A293143 Number of vertex points in a Sierpinski Carpet grid subdivided into squares: a(n+1) = 8*a(n) - 8*(3^n+1), a(0) = 4.

4, 16, 96, 688, 5280, 41584, 330720, 2639920, 21101856, 168762352, 1349941344, 10799058352, 86391049632, 691124145520, 5528980409568, 44231805012784, 353854325311008, 2830834258114288, 22646673031792992, 181173381154980016

Offset: 0

Author

Steven Beard, Oct 01 2017

Keywords

• nonn
• easy

Comments

Figurate number sequence for the Sierpinski Carpet lattice. See the faces of the cubes in "Image 2" in the Wikipedia link of A293144 for an example of the construction grid of the Sierpinski Carpet.

Examples

			The carpet is formed by squares within a square grid. The initial term is a(0) = 4 for the corners of the unit square. For n = 1 there are 3 X 3 squares, the middle one being open (empty), with 16 vertex points. At the next stage of recursion, these become eight squares with open center, now based on a square of 10 X 10 points. The remaining center square is empty, missing 4 points, thus the next term is 100 - 4 = 96 for a(2). In the next stage there are 8 squares missing 4 points and the new center is missing 64, thus the 28 square grid now has 784 - 32 - 64 = 688 for a(3). This carpet sequence becomes the faces for the cubes in the Menger Sponge recursion of A293144.
		

Links

• Colin Barker, Table of n, a(n) for n = 0..1000 (Recomputed by M. F. Hasler to include the initial term 4.)
• Eric Weisstein's World of Mathematics, Sierpinski Carpet.
• Wikipedia, Sierpinski carpet.
• Index entries for linear recurrences with constant coefficients, signature (12,-35,24).

Crossrefs

Cf. A293144, A034472.

Programs

Mathematica

FoldList[8 #1 - 8 (3^(#2-1) + 1) &, 4, Range@ 18] (* Michael De Vlieger, Oct 02 2017 *)

PARI

prev=4; concat(prev, vector(20, n, prev=8*prev-8*(3^(n-1)+1))) \\ Colin Barker, Oct 08 2017

PARI

Vec(4*(1 - 8*x + 11*x^2)/((1 - x)*(1 - 3*x)*(1 - 8*x)) + O(x^30)) \\ Colin Barker, Oct 09 2017

PARI

A293143(n)=8*(5+11*2^(3*n-1)+7*3^n)/35 \\ M. F. Hasler, Oct 16 2017

Formula

From Colin Barker, Oct 02 2017, corrected for a(0) = 4 by M. F. Hasler, Oct 16 2017: (Start)

G.f.: 4*(1 - 8*x + 11*x^2) / ((1 - x)*(1 - 3*x)*(1 - 8*x)).

a(n) = 8*(5 + 11*2^(3*n-1) + 7*3^n) / 35.

a(n) = 12*a(n-1) - 35*a(n-2) + 24*a(n-3) for n > 2. (End)

Extensions

Edited to start with a(0) = 4 by M. F. Hasler, Oct 16 2017