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%I #23 Sep 03 2021 12:43:18
%S 1,1,1010,1000011,1110100,11010000111,11101000,111110100001111,
%T 111010000,1111111101000011111,1110100000,11111111111010000111111,
%U 11101000000,111111111111110100001111111,111010000000,1111111111111111101000011111111,1110100000000
%N Binary representation of the n-th iteration of the "Rule 25" elementary cellular automaton starting with a single ON (black) cell.
%H Robert Price, <a href="/A266442/b266442.txt">Table of n, a(n) for n = 0..1000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://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 (0,10011,0,-110010,0,100000).
%F From _Colin Barker_, Jan 01 2016 and Apr 16 2019: (Start)
%F a(n) = 10011*a(n-2) - 110010*a(n-4) + 100000*a(n-6) for n>9.
%F G.f.: (1 +x -9001*x^2 +990000*x^3 -8891000*x^4 +999000000*x^5 -10991100000*x^6 +999000000000*x^7 +11000000000*x^8 -1000000000000*x^9) / ((1 -x)*(1 +x)*(1 -100*x)*(1 +100*x)*(1 -10*x^2)).
%F (End)
%F a(n) = (10*100^n - 909999*10^((n+1)/2) - 1)/9 for odd n>3; a(n) = 11101*10^(n/2) for even n>2. - _Karl V. Keller, Jr._, Sep 02 2021
%t rule=25; rows=20; ca=CellularAutomaton[rule,{{1},0},rows-1,{All,All}]; (* Start with single black cell *) catri=Table[Take[ca[[k]],{rows-k+1,rows+k-1}],{k,1,rows}]; (* Truncated list of each row *) Table[FromDigits[catri[[k]]],{k,1,rows}] (* Binary Representation of Rows *)
%o (Python) print([1, 1, 10, 1000011] + [(10*100**n - 909999*10**((n+1)//2) - 1)//9 if n%2 else 11101*10**(n//2) for n in range(4,50)]) # _Karl V. Keller, Jr._, Sep 02 2021
%Y Cf. A266441, A266443.
%K nonn,easy
%O 0,3
%A _Robert Price_, Dec 29 2015