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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

%I #51 Oct 18 2017 13:34:02

%S 4,16,96,688,5280,41584,330720,2639920,21101856,168762352,1349941344,

%T 10799058352,86391049632,691124145520,5528980409568,44231805012784,

%U 353854325311008,2830834258114288,22646673031792992,181173381154980016

%N 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.

%C 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.

%H Colin Barker, <a href="/A293143/b293143.txt">Table of n, a(n) for n = 0..1000</a> (Recomputed by M. F. Hasler to include the initial term 4.)

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/SierpinskiCarpet.html">Sierpinski Carpet</a>.

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Sierpinski_carpet">Sierpinski carpet</a>.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (12,-35,24).

%F From _Colin Barker_, Oct 02 2017, corrected for a(0) = 4 by _M. F. Hasler_, Oct 16 2017: (Start)

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

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

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

%e 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.

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

%o (PARI) prev=4; concat(prev, vector(20, n, prev=8*prev-8*(3^(n-1)+1))) \\ _Colin Barker_, Oct 08 2017

%o (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

%o (PARI) A293143(n)=8*(5+11*2^(3*n-1)+7*3^n)/35 \\ _M. F. Hasler_, Oct 16 2017

%Y Cf. A293144, A034472.

%K nonn,easy

%O 0,1

%A _Steven Beard_, Oct 01 2017

%E Edited to start with a(0) = 4 by _M. F. Hasler_, Oct 16 2017

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Last modified April 19 21:09 EDT 2024. Contains 371798 sequences. (Running on oeis4.)