A338149 - OEIS

%I #8 Oct 14 2020 10:50:37

%S 1,1,4,3,0,1,68,1200,7268,20025,27750,18900,5040,0,0,0,0,1,8198,

%T 9055962,1467050480,74035775370,1679679306420,20864180531565,

%U 159341117375160,804216787965360,2808560520334800,6981656802951600

%N 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.

%C 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 an octahedron (cube) with 12 edges. For n>1, the number of edges (ridges) is 2*n*(n-1). The Schläfli symbols for the n-D orthotope (hypercube) and the n-D orthoplex (hyperoctahedron, cross polytope) are {4,3,...,3,3} and {3,3,...,3,4} respectively, with n-2 3's in each case. The figures are mutually dual.

%C 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).

%H K. Balasubramanian, <a href="https://doi.org/10.33187/jmsm.471940">Computational enumeration of colorings of hyperplanes of hypercubes for all irreducible representations and applications</a>, J. Math. Sci. & Mod. 1 (2018), 158-180.

%F For n>1, A337414(n,k) = Sum_{j=1..2*n*(n-1)} T(n,j) * binomial(k,j).

%F T(n,k) = 2*A338147(n,k) - A338146(n,k) = A338146(n,k) - 2*A338148(n,k) = A338147(n,k) - A338148(n,k).

%F T(2,k) = A338145(2,k) = A325019(2,k) = A325011(2,k); T(3,k) = A338145(3,k).

%e Triangle begins with T(1,1):

%e   1

%e   1    4       3          0

%e   1   68    1200       7268       20025         27750 18900 5040 0 0 0 0

%e   1 8198 9055962 1467050480 74035775370 1679679306420 ...

%e   ...

%e 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.

%t m=1; (* dimension of color element, here an edge *)

%t 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]];

%t FiSum[] := (Do[Fi2[k2] = Fi1[k2], {k2, Divisors[per]}]; DivisorSum[per, DivisorSum[d1 = #, MoebiusMu[d1/#] Fi2[#] &]/# &]);

%t 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[])]);

%t PartPol[p_List] := (cs = Count[p, #]&/@ Range[n]; Total[CCPol[#]&/@ Tuples[Range[0, cs]]]);

%t pc[p_List] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #]&/@ mb; n!/(Times@@(ci!) Times@@(mb^ci))] (*partition count*)

%t row[m]=b; row[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^(n-1))]

%t array[n_, k_] := row[n] /. b -> k

%t 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

%Y Cf. A338146 (oriented), A338147 (unoriented), A338148 (chiral), A337414 (k or fewer colors), A325011 (orthoplex vertices, orthotope facets).

%Y Cf. A327090 (simplex), A338145 (orthotope edges, orthoplex ridges).

%K nonn,tabf

%O 1,3

%A _Robert A. Russell_, Oct 12 2020