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%I #6 Oct 14 2020 10:38:46
%S 1,1,4,6,3,1,142,11682,310536,3460725,19870590,65886660,133585200,
%T 168399000,128898000,54885600,9979200,1,11251320,4825713121719,
%U 48019143606137456,60392840368910627325
%N 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.
%C 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.
%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 A337408(n,k) = Sum_{j=1..n*2^(n-1)} T(n,j) * binomial(k,j).
%F T(n,k) = A338142(n,k) - A338144(n,k) = (A338142(n,k) + A338145(n,k)) / 2 = A338144(n,k) + A338145(n,k).
%F T(2,k) = A338147(2,k) = A325017(2,k) = A325009(2,k); T(3,k) = A338147(3,k).
%e Triangle begins with T(1,1):
%e 1
%e 1 4 6 3
%e 1 142 11682 310536 3460725 19870590 65886660 133585200 168399000
%e ...
%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, n - m]];
%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}]; 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 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
%Y Cf. A338142 (oriented), A338144 (chiral), A338145 (achiral), A337408 (k or fewer colors), A325017 (orthotope vertices, orthoplex facets).
%Y Cf. A327088 (simplex), A338147 (orthoplex edges, orthotope ridges).
%K nonn,tabf
%O 1,3
%A _Robert A. Russell_, Oct 12 2020
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