A266323 Binary representation of the n-th iteration of the "Rule 19" elementary cellular automaton starting with a single ON (black) cell.
1, 101, 0, 1111111, 0, 11111111111, 0, 111111111111111, 0, 1111111111111111111, 0, 11111111111111111111111, 0, 111111111111111111111111111, 0, 1111111111111111111111111111111, 0, 11111111111111111111111111111111111, 0, 111111111111111111111111111111111111111
Offset: 0
Links
- Robert Price, Table of n, a(n) for n = 0..500
- Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
- Stephen Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.
- Index entries for sequences related to cellular automata
- Index to Elementary Cellular Automata
- Index entries for linear recurrences with constant coefficients, signature (0,10001,0,-10000).
Programs
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Mathematica
rule=19; 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 *) -
Python
print([(10*100**n - 1)//9*(n%2) + 0**n - 10*0**abs(n-1) for n in range(50)]) # Karl V. Keller, Jr., Sep 02 2021
Formula
From Colin Barker, Dec 28 2015 and Apr 15 2019: (Start)
a(n) = 10001*a(n-2) - 10000*a(n-4) for n>5.
G.f.: (1+101*x-10001*x^2+101010*x^3+10000*x^4-100000*x^5) / ((1-x)*(1+x)*(1-100*x)*(1+100*x)).
(End)
a(n) = (10*100^n - 1)/9*(n mod 2) + 0^n - 10*0^abs(n-1). - Karl V. Keller, Jr., Sep 02 2021