%I #45 Sep 09 2016 00:03:17
%S 1,6,6,30,6,36,30,138,6,36,36,180,30,180,138,606,6,36,36,180,36,216,
%T 180,828,30,180,180,900,138,828,606,2586,6,36,36,180,36,216,180,828,
%U 36,216,216,1080,180,1080,828,3636,30,180,180,900,180,1080,900,4140,138,828,828,4140
%N Number of ON cells after n generations of the odd-rule cellular automaton defined by OddRule 365 when started with a single ON cell.
%H Paul Tek, <a href="/A255462/b255462.txt">Table of n, a(n) for n = 0..10000</a>
%H Shalosh B. Ekhad, N. J. A. Sloane, and Doron Zeilberger, <a href="http://arxiv.org/abs/1503.01796">A Meta-Algorithm for Creating Fast Algorithms for Counting ON Cells in Odd-Rule Cellular Automata</a>, arXiv:1503.01796 [math.CO], 2015; see also the <a href="http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/CAcount.html">Accompanying Maple Package</a>.
%H Shalosh B. Ekhad, N. J. A. Sloane, and Doron Zeilberger, <a href="http://arxiv.org/abs/1503.04249">Odd-Rule Cellular Automata on the Square Grid</a>, arXiv:1503.04249 [math.CO], 2015.
%H N. J. A. Sloane, On the No. of ON Cells in Cellular Automata, Video of talk in Doron Zeilberger's Experimental Math Seminar at Rutgers University, Feb. 05 2015: <a href="https://vimeo.com/119073818">Part 1</a>, <a href="https://vimeo.com/119073819">Part 2</a>
%H N. J. A. Sloane, <a href="http://arxiv.org/abs/1503.01168">On the Number of ON Cells in Cellular Automata</a>, arXiv:1503.01168 [math.CO], 2015
%H N. J. A. Sloane, <a href="/A255462/a255462.png">Illustration of generations 0 to 15</a>
%H N. J. A. Sloane, <a href="/A255462/a255462_3.png">Illustration of generations 0 to 35</a>
%H N. J. A. Sloane, <a href="/A255462/a255462_1.png">Illustration of generation 7</a>
%H N. J. A. Sloane, <a href="/A255462/a255462_2.png">Illustration of generation 15</a>
%H N. J. A. Sloane, <a href="/A255462/a255462.txt">Mathematica notebook to generate this cellular automaton</a>
%H <a href="/index/Ce#cell">Index entries for sequences related to cellular automata</a>
%F It follows from Theorem 3 of the Fredkin.pdf (2015) paper that this satisfies the recurrence a(2t)=a(t), a(4t+1)=6*a(t), and a(4t+3)=7*a(2t+1)-12*a(t) for t>0, with a(0)=1. - _N. J. A. Sloane_, Mar 10 2015
%e From _Omar E. Pol_, Sep 08 2016: (Start)
%e Written as an irregular triangle in which the row lengths are the terms of A011782 the sequence begins:
%e 1;
%e 6;
%e 6, 30;
%e 6, 36, 30, 138;
%e 6, 36, 36, 180, 30, 180, 138, 606;
%e 6, 36, 36, 180, 36, 216, 180, 828, 30, 180, 180, 900, 138, 828, 606, 2586;
%e ...
%e Right border gives A255463. (End)
%t (* See Mathematica notebook in link *)
%t (* or *)
%t A255462[n_] := Total[CellularAutomaton[{42, {2, {{0, 1, 1}, {1, 1, 0}, {1, 0, 1}}}, {1, 1}}, {{{1}}, 0}, {{{n}}}], 2]; Array[A255462, 60, 0] (* _JungHwan Min_, Sep 06 2016 *)
%t A255462L[n_] := Total[#, 2] & /@ CellularAutomaton[{42, {2, {{0, 1, 1}, {1, 1, 0}, {1, 0, 1}}}, {1, 1}}, {{{1}}, 0}, n]; A255462L[59] (* _JungHwan Min_, Sep 06 2016 *)
%Y Run length transform of A255463.
%K nonn,tabf,look
%O 0,2
%A _N. J. A. Sloane_ and _Doron Zeilberger_, Feb 23 2015