A171588 The Pell word: Fixed point of the morphism 0->001, 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, 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

Author

Alexis Monnerot-Dumaine, Dec 12 2009

Keywords

• nonn
• easy

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

Examples

			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)
		

References

• J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 284.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..5000
• 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.
• M. Lothaire, Combinatorics on Words.
• Wikipedia, Sturmian word

Crossrefs

Cf. A000129, A001333, A001951, A001952, A003849, A080764, A159684, A289001.

Programs

Magma

[Floor((n+1)*(1-1/Sqrt(2))-Floor(n*(1-1/Sqrt(2)))): n in [1..100]]; // Vincenzo Librandi, Jan 31 2017

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 *)

Python

from math import isqrt
def A171588(n): return 1+isqrt(n**2>>1)-isqrt((n+1)**2>>1) # Chai Wah Wu, May 24 2025

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