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A338144
Triangle read by rows: T(n,k) is the number of chiral pairs of 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.
4
0, 0, 0, 3, 3, 0, 74, 10482, 303268, 3440700, 19842840, 65867760, 133580160, 168399000, 128898000, 54885600, 9979200, 0, 11158298, 4825419243699, 48019052798280376, 60392832865887732525, 20362602448352682660450
OFFSET
1,4
COMMENTS
Chiral colorings come in pairs, each the reflection of the other. 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).
FORMULA
A337409(n,k) = Sum_{j=1..n*2^(n-1)} T(n,j) * binomial(k,j).
T(n,k) = A338142(n,k) - A338143(n,k) = (A338142(n,k) - A338145(n,k)) / 2 = A338143(n,k) - A338145(n,k).
T(2,k) = A338148(2,k) = A325018(2,k) = A325010(2,k); T(3,k) = A338148(3,k).
EXAMPLE
Triangle begins with T(1,1):
0
0 0 3 3
0 74 10482 303268 3440700 19842840 65867760 133580160 168399000
...
For T(2,3)=3, the chiral pairs are AABC-AACB, ABBC-ACBB, and ABCC-ACCB. For T(2,4)=3, the chiral pairs are ABCD-ADCB, ACBD-ADBC, and ABDC-ACDB.
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+2x^(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}]; If[EvenQ[Sum[If[EvenQ[j3], r1[[j3]], r2[[j3]]], {j3, n}]], 1, -1]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
CROSSREFS
Cf. A338142 (oriented), A338143 (unoriented), A338145 (achiral), A337409 (k or fewer colors), A325018 (orthotope vertices, orthoplex facets).
Cf. A327089 (simplex), A338148 (orthoplex edges, orthotope ridges).
Sequence in context: A111843 A119537 A338148 * A031438 A096964 A350483
KEYWORD
nonn,tabf
AUTHOR
Robert A. Russell, Oct 12 2020
STATUS
approved