A290396 a(n) = 3*2^n + 3*4^n + 6^(n+1) + 1.

13, 55, 277, 1513, 8593, 49825, 292417, 1729153, 10275073, 61254145, 365945857, 2189371393, 13111037953, 78565515265, 470990340097, 2824331231233, 16939544543233, 101611496669185, 609565899227137, 3656983075356673, 21940249178406913, 131634897988091905, 789782999624318977

Offset: 0

Author

Steven Beard, Jul 29 2017

Keywords

• nonn
• easy

Comments

One of several types of cuboctahedra with Sierpinski recursion. This Sierpinski cuboctahedron is a truncation of the six corners of the Sierpinski octahedron (A279512). The resulting Sierpinski cuboctahedron contains six square pyramids joined along the edges thus it can also be constructed by joining Sierpinski square pyramids (A281698).

Each face of the Sierpinski octahedron is a Sierpinski sieve with all triangular spaces completely formed. The triangles opening within the sieve, on each face, become tetrahedral excavations (cf. A067771). The third term above, for example, begins with an octahedron of 489 counters from A005900. From each of the eight faces a tetrahedron of four counters is removed. Thus the corresponding Sierpinski octahedron has 457 counters at this stage of recursion. Next, to construct the Sierpinski cuboctahedron, the truncation of the six corners is 30 counters each, thus 457 minus 180 equals 277, the third term of the Sierpinski cuboctahedral sequence shown above. For the next term in the sequence, two different sizes of tetrahedra are removed (and not only seen on the outer eight faces but also on the newly opened internal faces, now excavated in like manner.

To arrive at the next term, 1513, there are eight tetrahedra of 56 counters AND forty-eight tetrahedra of 4 counters each excavated from the octahedron, and then the six corners of 188 counters each are truncated thus: (56*8) + (4*48) = 640 total counters removed from the octahedral number 3281 (in A005900) to yield the Sierpinski octahedral number 2641 (in A279512, note that for subsequent terms the number of different size excavations of the octahedron increases, thus the next term would require three different sized excavations for the associated octahedron: 8 tetrahedra of 560 counters, 48 tetrahedra of 56 counters, and 288 tetrahedra of 4 counters).

Next truncate the six corners of 188 counters each: (6*188) = 1128. Sierpinski octahedron 2641 - 1128 = 1513, the fourth term in the Sierpinski cuboctahedral sequence above. The square pyramid of 188 counters truncated from each corner is found in A281698 by, in this case, taking the fourth term of 269 and subtracting its base of 81 counters.

The base of the square pyramid that is not counted is actually the square face of the remaining cuboctahedron. It is therefore easier to construct this Sierpinski cuboctahedral geometry by taking the six square pyramids of 269 counters each, subtracting 5 for the overlapping apices and then subtracting 96 to account for the edges where the square pyramids are overlapped: 6*269 = 1614 - 5 = 1609, - 96 = 1513, the fourth term shown in the Sierpinski cuboctahedral sequence above.

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Wikipedia, , Octahedron flake and Sierpinski tetrahedron.
• Index entries for linear recurrences with constant coefficients, signature (13,-56,92,-48).

Crossrefs

Cf. A005902, A279512, A274973, A281698, A067771.

Programs

Mathematica

Table[3*2^n + 3*4^n + 6^(n + 1) + 1, {n, 0, 22}] (* Michael De Vlieger, Jul 29 2017 *)

PARI

Vec((13 - 114*x + 290*x^2 - 204*x^3) / ((1 - x)*(1 - 2*x)*(1 - 4*x)*(1 - 6*x)) + O(x^30)) \\ Colin Barker, Jul 29 2017

PARI

a(n) = 3*2^n + 3*4^n + 6^(n+1) + 1 \\ Charles R Greathouse IV, Nov 03 2017

Formula

a(n) = 3*2^n + 3*4^n + 6^(n+1) + 1.

From Colin Barker, Jul 29 2017: (Start)

G.f.: (13 - 114*x + 290*x^2 - 204*x^3) / ((1 - x)*(1 - 2*x)*(1 - 4*x)*(1 - 6*x)).

a(n) = 13*a(n-1) - 56*a(n-2) + 92*a(n-3) - 48*a(n-4) for n>3.

(End)