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%I #9 Oct 14 2020 10:50:29
%S 0,0,0,3,3,0,74,10482,303268,3440700,19842840,65867760,133580160,
%T 168399000,128898000,54885600,9979200,0,40927,731157018,729348051686,
%U 151526009158620,11418355290999750,415756294427389020,8643340000393019040
%N Triangle read by rows: T(n,k) is the number of chiral pairs of 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.
%C 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 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.
%C 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).
%H K. Balasubramanian, <a href="https://doi.org/10.33187/jmsm.471940">Computational enumeration of colorings of hyperplanes of hypercubes for all irreducible representations and applications</a>, J. Math. Sci. & Mod. 1 (2018), 158-180.
%F For n>1, A337413(n,k) = Sum_{j=1..2*n*(n-1)} T(n,j) * binomial(k,j).
%F T(n,k) = A338146(n,k) - A338147(n,k) = (A338146(n,k) - A338149(n,k)) / 2 = A338147(n,k) - A338149(n,k).
%F T(2,k) = A338144(2,k) = A325018(2,k) = A325010(2,k); T(3,k) = A338144(3,k).
%e Triangle begins with T(1,1):
%e 0
%e 0 0 3 3
%e 0 74 10482 303268 3440700 19842840 65867760 133580160 168399000
%e ...
%e 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.
%t m=1; (* dimension of color element, here an edge *)
%t 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]];
%t FiSum[] := (Do[Fi2[k2] = Fi1[k2], {k2, Divisors[per]}]; DivisorSum[per, DivisorSum[d1 = #, MoebiusMu[d1/#] Fi2[#] &]/# &]);
%t 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[]);
%t PartPol[p_List] := (cs = Count[p, #]&/@ Range[n]; Total[CCPol[#]&/@ Tuples[Range[0, cs]]]);
%t pc[p_List] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #]&/@ mb; n!/(Times@@(ci!) Times@@(mb^ci))] (*partition count*)
%t row[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^n)]
%t array[n_, k_] := row[n] /. b -> k
%t Join[{{0}},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
%Y Cf. A338146 (oriented), A338147 (unoriented), A338149 (achiral), A337413 (k or fewer colors), A325010 (orthoplex vertices, orthotope facets).
%Y Cf. A327089 (simplex), A338144 (orthotope edges, orthoplex ridges).
%K nonn,tabf
%O 1,4
%A _Robert A. Russell_, Oct 12 2020
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