A010059 Another version of the Thue-Morse sequence: let A_k denote the first 2^k terms; then A_0 = 1 and for k >= 0, A_{k+1} = A_k B_k, where B_k is obtained from A_k by interchanging 0's and 1's.
1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0
Offset: 0
Examples
The evolution starting at 1 is: .1 .1, 0 .1, 0, 0, 1, .1, 0, 0, 1, 0, 1, 1, 0 .1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1 .1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0 ...........
References
- W. H. Gottschalk and G. A. Hedlund, Topological Dynamics. American Mathematical Society, Colloquium Publications, Vol. 36, Providence, RI, 1955, p. 105.
- M. Lothaire, Combinatorics on Words. Addison-Wesley, Reading, MA, 1983, p. 23.
- A. Salomaa, Jewels of Formal Language Theory. Computer Science Press, Rockville, MD, 1981, p. 6.
Links
- Robert Israel, Table of n, a(n) for n = 0..10000
- J.-P. Allouche and Jeffrey Shallit, The Ubiquitous Prouhet-Thue-Morse Sequence, in C. Ding. T. Helleseth and H. Niederreiter, eds., Sequences and Their Applications: Proceedings of SETA '98, Springer-Verlag, 1999, pp. 1-16.
- Scott Balchin and Dan Rust, Computations for Symbolic Substitutions, Journal of Integer Sequences, Vol. 20 (2017), Article 17.4.1.
- Françoise Dejean, Sur un Théorème de Thue, J. Combinatorial Theory, vol. 13 A, iss. 1 (1972) 90-99.
- F. Michel Dekking, Morphisms, Symbolic Sequences, and Their Standard Forms, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.
- Michael Gilleland, Some Self-Similar Integer Sequences
- G. A. Hedlund, Remarks on the work of Axel Thue on sequences, Nordisk Mat. Tid., 15 (1967), 148-150.
- Tanya Khovanova, There are no coincidences, arXiv preprint 1410.2193 [math.CO], 2014.
- H. M. Morse, Recurrent geodesics on a surface of negative curvature, Trans. Amer. Math. Soc., 22 (1921), 84-100.
- Stephen Wolfram, A New Kind Of Science | Online.
- Index entries for sequences related to binary expansion of n
- Index entries for characteristic functions
Crossrefs
Programs
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Haskell
a010059 = (1 -) . a010060 -- Reinhard Zumkeller, Feb 04 2013
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Maple
A010059 := n->1-A010060(n); map(t->49-t,convert(StringTools[ThueMorse](1000),bytes)); # Robert Israel, Feb 02 2016 # second Maple program: a := n -> ifelse(type(add(convert(n, base, 2)), even), 1, 0): seq(a(n), n = 0 .. 104); # Peter Luschny, Mar 11 2024
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Mathematica
Mod[ CoefficientList[Series[(1 + Sqrt[(1 - 3x)/(1 + x)])/(2(1 + x)), {x, 0, 111}], x], 2] (* Stephan Wolfram *) CoefficientList[ Series[1/(1 - x) + Product[1 - x^2^k, {k, 0, 10}], {x, 0, 111}]/2, x] (* Robert G. Wilson v, Jul 16 2004 *) Nest[ Flatten[ # /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {1}, 7] (* Robert G. Wilson v Sep 26 2006 *) od = Select[ Range[0, 129], OddQ@ DigitCount[ #, 2, 1] &]; ev = Select[ Range[0, 129], EvenQ@ DigitCount[ #, 2, 1] &]; Mod[ Flatten@ Transpose[{od, ev}], 2] (* Robert G. Wilson v, Apr 14 2009 *) Nest[ Join[ #, Mod[2# + 1, 3]] &, {1}, 7] (* Robert G. Wilson v, Jul 27 2014 *) {{1}}~Join~SubstitutionSystem[{0 -> {0, 1}, 1 -> {1, 0}}, {0}, 6] // Flatten (* Michael De Vlieger, Aug 15 2016, Version 10.2 *) 1 - ThueMorse[Range[0, 100]] (* Paolo Xausa, Oct 25 2024 *)
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PARI
a(n)=!(hammingweight(n)%2); \\ Charles R Greathouse IV, Mar 29 2013
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Python
def A010059(n): return n.bit_count()&1^1 # Chai Wah Wu, Mar 01 2023
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R
maxrow <- 8 # by choice b01 <- 0 for(m in 0:maxrow) for(k in 0:(2^m-1)){ b01[2^(m+1)+ k] <- b01[2^m+k] b01[2^(m+1)+2^m+k] <- 1-b01[2^m+k] } (b01 <- c(1,b01)) # Yosu Yurramendi, Apr 10 2017
Formula
G.f.: (1/2) * (1/(1-x) + Product_{k>=0} (1 - x^2^k)). - Ralf Stephan, Jun 20 2003
a(n) = A143579(n) mod 2. - Gary W. Adamson, Aug 24 2008
a(n) + A010060(n) = 1 for all n.
a(n) + A026147(n-1) = 2n for n >= 1. - Clark Kimberling, Oct 06 2014
a(n) = A000069(n+1) (mod 2). - John M. Campbell, Jun 30 2016
a(n) = A059448(A054429(n)) = (A106400(n)+1)/2 = (1+A008836(A005940(1+n)))/2. - Antti Karttunen, May 30 2017
If A(n)=(a(0),a(2),...,a(2^n-1)), then A(n+1)=(A(n),1-A(n)). - Arie Bos, Jul 27 2022
a(n) = (1 + (-1)^A000120(n))/2. - Lorenzo Sauras Altuzarra, Mar 10 2024
Comments