|
| |
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
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
|
|
|
|
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
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
