A193228 Truncated octahedron with faces of centered polygons.

1, 39, 185, 511, 1089, 1991, 3289, 5055, 7361, 10279, 13881, 18239, 23425, 29511, 36569, 44671, 53889, 64295, 75961, 88959, 103361, 119239, 136665, 155711, 176449, 198951, 223289, 249535, 277761, 308039, 340441, 375039, 411905, 451111, 492729, 536831, 583489

Offset: 1

Author

Craig Ferguson, Jul 18 2011

Keywords

• nonn
• easy

Comments

The sequence starts with a central dot and expands outward with (n-1) centered polygonal pyramids producing a truncated octahedron. Each iteration requires the addition of (n-2) edges and (n-1) vertices to complete the centered polygon of each face. [centered squares (A001844) and centered hexagons (A003215)]

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..10000
• OEIS, (Centered_polygons) pyramidal numbers
• Wikipedia, Tetrahedral number
• Wikipedia, Triangular number
• Wikipedia, Centered polygonal number
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Programs

Excel

(copy and paste the following formula =12*ROW()^3-18*ROW()^2+8*ROW()-1 fill down to desired size.)

Magma

[12*n^3-18*n^2+8*n-1: n in [1..50]]; // Vincenzo Librandi, Aug 30 2011

Mathematica

Table[12n^3-18n^2+8n-1,{n,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{1,39,185,511},40] (* Harvey P. Dale, Aug 27 2011 *)

PARI

vector(40, n, 12*n^3 - 18*n^2 + 8*n - 1) \\ G. C. Greubel, Nov 10 2018

Formula

a(n) = 12*n^3 - 18*n^2 + 8*n - 1.

G.f.: x*(1+x)*(x^2 + 34*x + 1) / (x-1)^4. - R. J. Mathar, Aug 26 2011

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(0)=1, a(1)=39, a(2)=185, a(3)=511. - Harvey P. Dale, Aug 27 2011

E.g.f.: 1 - (1 - 2*x - 18*x^2 - 12*x^3)*exp(x). - G. C. Greubel, Nov 10 2018