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Binary representation of the n-th iteration of the "Rule 1" elementary cellular automaton starting with a single ON (black) cell.
5

%I #37 Feb 16 2025 08:33:28

%S 1,0,100,1100011,10000,11110001111,1000000,111111000111111,100000000,

%T 1111111100011111111,10000000000,11111111110001111111111,

%U 1000000000000,111111111111000111111111111,100000000000000,1111111111111100011111111111111,10000000000000000

%N Binary representation of the n-th iteration of the "Rule 1" elementary cellular automaton starting with a single ON (black) cell.

%C Rule 33 also generates this sequence.

%H Robert Price, <a href="/A265720/b265720.txt">Table of n, a(n) for n = 0..999</a>

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/ElementaryCellularAutomaton.html">Elementary Cellular Automaton</a>

%H Stephen Wolfram, <a href="http://wolframscience.com/">A New Kind of Science</a>, Wolfram Media, 2002; p. 55.

%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>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,10101,0,-1010100,0,1000000).

%F From _Colin Barker_, Dec 14 2015 and Apr 16 2019: (Start)

%F a(n) = 10101*a(n-2) - 1010100*a(n-4) + 1000000*a(n-6) for n > 5.

%F G.f.: (1-10001*x^2+1100011*x^3+10000*x^4-1210000*x^5) / ((1-x)*(1+x)*(1-10*x)*(1+10*x)*(1-100*x)*(1+100*x)).

%F (End)

%F a(n) = (10*100^n - 999*10^(n-1) - 1)/9 for odd n; a(n) = 10^n for even n. - _Karl V. Keller, Jr._, Aug 25 2021

%e From _Michael De Vlieger_, Dec 14 2015: (Start)

%e First 10 rows, replacing leading zeros with ".", the row converted to its binary equivalent at right:

%e 1 = 1

%e . . 0 = 0

%e . . 1 0 0 = 100

%e 1 1 0 0 0 1 1 = 1100011

%e . . . . 1 0 0 0 0 = 10000

%e 1 1 1 1 0 0 0 1 1 1 1 = 11110001111

%e . . . . . . 1 0 0 0 0 0 0 = 1000000

%e 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 = 111111000111111

%e . . . . . . . . 1 0 0 0 0 0 0 0 0 = 100000000

%e 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 = 1111111100011111111

%e (End)

%t rule = 1; rows = 20; Table[FromDigits[Table[Take[CellularAutomaton[rule, {{1}, 0}, rows - 1, {All, All}][[k]], {rows - k + 1, rows + k - 1}], {k, 1, rows}][[k]]], {k, 1, rows}]

%o (Python) print([(10*100**n - 999*10**(n-1) - 1)//9 if n%2 else 10**n for n in range(50)]) # _Karl V. Keller, Jr._, Aug 25 2021

%Y Cf. A265718, A265721.

%K nonn,easy,changed

%O 0,3

%A _Robert Price_, Dec 14 2015