A338149 Triangle read by rows: T(n,k) is the number of achiral colorings of the edges of a regular n-D orthoplex (or ridges of a regular n-D orthotope) using exactly k colors. Row 1 has 1 column; row n>1 has 2*n*(n-1) columns.

1, 1, 4, 3, 0, 1, 68, 1200, 7268, 20025, 27750, 18900, 5040, 0, 0, 0, 0, 1, 8198, 9055962, 1467050480, 74035775370, 1679679306420, 20864180531565, 159341117375160, 804216787965360, 2808560520334800, 6981656802951600

Offset

1,3

Comments

An achiral coloring is identical to its reflection. A ridge is an (n-2)-face of an n-D polytope. For n=1, the figure is a line segment with one edge. For n=2, the figure is a square with 4 edges (vertices). For n=3, the figure is an octahedron (cube) with 12 edges. For n>1, the number of edges (ridges) is 2*n*(n-1). The Schläfli symbols for the n-D orthotope (hypercube) and the n-D orthoplex (hyperoctahedron, cross polytope) are {4,3,...,3,3} and {3,3,...,3,4} respectively, with n-2 3's in each case. The figures are mutually dual.

The algorithm used in the Mathematica program below assigns each permutation of the axes to a partition of n and then considers separate conjugacy classes for axis reversals. It uses the formulas in Balasubramanian's paper. If the value of m is increased, one can enumerate colorings of higher-dimensional elements beginning with T(m,1).

Links

Table of n, a(n) for n=1..28.

K. Balasubramanian, Computational enumeration of colorings of hyperplanes of hypercubes for all irreducible representations and applications, J. Math. Sci. & Mod. 1 (2018), 158-180.

Formula

For n>1, A337414(n,k) = Sum_{j=1..2*n*(n-1)} T(n,j) * binomial(k,j).

T(n,k) = 2*A338147(n,k) - A338146(n,k) = A338146(n,k) - 2*A338148(n,k) = A338147(n,k) - A338148(n,k).

T(2,k) = A338145(2,k) = A325019(2,k) = A325011(2,k); T(3,k) = A338145(3,k).

Example

Triangle begins with T(1,1):
  1
  1    4       3          0
  1   68    1200       7268       20025         27750 18900 5040 0 0 0 0
  1 8198 9055962 1467050480 74035775370 1679679306420 ...
  ...
For T(2,2)=4, the achiral colorings are AAAB, AABB, ABAB, and ABBB. For T(2,3)=3, the 3 achiral colorings are ABAC, ABCB, and ACBC.

Mathematica

m=1; (* dimension of color element, here an edge *)
Fi1[p1_] := Module[{g, h}, Coefficient[Product[g = GCD[k1, p1]; h = GCD[2 k1, p1]; (1 + 2 x^(k1/g))^(r1[[k1]] g) If[Divisible[k1, h], 1, (1+2x^(2 k1/h))^(r2[[k1]] h/2)], {k1, Flatten[Position[cs, n1_ /; n1 > 0]]}], x, m+1]];
FiSum[] := (Do[Fi2[k2] = Fi1[k2], {k2, Divisors[per]}]; DivisorSum[per, DivisorSum[d1 = #, MoebiusMu[d1/#] Fi2[#] &]/# &]);
CCPol[r_List] := (r1 = r; r2 = cs - r1; If[EvenQ[Sum[If[EvenQ[j3], r1[[j3]], r2[[j3]]], {j3, n}]], 0, (per = LCM @@ Table[If[cs[[j2]] == r1[[j2]], If[0 == cs[[j2]], 1, j2], 2j2], {j2, n}]; Times @@ Binomial[cs, r1] 2^(n-Total[cs]) b^FiSum[])]);
PartPol[p_List] := (cs = Count[p, #]&/@ Range[n]; Total[CCPol[#]&/@ Tuples[Range[0, cs]]]);
pc[p_List] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #]&/@ mb; n!/(Times@@(ci!) Times@@(mb^ci))] (*partition count*)
row[m]=b; row[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^(n-1))]
array[n_, k_] := row[n] /. b -> k
Join[{{1}}, Table[LinearSolve[Table[Binomial[i, j], {i, 2^(m+1)Binomial[n, m+1]}, {j, 2^(m+1)Binomial[n, m+1]}], Table[array[n, k], {k, 2^(m+1)Binomial[n, m+1]}]], {n, m+1, m+4}]] // Flatten

Crossrefs

Cf. A338146 (oriented), A338147 (unoriented), A338148 (chiral), A337414 (k or fewer colors), A325011 (orthoplex vertices, orthotope facets).

Cf. A327090 (simplex), A338145 (orthotope edges, orthoplex ridges).

Sequence in context: A270708 A198261 A284056 * A338145 A296002 A227423

Adjacent sequences: A338146 A338147 A338148 * A338150 A338151 A338152

Keyword

nonn,tabf

Author

Robert A. Russell, Oct 12 2020

Status

approved