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A267684
Binary representation of the n-th iteration of the "Rule 203" elementary cellular automaton starting with a single ON (black) cell.
3
1, 100, 11011, 1110111, 111101111, 11111011111, 1111110111111, 111111101111111, 11111111011111111, 1111111110111111111, 111111111101111111111, 11111111111011111111111, 1111111111110111111111111, 111111111111101111111111111, 11111111111111011111111111111
OFFSET
0,2
REFERENCES
Stephen Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.
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
E.g.f.: 1 - x + exp(x)*(-1 - 9*exp(9*x) + 10*exp(99*x))/9. - Elmo R. Oliveira, Apr 13 2026
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 *)
PROG
(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
CROSSREFS
Essentially the same as A138148.
Sequence in context: A272681 A267674 A267846 * A267885 A307810 A394861
KEYWORD
nonn,easy
AUTHOR
Robert Price, Jan 19 2016
STATUS
approved