A337958 Number of achiral colorings of the 8 cubic facets of a tesseract or of the 8 vertices of a hyperoctahedron.

1, 15, 126, 700, 2850, 9261, 25480, 61776, 135675, 275275, 523446, 943020, 1623076, 2686425, 4298400, 6677056, 10104885, 14942151, 21641950, 30767100, 43008966, 59208325, 80378376, 107730000, 142699375, 186978051, 242545590

Offset: 1

Author

Robert A. Russell, Oct 03 2020

Keywords

• nonn
• easy

Comments

An achiral coloring is identical to its reflection. The Schläfli symbols for the tesseract and the hyperoctahedron are {4,3,3} and {3,3,4} respectively. Both figures are regular 4-D polyhedra and they are mutually dual.

Links

• Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Crossrefs

Cf. A337956 (oriented), A337956 (unoriented), A234249(n+1) (chiral).

Other elements: A331357 (hyperoctahedron edges, tesseract faces), A331361 (hyperoctahedron faces, tesseract edges), A337955 (hyperoctahedron facets, tesseract vertices).

Other polychora: A132366(n-1) (5-cell), A338951 (24-cell), A338967 (120-cell, 600-cell).

Row 4 of A325007 (orthotope facets, orthoplex vertices).

Programs

Mathematica

Table[Binomial[Binomial[n+1,2]+3,4] - Binomial[Binomial[n,2],4],{n,30}]

Formula

a(n) = binomial(binomial(n+1,2)+3,4) - binomial(binomial(n,2),4).

a(n) = n^2 * (n+1)^2 * (n+3) * (n^2 -2n +4) / 48.

a(n) = 1*C(n,1) + 13*C(n,2) + 84*C(n,3) + 282*C(n,4) + 465*C(n,5) + 360*C(n,6) + 105*C(n,7), where the coefficient of C(n,k) is the number of achiral colorings using exactly k colors.

a(n) = 2*A337957(n) - A337956(n) = A337956(n) - 2 * A234249(n+1) = A337957(n) - A234249(n+1).

From Stefano Spezia, Oct 04 2020: (Start)

G.f.: x*(1 + 7*x + 34*x^2 + 56*x^3 + 8*x^4 - x^5)/(1 - x)^8.

a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8) for n > 8.

(End)