%I #18 May 29 2022 18:00:09
%S 0,0,0,0,0,0,0,3,74,0,0,15,10704,40927,0,0,45,345640,731279799,
%T 280317324,0,0,105,5062600,732272925320,3163614120031068,
%U 24869435516248,0,0,210,45246810,155180061396500,314800331906964016128,919853357924272852197243,29931599129719666392,0
%N Array read by descending antidiagonals: T(n,k) is the number of chiral pairs of colorings of the edges of a regular n-dimensional orthoplex (cross polytope) using k or fewer colors.
%C Each member of a chiral pair is a reflection, but not a rotation, of the other. 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 an octahedron with 12 edges. The number of edges is 2n*(n-1) for n>1.
%C Also the number of chiral pairs of colorings of the regular (n-2)-dimensional orthotopes (hypercubes) in a regular n-dimensional orthotope.
%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) = A337411(n,k) - A337412(n,k) = (A337411(n,k) - A337414(n,k)) / 2 = A337412(n,k) - A337414(n,k).
%e Table begins with T(1,1):
%e 0 0 0 0 0 0 0 0 0 ...
%e 0 0 3 15 45 105 210 378 630 ...
%e 0 74 10704 345640 5062600 45246810 288005144 1430618784 5881281480 ...
%e For T(2,3)=3, the chiral arrangements are AABC-AACB, ABBC-ACBB, and ABCC-ACCB.
%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 Table[array[n,d+m-n], {d,8}, {n,m,d+m-1}] // Flatten
%Y Cf. A337411 (oriented), A337412 (unoriented), A337414 (achiral).
%Y Rows 2-4 are A050534, A337406, A331356.
%Y Cf. A327085 (simplex edges), A337409 (orthotope edges), A325006 (orthoplex vertices).
%K nonn,tabl
%O 1,8
%A _Robert A. Russell_, Aug 26 2020