A274974 - OEIS

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A274974

Centered octahemioctahedral numbers: a(n) = (4*n^3+24*n^2+8*n+3)/3.

1, 13, 49, 117, 225, 381, 593, 869, 1217, 1645, 2161, 2773, 3489, 4317, 5265, 6341, 7553, 8909, 10417, 12085, 13921, 15933, 18129, 20517, 23105, 25901, 28913, 32149, 35617, 39325, 43281, 47493, 51969, 56717, 61745, 67061, 72673, 78589, 84817, 91365, 98241

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

Related to a faceting of the cuboctahedron, sharing the same triangular faces. The octahemioctahedron has the same edge and vertex arrangement as the cuboctahedron (as does A274973). Beginning with the third term, the six square faces are each now "missing" a square pyramid of size 1, 5, 14, 30, 55, 91...(A000330). See A274973 centered cubohemioctahedron for similar cuboctahedral faceting but without the triangular faces.

LINKS

Table of n, a(n) for n=0..40.

Steven Beard, Music track made with this sequence

Wikipedia, Octahemioctahedron

FORMULA

a(n) = (4*n^3+24*n^2+8*n+3)/3.

G.f.: (-5*x^3+3*x^2+9*x+1)/(x-1)^4.

MATHEMATICA

CoefficientList[Series[(-5 x^3 + 3 x^2 + 9 x + 1)/(x - 1)^4, {x, 0, 40}], x] (* or *)

Table[(4 n^3 + 24 n^2 + 8 n+3)/3, {n, 41}] (* Michael De Vlieger, Jul 13 2016 *)

PROG

(PARI) a(n)=(4*n^3+24*n^2+8*n+3)/3 \\ Charles R Greathouse IV, Nov 03 2017