A265720 Binary representation of the n-th iteration of the "Rule 1" elementary cellular automaton starting with a single ON (black) cell.
1, 0, 100, 1100011, 10000, 11110001111, 1000000, 111111000111111, 100000000, 1111111100011111111, 10000000000, 11111111110001111111111, 1000000000000, 111111111111000111111111111, 100000000000000, 1111111111111100011111111111111, 10000000000000000
Offset: 0
Examples
From _Michael De Vlieger_, Dec 14 2015: (Start)
First 10 rows, replacing leading zeros with ".", the row converted to its binary equivalent at right:
1 = 1
. . 0 = 0
. . 1 0 0 = 100
1 1 0 0 0 1 1 = 1100011
. . . . 1 0 0 0 0 = 10000
1 1 1 1 0 0 0 1 1 1 1 = 11110001111
. . . . . . 1 0 0 0 0 0 0 = 1000000
1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 = 111111000111111
. . . . . . . . 1 0 0 0 0 0 0 0 0 = 100000000
1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 = 1111111100011111111
(End)
Links
- Robert Price, Table of n, a(n) for n = 0..999
- 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 (0,10101,0,-1010100,0,1000000).
Programs
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Mathematica
rule = 1; rows = 20; Table[FromDigits[Table[Take[CellularAutomaton[rule, {{1}, 0}, rows - 1, {All, All}][[k]], {rows - k + 1, rows + k - 1}], {k, 1, rows}][[k]]], {k, 1, rows}] -
Python
print([(10*100**n - 999*10**(n-1) - 1)//9 if n%2 else 10**n for n in range(50)]) # Karl V. Keller, Jr., Aug 25 2021
Formula
From Colin Barker, Dec 14 2015 and Apr 16 2019: (Start)
a(n) = 10101*a(n-2) - 1010100*a(n-4) + 1000000*a(n-6) for n > 5.
G.f.: (1-10001*x^2+1100011*x^3+10000*x^4-1210000*x^5) / ((1-x)*(1+x)*(1-10*x)*(1+10*x)*(1-100*x)*(1+100*x)).
(End)
a(n) = (10*100^n - 999*10^(n-1) - 1)/9 for odd n; a(n) = 10^n for even n. - Karl V. Keller, Jr., Aug 25 2021
Comments