frac{1}{2}\left \lfloor
  \frac{n}{2}\right \rfloor \left \lfloor
  \frac{n-2}{2}\right \rfloor \left ( 2n-2(-1)^{n}-3\right )\\
  &=&   \binom{n-1}{4}+ \binom{n-1}{2}+\frac{4n^3-18n^2+20n-3}{16}+(-1)^n\frac{3}{16}\\
  &=& \frac{2n^4-8n^3+40n^2-112n+87}{48}+(-1)^n\frac{3}{16}
  \end{eqnarray*}
For $t_{N0}(n,7))$, we have $\binom{n-1}{4}$) chain topologies. Insert the graph in Figure 1 into a
chain of 4 elements (the number is counted 4 times): this gives the number

$$ 4\sum_{k=2}^{n-3}\binom{n-k-1}{2}\left \lfloor \frac{k}{2}\right
\rfloor.$$ By Lemma~\ref{lem32}, this number is

\begin{eqnarray*}
4\sum_{k=2}^{n-3}\binom{n-k-1}{2}\left \lfloor \frac{k}{2}\right
\rfloor &=& 2\sum_{k=2}^{n-3}(n-k-1)(n-k-2) \left \lfloor \frac{k}{2}\right
\rfloor \\
&=& 2(n^2-3n+2)\sum_{k=2}^{n-3}\left \lfloor \frac{k}{2}\right \rfloor-2(n-3)\sum_{k=2}^{n-3}\left \lfloor \frac{k}{2}\right
\rfloor+2\sum_{k=2}^{n-3}(k^2+k)\left \lfloor \frac{k}{2}\right
\rfloor.\\
&=& \frac{1}{3} \left \lfloor \frac{n}{2}\right \rfloor \left \lfloor \frac{n-2}{2}\right \rfloor \left ( n^2-4n-(1+(-1)^{n})(n-4)+ \frac{1-(-1)^{n\underline{}}}{2}\right )
\end{eqnarray*}

Again, replacing $\displaystyle \left \lfloor \frac{l}{2}\right \rfloor $ by its value, we obtain
$$ \frac{1}{3}\left \lfloor \frac{n}{2}\right \rfloor \left \lfloor \frac{n-2}{2}\right \rfloor \left ( n^2-4n-(1+(-1)^{n})(n-4)+ \frac{1-(-1)^{n\underline{}}}{2}\right )=\frac{4n^4-32n^3+80n^2-64n+6}{48}-\frac{1}{8}(-1)^n $$
The number corresponding to the graph below is is evaluated as
follows:
\par
\centerline{
\begin{picture}(100,50)(-5,-5)
%% the frist plot
\put(15,15){\line(1,1){15}} \put(30,15){\line(-1,1){15}}
  \put(15,15){\line(1,-1){7.5}}\put(22.5,7.5){\line(1,1){7.5}}
  \put(15,30){\line(1,1){7.5}}\put(30,30){\line(-1,1){7.5}}
\put(14.5,14){$\bullet$}  \put(21.6,21.5){$\bullet$}
   \put(29,29){$\bullet$}\put(29,14){$\bullet$}
 \put(14.5,29){$\bullet$}\put(21.6,36.5){$\bullet$}
  \put(21.6,7){$\bullet$} %\put(21.6,47){$\bullet$}
\put(18, -3) {Figure 4}
\end{picture} }
 Take a subset of cardinality $ 2\leq k\leq n-2$, then partition it
into two parts, this is done by $\lfloor \frac{k}{2} \rfloor$ and
then put over it two subsets containing it, such that their union is
the whole set, this can be done by $ \lfloor
  \frac{n-k}{2} \rfloor$, so the wanted number is
  \begin{eqnarray*}
\sum_{k=2}^{n-2} \left \lfloor \frac{k}{2}\right \rfloor
    \left \lfloor \frac{n-k}{2} \right
\rfloor&=& \left \lfloor
  \frac{n}{2}\right \rfloor \left \lfloor
  \frac{n-2}{2}\right \rfloor \frac{\left ( n-(-1)^{n}\right)}{6}\\
  &=&\frac{2n^3-6n^2+n+3}{48}+\frac{3(n-1)}{48}(-1)^n
    \end{eqnarray*}

  The number corresponding to the last case in  $t_{N0}(n,6)$, with a least or a greatest
element is (this is counted twice):
   $$2\sum_{k=3}^{n-1}\binom{k-1}{2}=2\binom{n-1}{3}. $$
   Summing all these numbers,  we obtain $t_{N0}(n,7)$ as claimed.

For $t_{N0}(n,8)$, we have six forms: chains ($ \binom{n-1}{6}$). Topologies such that their
diagram is formed by three copies of the graph in Figure 1. Their number is $2\binom{n-1}{3}$.
Topologies in the figure corresponding to the last case in $t_{N0}(n,6)$, with a least and a
greatest element added (counted three times) is  $3 \binom{n-1}{3}$. The number of different and
non homeomorphic topologies corresponding to
the graph below, is counted as follows: \\
\begin{picture}(100,60)(0,0)
  %%% the second plot
  \put(55,15){\line(1,1){15}} \put(70,15){\line(-1,1){15}}
  \put(55,15){\line(1,-1){7.5}}\put(62.5,7.5){\line(1,1){7.5}}
 \put(55,30){\line(1,1){7.5}}\put(70,30){\line(-1,1){7.5}}
  \put(54.5,14){$\bullet$} \put(69,14){$\bullet$} \put(61.6,21.5){$\bullet$}
  \put(61.6,7){$\bullet$} \put(61.6,36.5){$\bullet$} \put(69,29){$\bullet$}
  \put(54.5,29){$\bullet$}
  \put(62.4,37.5){\line(0,1){10}}  \put(61.6,47){$\bullet$} \put
(55,0){Figure 5}
\end{picture}
\\ \\
The number of topologies with $7$ open sets on a set of $k>3$
elements, having the form in Figure 4, is
$$ \left \lfloor
  \frac{k}{2}\right \rfloor \left \lfloor
  \frac{k-2}{2}\right \rfloor \frac{\left ( k-(-1)^{k}\right)}{6}.  $$
In order to obtain a topology on $E$, with 8 open sets, we adjoin the whole set. Thus, the number
is $$ \sum_{k=4}^{n-1} \left \lfloor
  \frac{k}{2}\right \rfloor \left \lfloor
  \frac{k-2}{2}\right \rfloor \frac{\left (
  k-(-1)^{k}\right)}{6}.$$ The last sum is counted three times.
  We  obtain
  \begin{eqnarray*}
  3\sum_{k=4}^{n-1} \left \lfloor
  \frac{k}{2}\right \rfloor \left \lfloor
  \frac{k-2}{2}\right \rfloor \frac{\left (
  k-(-1)^{k}\right)}{6}&=&\frac{1}{2}\left \lfloor
  \frac{n}{2}\right \rfloor \left \lfloor
  \frac{n-2}{2}\right  \rfloor \left \lfloor
  \frac{n-3}{2}\right  \rfloor \left ( \frac{n}{2}+\frac{(1-(-1)^{n})}{4}\right )\\
   &=&\frac{2n^4-12n^3+16n^2+6n-9}{64}-\frac{3(2n-3)}{64}(-1)^n.
  \end{eqnarray*}


Topologies formed by a chain and a copy of the graph in Figure 1 (a
chain with four elements over the graph). These topologies are
counted 5 times. This is counted as follows: For each $2\leq k\leq
n-4$, we may construct
$$\binom{n-k-1}{3}\left \lfloor
  \frac{k}{2}\right \rfloor$$ different topologies. Using the summation formulas of Lemma~\ref{lem32}, the total
  number is
  \begin{eqnarray*}
  5\sum_{k=2}^{n-4}\binom{n-k-1}{3}\left \lfloor
  \frac{k}{2}\right \rfloor&=&\frac{1}{12}\left \lfloor
  \frac{n}{2}\right \rfloor \left \lfloor
  \frac{n-2}{2}\right  \rfloor \left \lfloor
  \frac{n-4}{2}\right  \rfloor \left ( 2n^2-(10+3(-1)^n)n+1+\frac{15}{2}(1+(-1)^{n})\right ) \\
  &=&\frac{4n^5-50n^4+220n^3-400n^2+256n-15}{192}+\frac{15}{192}(-1)^n
  \end{eqnarray*}
The number of topologies corresponding to the boolean algebra
(the figure below) is $P(n,3)$, the number of partitions of the integer $n$ into 3 summands. Summing all these quantities, we obtain the desired formula.

%\begin{center}
%\begin{minipage}{6truecm}
%\par\noindent
 %\unitlength=1truemm
\begin{center}
 $$
\begin{picture}(50,60)(-10,-10)
  \put(9.25,15){$\bullet$}
  \put(9.25,25){$\bullet$}
  \put(-10.55,25){$\bullet$}
   \put(-10.55,15){$\bullet$}
\put(-0.75,5){$\bullet$}
 \put(0,6){\line(1,1){10}}
 \put(0,6){\line(0,1){10}}
 \put(0,26){\line(0,1){10}}
  \put(0,6){\line(-1,1){10}}
   \put(0,16){\line(-1,1){10}}
    \put(0,16){\line(1,1){10}}
    \put(0,26){\line(-1,-1){10}}
    \put(0,36){\line(1,-1){10}}
    \put(0,36){\line(-1,-1){10}}
    \put(0,26){\line(1,-1){10}}
 \put(10,16){\line(0,1){10}}
 \put(-10,16){\line(0,1){10}}
 \put(-0.75,15){$\bullet$}
 \put(-0.75,35){$\bullet$}
 \put(-0.75,24.5){$\bullet$}
 %\put(-8,-8){Figure X}
 \end{picture} \put(-45,-5) { Figure 6}
%\end{minipage}
$$
\end{center}
\end{proof}

\section{ Conclusion}

In the first part of this paper, we computed  $T(n,k)$  for every
$n$ and $k \leq 17$. Our method does not use the powerful tool of
poset theory. It is somewhat direct and  rests heavily on some
elementary and simple diophantine equations.  Calculations may be
pushed till a reasonable range of $k$,  ($k \leq 30$ for example),
and if refined, we may go far beyond this range.

In the second part of the paper, we encounter a difficulty  to compute
the numbers $t_{N0}(n,k)$, since there is no method to deal with them,
and there is no analogous formula, as for  $T_0(n,k)$. Here too, we
may reach the range of $k=10 $ or $k=11$. This is the project of a
forthcoming paper.


\section{Acknowledgement}

We are very indebted to the anonymous referee for her (his) careful
reading of the manuscript, corrections, valuable suggestions and
comments which greatly improved the paper.

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\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:
Primary 05A15; Secondary 06A07, 06A99.

\noindent \emph{Keywords: } binary relation, enumeration, finite set,
finite topology, partial order, posets.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000798},
\seqnum{A001930},
\seqnum{A008277}, and
\seqnum{A122934}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received September 11 2007;
revised versions received  October 18 2007; July 7 2009; January 19 2010;
March 6 2010.
Published in {\it Journal of Integer Sequences}, March 10 2010.
Minor revision, March 18 2010.
Added author's new address, December 1
2010.


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