A337887 Array read by descending antidiagonals: T(n,k) is the number of oriented colorings of the square faces of a regular n-dimensional orthotope (hypercube) using k or fewer colors.

1, 2, 1, 3, 10, 1, 4, 57, 90054, 1, 5, 240, 1471640157, 629648865588086369152, 1, 6, 800, 1466049174160, 76983765319971901895960429658208179, 76686070519895153193719509580895099970955878067526648007224125292544, 1

Offset: 2

Author

Robert A. Russell, Sep 28 2020

Keywords

• nonn
• tabl

Comments

Each chiral pair is counted as two when enumerating oriented arrangements. Each face is a square bounded by four edges. For n=2, the figure is a square with one face. For n=3, the figure is a cube with 6 faces. For n=4, the figure is a tesseract with 24 faces. The number of faces is 2^(n-2)*C(n,2).

Also the number of oriented colorings of peaks of an n-dimensional orthoplex. A peak is an (n-3)-dimensional simplex.

Examples

			Array begins with T(2,1):
 1     2          3             4               5                 6 ...
 1    10         57           240             800              2226 ...
 1 90054 1471640157 1466049174160 310441584462375 24679078461920106 ...
		

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

Cf. A337888 (unoriented), A337889 (chiral), A337890 (achiral).

Other elements: A325012 (vertices), A337407 (edges).

Other polytopes: A337883 (simplex), A337891 (orthoplex).

Rows 2-4 are A000027, A047780, A331354.

Programs

Mathematica

m = 2;(* dimension of color element, here a square face *)
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}]], (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[]),0]);
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[array[n,d+m-n], {d,6}, {n,m,d+m-1}] // Flatten

Formula

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).

T(n,k) = A337888(n,k) + A337889(n,k) = 2*A337888(n,k) - A337890(n,k) = 2*A337889(n,k) + A337890(n,k).