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A160722 Number of "ON" cells at n-th stage in a certain 2-dimensional cellular automaton based on Sierpinski triangles (see Comments for precise definition). 8

%I #19 Nov 02 2022 07:45:48

%S 0,1,5,9,19,23,33,43,65,69,79,89,111,121,143,165,211,215,225,235,257,

%T 267,289,311,357,367,389,411,457,479,525,571,665,669,679,689,711,721,

%U 743,765,811,821,843,865,911,933,979,1025,1119,1129,1151,1173,1219,1241

%N Number of "ON" cells at n-th stage in a certain 2-dimensional cellular automaton based on Sierpinski triangles (see Comments for precise definition).

%C This cellular automata is formed by the concatenation of three Sierpinski triangles, starting from a central vertex. Adjacent polygons are fused. The ON cells are triangles, but we only count after fusion. The sequence gives the number of polygons at the n-th round.

%C If instead we start from four Sierpinski triangles we get A160720.

%H Michael De Vlieger, <a href="/A160722/b160722.txt">Table of n, a(n) for n = 0..9999</a>

%H David Applegate, Omar E. Pol and N. J. A. Sloane, <a href="/A000695/a000695_1.pdf">The Toothpick Sequence and Other Sequences from Cellular Automata</a>, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.]

%H Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, <a href="https://arxiv.org/abs/2210.10968">Identities and periodic oscillations of divide-and-conquer recurrences splitting at half</a>, arXiv:2210.10968 [cs.DS], 2022, p. 30.

%H Omar E. Pol, <a href="http://www.polprimos.com/imagenespub/polca722.jpg">Illustration if initial terms</a>

%H N. J. A. Sloane, <a href="/wiki/Catalog_of_Toothpick_and_CA_Sequences_in_OEIS">Catalog of Toothpick and Cellular Automata Sequences in the OEIS</a>

%F a(n) = 3*A006046(n) - 2*n. - _Max Alekseyev_, Jan 21 2010

%e We start at round 0 with no polygons, a(0) = 0.

%e At round 1 we turn ON the first triangle in each of the three Sierpinski triangles. After fusion we have a concave pentagon, so a(1) = 1.

%e At round 2 we turn ON two triangles in each the three Sierpinski triangles. After fusions we have the concave pentagon and four triangles. So a(2) = 1 + 4 = 5.

%t a[0] = 0; a[1] = 1; a[n_] := a[n] = 2 a[Floor[#]] + a[Ceiling[#]] &[n/2]; Array[3 a[#] - 2 # &, 54, 0] (* _Michael De Vlieger_, Nov 01 2022 *)

%Y A160723 gives the first differences.

%Y Cf. A139250, A160720.

%K nonn

%O 0,3

%A _Omar E. Pol_, May 25 2009, Jan 03 2010

%E Extended by _Max Alekseyev_, Jan 21 2010

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Last modified March 28 16:34 EDT 2024. Contains 371254 sequences. (Running on oeis4.)