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A337409
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Array read by descending antidiagonals: T(n,k) is the number of chiral pairs of colorings of the edges of a regular n-dimensional orthotope (hypercube) using k or fewer colors.
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9
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0, 0, 0, 0, 0, 0, 0, 3, 74, 0, 0, 15, 10704, 11158298, 0, 0, 45, 345640, 4825452718593, 314824408633217132928, 0, 0, 105, 5062600, 48038354542204960, 38491882659952177472606694634030116, 136221825854745676076981182469325427379054390050209792, 0
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OFFSET
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1,8
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COMMENTS
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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 a cube with 12 edges. The number of edges is n*2^(n-1).
Also the number of chiral pairs of colorings of the regular (n-2)-dimensional simplexes in a regular n-dimensional orthoplex.
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LINKS
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FORMULA
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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).
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EXAMPLE
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Table begins with T(1,1):
0 0 0 0 0 0 0 0 0 ...
0 0 3 15 45 105 210 378 630 ...
0 74 10704 345640 5062600 45246810 288005144 1430618784 5881281480 ...
For T(2,3)=3, the chiral arrangements are AABC-AACB, ABBC-ACBB, and ABCC-ACCB.
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MATHEMATICA
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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[array[n, d+m-n], {d, 7}, {n, m, d+m-1}] // Flatten
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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