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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.

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    $\begingroup$ There is most likely an exercise in one of R. Stanley books on Enumerative Combinatorics that solves this. $\endgroup$
    – Per Alexandersson
    Commented Aug 14, 2014 at 12:52

2 Answers 2

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This is http://oeis.org/A006602 . I checked the numbers there.

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

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    $\begingroup$ This is also A003182(n)-A003182(n-1) (oeis.org/A003182) which has many more references $\endgroup$
    – Philippe Nadeau
    Commented Aug 14, 2014 at 13:20
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Since Brendan has identified the sequence and provided values for small $n$, let me point out that the asymptotic behavior of your sequence $s(n)$ will be $$s(n) \sim \frac{1}{n!}d(n)$$ where $d(n)$ is the $n$-th Dedekind number.

To see why, note that a simplicial automorphism must at least induce a graph automorphism of the 1-skeleton, but the number of graph automorphisms of a graph on $n$ vertices are almost surely zero unless the number of edges is within $\log(n)$ of the extreme values $0$ and $\binom{n}{2}$. See Chapter 9 of Bollobas' book on Random graphs for a proof.

To answer your post-script -- this is very different from my question on homotopy classes of simplicial complexes in the sense that you are seeking isomorphism classes whereas I was only seeking homotopy classes. In particular, I would expect (at least for small $n$) for your sequence to be much larger than mine since there are many homotopy-equivalent but non-isomorphic simplicial complexes on $n$ vertices.

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