A187181 Parse the infinite string 012012012012... into distinct phrases 0, 1, 2, 01, 20, 12, 012, ...; a(n) = length of n-th phrase.
1, 1, 1, 2, 2, 2, 3, 4, 3, 4, 3, 4, 5, 5, 5, 6, 7, 6, 7, 6, 7, 8, 8, 8, 9, 10, 9, 10, 9, 10, 11, 11, 11, 12, 13, 12, 13, 12, 13, 14, 14, 14, 15, 16, 15, 16, 15, 16, 17, 17, 17, 18, 19, 18, 19, 18, 19, 20, 20, 20, 21, 22, 21, 22, 21, 22, 23, 23, 23, 24, 25, 24, 25, 24, 25, 26, 26, 26, 27, 28, 27, 28, 27, 28, 29, 29, 29, 30, 31, 30, 31, 30, 31, 32, 32, 32, 33, 34, 33, 34
Offset: 1
Examples
The sequence begins 1 1 1 2 2 2 3 4 3 4 3 4 5 5 5 6 7 6 7 6 7 8 8 8 9 10 9 10 9 10 11 11 11 12 13 12 13 12 13 14 14 14 15 16 15 16 15 16 ...
Links
- Ray Chandler, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (1, 0, 0, 0, 0, 0, 0, 0, 1, -1).
Programs
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Mathematica
Join[{1, 1},LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, 1, -1},{1, 2, 2, 2, 3, 4, 3, 4, 3, 4},98]] (* Ray Chandler, Aug 26 2015 *) -
PARI
Vec(x*(x^11-x^10-x^8+x^7+x^6+x^3+1)/((x-1)^2*(x^2+x+1)*(x^6+x^3+1)) + O(x^100)) \\ Colin Barker, Nov 05 2015
Formula
After the initial block of three 1's, the sequence is quasi-periodic with period 9, increasing by 3 after each block.
From Colin Barker, Nov 05 2015: (Start)
a(n) = a(n-1) + a(n-9) - a(n-10) for n>12.
G.f.: x*(x^11-x^10-x^8+x^7+x^6+x^3+1) / ((x-1)^2*(x^2+x+1)*(x^6+x^3+1)).
(End)
Comments