My question is about (abstract) simplicial complices.  

In particular, how many are they if I consider $n$ unlabelled vertices?


For example, if $n=4$, the two complices
$$
\{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{3, 4\}\}
$$
and
$$
\{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{2, 3\}, \{1, 4\}\}
$$
are the same, but not
$$
\{\varnothing, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}\}
$$
(since the last two sides of this one intersect in one vertex).


If $n=3$, there are 5 of them (while the Dedekind number for 3 is 20).  

They are:  

- dim=2
$$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}\}$$
- dim=1
$$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}\}$$
$$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}\}$$
$$\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}\}$$
- dim 0
$$\{\varnothing, \{1\}, \{2\}, \{3\}\}$$


Since this last observation, I think that the answer is not the Dedekind number, but please prove me wrong if you think it is.


Thank you in advance,
Davide


PS: I am not sure whether or not this question is related to [this other one](https://mathoverflow.net/q/102587/42995). If so, please can you explain why?  

PPS: I posted this question also [on Math.SE](https://math.stackexchange.com/q/896042/110046), but no one answered.



Simplicial complices on unlabelled vertices
This is <http://oeis.org/A006602> . I checked the numbers there.


The correspondence is because the maximal elements form an antichain cover.


