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

Author

Robert Price, Jan 19 2016

Keywords

• nonn
• easy

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).

Crossrefs

Cf. A267683, A267685.

Essentially the same as A138148.

Programs

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