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A338145
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Triangle read by rows: T(n,k) is the number of achiral 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.
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4
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1, 1, 4, 3, 0, 1, 68, 1200, 7268, 20025, 27750, 18900, 5040, 0, 0, 0, 0, 1, 93022, 293878020, 90807857080, 7503022894800, 258528829444320, 4681671089961600, 50981530073846400, 363246007692204000
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OFFSET
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1,3
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COMMENTS
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An achiral coloring is identical to its reflection. 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.
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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LINKS
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FORMULA
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A337410(n,k) = Sum_{j=1..n*2^(n-1)} T(n,j) * binomial(k,j).
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EXAMPLE
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Triangle begins with T(1,1):
1
1 4 3 0
1 68 1200 7268 20025 27750 18900 5040 0 0 0 0
...
For T(2,2)=4, the achiral colorings are AAAB, AABB, ABAB, and ABBB. For T(2,3)=3, the achiral colorings are ABAC, ABCB, and ACBC.
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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; 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[])]);
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-1))]
array[n_, k_] := row[n] /. b -> k
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
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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