OFFSET
1,3
COMMENTS
Each chiral pair is counted as two when enumerating oriented 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 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
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
EXAMPLE
Triangle begins with T(1,1):
1
1 4 9 6
1 216 22164 613804 6901425 39713430 131754420 267165360 336798000
...
For T(2,3)=9, the 3 achiral colorings are ABAC, ABCB, and ACBC. The three chiral pairs are AABC-AACB, ABBC-ACBB, and ABCC-ACCB.
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}]], (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[]), 0]);
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
KEYWORD
nonn,tabf
AUTHOR
Robert A. Russell, Oct 12 2020
STATUS
approved