Verified answer

My guess would be the number of regular convex polytopes (counted up to similarity) in R^n.

There is only one (up to similarity) in R^1.

There are infinitely many regular convex polygons in R^2 (for each positive integer n >= 3, there is a regular convex n-gon, and the regular n-gon and regular m-gon are not similar unless m = n).

There are exactly five regular convex polytopes in R^3 (corresponding to the five "platonic solids", familiar to people who play role playing games as the four, six, eight, twelve, and twenty-sided dice). The ancient Greeks proved that there are only 5 (although they were not using modern definitions or modern notions of rigorous proof, it is not hard to translate their work into this form).

That there are exactly six convex polytopes in R^4 up to similarity. According to Wikipedia this was first proved by Schlafli in the 1800s (and I think that is even right--- unlike a lot of historical math stuff on Wikipedia).

And for each n >= 5 there are exactly three regular convex polytopes (the higher dimensional analogues of the tetrahedron, cube, and octahedron). I don't know who proved this first but it is certainly covered well in Coxeter's book on convex polytopes.

I honestly don't know why R^0 (usually defined to be a single point) is said to have exactly one regular convex polytope (which I guess would be that point). But it doesn't bother me very much. Presumably it comes from putting n = 0 into somebody's favorite definition of "convex polytope in R^n."