A338143 Triangle read by rows: T(n,k) is the number of unoriented colorings of the edges of a regular n-D orthotope (or ridges of a regular n-D orthoplex) using exactly k colors. Row n has n*2^(n-1) columns.

1, 1, 4, 6, 3, 1, 142, 11682, 310536, 3460725, 19870590, 65886660, 133585200, 168399000, 128898000, 54885600, 9979200, 1, 11251320, 4825713121719, 48019143606137456, 60392840368910627325

Offset: 1

Author

Robert A. Russell, Oct 12 2020

Keywords

• nonn
• tabf

Comments

Each chiral pair is counted as one when enumerating unoriented arrangements. 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 a cube (octahedron) with 12 edges. The number of edges (ridges) is n*2^(n-1). The Schläfli symbols for the n-D orthotope (hypercube) and the n-D orthoplex (hyperoctahedron, cross polytope) are {4,...,3,3} and {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).

Examples

			Triangle begins with T(1,1):
  1
  1   4     6      3
  1 142 11682 310536 3460725 19870590 65886660 133585200 168399000
  ...
		

Links

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

Crossrefs

Cf. A338142 (oriented), A338144 (chiral), A338145 (achiral), A337408 (k or fewer colors), A325017 (orthotope vertices, orthoplex facets).

Cf. A327088 (simplex), A338147 (orthoplex edges, orthotope ridges).

Programs

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, n - m]];
FiSum[] := (Do[Fi2[k2] = Fi1[k2], {k2, Divisors[per]}]; DivisorSum[per, DivisorSum[d1 = #, MoebiusMu[d1/#] Fi2[#] &]/# &]);
CCPol[r_List] := (r1 = r; r2 = cs - r1; 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[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^n)]
array[n_, k_] := row[n] /. b -> k
Table[LinearSolve[Table[Binomial[i,j],{i,2^(n-m)Binomial[n,m]},{j,2^(n-m)Binomial[n,m]}], Table[array[n,k],{k,2^(n-m)Binomial[n,m]}]], {n,m,m+4}] // Flatten

Formula

A337408(n,k) = Sum_{j=1..n*2^(n-1)} T(n,j) * binomial(k,j).

T(n,k) = A338142(n,k) - A338144(n,k) = (A338142(n,k) + A338145(n,k)) / 2 = A338144(n,k) + A338145(n,k).

T(2,k) = A338147(2,k) = A325017(2,k) = A325009(2,k); T(3,k) = A338147(3,k).