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. If so, please can you explain why?

PPS: I posted this question also on Math.SE, but no one answered.