Vertex configurations are a topic of geometry as sequences of geometrical faces around a vertex. They give rise to uniform polyhedra (i.e. tilings of the sphere) or periodic tilings of the Euclidean or hyperbolic plane ("uniform tilings" in short).
Firstly, I don't know if for an arbitrary vertex configuration $(n_1.n_2.\dots.n_k)^m$ there exists either
a uniform polyhedron
a periodic tiling of the Euclidean plane or
a periodic tiling of the hyperbolic plane
or possibly none of these.
In case of the latter, I wonder:
Is there for every vertex configuration some geometry (possibly different from spherical, Euclidean, or hyperbolic) in which it defines a uniform tiling?
In any case: Each vertex configuration that defines a uniform tiling defines also an abstract vertex-transitive graph: its skeleton graph.
The question then arises:
In case there is a vertex configuration that doesn't define a uniform tiling (in any geometry) could it nevertheless define an abstract vertex-transitive graph (or a set of)?
In terms of abstract graphs, the vertex configuration is a sequence of lengths of minimal cycles a vertex is contained in (these cycles forming "faces"). It defines the neighbourhood of a vertex, including shortest paths between its immediate neighbours.
Can from the vertex configuration be told if the corresponding graph(s) have to be finite or can (or must) be infinite? For (geometrical) tilings the angle defect can be used to tell how many vertices the tiling (and thus the graph) must have (by Descartes's theorem), which may be a finite or an infinite number. But what if no geometrical argument is at hand? (To make it clear: In case that every vertex configuration does define a uniform tiling, the angle defect is always at hand.)