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A180916
Number of convex polyhedra with n faces that are all regular polygons.
5
0, 0, 0, 1, 2, 3, 2, 7, 3, 6, 4, 7, 3, 13, 2, 5, 4, 6, 1, 9, 2, 6, 1, 4, 1, 8, 4, 2, 1, 3, 1, 10, 1, 3, 1, 2, 4, 3, 1, 2, 1, 9, 1, 2, 1, 2, 2, 2, 1, 2, 1, 9, 1, 2, 1, 2, 1, 2, 1, 2, 1, 9, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 2, 1, 2
OFFSET
1,5
COMMENTS
For all n > 92, the sequence is identical to A000034 because for large n only prisms (even and odd n) and antiprisms (even n) are convex and have regular polygonal faces. The MathWorld article about Johnson Solids is very informative about this topic.
In a regular-faced polyhedron, any two faces with the same number of edges are congruent. (Proof: As the two faces are regular polygons, it suffices to show their edges have the same length. But as all faces are regular polygons and the polyhedron is connected, all edges have the same length.) - Jonathan Sondow, Feb 11 2018
LINKS
FORMULA
a(A296602(n)) = 1. - Jonathan Sondow, Jan 29 2018
EXAMPLE
a(6) = 3 because the cube, pentagonal pyramid, and triangular bipyramid all qualify. a(7) = 2 because only the pentagonal prism and elongated triangular pyramid qualify; the hexagonal pyramid is impossible with equilateral triangles
MATHEMATICA
f = Tally[Join[PolyhedronData["Platonic", "FaceCount"], PolyhedronData["Archimedean", "FaceCount"], PolyhedronData["Johnson", "FaceCount"], {PolyhedronData[{"Prism", 3}, "FaceCount"]}]]; f2 = Transpose[f]; cnt = Table[0, {n, 100}]; cnt[[f2[[1]]]] = f2[[2]]; Do[cnt[[n]]++, {n, 7, 100}] (* add prisms *); Do[ cnt[[n]]++, {n, 10, 100, 2}] (* add antiprisms *); cnt (* T. D. Noe, Mar 04 2011 *)
CROSSREFS
KEYWORD
nice,nonn
AUTHOR
J. Lowell, Sep 23 2010
EXTENSIONS
More terms from J. Lowell, Feb 28 2011
Corrected by T. D. Noe, Mar 04 2011
STATUS
approved