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%I #47 Oct 21 2021 01:10:58
%S 1,4,6,12,15,24,28,40,45,60,66,84,91,112,120,144,153,180,190,220,231,
%T 264,276,312,325,364,378,420,435,480,496,544,561,612,630,684,703,760,
%U 780,840,861,924,946,1012,1035,1104,1128,1200,1225,1300,1326,1404,1431
%N Total number of ON (black) cells after n iterations of the "Rule 54" elementary cellular automaton starting with a single ON (black) cell.
%C Take the first 2n positive integers and choose n of them such that their sum: a) is divisible by n, and b) is minimal. It seems their sum equals a(n). - _Ivan N. Ianakiev_, Feb 16 2019
%D S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.
%H Robert Price, <a href="/A265225/b265225.txt">Table of n, a(n) for n = 0..999</a>
%H Emanuele Munarini, <a href="https://doi.org/10.4418/2021.76.1.14">Topological indices for the antiregular graphs</a>, Le Mathematiche (2021) Vol. 76, No. 1, see p. 301.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/ElementaryCellularAutomaton.html">Elementary Cellular Automaton</a>
%H <a href="/index/Ce#cell">Index entries for sequences related to cellular automata</a>
%H <a href="https://oeis.org/wiki/Index_to_Elementary_Cellular_Automata">Index to Elementary Cellular Automata</a>
%F Conjectures from _Colin Barker_, Dec 08 2015 and Apr 20 2019: (Start)
%F a(n) = (n+1)*(2*n -(-1)^n +5)/4.
%F a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n>4.
%F G.f.: (1+3*x) / ((1-x)^3*(1+x)^2).
%F (End)
%F a(n) = n + 1 + (n+1) * floor((n+1)/2), conjectured. - _Wesley Ivan Hurt_, Dec 25 2016
%F a(n) = A093353(n) + n + 1, conjectured. - _Matej Veselovac_, Jan 21 2020
%e From _Michael De Vlieger_, Dec 14 2015: (Start)
%e First 12 rows, replacing "0" with "." for better visibility of ON cells, followed by the total number of 1's per row, and the running total up to that row:
%e 1 = 1 -> 1
%e 1 1 1 = 3 -> 4
%e 1 . . . 1 = 2 -> 6
%e 1 1 1 . 1 1 1 = 6 -> 12
%e 1 . . . 1 . . . 1 = 3 -> 15
%e 1 1 1 . 1 1 1 . 1 1 1 = 9 -> 24
%e 1 . . . 1 . . . 1 . . . 1 = 4 -> 28
%e 1 1 1 . 1 1 1 . 1 1 1 . 1 1 1 = 12 -> 40
%e 1 . . . 1 . . . 1 . . . 1 . . . 1 = 5 -> 45
%e 1 1 1 . 1 1 1 . 1 1 1 . 1 1 1 . 1 1 1 = 15 -> 60
%e 1 . . . 1 . . . 1 . . . 1 . . . 1 . . . 1 = 6 -> 66
%e 1 1 1 . 1 1 1 . 1 1 1 . 1 1 1 . 1 1 1 . 1 1 1 = 18 -> 84
%e 1 . . . 1 . . . 1 . . . 1 . . . 1 . . . 1 . . . 1 = 7 -> 91
%e (End)
%p A265225:=n->1/4*(n+1)*(2*n-(-1)^n+5): seq(A265225(n), n=0..60); # _Wesley Ivan Hurt_, Dec 25 2016
%t rule = 54; rows = 30; Table[Total[Take[Table[Total[Table[Take[CellularAutomaton[rule,{{1},0},rows-1,{All,All}][[k]],{rows-k+1,rows+k-1}],{k,1,rows}][[k]]],{k,1,rows}],k]],{k,1,rows}]
%t Accumulate[Total /@ CellularAutomaton[54, {{1}, 0}, 52]]
%Y Cf. A071030, A118108, A118109, A133872.
%K nonn,easy
%O 0,2
%A _Robert Price_, Dec 05 2015