cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A010060 Thue-Morse sequence: let A_k denote the first 2^k terms; then A_0 = 0 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.

Original entry on oeis.org

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, 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, 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
Offset: 0

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Author

Keywords

Comments

Named after Axel Thue, whose name is pronounced as if it were spelled "Tü" where the ü sound is roughly as in the German word üben. (It is incorrect to say "Too-ee" or "Too-eh".) - N. J. A. Sloane, Jun 12 2018
Also called the Thue-Morse infinite word, or the Morse-Hedlund sequence, or the parity sequence.
Fixed point of the morphism 0 --> 01, 1 --> 10, see example. - Joerg Arndt, Mar 12 2013
The sequence is cubefree (does not contain three consecutive identical blocks) [see Offner for a direct proof] and is overlap-free (does not contain XYXYX where X is 0 or 1 and Y is any string of 0's and 1's).
a(n) = "parity sequence" = parity of number of 1's in binary representation of n.
To construct the sequence: alternate blocks of 0's and 1's of successive lengths A003159(k) - A003159(k-1), k = 1, 2, 3, ... (A003159(0) = 0). Example: since the first seven differences of A003159 are 1, 2, 1, 1, 2, 2, 2, the sequence starts with 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0. - Emeric Deutsch, Jan 10 2003
Characteristic function of A000069 (odious numbers). - Ralf Stephan, Jun 20 2003
a(n) = S2(n) mod 2, where S2(n) = sum of digits of n, n in base-2 notation. There is a class of generalized Thue-Morse sequences: Let Sk(n) = sum of digits of n; n in base-k notation. Let F(t) be some arithmetic function. Then a(n)= F(Sk(n)) mod m is a generalized Thue-Morse sequence. The classical Thue-Morse sequence is the case k=2, m=2, F(t)= 1*t. - Ctibor O. Zizka, Feb 12 2008 (with correction from Daniel Hug, May 19 2017)
More generally, the partial sums of the generalized Thue-Morse sequences a(n) = F(Sk(n)) mod m are fractal, where Sk(n) is sum of digits of n, n in base k; F(t) is an arithmetic function; m integer. - Ctibor O. Zizka, Feb 25 2008
Starting with offset 1, = running sums mod 2 of the kneading sequence (A035263, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, ...); also parity of A005187: (1, 3, 4, 7, 8, 10, 11, 15, 16, 18, 19, ...). - Gary W. Adamson, Jun 15 2008
Generalized Thue-Morse sequences mod n (n>1) = the array shown in A141803. As n -> infinity the sequences -> (1, 2, 3, ...). - Gary W. Adamson, Jul 10 2008
The Thue-Morse sequence for N = 3 = A053838, (sum of digits of n in base 3, mod 3): (0, 1, 2, 1, 2, 0, 2, 0, 1, 1, 2, ...) = A004128 mod 3. - Gary W. Adamson, Aug 24 2008
For all positive integers k, the subsequence a(0) to a(2^k-1) is identical to the subsequence a(2^k+2^(k-1)) to a(2^(k+1)+2^(k-1)-1). That is to say, the first half of A_k is identical to the second half of B_k, and the second half of A_k is identical to the first quarter of B_{k+1}, which consists of the k/2 terms immediately following B_k.
Proof: The subsequence a(2^k+2^(k-1)) to a(2^(k+1)-1), the second half of B_k, is by definition formed from the subsequence a(2^(k-1)) to a(2^k-1), the second half of A_k, by interchanging its 0's and 1's. In turn, the subsequence a(2^(k-1)) to a(2^k-1), the second half of A_k, which is by definition also B_{k-1}, is by definition formed from the subsequence a(0) to a(2^(k-1)-1), the first half of A_k, which is by definition also A_{k-1}, by interchanging its 0's and 1's. Interchanging the 0's and 1's of a subsequence twice leaves it unchanged, so the subsequence a(2^k+2^(k-1)) to a(2^(k+1)-1), the second half of B_k, must be identical to the subsequence a(0) to a(2^(k-1)-1), the first half of A_k.
Also, the subsequence a(2^(k+1)) to a(2^(k+1)+2^(k-1)-1), the first quarter of B_{k+1}, is by definition formed from the subsequence a(0) to a(2^(k-1)-1), the first quarter of A_{k+1}, by interchanging its 0's and 1's. As noted above, the subsequence a(2^(k-1)) to a(2^k-1), the second half of A_k, which is by definition also B_{k-1}, is by definition formed from the subsequence a(0) to a(2^(k-1)-1), which is by definition A_{k-1}, by interchanging its 0's and 1's, as well. If two subsequences are formed from the same subsequence by interchanging its 0's and 1's then they must be identical, so the subsequence a(2^(k+1)) to a(2^(k+1)+2^(k-1)-1), the first quarter of B_{k+1}, must be identical to the subsequence a(2^(k-1)) to a(2^k-1), the second half of A_k.
Therefore the subsequence a(0), ..., a(2^(k-1)-1), a(2^(k-1)), ..., a(2^k-1) is identical to the subsequence a(2^k+2^(k-1)), ..., a(2^(k+1)-1), a(2^(k+1)), ..., a(2^(k+1)+2^(k-1)-1), QED.
According to the German chess rules of 1929 a game of chess was drawn if the same sequence of moves was repeated three times consecutively. Euwe, see the references, proved that this rule could lead to infinite games. For his proof he reinvented the Thue-Morse sequence. - Johannes W. Meijer, Feb 04 2010
"Thue-Morse 0->01 & 1->10, at each stage append the previous with its complement. Start with 0, 1, 2, 3 and write them in binary. Next calculate the sum of the digits (mod 2) - that is divide the sum by 2 and use the remainder." Pickover, The Math Book.
Let s_2(n) be the sum of the base-2 digits of n and epsilon(n) = (-1)^s_2(n), the Thue-Morse sequence, then prod(n >= 0, ((2*n+1)/(2*n+2))^epsilon(n) ) = 1/sqrt(2). - Jonathan Vos Post, Jun 06 2012
Dekking shows that the constant obtained by interpreting this sequence as a binary expansion is transcendental; see also "The Ubiquitous Prouhet-Thue-Morse Sequence". - Charles R Greathouse IV, Jul 23 2013
Drmota, Mauduit, and Rivat proved that the subsequence a(n^2) is normal--see A228039. - Jonathan Sondow, Sep 03 2013
Although the probability of a 0 or 1 is equal, guesses predicated on the latest bit seen produce a correct match 2 out of 3 times. - Bill McEachen, Mar 13 2015
From a(0) to a(2n+1), there are n+1 terms equal to 0 and n+1 terms equal to 1 (see Hassan Tarfaoui link, Concours Général 1990). - Bernard Schott, Jan 21 2022

Examples

			The evolution starting at 0 is:
  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
  .......
A_2 = 0 1 1 0, so B_2 = 1 0 0 1 and A_3 = A_2 B_2 = 0 1 1 0 1 0 0 1.
From _Joerg Arndt_, Mar 12 2013: (Start)
The first steps of the iterated substitution are
Start: 0
Rules:
  0 --> 01
  1 --> 10
-------------
0:   (#=1)
  0
1:   (#=2)
  01
2:   (#=4)
  0110
3:   (#=8)
  01101001
4:   (#=16)
  0110100110010110
5:   (#=32)
  01101001100101101001011001101001
6:   (#=64)
  0110100110010110100101100110100110010110011010010110100110010110
(End)
From _Omar E. Pol_, Oct 28 2013: (Start)
Written as an irregular triangle in which row lengths is A011782, the sequence begins:
  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;
  1,0,0,1,0,1,1,0,0,1,1,0,1,0,0,1,0,1,1,0,1,0,0,1,1,0,0,1,0,1,1,0;
It appears that: row j lists the first A011782(j) terms of A010059, with j >= 0; row sums give A166444 which is also 0 together with A011782; right border gives A000035.
(End)
		

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 15.
  • Jason Bell, Michael Coons, and Eric Rowland, "The Rational-Transcendental Dichotomy of Mahler Functions", Journal of Integer Sequences, Vol. 16 (2013), #13.2.10.
  • J. Berstel and J. Karhumaki, Combinatorics on words - a tutorial, Bull. EATCS, #79 (2003), pp. 178-228.
  • B. Bollobas, The Art of Mathematics: Coffee Time in Memphis, Cambridge, 2006, p. 224.
  • S. Brlek, Enumeration of factors in the Thue-Morse word, Discrete Applied Math., 24 (1989), 83-96. doi:10.1016/0166-218X(92)90274-E.
  • Yann Bugeaud and Guo-Niu Han, A combinatorial proof of the non-vanishing of Hankel determinants of the Thue-Morse sequence, Electronic Journal of Combinatorics 21(3) (2014), #P3.26.
  • Y. Bugeaud and M. Queffélec, On Rational Approximation of the Binary Thue-Morse-Mahler Number, Journal of Integer Sequences, 16 (2013), #13.2.3.
  • Currie, James D. "Non-repetitive words: Ages and essences." Combinatorica 16.1 (1996): 19-40
  • Colin Defant, Anti-Power Prefixes of the Thue-Morse Word, Journal of Combinatorics, 24(1) (2017), #P1.32
  • F. M. Dekking, Transcendance du nombre de Thue-Morse, Comptes Rendus de l'Academie des Sciences de Paris 285 (1977), pp. 157-160.
  • F. M. Dekking, On repetitions of blocks in binary sequences. J. Combinatorial Theory Ser. A 20 (1976), no. 3, pp. 292-299. MR0429728(55 #2739)
  • Dekking, Michel, Michel Mendès France, and Alf van der Poorten. "Folds." The Mathematical Intelligencer, 4.3 (1982): 130-138 & front cover, and 4:4 (1982): 173-181 (printed in two parts).
  • Dubickas, Artūras. On a sequence related to that of Thue-Morse and its applications. Discrete Math. 307 (2007), no. 9-10, 1082--1093. MR2292537 (2008b:11086).
  • Fabien Durand, Julien Leroy, and Gwenaël Richomme, "Do the Properties of an S-adic Representation Determine Factor Complexity?", Journal of Integer Sequences, Vol. 16 (2013), #13.2.6.
  • M. Euwe, Mengentheoretische Betrachtungen Über das Schachspiel, Proceedings Koninklijke Nederlandse Akademie van Wetenschappen, Amsterdam, Vol. 32 (5): 633-642, 1929.
  • S. Ferenczi, Complexity of sequences and dynamical systems, Discrete Math., 206 (1999), 145-154.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 6.8.
  • W. H. Gottschalk and G. A. Hedlund, Topological Dynamics. American Mathematical Society, Colloquium Publications, Vol. 36, Providence, RI, 1955, p. 105.
  • J. Grytczuk, Thue type problems for graphs, points and numbers, Discrete Math., 308 (2008), 4419-4429.
  • A. Hof, O. Knill and B. Simon, Singular continuous spectrum for palindromic Schroedinger operators, Commun. Math. Phys. 174 (1995), 149-159.
  • Mari Huova and Juhani Karhumäki, "On Unavoidability of k-abelian Squares in Pure Morphic Words", Journal of Integer Sequences, Vol. 16 (2013), #13.2.9.
  • B. Kitchens, Review of "Computational Ergodic Theory" by G. H. Choe, Bull. Amer. Math. Soc., 44 (2007), 147-155.
  • Le Breton, Xavier, Linear independence of automatic formal power series. Discrete Math. 306 (2006), no. 15, 1776-1780.
  • M. Lothaire, Combinatorics on Words. Addison-Wesley, Reading, MA, 1983, p. 23.
  • Donald MacMurray, A mathematician gives an hour to chess, Chess Review 6 (No. 10, 1938), 238. [Discusses Marston's 1938 article]
  • Mauduit, Christian. Multiplicative properties of the Thue-Morse sequence. Period. Math. Hungar. 43 (2001), no. 1-2, 137--153. MR1830572 (2002i:11081)
  • C. A. Pickover, Wonders of Numbers, Adventures in Mathematics, Mind and Meaning, Chapter 17, 'The Pipes of Papua,' Oxford University Press, Oxford, England, 2000, pages 34-38.
  • C. A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 60.
  • Clifford A. Pickover, The Math Book, From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics, Sterling Publ., NY, 2009, page 316.
  • Narad Rampersad and Elise Vaslet, "On Highly Repetitive and Power Free Words", Journal of Integer Sequences, Vol. 16 (2013), #13.2.7.
  • G. Richomme, K. Saari, L. Q. Zamboni, Abelian complexity in minimal subshifts, J. London Math. Soc. 83(1) (2011) 79-95.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
  • M. Rigo, P. Salimov, and E. Vandomme, "Some Properties of Abelian Return Words", Journal of Integer Sequences, Vol. 16 (2013), #13.2.5.
  • Benoit Rittaud, Elise Janvresse, Emmanuel Lesigne and Jean-Christophe Novelli, Quand les maths se font discrètes, Le Pommier, 2008 (ISBN 978-2-7465-0370-0).
  • A. Salomaa, Jewels of Formal Language Theory. Computer Science Press, Rockville, MD, 1981, p. 6.
  • Shallit, J. O. "On Infinite Products Associated with Sums of Digits." J. Number Th. 21, 128-134, 1985.
  • Ian Stewart, "Feedback", Mathematical Recreations Column, Scientific American, 274 (No. 3, 1996), page 109 [Historical notes on this sequence]
  • Thomas Stoll, On digital blocks of polynomial values and extractions in the Rudin-Shapiro sequence, RAIRO - Theoretical Informatics and Applications (RAIRO: ITA), EDP Sciences, 2016, 50, pp. 93-99. .
  • A. Thue. Über unendliche Zeichenreihen, Norske Vid. Selsk. Skr. I. Mat. Nat. Kl. Christiania, No. 7 (1906), 1-22.
  • A. Thue, Über die gegenseitige Lage gleicher Teile gewisser Zeichenreihen, Norske Vid. Selsk. Skr. I. Mat. Nat. Kl. Christiania, 1 (1912), 1-67.
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 890.

Crossrefs

Cf. A001285 (for 1, 2 version), A010059 (for 1, 0 version), A106400 (for +1, -1 version), A048707. A010060(n)=A000120(n) mod 2.
Cf. A007413, A080813, A080814, A036581, A108694. See also the Thue (or Roth) constant A014578, also A014571.
Run lengths give A026465. Backward first differences give A029883.
Cf. A004128, A053838, A059448, A171900, A161916, A214212, A005942 (subword complexity), A010693 (Abelian complexity), A225186 (squares), A228039 (a(n^2)), A282317.
Sequences mentioned in the Allouche et al. "Taxonomy" paper, listed by example number: 1: A003849, 2: A010060, 3: A010056, 4: A020985 and A020987, 5: A191818, 6: A316340 and A273129, 18: A316341, 19: A030302, 20: A063438, 21: A316342, 22: A316343, 23: A003849 minus its first term, 24: A316344, 25: A316345 and A316824, 26: A020985 and A020987, 27: A316825, 28: A159689, 29: A049320, 30: A003849, 31: A316826, 32: A316827, 33: A316828, 34: A316344, 35: A043529, 36: A316829, 37: A010060.

Programs

  • Haskell
    a010060 n = a010060_list !! n
    a010060_list =
       0 : interleave (complement a010060_list) (tail a010060_list)
       where complement = map (1 - )
             interleave (x:xs) ys = x : interleave ys xs
    -- Doug McIlroy (doug(AT)cs.dartmouth.edu), Jun 29 2003
    -- Edited by Reinhard Zumkeller, Oct 03 2012
    
  • Maple
    s := proc(k) local i, ans; ans := [ 0,1 ]; for i from 0 to k do ans := [ op(ans),op(map(n->(n+1) mod 2, ans)) ] od; return ans; end; t1 := s(6); A010060 := n->t1[n]; # s(k) gives first 2^(k+2) terms.
    a := proc(k) b := [0]: for n from 1 to k do b := subs({0=[0,1], 1=[1,0]},b) od: b; end; # a(k), after the removal of the brackets, gives the first 2^k terms. # Example: a(3); gives [[[[0, 1], [1, 0]], [[1, 0], [0, 1]]]]
    A010060:=proc(n)
        add(i,i=convert(n, base, 2)) mod 2 ;
    end proc:
    seq(A010060(n),n=0..104); # Emeric Deutsch, Mar 19 2005
    map(`-`,convert(StringTools[ThueMorse](1000),bytes),48); # Robert Israel, Sep 22 2014
  • Mathematica
    Table[ If[ OddQ[ Count[ IntegerDigits[n, 2], 1]], 1, 0], {n, 0, 100}];
    mt = 0; Do[ mt = ToString[mt] <> ToString[(10^(2^n) - 1)/9 - ToExpression[mt] ], {n, 0, 6} ]; Prepend[ RealDigits[ N[ ToExpression[mt], 2^7] ] [ [1] ], 0]
    Mod[ Count[ #, 1 ]& /@Table[ IntegerDigits[ i, 2 ], {i, 0, 2^7 - 1} ], 2 ] (* Harlan J. Brothers, Feb 05 2005 *)
    Nest[ Flatten[ # /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 7] (* Robert G. Wilson v Sep 26 2006 *)
    a[n_] := If[n == 0, 0, If[Mod[n, 2] == 0, a[n/2], 1 - a[(n - 1)/2]]] (* Ben Branman, Oct 22 2010 *)
    a[n_] := Mod[Length[FixedPointList[BitAnd[#, # - 1] &, n]], 2] (* Jan Mangaldan, Jul 23 2015 *)
    Table[2/3 (1 - Cos[Pi/3 (n - Sum[(-1)^Binomial[n, k], {k, 1, n}])]), {n, 0, 100}] (* or, for version 10.2 or higher *) Table[ThueMorse[n], {n, 0, 100}] (* Vladimir Reshetnikov, May 06 2016 *)
    ThueMorse[Range[0, 100]] (* The program uses the ThueMorse function from Mathematica version 11 *) (* Harvey P. Dale, Aug 11 2016 *)
    Nest[Join[#, 1 - #] &, {0}, 7] (* Paolo Xausa, Oct 25 2024 *)
  • PARI
    a(n)=if(n<1,0,sum(k=0,length(binary(n))-1,bittest(n,k))%2)
    
  • PARI
    a(n)=if(n<1,0,subst(Pol(binary(n)), x,1)%2)
    
  • PARI
    default(realprecision, 6100); x=0.0; m=20080; for (n=1, m-1, x=x+x; x=x+sum(k=0, length(binary(n))-1, bittest(n, k))%2); x=2*x/2^m; for (n=0, 20000, d=floor(x); x=(x-d)*2; write("b010060.txt", n, " ", d)); \\ Harry J. Smith, Apr 28 2009
    
  • PARI
    a(n)=hammingweight(n)%2 \\ Charles R Greathouse IV, Mar 22 2013
    
  • Python
    A010060_list = [0]
    for _ in range(14):
        A010060_list += [1-d for d in A010060_list] # Chai Wah Wu, Mar 04 2016
    
  • Python
    def A010060(n): return n.bit_count()&1 # Chai Wah Wu, Mar 01 2023
    
  • R
    maxrow <- 8 # by choice
    b01 <- 1
    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(0,b01))
    # Yosu Yurramendi, Apr 10 2017

Formula

a(2n) = a(n), a(2n+1) = 1 - a(n), a(0) = 0. Also, a(k+2^m) = 1 - a(k) if 0 <= k < 2^m.
If n = Sum b_i*2^i is the binary expansion of n then a(n) = Sum b_i (mod 2).
Let S(0) = 0 and for k >= 1, construct S(k) from S(k-1) by mapping 0 -> 01 and 1 -> 10; sequence is S(infinity).
G.f.: (1/(1 - x) - Product_{k >= 0} (1 - x^(2^k)))/2. - Benoit Cloitre, Apr 23 2003
a(0) = 0, a(n) = (n + a(floor(n/2))) mod 2; also a(0) = 0, a(n) = (n - a(floor(n/2))) mod 2. - Benoit Cloitre, Dec 10 2003
a(n) = -1 + (Sum_{k=0..n} binomial(n,k) mod 2) mod 3 = -1 + A001316(n) mod 3. - Benoit Cloitre, May 09 2004
Let b(1) = 1 and b(n) = b(ceiling(n/2)) - b(floor(n/2)) then a(n-1) = (1/2)*(1 - b(2n-1)). - Benoit Cloitre, Apr 26 2005
a(n) = 1 - A010059(n) = A001285(n) - 1. - Ralf Stephan, Jun 20 2003
a(n) = A001969(n) - 2n. - Franklin T. Adams-Watters, Aug 28 2006
a(n) = A115384(n) - A115384(n-1) for n > 0. - Reinhard Zumkeller, Aug 26 2007
For n >= 0, a(A004760(n+1)) = 1 - a(n). - Vladimir Shevelev, Apr 25 2009
a(A160217(n)) = 1 - a(n). - Vladimir Shevelev, May 05 2009
a(n) == A000069(n) (mod 2). - Robert G. Wilson v, Jan 18 2012
a(n) = A000035(A000120(n)). - Omar E. Pol, Oct 26 2013
a(n) = A000035(A193231(n)). - Antti Karttunen, Dec 27 2013
a(n) + A181155(n-1) = 2n for n >= 1. - Clark Kimberling, Oct 06 2014
G.f. A(x) satisfies: A(x) = x / (1 - x^2) + (1 - x) * A(x^2). - Ilya Gutkovskiy, Jul 29 2021
From Bernard Schott, Jan 21 2022: (Start)
a(n) = a(n*2^k) for k >= 0.
a((2^m-1)^2) = (1-(-1)^m)/2 (see Hassan Tarfaoui link, Concours Général 1990). (End)

A035263 Trajectory of 1 under the morphism 0 -> 11, 1 -> 10; parity of 2-adic valuation of 2n: a(n) = A000035(A001511(n)).

Original entry on oeis.org

1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1
Offset: 1

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Author

Keywords

Comments

First Feigenbaum symbolic (or period-doubling) sequence, corresponding to the accumulation point of the 2^{k} cycles through successive bifurcations.
To construct the sequence: start with 1 and concatenate: 1,1, then change the last term (1->0; 0->1) gives: 1,0. Concatenate those 2 terms: 1,0,1,0, change the last term: 1,0,1,1. Concatenate those 4 terms: 1,0,1,1,1,0,1,1 change the last term: 1,0,1,1,1,0,1,0, etc. - Benoit Cloitre, Dec 17 2002
Let T denote the present sequence. Here is another way to construct T. Start with the sequence S = 1,0,1,,1,0,1,,1,0,1,,1,0,1,,... and fill in the successive holes with the successive terms of the sequence T (from paper by Allouche et al.). - Emeric Deutsch, Jan 08 2003 [Note that if we fill in the holes with the terms of S itself, we get A141260. - N. J. A. Sloane, Jan 14 2009]
From N. J. A. Sloane, Feb 27 2009: (Start)
In more detail: define S to be 1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1,0,1___...
If we fill the holes with S we get A141260:
1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0,
........1.........0.........1.........1.........0.......1.........1.........0...
- the result is
1..0..1.1.1..0..1.0.1..0..1.1.1..0..1.1.1..0..1.0.1.... = A141260.
But instead, if we define T recursively by filling the holes in S with the terms of T itself, we get A035263:
1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0,
........1.........0.........1.........1.........1.......0.........1.........0...
- the result is
1..0..1.1.1..0..1.0.1..0..1.1.1..0..1.1.1..0..1.1.1.0.1.0.1..0..1.1.1..0..1.0.1.. = A035263. (End)
Characteristic function of A003159, i.e., A035263(n)=1 if n is in A003159 and A035263(n)=0 otherwise (from paper by Allouche et al.). - Emeric Deutsch, Jan 15 2003
This is the sequence of R (=1), L (=0) moves in the Towers of Hanoi puzzle: R, L, R, R, R, L, R, L, R, L, R, R, R, ... - Gary W. Adamson, Sep 21 2003
Manfred Schroeder, p. 279 states, "... the kneading sequences for unimodal maps in the binary notation, 0, 1, 0, 1, 1, 1, 0, 1..., are obtained from the Morse-Thue sequence by taking sums mod 2 of adjacent elements." On p. 278, in the chapter "Self-Similarity in the Logistic Parabola", he writes, "Is there a closer connection between the Morse-Thue sequence and the symbolic dynamics of the superstable orbits? There is indeed. To see this, let us replace R by 1 and C and L by 0." - Gary W. Adamson, Sep 21 2003
Partial sums modulo 2 of the sequence 1, a(1), a(1), a(2), a(2), a(3), a(3), a(4), a(4), a(5), a(5), a(6), a(6), ... . - Philippe Deléham, Jan 02 2004
Parity of A007913, A065882 and A065883. - Philippe Deléham, Mar 28 2004
The length of n-th run of 1's in this sequence is A080426(n). - Philippe Deléham, Apr 19 2004
Also parity of A005043, A005773, A026378, A104455, A117641. - Philippe Deléham, Apr 28 2007
Equals parity of the Towers of Hanoi, or ruler sequence (A001511), where the Towers of Hanoi sequence (1, 2, 1, 3, 1, 2, 1, 4, ...) denotes the disc moved, labeled (1, 2, 3, ...) starting from the top; and the parity of (1, 2, 1, 3, ...) denotes the direction of the move, CW or CCW. The frequency of CW moves converges to 2/3. - Gary W. Adamson, May 11 2007
A conjectured identity relating to the partition sequence, A000041: p(x) = A(x) * A(x^2) when A(x) = the Euler transform of A035263 = polcoeff A174065: (1 + x + x^2 + 2x^3 + 3x^4 + 4x^5 + ...). - Gary W. Adamson, Mar 21 2010
a(n) is 1 if the number of trailing zeros in the binary representation of n is even. - Ralf Stephan, Aug 22 2013
From Gary W. Adamson, Mar 25 2015: (Start)
A conjectured identity relating to the partition sequence, A000041 as polcoeff p(x); A003159, and its characteristic function A035263: (1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, ...); and A036554 indicating n-th terms with zeros in A035263: (2, 6, 8, 10, 14, 18, 22, ...).
The conjecture states that p(x) = A(x) = A(x^2) when A(x) = polcoeffA174065 = the Euler transform of A035263 = 1/(1-x)*(1-x^3)*(1-x^4)*(1-x^5)*... = (1 + x + x^2 + 2x^3 + 3x^4 + 4x^5 + ...) and the aerated variant = the Euler transform of the complement of A035263: 1/(1-x^2)*(1-x^6)*(1-x^8)*... = (1 + x^2 + x^4 + 2x^6 + 3x^8 + 4x^10 + ...).
(End)
The conjecture above was proved by Jean-Paul Allouche on Dec 21 2013.
Regarded as a column vector, this sequence is the product of A047999 (Sierpinski's gasket) regarded as an infinite lower triangular matrix and A036497 (the Fredholm-Rueppel sequence) where the 1's have alternating signs, 1, -1, 0, 1, 0, 0, 0, -1, .... - Gary W. Adamson, Jun 02 2021
The numbers of 1's through n (A050292) can be determined by starting with the binary (say for 19 = 1 0 0 1 1) and writing: next term is twice current term if 0, otherwise twice plus 1. The result is 1, 2, 4, 9, 19. Take the difference row, = 1, 1, 2, 5, 10; and add the odd-indexed terms from the right: 5, 4, 3, 2, 1 = 10 + 2 + 1 = 13. The algorithm is the basis for determining the disc configurations in the tower of Hanoi game, as shown in the Jul 24 2021 comment of A060572. - Gary W. Adamson, Jul 28 2021

References

  • Karamanos, Kostas. "From symbolic dynamics to a digital approach." International Journal of Bifurcation and Chaos 11.06 (2001): 1683-1694. (Full version. See p. 1685)
  • Karamanos, K. (2000). From symbolic dynamics to a digital approach: chaos and transcendence. In Michel Planat (Ed.), Noise, Oscillators and Algebraic Randomness (Lecture Notes in Physics, pp. 357-371). Springer, Berlin, Heidelberg. (Short version. See p. 359)
  • Manfred R. Schroeder, "Fractals, Chaos, Power Laws", W. H. Freeman, 1991
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 892, column 2, Note on p. 84, part (a).

Crossrefs

Parity of A001511. Anti-parity of A007814.
Absolute values of first differences of A010060. Apart from signs, same as A029883. Essentially the same as A056832.
Swapping 0 and 1 gives A096268.
Cf. A033485, A050292 (partial sums), A089608, A088172, A019300, A039982, A073675, A121701, A141260, A000041, A174065, A220466, A154269 (Mobius transform).
Limit of A317957(n) for large n.

Programs

  • Haskell
    import Data.Bits (xor)
    a035263 n = a035263_list !! (n-1)
    a035263_list = zipWith xor a010060_list $ tail a010060_list
    -- Reinhard Zumkeller, Mar 01 2012
    
  • Maple
    nmax:=105: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 1 to ceil(nmax/(p+2)) do a((2*n-1)*2^p) := (p+1) mod 2 od: od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Feb 07 2013
    A035263 := n -> 1 - padic[ordp](n, 2) mod 2:
    seq(A035263(n), n=1..105); # Peter Luschny, Oct 02 2018
  • Mathematica
    a[n_] := a[n] = If[ EvenQ[n], 1 - a[n/2], 1]; Table[ a[n], {n, 1, 105}] (* Or *)
    Rest[ CoefficientList[ Series[ Sum[ x^(2^k)/(1 + (-1)^k*x^(2^k)), {k, 0, 20}], {x, 0, 105}], x]]
    f[1] := True; f[x_] := Xor[f[x - 1], f[Floor[x/2]]]; a[x_] := Boole[f[x]] (* Ben Branman, Oct 04 2010 *)
    a[n_] := If[n == 0, 0, 1 - Mod[ IntegerExponent[n, 2], 2]]; (* Jean-François Alcover, Jul 19 2013, after Michael Somos *)
    Nest[ Flatten[# /. {0 -> {1, 1}, 1 -> {1, 0}}] &, {0}, 7] (* Robert G. Wilson v, Jul 23 2014 *)
    SubstitutionSystem[{0->{1,1},1->{1,0}},1,{7}][[1]] (* Harvey P. Dale, Jun 06 2022 *)
  • PARI
    {a(n) = if( n==0, 0, 1 - valuation(n, 2)%2)}; /* Michael Somos, Sep 04 2006 */
    
  • PARI
    {a(n) = if( n==0, 0, n = abs(n); subst( Pol(binary(n)) - Pol(binary(n-1)), x, 1)%2)}; /* Michael Somos, Sep 04 2006 */
    
  • PARI
    {a(n) = if( n==0, 0, n = abs(n); direuler(p=2, n, 1 / (1 - X^((p<3) + 1)))[n])}; /* Michael Somos, Sep 04 2006 */
    
  • Python
    def A035263(n): return (n&-n).bit_length()&1 # Chai Wah Wu, Jan 09 2023
  • Scheme
    (define (A035263 n) (let loop ((n n) (i 1)) (cond ((odd? n) (modulo i 2)) (else (loop (/ n 2) (+ 1 i)))))) ;; (Use mod instead of modulo in R6RS) Antti Karttunen, Sep 11 2017
    

Formula

Absolute values of first differences (A029883) of Thue-Morse sequence (A001285 or A010060). Self-similar under 10->1 and 11->0.
Series expansion: (1/x) * Sum_{i>=0} (-1)^(i+1)*x^(2^i)/(x^(2^i)-1). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 17 2003
a(n) = Sum_{k>=0} (-1)^k*(floor((n+1)/2^k)-floor(n/2^k)). - Benoit Cloitre, Jun 03 2003
Another g.f.: Sum_{k>=0} x^(2^k)/(1+(-1)^k*x^(2^k)). - Ralf Stephan, Jun 13 2003
a(2*n) = 1-a(n), a(2*n+1) = 1. - Ralf Stephan, Jun 13 2003
a(n) = parity of A033485(n). - Philippe Deléham, Aug 13 2003
Equals A088172 mod 2, where A088172 = 1, 2, 3, 7, 13, 26, 53, 106, 211, 422, 845, ... (first differences of A019300). - Gary W. Adamson, Sep 21 2003
a(n) = a(n-1) - (-1)^n*a(floor(n/2)). - Benoit Cloitre, Dec 02 2003
a(1) = 1 and a(n) = abs(a(n-1) - a(floor(n/2))). - Benoit Cloitre, Dec 02 2003
a(n) = 1 - A096268(n+1); A050292 gives partial sums. - Reinhard Zumkeller, Aug 16 2006
Multiplicative with a(2^k) = 1 - (k mod 2), a(p^k) = 1, p > 2. Dirichlet g.f.: Product_{n = 4 or an odd prime} (1/(1-1/n^s)). - Christian G. Bower, May 18 2005
a(-n) = a(n). a(0)=0. - Michael Somos, Sep 04 2006
Dirichlet g.f.: zeta(s)*2^s/(2^s+1). - Ralf Stephan, Jun 17 2007
a(n+1) = a(n) XOR a(ceiling(n/2)), a(1) = 1. - Reinhard Zumkeller, Jun 11 2009
Let D(x) be the generating function, then D(x) + D(x^2) == x/(1-x). - Joerg Arndt, May 11 2010
a(n) = A010060(n) XOR A010060(n+1); a(A079523(n)) = 0; a(A121539(n)) = 1. - Reinhard Zumkeller, Mar 01 2012
a((2*n-1)*2^p) = (p+1) mod 2, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 07 2013
a(n) = A000035(A001511(n)). - Omar E. Pol, Oct 29 2013
a(n) = 2-A056832(n) = (5-A089608(n))/4. - Antti Karttunen, Sep 11 2017, after Benoit Cloitre
For n >= 0, a(n+1) = M(2n) mod 2 where M(n) is the Motzkin number A001006 (see Deutsch and Sagan 2006 link). - David Callan, Oct 02 2018
a(n) = A038712(n) mod 3. - Kevin Ryde, Jul 11 2019
Given any n in the form (k * 2^m, k odd), extract k and m. Categorize the results into two outcomes of (k, m, even or odd). If (k, m) is (odd, even) substitute 1. If (odd, odd), denote the result 0. Example: 5 = (5 * 2^0), (odd, even, = 1). (6 = 3 * 2^1), (odd, odd, = 0). - Gary W. Adamson, Jun 23 2021

Extensions

Alternative description added to the name by Antti Karttunen, Sep 11 2017

A007413 A squarefree (or Thue-Morse) ternary sequence: closed under 1->123, 2->13, 3->2. Start with 1.

Original entry on oeis.org

1, 2, 3, 1, 3, 2, 1, 2, 3, 2, 1, 3, 1, 2, 3, 1, 3, 2, 1, 3, 1, 2, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1, 2, 3, 1, 3, 2, 1, 3, 1, 2, 3, 1, 3, 2, 1, 2, 3, 2, 1, 3, 1, 2, 3, 1, 3, 2, 1, 3, 1, 2, 3, 2, 1, 2, 3, 1, 3, 2, 1, 3, 1, 2, 3, 1, 3, 2, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3
Offset: 1

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Author

Keywords

Comments

a(n)=2 if and only if n-1 is in A079523. - Benoit Cloitre, Mar 10 2003
Partial sums modulo 4 of the sequence 1, a(1), a(1), a(2), a(2), a(3), a(3), a(4), a(4), a(5), a(5), a(6), a(6), ... - Philippe Deléham, Mar 04 2004
To construct the sequence: start with 1 and concatenate 4 -1 = 3: 1, 3, then change the last term (2 -> 1, 3 ->2 ) gives 1, 2. Concatenate 1, 2 with 4 -1 = 3, 4 - 2 = 2: 1, 2, 3, 2 and change the last term: 1, 2, 3, 1. Concatenate 1, 2, 3, 1 with 4 - 1 = 3, 4 - 2 = 2, 4 - 3 = 1, 4 - 1 = 3: 1, 2, 3, 1, 3, 2, 1, 3 and change the last term: 1, 2, 3, 1, 3, 2, 1, 2 etc. - Philippe Deléham, Mar 04 2004
To construct the sequence: start with the Thue-Morse sequence A010060 = 0, 1, 1, 0, 1, 0, 0, 1, ... Then change 0 -> 1, 2, 3, and 1 -> 3, 2, 1, gives: 1, 2, 3, , 3, 2, 1, ,3, 2, 1, , 1, 2, 3, , 3, 2, 1, , ... and fill in the successive holes with the successive terms of the sequence itself. - _Philippe Deléham, Mar 04 2004
To construct the sequence: to insert the number 2 between the A003156(k)-th term and the (1 + A003156(k))-th term of the sequence 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, ... - Philippe Deléham, Mar 04 2004
Conjecture. The sequence is formed by the numbers of 1's between every pair of consecutive 2's in A076826. - Vladimir Shevelev, May 31 2009

Examples

			Here are the first 5 stages in the construction of this sequence, together with Mma code, taken from Keranen's article. His alphabet is a,b,c rather than 1,2,3.
productions = {"a" -> "abc ", "b" -> "ac ", "c" -> "b ", " " -> ""};
NestList[g, "a", 5] // TableForm
a
abc
abc ac b
abc ac b abc b ac
abc ac b abc b ac abc ac b ac abc b
abc ac b abc b ac abc ac b ac abc b abc ac b abc b ac abc b abc ac b ac
		

References

  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 18.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • A. Thue, Über unendliche Zeichenreihen, Norske Vid. Selsk. Skr. I. Mat. Nat. Kl. Christiania, No. 7 (1906), 1-22.

Crossrefs

First differences of A000069.
Equals A036580(n-1) + 1.

Programs

  • Mathematica
    Nest[ Flatten[ # /. {1 -> {1, 2, 3}, 2 -> {1, 3}, 3 -> {2}}] &, {1}, 7] (* Robert G. Wilson v, May 07 2005 *)
    2 - Differences[ThueMorse[Range[0, 100]]] (* Paolo Xausa, Oct 25 2024 *)
  • PARI
    {a(n) = if( n<1 || valuation(n, 2)%2, 2, 2 + (-1)^subst( Pol(binary(n)), x,1))};
    
  • Python
    def A007413(n): return 2-(n.bit_count()&1)+((n-1).bit_count()&1) # Chai Wah Wu, Mar 03 2023

Formula

a(n) modulo 2 = A035263(n). a(A036554(n)) = 2. a(A003159(n)) = 1 if n odd. a(A003159(n)) = 3 if n even. a(n) = A033485(n) mod 4. a(n) = 4 - A036585(n-1). - Philippe Deléham, Mar 04 2004
a(n) = 2 - A029883(n) = 3 - A036577(n). - Philippe Deléham, Mar 20 2004
For n>=1, we have: 1) a(A108269(n))=A010684(n-1); 2) a(A079523(n))=A010684(n-1); 3) a(A081706(2n))=A010684(n). - Vladimir Shevelev, Jun 22 2009

A036577 Ternary Thue-Morse sequence: closed under a->abc, b->ac, c->b.

Original entry on oeis.org

2, 1, 0, 2, 0, 1, 2, 1, 0, 1, 2, 0, 2, 1, 0, 2, 0, 1, 2, 0, 2, 1, 0, 1, 2, 1, 0, 2, 0, 1, 2, 1, 0, 1, 2, 0, 2, 1, 0, 1, 2, 1, 0, 2, 0, 1, 2, 0, 2, 1, 0, 2, 0, 1, 2, 1, 0, 1, 2, 0, 2, 1, 0, 2, 0, 1, 2, 0, 2, 1, 0, 1, 2, 1, 0, 2, 0, 1, 2, 0, 2, 1, 0, 2, 0, 1, 2, 1, 0, 1, 2, 0, 2, 1, 0, 1, 2, 1, 0, 2, 0, 1, 2, 1, 0
Offset: 1

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Author

Keywords

Comments

Number of 1's between successive 0's in A010060.
The infinite sequence is abcacbabcbac... which is encoded with a->2, b->1, c->0 to produce this integer sequence.
From Jeffrey Shallit, Dec 07 2019: (Start)
This word is sometimes called 'vtm'; see, for example, see the Blanchet-Sadri et al. reference.
It is a squarefree word containing no instances of the factor 010 or 212 (or cbc, aba in the encoding).
Berstel proved many different definitions (e.g., Braunholtz, Istrail) of the word are equivalent. (End)

Examples

			2*x + x^2 + 2*x^4 + x^6 + 2*x^7 + x^8 + x^10 + 2*x^11 + 2*x^13 + x^14 + ...
		

References

  • M. Lothaire, Combinatorics on Words. Addison-Wesley, Reading, MA, 1983, p. 26.

Crossrefs

See A007413, A036580 for other versions.

Programs

  • Mathematica
    (* ThueMorse is built-in since version 10.2, for lower versions it needs to be defined manually *) ThueMorse[n_] := Mod[DigitCount[n, 2, 1], 2]; Table[1 + ThueMorse[n] - ThueMorse[n-1], {n, 1, 100}]  (* Vladimir Reshetnikov, May 17 2016 *)
    Nest[Flatten[# /. {2 -> {2, 1, 0}, 1 -> {2, 0}, 0 -> {1}}] &, {2, 1, 0}, 7] (* Robert G. Wilson v, Jul 30 2018 *)
    Differences[ThueMorse[Range[0, 100]]] + 1 (* Paolo Xausa, Jul 17 2025 *)
  • PARI
    {a(n) = if( n<1, 0, if( valuation( n, 2)%2, 1, 1 - (-1)^subst( Pol( binary(n)), x, 1)))} /* Michael Somos, Aug 03 2011 */
    
  • Python
    def A036577(n): return (n.bit_count()&1)+((n-1).bit_count()&1^1) # Chai Wah Wu, Mar 03 2023

Formula

a(n) = A036585(n) - 1 = A029883(n) + 1.
a(n) = 3 - A007413(n). a(A036554(n)) = 1; a(A091785(n)) = 0; a(A091855(n)) = 2. - Philippe Deléham, Mar 20 2004
a(4*n + 2) = 1. a(2*n + 1) = 2 * A010059(n). a(4*n + 3) = 2 * A010060(n). - Michael Somos, Aug 03 2011
a(n) = A010060(2*n - 1) + A010060(2*n) = A115384(2*n) - A115384(2*n - 2). - Zhuorui He, Jul 11 2025

A091855 Odious numbers (see A000069) in A003159.

Original entry on oeis.org

1, 4, 7, 11, 13, 16, 19, 21, 25, 28, 31, 35, 37, 41, 44, 47, 49, 52, 55, 59, 61, 64, 67, 69, 73, 76, 79, 81, 84, 87, 91, 93, 97, 100, 103, 107, 109, 112, 115, 117, 121, 124, 127, 131, 133, 137, 140, 143, 145, 148, 151, 155, 157, 161, 164, 167, 171, 173, 176, 179, 181
Offset: 1

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Author

Philippe Deléham, Mar 16 2004

Keywords

Comments

Also n such that A033485(n) == 1 (mod 4); see A007413.
Also n such that A029883(n-1) = 1, A036577(n-1) = 2, A036585(n-1) = 3. - Philippe Deléham, Mar 25 2004
The number of odd numbers before the n-th even number in this sequence is a(n). - Philippe Deléham, Mar 27 2004
Numbers n such that {A010060(n-1), A010060(n)}={0,1} where A010060 is the Thue-Morse sequence. - Benoit Cloitre, Jun 16 2006
Positive integers not of the form n+A010060(n). - Jeffrey Shallit, Feb 13 2024

Programs

Formula

a(n) = A003159(2n-1) = A036554(2n-1)/2.
a(n) is asymptotic to 3*n - Benoit Cloitre, Mar 22 2004
A050292(a(n)) = 2n - 1. - Philippe Deléham, Mar 26 2004

Extensions

More terms from Benoit Cloitre, Mar 22 2004

A003156 A self-generating sequence (see Comments for definition).

Original entry on oeis.org

1, 4, 5, 6, 9, 12, 15, 16, 17, 20, 21, 22, 25, 26, 27, 30, 33, 36, 37, 38, 41, 44, 47, 48, 49, 52, 55, 58, 59, 60, 63, 64, 65, 68, 69, 70, 73, 76, 79, 80, 81, 84, 85, 86, 89, 90, 91, 94, 97, 100, 101, 102, 105, 106, 107, 110, 111, 112, 115, 118, 121, 122, 123, 126, 129, 132
Offset: 1

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Comments

From N. J. A. Sloane, Dec 26 2020: (Start)
The best definitions of the triple [this sequence, A003157, A003158] are as the rows a(n), b(n), c(n) of the table:
n: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ...
a: 1, 4, 5, 6, 9, 12, 15, 16, 17, 20, 21, 22, ...
b: 3, 8, 11, 14, 19, 24, 29, 32, 35, 40, 43, 46, ...
c: 2, 7, 10, 13, 18, 23, 28, 31, 34, 39, 42, 45, ...
where a(1)=1, b(1)=3, c(1)=2, and thereafter
a(n) = mex{a(i), b(i), c(i), i
b(n) = a(n) + 2*n,
c(n) = b(n) - 1.
Then a,b,c form a partition of the positive integers.
Note that there is another triple of sequences (A003144, A003145, A003146) also called a, b, c and also a partition of the positive integers, in a different paper by the same authors (Carlitz-Scovelle-Hoggatt) in the same volume of the same journal.
(End)
a(n) is the number of ones before the n-th zero in the Feigenbaum sequence A035263. - Philippe Deléham, Mar 27 2004
Number of odd numbers before the n-th even number in A007413, A007913, A001511, A029883, A033485, A035263, A036585, A065882, A065883, A088172, A092412. - Philippe Deléham, Apr 03 2004
Indices of a in the sequence closed under a -> abc, b -> a, c -> a, starting with a(1) = a; see A092606 where a = 0, b = 2, c = 1. - Philippe Deléham, Apr 12 2004

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Programs

  • Haskell
    following Deléham
    a003156 n = a003156_list !! (n-1)
    a003156_list = scanl1 (+) a080426_list
    -- Reinhard Zumkeller, Oct 27 2014
    
  • Maple
    a:= proc(n) global l; while nops(l) [1, 3$d, 1][], l) od; `if` (n=1, 1, a(n-1) +l[n]) end: l:= [1]: seq (a(n), n=1..80); # Alois P. Heinz, Oct 31 2009
  • Mathematica
    Position[Nest[Flatten[# /. {0 -> {0, 2, 1}, 1 -> {0}, 2 -> {0}}]&, {0}, 7], 0] // Flatten (* Jean-François Alcover, Mar 14 2014 *)
  • Python
    def A003156(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x):
            c, s = n+x, bin(x)[2:]
            l = len(s)
            for i in range(l&1,l,2):
                c -= int(s[i])+int('0'+s[:i],2)
            return c
        return bisection(f,n,n)-n # Chai Wah Wu, Jan 29 2025

Formula

a(n) = A079523(n) - n + 1 = A003157(n) - 2n = A003158(n) - 2n + 1. - Philippe Deléham, Feb 28 2004
a(n) = A036554(n) - n = A072939(n) - n - 1 = 2*A003159(n) - n. - Philippe Deléham, Apr 10 2004
a(n) = Sum_{k = 1..n} A080426(k). - Philippe Deléham, Apr 16 2004

Extensions

More terms from Alois P. Heinz, Oct 31 2009
Incorrect equation removed from formula by Peter Munn, Dec 11 2020

A036585 Ternary Thue-Morse sequence: closed under a->abc, b->ac, c->b.

Original entry on oeis.org

3, 2, 1, 3, 1, 2, 3, 2, 1, 2, 3, 1, 3, 2, 1, 3, 1, 2, 3, 1, 3, 2, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 2, 1, 3, 1, 2, 3, 1, 3, 2, 1, 3, 1, 2, 3, 2, 1, 2, 3, 1, 3, 2, 1, 3, 1, 2, 3, 1, 3, 2, 1, 2, 3, 2, 1, 3, 1, 2, 3, 1, 3, 2, 1, 3, 1, 2, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1
Offset: 1

Keywords

Comments

First differences of A001969. Observed by Franklin T. Adams-Watters, proved by Max Alekseyev, Aug 30 2006

References

  • M. Lothaire, Combinatorics on Words. Addison-Wesley, Reading, MA, 1983, p. 26.

Crossrefs

Programs

  • Haskell
    a036585 n = a036585_list !! (n-1)
    a036585_list = 3 : concat (map f a036585_list)
    where f 1 = [1,2,3]; f 2 = [1,3]; f 3 = [2]
    -- Reinhard Zumkeller, Oct 31 2012
    
  • Mathematica
    Differences[ThueMorse[Range[0, 100]]] + 2 (* Paolo Xausa, Oct 25 2024 *)
  • PARI
    a(n)=if(n<1 || valuation(n,2)%2,2,2-(-1)^subst(Pol(binary(n)),x,1))
    
  • Python
    def A036585(n): return 2+(n.bit_count()&1)-((n-1).bit_count()&1) # Chai Wah Wu, Mar 03 2023

Formula

a(n) = A001969(n+1) - A001969(n). - Franklin T. Adams-Watters, Aug 28 2006
a(n) = A029883(n) + 2 = A036577(n) + 1.

A091785 Evil numbers (see A001969) in A003159.

Original entry on oeis.org

3, 5, 9, 12, 15, 17, 20, 23, 27, 29, 33, 36, 39, 43, 45, 48, 51, 53, 57, 60, 63, 65, 68, 71, 75, 77, 80, 83, 85, 89, 92, 95, 99, 101, 105, 108, 111, 113, 116, 119, 123, 125, 129, 132, 135, 139, 141, 144, 147, 149, 153, 156, 159, 163, 165, 169, 172, 175, 177, 180, 183
Offset: 1

Author

Philippe Deléham, Mar 16 2004

Keywords

Comments

Also n such that A033485(n) == 3 (mod 4); see A007413.
Also n such that A029883(n-1) = -1, A036577(n-1) = 0, A036585(n-1) = 1. - Philippe Deléham, Mar 25 2004
The number of odd numbers before the n-th even number in this sequence is a(n). - Philippe Deléham, Mar 27 2004
Numbers n such that {A010060(n-1), A010060(n)}={1,0} where A010060 is the Thue-Morse sequence. - Benoit Cloitre, Jun 16 2006

Programs

Formula

a(n) = A003159(2*n) = A036554(2*n)/2.
a(n) is asymptotic to 3*n. - Benoit Cloitre, Mar 22 2004
A050292(a(n)) = 2n. - Philippe Deléham, Mar 26 2004

Extensions

More terms from Benoit Cloitre, Mar 22 2004

A285951 Positions of 1's in A285949; complement of A285950.

Original entry on oeis.org

2, 6, 9, 11, 15, 17, 20, 24, 27, 29, 32, 36, 38, 42, 45, 47, 51, 53, 56, 60, 62, 66, 69, 71, 74, 78, 81, 83, 87, 89, 92, 96, 99, 101, 104, 108, 110, 114, 117, 119, 122, 126, 129, 131, 135, 137, 140, 144, 146, 150, 153, 155, 159, 161, 164, 168, 171, 173, 176
Offset: 1

Author

Clark Kimberling, May 02 2017

Keywords

Comments

Conjecture: 3n - a(n) is in {0, 1} for all n >= 1.
From Michel Dekking, Sep 03 2019: (Start)
Proof of the conjecture by Kimberling: more is true. Here follows a proof of the formula below. Let T be the transform T(01)=0, T(1)=0.
Consider the return word structure of A285949 for the word 1:
A285949 = 0|1000|100|10|1000|10|100| ....
[See Justin & Vuillon (2000) for definition of return word. - N. J. A. Sloane, Sep 23 2019]
The three return words are u:=10, v:=100 and w:=1000. These words uniquely correspond to the conjugated three words u'=01, v'=010, w'=0100 in A285949, which are the unique images u'=T(0), v'=T(01) and w'=T(011) of the words 0, 01 and 011 in the Thue-Morse word A010060. The images of these three words under the Thue-Morse morphism 0->01, 1->10 are 01, 0110 and 011010, and we have
T(01)=010, T(0110)=010001, T(011010)=010001001.
Shifting by 1 in A285949, these correspond uniquely to the conjugated words 100, 100010, and 100010010. It follows that the Thue-Morse morphism induces the morphism u->v, v->wu, w->wvu on the return words.
This morphism is modulo a change of alphabet equal to the ternary Thue-Morse morphism with fixed point A007413.
Note that on the alphabet {4,3,2} of the respective lengths of w, v, and u we obtain the sequence (a(n+1)-a(n)) = 4,3,2,4,2,3,4,3,2,... of first differences of the positions of the 1's in A285949.
To prove the formula a(n) = A010060(n)+ 3n-1, it suffices to show that a(n+1)-a(n) = A010060(n+1)-A010060(n)+3.
That this indeed is true: see the Comments of A029883, the first differences of the standard form of the Thue-Morse sequence A001285.
(End)

Examples

			As a word, A285949 = 0100010010100010100100010..., in which 1 is in positions  2,6,9,11,...
		

Crossrefs

Programs

  • Mathematica
    s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 7]  (* Thue-Morse, A010060 *)
    w = StringJoin[Map[ToString, s]]
    w1 = StringReplace[w, {"0" -> "01", "1" -> "0"}]  (* A284949, word *)
    st = ToCharacterCode[w1] - 48 (* A284949, sequence *)
    Flatten[Position[st, 0]] (* A285950 *)
    Flatten[Position[st, 1]] (* A285951 *)
  • Python
    def A285951(n): return ((n-1).bit_count()&1)+3*n-1 # Chai Wah Wu, May 21 2025

Formula

a(n) = A010060(n) + 3n-1. - Michel Dekking, Sep 03 2019

A088172 First differences of A019300.

Original entry on oeis.org

1, 2, 3, 7, 13, 26, 53, 106, 211, 422, 845, 1689, 3379, 6758, 13515, 27031, 54061, 108122, 216245, 432489, 864979, 1729958, 3459915, 6919830, 13839661, 27679322, 55358643, 110717287, 221434573, 442869146, 885738293, 1771476586, 3542953171
Offset: 1

Author

Gary W. Adamson, Sep 21 2003

Keywords

Comments

Partial sums give A019300, while partial sums mod 2 = the Thue-Morse sequence, A010060. a(n) mod 2 = A035263: 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, ...

Crossrefs

Programs

  • Mathematica
    tm = Table[ If[ OddQ[ Count[ IntegerDigits[n, 2], 1]], 1, 0], {n, 0, 35}]; Table[ FromDigits[ Take[ tm, n + 1], 2] - FromDigits[ Take[ tm, n], 2], {n, 1, 33}] (* Robert G. Wilson v *)

Formula

a(n+1) = 2*a(n) + 0 or +-1. - Robert G. Wilson v, Sep 27 2003
a(n+1) = 2*a(n) + A029883(n). - Philippe Deléham, Mar 21 2004

Extensions

More terms from Robert G. Wilson v, Sep 27 2003
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