A159684 Sturmian word: limit S(infinity) where S(0) = 0, S(1) = 0,1 and for n>=1, S(n+1) = S(n)S(n)S(n-1).

0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0

Offset: 0

Author

Philippe Deléham, Apr 19 2009

Keywords

• nonn
• easy

Comments

Fixed point of morphism 0 -> 0,1; 1 -> 0,1,0.

This sequence corresponds to the case k = 1 of the Sturmian word S_k(infinity) as defined in A080764. See A171588 for the case k = 2. - Peter Bala, Nov 22 2013

This sequence is the {1->01}-transform of the Sturmian word A080764. - Clark Kimberling, May 17 2017

Also, sequence (1 if x/sqrt(2) is integer, 0 else) as x runs over the elements of N U N*sqrt(2) in increasing order, N = {0, 1, 2, ...}. See A144612 for the sqrt(3) analog. - M. F. Hasler, Feb 06 2025

Examples

			0 -> 0,1 -> 0,1,0,1,0 -> 0,1,0,1,0,0,1,0,1,0,0,1 ->...
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
• W. W. Adams and J. L. Davison, A remarkable class of continued fractions, Proc. Amer. Math. Soc. 65 (1977), 194-198.
• P. G. Anderson, T. C. Brown, and P. J.-S. Shiue, A simple proof of a remarkable continued fraction identity Proc. Amer. Math. Soc. 123 (1995), 2005-2009.
• Jeffrey Shallit, Characteristic words as fixed points of homomorphisms, University of Waterloo Technical Report CS-91-72, 1991. See Example 1.
• Jeffrey Shallit, Characteristic words as fixed points of homomorphisms. See Example 1. [Cached copy, with permission]
• N. J. A. Sloane, Families of Essentially Identical Sequences, Mar 24 2021 (Includes this sequence)
• Vincent Van Dongen, Tileset As: 3 squares with local rules for non-periodic tiling, arXiv:2503.11682 [math.GM], 2025. See p. 5.
• Wikipedia, Sturmian word

Crossrefs

See A188037 for another version of this sequence. - N. J. A. Sloane, Mar 22 2011

Cf. A000129, A001333, A005614, A080764, A119812, A171588, A221150, A221151, A221152, A096270.

The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A003151 as the parent: A003151, A001951, A001952, A003152, A006337, A080763, A082844 (conjectured), A097509, A159684, A188037, A245219 (conjectured), A276862. - N. J. A. Sloane, Mar 09 2021

Programs

Haskell

a159684 n = a159684_list !! n
a159684_list = 0 : concat (iterate (concatMap s) [1])
   where s 0 = [0,1]; s 1 = [0,1,0]
-- Reinhard Zumkeller, Oct 26 2013

Mathematica

Nest[ Flatten[ # /. {0 -> {0, 1}, 1 -> {0, 1, 0}}] &, {1}, 6] (* Robert G. Wilson v, May 02 2009 *)
SubstitutionSystem[{0->{0,1},1->{0,1,0}},{1},{6}][[1]] (* Harvey P. Dale, Dec 25 2021 *)

Python

def aupto(nn):
    Snm1, Sn = [0], [0, 1]
    while len(Sn) < nn+1: Snm1, Sn = Sn, Sn + Sn + Snm1
    return Sn[:nn+1]
print(aupto(104)) # Michael S. Branicky, Jul 23 2022

Python

from math import isqrt
def A159684(n): return -isqrt(m:=(n+1)**2<<1)+isqrt(m+(n<<2)+6)-1 # Chai Wah Wu, Aug 03 2022

Formula

From Peter Bala, Nov 22 2013: (Start)

a(n) = floor((n + 2)*(sqrt(2) - 1)) - floor((n + 1)*(sqrt(2) - 1)).

If we read the sequence as the decimal constant C = 0.01010 01010 01010 10010 10010 ... then C = sum {n >= 1} 1/10^floor(n*(1 + sqrt(2))).

The real number 9*C has the simple continued fraction expansion [0; 11, 1010, 10000100, 100000000000100000, 100000000000000000000000000001000000000000, ...], the partial quotients having the form 10^Pell(n)*(1 + 10^Pell(n+1)) = 10^A001333(n+1) + 10^A000129(n) (see Adams and Davison).

A rapidly converging series for C is C = 9*sum {n >= 1} 10^Pell(2*n-1)*(1 + 10^Pell(2*n))/( (10^Pell(2*n-1) - 1)*(10^Pell(2*n+1) - 1) ): for example, the first 10 terms of the series give a rational approximation to C accurate to more than 130 million decimal places. Compare with the Fibonacci words A005614 and A221150. (End)

Extensions

More terms from Robert G. Wilson v, May 02 2009