A267684 Binary representation of the n-th iteration of the "Rule 203" elementary cellular automaton starting with a single ON (black) cell.
1, 100, 11011, 1110111, 111101111, 11111011111, 1111110111111, 111111101111111, 11111111011111111, 1111111110111111111, 111111111101111111111, 11111111111011111111111, 1111111111110111111111111, 111111111111101111111111111, 11111111111111011111111111111
Offset: 0
References
- S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.
Links
- Robert Price, Table of n, a(n) for n = 0..1000
- Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
- S. Wolfram, A New Kind of Science
- Index entries for sequences related to cellular automata
- Index to Elementary Cellular Automata
- Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).
Programs
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Mathematica
rule=203; 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 *) LinearRecurrence[{111, -1110, 1000}, {1, 100, 11011, 1110111, 111101111}, 20] (* Paolo Xausa, Aug 07 2025 *) -
Python
print([1, 100]+[(10*100**n - 9*10**n - 1)//9 for n in range(2, 50)]) # Karl V. Keller, Jr., Jun 07 2022
Formula
From Colin Barker, Jan 19 2016 and Apr 17 2019: (Start)
a(n) = 111*a(n-1)-1110*a(n-2)+1000*a(n-3) for n>4.
G.f.: (1-11*x+1021*x^2-2110*x^3+1000*x^4) / ((1-x)*(1-10*x)*(1-100*x)).
(End)
The above conjectures are correct. Also a(n) = (10*100^n - 9*10^n - 1)/9 for n > 1. - Karl V. Keller, Jr., Jun 07 2022