Counting the topologies of a fixed number of open sets Let $\ X\ $ be a finite set, and $\ n:= |X| > 0.\ $ Let $\ Top(X)\ $ be the set of all topologies in $\ X,\ $ and $\ top(n) := \left|Top(X)\right|.\ $ Define:
$$ Top(X;t)\ :=\ \left\{T\in Top\left(X\right)\ :\ |T| = t \right\} $$
and
$$ top(n;t)\ :=\ \left|Top\left(X;t\right)\right| $$
for all integers $\ t\ $ such that $ 2\le t\le 2^n.\ $ Thus now there are some small theorems and a number of tough open questions about $\ top(n;t).\ $ Let me formulate one (approximate answers are most welcome too):

QUESTION:   What are $\ t_n\ \&\ v_n\ \,(2\le t_n\le 2^n)\ $ such that
  $$ v_n = top(n;t_n)\ =\ \max_{t\ :\ 2\le t\le 2^n}\,\ top(n;t) $$
  for positive integers $\ n.$

Of course $\ t_n\ $ does not have to be unique.
About Notation: we have $\ top(2)=4\ $ while there are only $\ 3\ $ different homeomorphic classes in a $2$-element set.

In view of the @RichardStanley's answer and of my addition below, let me ask a simpler and still a specific question from in-between:


QUESTION   Compute or estimate $\ top\left(n,\,\ 2^{\lfloor\frac n2   \rfloor}\right),\ $ also $\ top(n\,\ 2^{n-1}),\ top(n\,\ 2^{n-2}\ )$ and $\ top(n\,\ 2^{n-3})\ $ for $\ n>3$.

In general, can one take an advantage of an eventual divisibility $\ k\,|\,n\ $ (or similar) when computing $\ top(n\ k)\,$?
 A: Information on this question is given in
https://arxiv.org/pdf/0802.2550.pdf. In particular,
$t(n,k)$ was computed for $k\leq 23$ by Erné and Stege, Ars
Combinatoria 40 (1995), 65--88. For $k\leq 5$ we have
$t(n,3)=2^n-2$, and for $n\geq 4$,
  $$ t(n,4) = \frac 12(2\cdot 3^n-5\cdot 2^n+4) $$
  $$ t(n,5) = 4^n-3\cdot 3^n+2^n-3. $$
A: In view of the @RichardStanley's answer (thank you!) let me add some information from the other end, I'll assume $\ n\ge 2$:


*

*$\ top(n\,\ 2^n)\ =\ 1 $

*$\ top(n\,\ k)\ =\ 0\quad $ for $\ 3\cdot 2^{n-2} < k < 2^n $

*$\ top(n\ \ 3\!\cdot\!2^{n-2})\,\ =\,\  n\cdot(n-1) $


After a couple more steps in this direction, the exact calculations get very tough.

We also get the following general property:
$$ top\left(n\ t\right) > 0\quad\Rightarrow\quad
      top\left(n\!+\!m\quad 2^m\!\cdot\! top\left(n\ t\right)
\right)\ >\ 0 $$
A: The @RichardStanley's approach to the topic of this thread (see: math.mit.edu/~rstan/pubs/pubfiles/3.pdf) was via orders (partial and quasi). I'd like to give a gist of a topological approach--this was a byproduct of my replacement of filters and Moore-Smith sequences by singular spaces. It's a question of elegance, that's all.
The combinatorics underneath is the same but the approaches feel different--I apply results for the general spaces to the finite spaces.
DEFINITION A topological space (and its topology) is called singular when it has exactly one non-isolated point.
Given a topological space $\ (X\ T),\ $ and $\ a\in X,\ $ let:
$$ T_a\ :=\ 2^{X\setminus \{a\}}\,\cup\, \{G\in T: a\in G\} $$
Then $\ T_a\ $ is a singular topology such that $\ T\subseteq T_a.\ $ It follows that every maximal topology is singular, where maximal means maximal non-discrete. Furthermore:
THEOREM 0   Topologies, maximal among the singular topologies, are the same as maximal topologies.
THEOREM 1   Every topology $\ T\ $ is the intersection of the singular topologies (in the same set of points) that contain $\ T$.
THEOREM 2   Every singular topology $\ S\ $ is the intersection of the maximal singular topologies that contain $\ S.\ $ Every topology $\ T\ $ is the intersection of the maximal (singular) topologies that contain $\ T$.
In the case of a finite space $\ (X\ T),\ $ the topology admits the minimal base function $\ \beta_T: a\mapsto \beta_T(a) := \bigcap\{G\in T: a\in G\}.\ $ We get a reverse ordering of these functions w.r. to topologies:
$$ S\subseteq T\quad\Leftrightarrow\quad \beta_T\le_X\beta_S $$
for arbitrary topologies $\ S\ T\ $ in $\ X,\ $ where
$$ \beta_T\le_X\beta_S\quad\Leftarrow:\Rightarrow\quad
        \forall_{x\in X}\,\ \beta_T(x)\subseteq\beta_S(x) $$
It's clear that in the finite case, a topology $\ T\ $ in $\ X\ $ is maximal $\ \Leftrightarrow $
$$ \exists_{a\ b\in X: a\ne b}\ \ \left(\left(\beta_T(a)\, =\, \{a\ b\}\right)
   \ \ \&\ \ 
  \left(\forall_{x\in X\setminus\{a\}}\ \beta_T(x)=\{x\}\right)\right) $$
Call such topology $\ T_{a\ b}.\ $ Obviously, $ |T_{a\ b}|=\frac 34\cdot 2^n,\ $ hence


*

*$\ \forall_{t:\ \frac 34\cdot 2^n\,<\,t\,<\,2^n}\ \ top(n\ t) = 0 $

*$\ top\left(n\ \ \frac 34\!\cdot\! 2^n\right)\ =\,\ n\cdot(n-1) $



Next step. Consider submaximal topologies, i.e. topologies which are maximal among non-discrete and non-maximal topologies. Obviously, each of them must be an intersection of exactly two maximal topologies (see the 2nd part of Theorem 2 above). This leads to $5$ cases (all points $a$ and $b$, and eventually $c$ or even $d$, are all different):


*

*$\ T := T_{a\ b}\cap T_{b\ a},\ \ |T| = 2^{n-1} $ -- remark: the non-$T_0$ case;

*$\ T := T_{a\ b}\cap T_{a c},\ \ |T| = 2^{n-1}+2^{n-3}\ $ -- remark: $\ \beta_T(a) = \{a\ b\ c\};$

*$\ T := T_{a\ c}\cap T_{b\ c},\ \ |T| = 2^{n-2}+2\cdot 2^{n-3}+2^{n-3}=2^{n-1}+2^{n-3} $

*$\ T := T_{a\ b}\cap T_{b\ c},\ \ |T| = 2^{n-2}+2^{n-3}+2^{n-3} = 2^{n-1} $ -- remark: $\ \beta_T(a) = \{a\ b\ c\};$

*$\ T:= T_{a\ b}\cup T_{c\ d},\ \ |T| = 2^{n-2}+2\cdot 2^{n-3}+2^{n-4} = 2^{n-1}+2^{n-4} $


Next, one may consider the subsubmaximal topologies $\ T\ $ (some of these cardinalities $\ T\ $ coincide with the previous submaximal case). Etc. etc.
