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A337408
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Array read by descending antidiagonals: T(n,k) is the number of unoriented colorings of the edges of a regular n-dimensional orthotope (hypercube) using k or fewer colors.
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8
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1, 2, 1, 3, 6, 1, 4, 21, 144, 1, 5, 55, 12111, 11251322, 1, 6, 120, 358120, 4825746875682, 314824456456819827136, 1, 7, 231, 5131650, 48038446526132256, 38491882660019692002988737797054040, 136221825854745676520058554256163406987047485113810944, 1
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OFFSET
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1,2
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COMMENTS
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Each chiral pair is counted as one when enumerating unoriented arrangements. 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 unoriented 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):
1 2 3 4 5 6 7 8 9 ...
1 6 21 55 120 231 406 666 1035 ...
1 144 12111 358120 5131650 45528756 288936634 1433251296 5887880415 ...
For T(2,2)=6, the arrangements are AAAA, AAAB, AABB, ABAB, ABBB, and BBBB.
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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 + 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]];
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}]; 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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AUTHOR
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STATUS
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approved
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