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

%I #14 Aug 12 2021 10:26:32

%S 1,15,126,715,3060,10626,31465,82251,194580,424270,864501,1663740,

%T 3049501,5359095,9078630,14891626,23738715,36890001,56031760,83369265,

%U 121747626,174792640,247073751,344291325,473490550,643304376

%N Number of unoriented colorings of the 8 cubic facets of a tesseract or of the 8 vertices of a hyperoctahedron.

%C Each chiral pair is counted as one when enumerating unoriented arrangements. 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.

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (9,-36,84,-126,126,-84,36,-9,1).

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

%F a(n) = n * (n+1) * (n^2 + n + 2) * (n^2 + n + 4) * (n^2 + n + 6) / 384.

%F a(n) = 1*C(n,1) + 13*C(n,2) + 84*C(n,3) + 297*C(n,4) + 600*C(n,5) + 690*C(n,6) + 420*C(n,7) + 105*C(n,8), where the coefficient of C(n,k) is the number of unoriented colorings using exactly k colors.

%F a(n) = A337956(n) - A234249(n+1) = (A337956(n) + A337958(n)) / 2 = A234249(n+1) + A337958(n).

%F From _Stefano Spezia_, Oct 04 2020: (Start)

%F G.f.: x*(1 + 6*x + 27*x^2 + 37*x^3 + 27*x^4 + 6*x^5 + x^6)/(1 - x)^9.

%F a(n) = 9*a(n-1)-36*a(n-2)+84*a(n-3)-126*a(n-4)+126*a(n-5)-84*a(n-6)+36*a(n-7)-9*a(n-8)+a(n-8) for n > 8.

%F (End)

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

%Y Cf. A337956 (oriented), A234249(n+1) (chiral), A337958 (achiral).

%Y Other elements: A331355 (hyperoctahedron edges, tesseract faces), A331359 (hyperoctahedron faces, tesseract edges), A128767 (hyperoctahedron facets, tesseract vertices).

%Y Other polychora: A000389(n+4) (5-cell), A338949 (24-cell), A338965 (120-cell, 600-cell).

%Y Row 4 of A325005 (orthotope facets, orthoplex vertices).

%K nonn,easy

%O 1,2

%A _Robert A. Russell_, Oct 03 2020