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A171588
The Pell word: Fixed point of the morphism 0->001, 1->0.
31
0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1
OFFSET
1,1
COMMENTS
From Peter Bala, Nov 22 2013: (Start)
This is a Sturmian word: equals the limit word S(infinity) where S(0) = 0, S(1) = 001 and for n >= 1, S(n+1) = S(n)S(n)S(n-1). See the examples below.
This sequence corresponds to the case k = 2 of the Sturmian word S_k(infinity) as defined in A080764. See A159684 for the case k = 1. (End)
Characteristic word with slope 1 - 1/sqrt(2). Since the characteristic word with slope 1-theta is the mirror image of the characteristic word with slope theta, a(n)= 1 - A080764(n) for all n. - Michel Dekking, Jan 31 2017
The positions of 0 comprise A001951 (Beatty sequence for sqrt(2)); the positions of 1 comprise A001952 (Beatty sequence for 2+sqrt(2)). - Clark Kimberling, May 11 2017
This is also the fixed point of the mapping 00->0010, 01->001, 10->010, starting with 00 [Dekking and Keane, 2022]. See A289001. - N. J. A. Sloane, Mar 09 2022
REFERENCES
J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 284.
LINKS
Scott Balchin and Dan Rust, Computations for Symbolic Substitutions, Journal of Integer Sequences, Vol. 20 (2017), Article 17.4.1.
Jean Berstel and Juhani Karhumäki, Combinatorics on words-a tutorial. Bull. Eur. Assoc. Theor. Comput. Sci. EATCS, 79:178-228, 2003.
Michel Dekking, Substitution invariant Sturmian words and binary trees, arXiv:1705.08607 [math.CO], 2017.
Michel Dekking, Substitution invariant Sturmian words and binary trees, Integers, Electronic Journal of Combinatorial Number Theory 18A (2018), #A17.
Michel Dekking and Mike Keane, Two-block substitutions and morphic words, arXiv:2202.13548 [math.CO], 2022.
Wikipedia, Sturmian word
FORMULA
a(n) = floor((n + 1)/(2 + sqrt(2))) - floor(n /(2 + sqrt(2))). - Peter Bala, Nov 22 2013
a(n) = floor((n+1)*(1 - 1/sqrt(2))) - floor(n*(1 - 1/sqrt(2))). - Michel Dekking, Jan 31 2017
EXAMPLE
From Peter Bala, Nov 22 2013: (Start)
The sequence of words S(n) begins
S(0) = 0
S(1) = 001
S(2) = 001 001 0
S(3) = 0010010 0010010 001
S(4) = 00100100010010001 00100100010010001 0010010.
The lengths of the words are [1, 3, 7, 17, 41, ...] = A001333 (apart from the initial term). (End)
MAPLE
Digits := 50: u := evalf(2 + sqrt(2)): A171588 := n->floor((n+1)/u) - floor(n/u): seq(A171588(n), n = 1..80); # Peter Bala, Nov 22 2013
MATHEMATICA
Table[Floor[(n + 1) (1 - 1/Sqrt[2]) - Floor[n (1 - 1/Sqrt[2])]], {n, 100}] (* Vincenzo Librandi, Jan 31 2017 *)
Nest[Flatten[# /. {0 -> {0, 0, 1}, 1 -> {0}}] &, {0}, 6] (* Clark Kimberling, May 11 2017 *)
PROG
(Magma) [Floor((n+1)*(1-1/Sqrt(2))-Floor(n*(1-1/Sqrt(2)))): n in [1..100]]; // Vincenzo Librandi, Jan 31 2017
KEYWORD
nonn,easy
AUTHOR
Alexis Monnerot-Dumaine (alexis.monnerotdumaine(AT)gmail.com), Dec 12 2009
STATUS
approved