

A193218


Number of vertices in truncated tetrahedron with faces that are centered polygons.


3



1, 21, 95, 259, 549, 1001, 1651, 2535, 3689, 5149, 6951, 9131, 11725, 14769, 18299, 22351, 26961, 32165, 37999, 44499, 51701, 59641, 68355, 77879, 88249, 99501, 111671, 124795, 138909, 154049, 170251, 187551, 205985, 225589, 246399, 268451, 291781, 316425
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OFFSET

1,2


COMMENTS

The sequence starts with a central vertex and expands outward with (n1) centered polygonal pyramids producing a truncated tetrahedron. Each iteration requires the addition of (n2) edges and (n1) vertices to complete the centered polygon in each face. For centered triangles see A005448 and centered hexagons A003215.
This sequence is the 18th in the series (1/12)*t*(2*n^33*n^2+n)+2*n1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496 and t = 36. While adjusting for offsets, the beginning sequence A049480 is generated by adding the square pyramidal numbers A000330 to the odd numbers A005408 and each subsequent sequence is found by adding another set of square pyramidals A000330. (T/2) * A000330(n) + A005408(n). At 30 * A000330 + A005408 = centered dodecahedral numbers, 36 * A000330 + A005408 = A193228 truncated octahedron and 90 * A000330 + A005408 = A193248 = truncated icosahedron and dodecahedron. All five of the "Centered Platonic Solids" numbers sequences are in this series of sequences. Also 4 out of five of the "truncated" platonic solid number sequences are in this series.  Bruce J. Nicholson, Jul 06 2018
It would be good to have a detailed description of how the sequence is constructed. Maybe in the Examples section?  N. J. A. Sloane, Sep 07 2018


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).


FORMULA

a(n) = 6*n^3  9*n^2 + 5*n  1.
G.f.: x*(1+x)*(x^2+16*x+1) / (1x)^4.  R. J. Mathar, Aug 26 2011
a(n) = 18 * A000330(n1) + A005408(n1) = A063496(n) + A006331(n1).  Bruce J. Nicholson, Jul 06 2018


MATHEMATICA

Table[6 n^3  9 n^2 + 5 n  1, {n, 35}] (* Alonso del Arte, Jul 18 2011 *)
CoefficientList[Series[(1+x)*(x^2+16*x+1)/(1x)^4, {x, 0, 50}], x] (* Stefano Spezia, Sep 04 2018 *)


PROG

(MAGMA) [6*n^39*n^2+5*n1: n in [1..40]]; // Vincenzo Librandi, Aug 30 2011


CROSSREFS

Cf. A260810 (partial sums).
Sequence in context: A123213 A296822 A174084 * A220157 A264239 A200255
Adjacent sequences: A193215 A193216 A193217 * A193219 A193220 A193221


KEYWORD

nonn,easy


AUTHOR

Craig Ferguson, Jul 18 2011


STATUS

approved



