How many simplicial complexes on n vertices up to homotopy equivalence?

Andrew Newman just posted a preprint to the arXiv showing that the answer is doubly exponential in $n$.

In particular, he showed that the number of homotopy types of simplicial complexes on $n$ vertices is at least $$\exp \left( \exp \left( 0.004n \right) \right),$$ for all large enough $n$.

This matches the upper bound from Dedekind numbers, up to the constant $0.004$.

Newman's result depends on showing that this many different torsion subgroups are possible for homology in dimension $d$, where $d \approx \delta n$ for some small constant $\delta > 0$. The existence proof is partly constructive and partly depends on the probabilistic method.

https://arxiv.org/abs/1804.06787

Fernando Muro's argument seems convincing that getting an exact formula is likely to be impossible. But we still might find lower and upper bounds that give us a sense of the asymptotics.

We can get a lower bound by restricting to a subset, like graphs up to homotopy equivalence. This has a pretty nice generating function.

$\sum_{n=0}^\infty \gamma(n) q^n=\frac{1}{(1-q)^2(1-q^3)(1-q^4)^2(1-q^5)^3(1-q^6)^4\dots}=\frac{1}{(1-q)^2}\prod_{n=1}^\infty \frac{1}{(1-q^{n+2})^n}$

The reason for this is that you can identify a complex by the number of connected graphs of each Euler characteristic it contains. Then each complex shows up at whatever the minimal $n$ is to express it, which is just a sum over the graphs, and at each larger $n$. Since the number of possible Euler characteristics is quadratic in $n$, the number of new types at each $n$ is linear. You have two extra $1-q$ terms, one to account for the 1-vertex graph, and one to account for homotopy types showing up after the minimal $n$.

I would expect that the answer for the Dedekind problem on the number of simplicial complexes on $n$ labelled vertices should essentially give the right behavior (doubly exponential with $n$) also for homotopy type. (The difference between labelled and unlabelled vertices is at most $n!$ so it is negligible.)