It is well-known since Plato that there are only 5 regular polyhedra which live in 3D Euclidean space. However abstract polytopes are defined solely by their incidences, and are not confined by the geometry of 3 dimensional Euclidean space, so there may be more of them. Wikipedia mentions one non-Euclidean example, the hemicube. Are there more? Is there an algorithm to enumerate them?
Or to be less ambitious, it seems to me that for any Schläfli symbol of rank 2 you can write down, $\{p,q\}$, there is a regular polyhedron of that type. For example a polyhedron with 13 heptogonal faces. Does that exist?
Thanks.
There are infinitely many abstract regular polyhedra. Several are discussed at https://en.wikipedia.org/wiki/Regular_polyhedron#Abstract_regular_polyhedra, including Petrials, spherical polyhedra, hosohedra, and dihedra.
Many abstract regular polyhedra are infinite; any tilings of the Euclidean or hyperbolic planes (as well as tilings of the 2-sphere) are considered abstract polyhedra. These include tilings of Schläfli type $\{p,q\}$ for any $p, q \geq 2$. You can take quotients of these abstract polyhedra by subgroups of their automorphism group to get more abstract regular polyhedra. In fact, every abstract regular polyhedron arises in this way, since any such polyhedron has a Schläfli type $\{p,q\}$, and is a quotient of the tiling of type $\{p,q\}$, which is called the "universal polytope $\{p,q\}$" for this reason. Sometimes these quotients are finite.
You may find data about finite abstract regular polyhedra, with at most 2000 flags (not including 1024 or 1536, which are apparently too numerous), at http://www.abstract-polytopes.com/atlas/r3. This includes 5946 nondegenerate and 993 degenerate polytopes, with 2250 different Schlafli types.
There is no polytope with 13 heptagonal faces, since the number of odd-length faces must be even, but there are three polytopes listed of type $\{7,13\}$. There is one with 14 heptagonal faces.