A325015 Array read by descending antidiagonals: T(n,k) is the number of achiral colorings of the facets of a regular n-dimensional orthoplex using up to k colors.

1, 2, 1, 3, 6, 1, 4, 18, 21, 1, 5, 40, 201, 308, 1, 6, 75, 1076, 34128, 180342, 1, 7, 126, 4025, 1056576, 2945136213, 366975285216, 1, 8, 196, 11901, 15303750, 2932338749408, 103863386269870076808, 10316179427644325573474464, 1

Offset

1,2

Comments

Also called cross polytope and hyperoctahedron. For n=1, the figure is a line segment with two vertices. For n=2 the figure is a square with four edges. For n=3 the figure is an octahedron with eight triangular faces. For n=4, the figure is a 16-cell with sixteen tetrahedral facets. The Schläfli symbol, {3,...,3,4}, of the regular n-dimensional orthoplex (n>1) consists of n-2 threes followed by a four. Each of its 2^n facets is an (n-1)-dimensional simplex. An achiral coloring is identical to its reflection.

Also the number of achiral colorings of the vertices of a regular n-dimensional orthotope (cube) using up to k colors.

Links

Robert A. Russell, Table of n, a(n) for n = 1..78

E. M. Palmer and R. W. Robinson, Enumeration under two representations of the wreath product, Acta Math., 131 (1973), 123-143.

Wikipedia, Cross-polytope

Formula

The algorithm used in the Mathematica program below assigns each permutation of the axes to a partition of n. It then determines the number of permutations for each partition and the cycle index for each partition.

T(n,k) = 2*A325013(n,k) - A325012(n,k) = A325012(n,k) - 2*A325014(n,k) = A325013(n,k) - A325014(n,k).

T(n,k) = Sum_{j=1..3*2^(n-2)} A325019(n,j) * binomial(k,j).

Example

Array begins with T(1,1):
1   2     3       4        5         6         7          8 ...
1   6    18      40       75       126       196        288 ...
1  21   201    1076     4025     11901     29841      66256 ...
1 308 34128 1056576 15303750 136236276 865711763 4296782848 ...
...
For T(2,2)=6, two squares have all edges the same color, two have three edges the same color, one has opposite edges the same color, and one has opposite edges different colors.

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, a48 /@ sub}]]] 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[(CI1[#] pc[#]) & /@ IntegerPartitions[n]])/(n! 2^(n - 1))] /. CI[l_List] :> j^(Total[l][[2]])
array[n_, k_] := row[n] /. j -> k
Table[array[n, d-n+1], {d, 1, 10}, {n, 1, d}] // Flatten

Crossrefs

Cf. A325012 (oriented), A325013 (unoriented), A325014 (chiral), A325019 (exactly k colors).

Other n-dimensional polytopes: A325001 (simplex), A325007 (orthotope).

Rows 1-2 are A000027, A002411.

Cf. A000048, A001037.

Sequence in context: A115597 A325007 A103371 * A337414 A337410 A337389

Adjacent sequences: A325012 A325013 A325014 * A325016 A325017 A325018

Keyword

nonn,tabl,easy

Author

Robert A. Russell, May 27 2019

Status

approved