A338149 Triangle read by rows: T(n,k) is the number of achiral colorings of the edges of a regular n-D orthoplex (or ridges of a regular n-D orthotope) using exactly k colors. Row 1 has 1 column; row n>1 has 2*n*(n-1) columns.
1, 1, 4, 3, 0, 1, 68, 1200, 7268, 20025, 27750, 18900, 5040, 0, 0, 0, 0, 1, 8198, 9055962, 1467050480, 74035775370, 1679679306420, 20864180531565, 159341117375160, 804216787965360, 2808560520334800, 6981656802951600
Offset: 1
Examples
Triangle begins with T(1,1): 1 1 4 3 0 1 68 1200 7268 20025 27750 18900 5040 0 0 0 0 1 8198 9055962 1467050480 74035775370 1679679306420 ... ... For T(2,2)=4, the achiral colorings are AAAB, AABB, ABAB, and ABBB. For T(2,3)=3, the 3 achiral colorings are ABAC, ABCB, and ACBC.
Links
- K. Balasubramanian, Computational enumeration of colorings of hyperplanes of hypercubes for all irreducible representations and applications, J. Math. Sci. & Mod. 1 (2018), 158-180.
Crossrefs
Programs
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Mathematica
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, m+1]]; 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[m]=b; row[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^(n-1))] array[n_, k_] := row[n] /. b -> k Join[{{1}},Table[LinearSolve[Table[Binomial[i,j],{i,2^(m+1)Binomial[n,m+1]},{j,2^(m+1)Binomial[n,m+1]}], Table[array[n,k],{k,2^(m+1)Binomial[n,m+1]}]], {n,m+1,m+4}]] // Flatten
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