A325016 Triangle read by rows: T(n,k) is the number of oriented colorings of the facets of a regular n-dimensional orthoplex using exactly k colors. Row n has 2^n columns.
1, 2, 1, 4, 9, 6, 1, 21, 267, 1718, 5250, 7980, 5880, 1680, 1, 494, 228591, 21539424, 685479375, 10257064650, 86151316860, 449772354360, 1551283253100, 3661969537800, 6015983173200, 6878457986400, 5371454088000, 2733402672000, 817296480000, 108972864000
Offset: 1
Examples
Triangle begins with T(1,1): 1 2 1 4 9 6 1 21 267 1718 5250 7980 5880 1680 For T(2,2)=4, two squares have three edges the same color, one has opposite edges the same color, and one has opposite edges different colors.
Links
- Robert A. Russell, Table of n, a(n) for n = 1..510, rows 1..8, flattened.
- E. M. Palmer and R. W. Robinson, Enumeration under two representations of the wreath product, Acta Math., 131 (1973), 123-143.
- Wikipedia, Cross-polytope
Crossrefs
Programs
-
Mathematica
a48[n_] := a48[n] = DivisorSum[NestWhile[#/2&, n, EvenQ], MoebiusMu[#]2^(n/#)&]/(2n); (* A000048 *) a37[n_] := a37[n] = DivisorSum[n, MoebiusMu[n/#]2^#&]/n; (* A001037 *) CI0[{n_Integer}] := CI0[{n}] = CI[Transpose[If[EvenQ[n], p2 = IntegerExponent[n, 2]; sub = Divisors[n/2^p2]; {2^(p2+1) sub, a48 /@ (2^p2 sub) }, sub = Divisors[n]; {sub, a37 /@ sub}]]] 2^(n-1); (* even perm. *) CI1[{n_Integer}] := CI1[{n}] = CI[sub = Divisors[n]; Transpose[If[EvenQ[n], {sub, a37 /@ sub}, {2 sub, (a37 /@ sub)/2}]]] 2^(n-1); (* odd perm. *) compress[x : {{, } ...}] := (s = Sort[x]; For[i = Length[s], i > 1, i -= 1, If[s[[i, 1]]==s[[i-1, 1]], s[[i-1, 2]] += s[[i, 2]]; s = Delete[s, i], Null]]; s) cix[{a_, b_}, {c_, d_}] := {LCM[a, c], (a b c d)/LCM[a, c]}; Unprotect[Times]; Times[CI[a_List], CI[b_List]] := (* combine *) CI[compress[Flatten[Outer[cix, a, b, 1], 1]]]; Protect[Times]; CI0[p_List] := CI0[p] = Expand[CI0[Drop[p, -1]] CI0[{Last[p]}] + CI1[Drop[p, -1]] CI1[{Last[p]}]] CI1[p_List] := CI1[p] = Expand[CI0[Drop[p, -1]] CI1[{Last[p]}] + CI1[Drop[p, -1]] CI0[{Last[p]}]] 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[(CI0[#] pc[#]) & /@ IntegerPartitions[n]])/(n! 2^(n - 1))] /. CI[l_List] :> j^(Total[l][[2]]) array[n_, k_] := row[n] /. j -> k (* A325012 *) Table[LinearSolve[Table[Binomial[i,j],{i,1,2^n},{j,1,2^n}],Table[array[n,k],{k,1,2^n}]],{n,1,6}] // Flatten
Comments