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%I #12 Aug 28 2020 06:05:32
%S 1,2,1,3,6,1,4,18,70,1,5,40,1407,93024,1,6,75,12480,294157089,
%T 47823602694208,1,7,126,69050,91983927296,67514530382043163023924,
%U 443077371786837979607993095063601152,1
%N Array read by descending antidiagonals: T(n,k) is the number of achiral colorings of the edges of a regular n-dimensional orthotope (hypercube) using k or fewer colors.
%C An achiral arrangement is identical to its reflection. For n=1, the figure is a line segment with one edge. For n=2, the figure is a square with 4 edges. For n=3, the figure is a cube with 12 edges. The number of edges is n*2^(n-1).
%C Also the number of achiral colorings of the regular (n-2)-dimensional simplexes in a regular n-dimensional orthoplex.
%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 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).
%F T(n,k) = 2*A337408(n,k) - A337407(n,k) = A337407(n,k) - 2*A337409(n,k) = A337408(n,k) - A337409(n,k).
%e Table begins with T(1,1):
%e 1 2 3 4 5 6 7 8 9 10 ...
%e 1 6 18 40 75 126 196 288 405 550 ...
%e 1 70 1407 12480 69050 281946 931490 2632512 6598935 15041950 ...
%e For T(2,2)=6, the arrangements are AAAA, AAAB, AABB, ABAB, ABBB, and BBBB.
%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; If[EvenQ[Sum[If[EvenQ[j3], r1[[j3]], r2[[j3]]], {j3,n}]],0,(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-1))]
%t array[n_, k_] := row[n] /. b -> k
%t Table[array[n,d+m-n], {d,7}, {n,m,d+m-1}] // Flatten
%Y Cf. A337407 (oriented), A337408 (unoriented), A337409 (chiral).
%Y Rows 1-4 are A000027, A002411, A331351, A331361.
%Y Cf. A327086 (simplex edges), A337414 (orthoplex edges), A325015 (orthotope vertices).
%K nonn,tabl
%O 1,2
%A _Robert A. Russell_, Aug 26 2020
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