OFFSET
1,1
COMMENTS
A 10-gon is the highest-sided polygon a Johnson solid can have as a face.
All terms in columns k = 7 and k = 9 are zero (there are no Johnson solids with heptagons or 9-gons as faces).
LINKS
Paolo Xausa, Table of n, a(n) for n = 1..356 (all 92 rows, flattened).
Wikipedia, List of Johnson solids.
EXAMPLE
Triangle begins:
n\k| 3 4 5 6 7 8 9 10
-------------------------------------
1 | 4, 1;
2 | 5, 0, 1;
3 | 4, 3, 0, 1;
4 | 4, 5, 0, 0, 0, 1;
5 | 5, 5, 1, 0, 0, 0, 0, 1;
6 | 10, 0, 6, 0, 0, 0, 0, 1;
7 | 4, 3;
8 | 4, 5;
9 | 5, 5, 1;
10 | 12, 1;
11 | 15, 0, 1;
12 | 6;
13 | 10;
14 | 6, 3;
15 | 8, 4;
16 | 10, 5;
17 | 16;
18 | 4, 9, 0, 1;
19 | 4, 13, 0, 0, 0, 1;
20 | 5, 15, 1, 0, 0, 0, 0, 1;
...
Johnson solid J_6 (pentagonal rotunda) has 10 triangular faces, 0 square faces, 6 pentagonal faces, 0 hexagonal faces, 0 heptagonal faces, 0 octagonal faces, 0 9-gonal faces and 1 10-gonal face.
MATHEMATICA
A394912row[n_] := BinCounts[#, {3, Max[#] + 1}] & [Map[Length, PolyhedronData["J" <> ToString[n], "FaceIndices"]]];
Array[A394912row, 92] (* Gives the full sequence *)
PROG
(Python)
"Return [T(n, k), k = 3 .. A394913(n)] if k is not given."
if (L := getattr(A394912, f"J{n}", 0))==0:
setattr(A394912, f"J{n}", L := [len(f) for f in JS[n-1]['faces']])
return sum(L==k for L in L) if k else [A394912(n, k)for k in range(3, max(L)+1)]
for n in range(1, 20): print(A394912(n)) # M. F. Hasler, Apr 10 2026
CROSSREFS
KEYWORD
nonn,tabf,fini,full
AUTHOR
Paolo Xausa, Apr 08 2026
STATUS
approved
