# Question about 3-regular graphs with a restriction (also fullerene and four color theorem)

(Crossposted to mathoverflow.net)

Studying all 3-regular graphs that have only faces with 5 edges or more (simplified), I empirically found (computer program) that many hypothetically possible graphs, that by Euler's identity may exist ($F5 = 12 + F7 + 2F8 + 3F9 + ...$), do not actually exist. Using a VF2 algorithm to filter out isomorphic maps being created, I also noticed that not so many graphs as I expected exist. And that one general category of graphs, that always represents a simplified 3-regular graph, is that of fullerenes (with 12 faces F5 and an arbitrary number of F6). Here is a list of what I found, so far, for each class of graphs, from 12 faces to 20 faces (surrounding area included).

The question is: Since the computation of maps with 17, 18, 19, 20 faces (simplified and not containing isomorphic graphs) it is taking very long time (days of CPU time on a PC), is this sequence already known?

• 12 faces: 1 (only 1 graph exists)
• On 3 dimentional space (sphere) it is a dodecahedron
• It is a fullerene: 20-fullerene Dodecahedral graph
• 13 faces: 0 (no simplified graphs exist with 13 faces)
• The hypothetical (by Euler's identity) map of 12 F5 and 1 F6 does not exist
• 14 faces: 1 (12 F5 + 2 F6)
• he hypothetical (by Euler's identity) map of 13 F5 and 1 F7 does not exist
• It is a fullerene: GP (12,2) Generalized Petersen graph
• 15 faces: 1 (12 F5 + 3 F6)
• The hypothetical (by Euler's identity) map of 14 F5 and 1 F8 does not exist
• It is a fullerene: 26-Fullerene
• 16 faces: 3 (Two graphs are 12 F5 + 4 F6. The other has 14 F5 + 2 F7)
• The hypothetical (by Euler's identity) map of 14 F5 and 2 F7 does exists
• The other two are Fullerenes
• 17 faces: ???
• 18 faces: ???
• 19 faces: ???
• 20 faces: ???