I was wondering how to count the number of abstract simplicial complexes there are on $\left[n\right]=\left\{1,…,n\right\}$. An abstract simplicial complex being a set $\Sigma\subset\wp\left(\left[n\right]\right)$ where $\emptyset\not\in \Sigma$ such that for any $\sigma\in \Sigma$ and any $\tau\subset \sigma$ with $\tau\neq\emptyset$, $\tau\in \Sigma$.
For $n=1$ there is one.
For $n=2$ there are 4 $\left\{\left\{1\right\}\right\}$, $\left\{\left\{2\right\}\right\}$, $\left\{\left\{1\right\},\left\{2\right\}\right\}$, and $\left\{\left\{1\right\},\left\{2\right\},\left\{1,2\right\}\right\}$.
It seems clear that for $n=3$ there should be $8$ "points", $8+3$ one dimensional simplices, one "filled in triangle" for a total of 20.
*In the above, I forgot the empty one, so add one to those answers, or exclude it from the definition.
Best Answer
This is an open problem; there is not believed to be any closed form and the exact value has not even been computed for $n=9$. It is known that the number $M(n)-1$ of abstract simplicial complexes on a set of size $n$ grows roughly like $2^{\binom{n}{\lfloor n/2 \rfloor}}$, or more precisely that $\frac{\log_2 M(n)}{\binom{n}{\lfloor n/2 \rfloor}}\to 1$ as $n\to\infty$ (see the link above for even more precise information about its asymptotics). The idea behind this estimate is that a simplicial complex is determined by picking its maximal faces which must form an antichain (no one is contained in any other) and when $n$ is large, the best way to get lots of antichains is to pick subsets of the same size and the best size to pick is $n/2$.