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

N. J. A. Sloane

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.

Links

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.
Cf. also A001969, A035263, A005187, A115384, A132680, A141803, A104248, A193231.
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)

A000201 Lower Wythoff sequence (a Beatty sequence): a(n) = floor(n*phi), where phi = (1+sqrt(5))/2 = A001622.

Original entry on oeis.org

1, 3, 4, 6, 8, 9, 11, 12, 14, 16, 17, 19, 21, 22, 24, 25, 27, 29, 30, 32, 33, 35, 37, 38, 40, 42, 43, 45, 46, 48, 50, 51, 53, 55, 56, 58, 59, 61, 63, 64, 66, 67, 69, 71, 72, 74, 76, 77, 79, 80, 82, 84, 85, 87, 88, 90, 92, 93, 95, 97, 98, 100, 101, 103, 105, 106, 108, 110
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

This is the unique sequence a satisfying a'(n)=a(a(n))+1 for all n in the set N of natural numbers, where a' denotes the ordered complement (in N) of a. - Clark Kimberling, Feb 17 2003
This sequence and A001950 may be defined as follows. Consider the maps a -> ab, b -> a, starting from a(1) = a; then A000201 gives the indices of a, A001950 gives the indices of b. The sequence of letters in the infinite word begins a, b, a, a, b, a, b, a, a, b, a, ... Setting a = 0, b = 1 gives A003849 (offset 0); setting a = 1, b = 0 gives A005614 (offset 0). - Philippe Deléham, Feb 20 2004
These are the numbers whose lazy Fibonacci representation (see A095791) includes 1; the complementary sequence (the upper Wythoff sequence, A001950) are the numbers whose lazy Fibonacci representation includes 2 but not 1.
a(n) is the unique monotonic sequence satisfying a(1)=1 and the condition "if n is in the sequence then n+(rank of n) is not in the sequence" (e.g. a(4)=6 so 6+4=10 and 10 is not in the sequence) - Benoit Cloitre, Mar 31 2006
Write A for A000201 and B for A001950 (the upper Wythoff sequence, complement of A). Then the composite sequences AA, AB, BA, BB, AAA, AAB,...,BBB,... appear in many complementary equations having solution A000201 (or equivalently, A001950). Typical complementary equations: AB=A+B (=A003623), BB=A+2B (=A101864), BBB=3A+5B (=A134864). - Clark Kimberling, Nov 14 2007
Cumulative sum of A001468 terms. - Eric Angelini, Aug 19 2008
The lower Wythoff sequence also can be constructed by playing the so-called Mancala-game: n piles of total d(n) chips are standing in a row. The piles are numbered from left to right by 1, 2, 3, ... . The number of chips in a pile at the beginning of the game is equal to the number of the pile. One step of the game is described as follows: Distribute the pile on the very left one by one to the piles right of it. If chips are remaining, build piles out of one chip subsequently to the right. After f(n) steps the game ends in a constant row of piles. The lower Wythoff sequence is also given by n -> f(n). - Roland Schroeder (florola(AT)gmx.de), Jun 19 2010
With the exception of the first term, a(n) gives the number of iterations required to reverse the list {1,2,3,...,n} when using the mapping defined as follows: remove the first term of the list, z(1), and add 1 to each of the next z(1) terms (appending 1's if necessary) to get a new list. See A183110 where this mapping is used and other references given. This appears to be essentially the Mancala-type game interpretation given by R. Schroeder above. - John W. Layman, Feb 03 2011
Also row numbers of A213676 starting with an even number of zeros. - Reinhard Zumkeller, Mar 10 2013
From Jianing Song, Aug 18 2022: (Start)
Numbers k such that {k*phi} > phi^(-2), where {} denotes the fractional part.
Proof: Write m = floor(k*phi).
If {k*phi} > phi^(-2), take s = m-k+1. From m < k*phi < m+1 we have k < (m-k+1)*phi < k + phi, so floor(s*phi) = k or k+1. If floor(s*phi) = k+1, then (see A003622) floor((k+1)*phi) = floor(floor(s*phi)*phi) = floor(s*phi^2)-1 = s+floor(s*phi)-1 = m+1, but actually we have (k+1)*phi > m+phi+phi^(-2) = m+2, a contradiction. Hence floor(s*phi) = k.
If floor(s*phi) = k, suppose otherwise that k*phi - m <= phi^(-2), then m < (k+1)*phi <= m+2, so floor((k+1)*phi) = m+1. Suppose that A035513(p,q) = k for p,q >= 1, then A035513(p,q+1) = floor((k+1)*phi) - 1 = m = A035513(s,1). But it is impossible for one number (m) to occur twice in A035513. (End)
The formula from Jianing Song above is a direct consequence of an old result by Carlitz et al. (1972). Their Theorem 11 states that (a(n)) consists of the numbers k such that {k*phi^(-2)} < phi^(-1). One has {k*phi^(-2)} = {k*(2-phi)} = {-k*phi}. Using that 1-phi^(-1) = phi^(-2), the Jianing Song formula follows. - Michel Dekking, Oct 14 2023
In the Fokkink-Joshi paper, this sequence is the Cloitre (1,1,2,1)-hiccup sequence, i.e., a(1) = 1; for m < n, a(n) = a(n-1)+2 if a(m) = n-1, else a(n) = a(n-1)+1. - Michael De Vlieger, Jul 28 2025

Examples

			From Roland Schroeder (florola(AT)gmx.de), Jul 13 2010: (Start)
Example for n = 5; a(5) = 8;
(Start: [1,2,3,4,5]; 8 steps until [5,4,3,2,1]):
[1,2,3,4,5]; [3,3,4,5]; [4,5,6]; [6,7,1,1]; [8,2,2,1,1,1]: [3,3,2,2,2,1,1,1]; [4,3,3,2,1,1,1]; [4,4,3,2,1,1]; [5,4,3,2,1]. (End)

References

  • Eric Friedman, Scott M. Garrabrant, Ilona K. Phipps-Morgan, A. S. Landsberg and Urban Larsson, Geometric analysis of a generalized Wythoff game, in Games of no Chance 5, MSRI publ. Cambridge University Press, date?
  • M. Gardner, Penrose Tiles to Trapdoor Ciphers, W. H. Freeman, 1989; see p. 107.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • I. M. Yaglom, Two games with matchsticks, pp. 1-7 of Qvant Selecta: Combinatorics I, Amer Math. Soc., 2001.

Links

Crossrefs

a(n) = least k such that s(k) = n, where s = A026242. Complement of A001950. See also A058066.
The permutation A002251 maps between this sequence and A001950, in that A002251(a(n)) = A001950(n), A002251(A001950(n)) = a(n).
First differences give A014675. a(n) = A022342(n) + 1 = A005206(n) + n + 1. a(2n)-a(n)=A007067(n). a(a(a(n)))-a(n) = A026274(n-1). - Benoit Cloitre, Mar 08 2003
A185615 gives values n such that n divides A000201(n)^m for some integer m>0.
Cf. A183110, A329825.
Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021
Bisections: A276854, A342279.

Programs

  • Haskell
    a000201 n = a000201_list !! (n-1)
a000201_list = f [1..] [1..] where
   f (x:xs) (y:ys) = y : f xs (delete (x + y) ys)
-- Reinhard Zumkeller, Jul 02 2015, Mar 10 2013
  • Maple
    Digits := 100; t := evalf((1+sqrt(5))/2); A000201 := n->floor(t*n);
  • Mathematica
    Table[Floor[N[n*(1+Sqrt[5])/2]], {n, 1, 75}]
Array[ Floor[ #*GoldenRatio] &, 68] (* Robert G. Wilson v, Apr 17 2010 *)
  • Maxima
    makelist(floor(n*(1+sqrt(5))/2),n,1,60); /* Martin Ettl, Oct 17 2012 */
  • PARI
    a(n)=floor(n*(sqrt(5)+1)/2)
  • PARI
    a(n)=(n+sqrtint(5*n^2))\2 \\ Charles R Greathouse IV, Feb 07 2013
  • Python
    def aupton(terms):
  alst, aset = [None, 1], {1}
  for n in range(1, terms):
    an = alst[n] + (1 if n not in aset else 2)
    alst.append(an); aset.add(an)
  return alst[1:]
print(aupton(68)) # Michael S. Branicky, May 14 2021
  • Python
    from math import isqrt
def A000201(n): return (n+isqrt(5*n**2))//2 # Chai Wah Wu, Jan 11 2022

Formula

Zeckendorf expansion of n (cf. A035517) ends with an even number of 0's.
Other properties: a(1)=1; for n>1, a(n) is taken to be the smallest integer greater than a(n-1) which is consistent with the condition "n is in the sequence if and only if a(n)+1 is not in the sequence".
a(1) = 1; for n>0, a(n+1) = a(n)+1 if n is not in the sequence, a(n+1) = a(n)+2 if n is in the sequence.
a(a(n)) = floor(n*phi^2) - 1 = A003622(n).
{a(k)} union {a(k)+1} = {1, 2, 3, 4, ...}. Hence a(1) = 1; for n>1, a(a(n)) = a(a(n)-1)+2, a(a(n)+1) = a(a(n))+1. - Benoit Cloitre, Mar 08 2003
{a(n)} is a solution to the recurrence a(a(n)+n) = 2*a(n)+n, a(1)=1 (see Barbeau et al.).
a(n) = A001950(n) - n. - Philippe Deléham, May 02 2004
a(0) = 0; a(n) = n + Max_{k : a(k) < n}. - Vladeta Jovovic, Jun 11 2004
a(Fibonacci(r-1)+j) = Fibonacci(r)+a(j) for 0 < j <= Fibonacci(r-2); 2 < r. - Paul Weisenhorn, Aug 18 2012
With 1 < k and A001950(k-1) < n <= A001950(k): a(n) = 2*n-k; A001950(n) = 3*n-k. - Paul Weisenhorn, Aug 21 2012

A001950 Upper Wythoff sequence (a Beatty sequence): a(n) = floor(n*phi^2), where phi = (1+sqrt(5))/2.

Original entry on oeis.org

2, 5, 7, 10, 13, 15, 18, 20, 23, 26, 28, 31, 34, 36, 39, 41, 44, 47, 49, 52, 54, 57, 60, 62, 65, 68, 70, 73, 75, 78, 81, 83, 86, 89, 91, 94, 96, 99, 102, 104, 107, 109, 112, 115, 117, 120, 123, 125, 128, 130, 133, 136, 138, 141, 143, 146, 149, 151, 154, 157
Offset: 1

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Author

N. J. A. Sloane

Keywords

Comments

Indices at which blocks (1;0) occur in infinite Fibonacci word; i.e., n such that A005614(n-2) = 0 and A005614(n-1) = 1. - Benoit Cloitre, Nov 15 2003
A000201 and this sequence may be defined as follows: Consider the maps a -> ab, b -> a, starting from a(1) = a; then A000201 gives the indices of a, A001950 gives the indices of b. The sequence of letters in the infinite word begins a, b, a, a, b, a, b, a, a, b, a, ... Setting a = 0, b = 1 gives A003849 (offset 0); setting a = 1, b = 0 gives A005614 (offset 0). - Philippe Deléham, Feb 20 2004
a(n) = n-th integer which is not equal to the floor of any multiple of phi, where phi = (1+sqrt(5))/2 = golden number. - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), May 09 2007
Write A for A000201 and B for the present sequence (the upper Wythoff sequence, complement of A). Then the composite sequences AA, AB, BA, BB, AAA, AAB, ..., BBB, ... appear in many complementary equations having solution A000201 (or equivalently, the present sequence). Typical complementary equations: AB=A+B (=A003623), BB=A+2B (=A101864), BBB=3A+5B (=A134864). - Clark Kimberling, Nov 14 2007
Apart from the initial 0 in A090909, is this the same as that sequence? - Alec Mihailovs (alec(AT)mihailovs.com), Jul 23 2007
If we define a base-phi integer as a positive number whose representation in the golden ratio base consists only of nonnegative powers of phi, and if these base-phi integers are ordered in increasing order (beginning 1, phi, ...), then it appears that the difference between the n-th and (n-1)-th base-phi integer is phi-1 if and only if n belongs to this sequence, and the difference is 1 otherwise. Further, if each base-phi integer is written in linear form as a + b*phi (for example, phi^2 is written as 1 + phi), then it appears that there are exactly two base-phi integers with b=n if and only if n belongs to this sequence, and exactly three base-phi integers with b=n otherwise. - Geoffrey Caveney, Apr 17 2014
Numbers with an odd number of trailing zeros in their Zeckendorf representation (A014417). - Amiram Eldar, Feb 26 2021
Numbers missing from A066096. - Philippe Deléham, Jan 19 2023

Examples

			From _Paul Weisenhorn_, Aug 18 2012 and Aug 21 2012: (Start)
a(14) = floor(14*phi^2) = 36; a'(14) = floor(14*phi)=22;
with r=9 and j=1: a(13+1) = 34 + 2 = 36;
with r=8 and j=1: a'(13+1) = 21 + 1 = 22.
k=6 and a(5)=13 < n <= a(6)=15
a(14) = 3*14 - 6 = 36; a'(14) = 2*14 - 6 = 22;
a(15) = 3*15 - 6 = 39; a'(15) = 2*15 - 6 = 24. (End)

References

  • Claude Berge, Graphs and Hypergraphs, North-Holland, 1973; p. 324, Problem 2.
  • Eric Friedman, Scott M. Garrabrant, Ilona K. Phipps-Morgan, A. S. Landsberg and Urban Larsson, Geometric analysis of a generalized Wythoff game, in Games of no Chance 5, MSRI publ. Cambridge University Press, 2019.
  • Martin Gardner, Penrose Tiles to Trapdoor Ciphers, W. H. Freeman, 1989; see p. 107.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • I. M. Yaglom, Two games with matchsticks, pp. 1-7 of Qvant Selecta: Combinatorics I, Amer Math. Soc., 2001.

Links

Crossrefs

a(n) = greatest k such that s(k) = n, where s = A026242.
Complement of A000201 or A066096.
A002251 maps between A000201 and A001950, in that A002251(A000201(n)) = A001950(n), A002251(A001950(n)) = A000201(n).
Cf. A001622, A026352, A004976, A004919.
Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864.
First differences give (essentially) A076662.
Bisections: A001962, A001966.
Cf. A014417, A329825.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

Programs

  • Haskell
    a001950 n = a000201 n + n  -- Reinhard Zumkeller, Mar 10 2013
  • Magma
    [Floor(n*((1+Sqrt(5))/2)^2): n in [1..80]]; // Vincenzo Librandi, Nov 19 2016
  • Maple
    A001950 := proc(n)
    floor(n*(3+sqrt(5))/2) ;
end proc:
seq(A001950(n),n=0..40) ; # R. J. Mathar, Jul 16 2024
  • Mathematica
    Table[Floor[N[n*(1+Sqrt[5])^2/4]], {n, 1, 75}]
Array[ Floor[ #*GoldenRatio^2] &, 60] (* Robert G. Wilson v, Apr 17 2010 *)
  • PARI
    a(n)=floor(n*(sqrt(5)+3)/2)
  • PARI
    A001950(n)=(sqrtint(n^2*5)+n*3)\2 \\ M. F. Hasler, Sep 17 2014
  • Python
    from math import isqrt
def A001950(n): return (n+isqrt(5*n**2)>>1)+n # Chai Wah Wu, Aug 10 2022

Formula

a(n) = n + floor(n*phi). In general, floor(n*phi^m) = Fibonacci(m-1)*n + floor(Fibonacci(m)*n*phi). - Benoit Cloitre, Mar 18 2003
a(n) = n + floor(n*phi) = n + A000201(n). - Paul Weisenhorn and Philippe Deléham
Append a 0 to the Zeckendorf expansion (cf. A035517) of n-th term of A000201.
a(n) = A003622(n) + 1. - Philippe Deléham, Apr 30 2004
a(n) = Min(m: A134409(m) = A006336(n)). - Reinhard Zumkeller, Oct 24 2007
If a'=A000201 is the ordered complement (in N) of {a(n)}, then a(Fib(r-2) + j) = Fib(r) + a(j) for 0 < j <= Fib(r-2), 3 < r; and a'(Fib(r-1) + j) = Fib(r) + a'(j) for 0 < j <= Fib(r-2), 2 < r. - Paul Weisenhorn, Aug 18 2012
With a(1)=2, a(2)=5, a'(1)=1, a'(2)=3 and 1 < k and a(k-1) < n <= a(k) one gets a(n)=3*n-k, a'(n)=2*n-k. - Paul Weisenhorn, Aug 21 2012

Extensions

Corrected by Michael Somos, Jun 07 2000

A003849 The infinite Fibonacci word (start with 0, apply 0->01, 1->0, take limit).

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane

Keywords

Comments

A Sturmian word.
Define strings S(0)=0, S(1)=01, S(n)=S(n-1)S(n-2); iterate; sequence is S(infinity). If the initial 0 is omitted from S(n) for n>0, we obtain A288582(n+1).
The 0's occur at positions in A022342 (i.e., A000201 - 1), the 1's at positions in A003622.
Replace each run (1;1) with (1;0) in the infinite Fibonacci word A005614 (and add 0 as prefix) A005614 begins: 1,0,1,1,0,1,0,1,1,0,1,1,... changing runs (1,1) with (1,0) produces 1,0,0,1,0,1,0,0,1,0,0,1,... - Benoit Cloitre, Nov 10 2003
Characteristic function of A003622. - Philippe Deléham, May 03 2004
The fraction of 0's in the first n terms approaches 1/phi (see for example Allouche and Shallit). - N. J. A. Sloane, Sep 24 2007
The limiting mean and variance of the first n terms are 2-phi and 2*phi-3, respectively. - Clark Kimberling, Mar 12 2014, Aug 16 2018
Let S(n) be defined as above. Then this sequence is S(1) + Sum_{n=0..} S(n), where the addition of strings represents concatenation. - Isaac Saffold, May 03 2019
The word is a concatenation of three runs: 0, 1, and 00. The limiting proportions of these are respectively 1 - phi/2, 1/2, and (phi - 1)/2. The mean runlength is (phi + 1)/2. - Clark Kimberling, Dec 26 2010
From Amiram Eldar, Mar 10 2021: (Start)
a(n) is the number of the trailing 0's in the dual Zeckendorf representation of (n+1) (A104326).
The asymptotic density of the occurrences of k (0 or 1) is 1/phi^(k+1), where phi is the golden ratio (A001622).
The asymptotic mean of this sequence is 1/phi^2 (A132338). (End)

Examples

			The word is 010010100100101001010010010100...
Over the alphabet {a,b} this is a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, ...

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
  • Jean Berstel, Fibonacci words—a survey, In The book of L, pp. 13-27. Springer Berlin Heidelberg, 1986.
  • J. C. Lagarias, Number Theory and Dynamical Systems, pp. 35-72 of S. A. Burr, ed., The Unreasonable Effectiveness of Number Theory, Proc. Sympos. Appl. Math., 46 (1992). Amer. Math. Soc. - see p. 64.
  • Wolfdieter Lang, The Wythoff and the Zeckendorf representations of numbers are equivalent, in G. E. Bergum et al. (edts.) Application of Fibonacci numbers vol. 6, Kluwer, Dordrecht, 1996, pp. 319-337. [See A317208 for a link.]
  • G. Melançon, Factorizing infinite words using Maple, MapleTech journal, vol. 4, no. 1, 1997, pp. 34-42, esp. p. 36.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

Links

Crossrefs

There are several versions of this sequence in the OEIS. This one and A003842 are probably the most important. See also A008352, A076662, A288581, A288582.
Binary complement of A005614. Cf. A014675, A036299, A003714, A014417, A096268, A096270, A133235, A182028, A213975.
Positions of 1's gives A003622.
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.
Cf. A001622, A104326, A132338.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

Programs

  • Haskell
    a003849 n = a003849_list !! n
a003849_list = tail $ concat fws where
   fws = [1] : [0] : (zipWith (++) fws $ tail fws)
-- Reinhard Zumkeller, Nov 01 2013, Apr 07 2012
  • Magma
    t1:=[ n le 2 select ["0","0,1"][n] else Self(n-1) cat "," cat Self(n-2) : n in [1..12]]; t1[12];
  • Maple
    z := proc(m) option remember; if m=0 then [0] elif m=1 then [0,1] else [op(z(m-1)),op(z(m-2))]; fi; end; z(12);
M:=19; S[0]:=`0`; S[1]:=`01`; for n from 2 to M do S[n]:=cat(S[n-1], S[n-2]); od:
t0:=S[M]: l:=length(t0); for i from 1 to l do lprint(i-1,substring(t0,i..i)); od: # N. J. A. Sloane, Nov 01 2006
  • Mathematica
    Nest[ Flatten[ # /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 10] (* Robert G. Wilson v, Mar 05 2005 *)
Flatten[Nest[{#, #[[1]]} &, {0, 1}, 9]] (* IWABUCHI Yu(u)ki, Oct 23 2013 *)
Table[Floor[(n + 2) #] - Floor[(n + 1) #], {n, 0, 120}] &[2 - GoldenRatio] (* Michael De Vlieger, Aug 15 2016 *)
SubstitutionSystem[{0->{0,1},1->{0}},{0},{10}][[1]] (* Harvey P. Dale, Dec 20 2021 *)
  • PARI
    a(n)=my(k=2);while(fibonacci(k)<=n,k++);while(n>1,while(fibonacci(k--)>n,); n-=fibonacci(k)); n==1 \\ Charles R Greathouse IV, Feb 03 2014
  • PARI
    M3849=[2,2,1,0]/*L(k),S(k),L(k-1),S(k-1)*/; A003849(n)={while(n>M3849[1],M3849=vecextract(M3849,[1,2,1,2])+[M3849[3],M3849[4]<M. F. Hasler, Apr 07 2021
  • Python
    def fib(n):
    """Return the concatenation of A003849(0..F-1) where F is the smallest
       Fibonacci number > n, so that the result contains a(n) at index n."""
    a, b = '10'
    while len(b)<=n:
        a, b = b, b + a
    return b # Robert FERREOL, Apr 15 2016, edited by M. F. Hasler, Apr 07 2021
  • Python
    from math import isqrt
def A003849(n): return 2-(n+2+isqrt(m:=5*(n+2)**2)>>1)+(n+1+isqrt(m-10*n-15)>>1) # Chai Wah Wu, Aug 25 2022

Formula

a(n) = floor((n+2)*r) - floor((n+1)*r) where r=phi/(1+2*phi) and phi is the Golden Ratio. - Benoit Cloitre, Nov 10 2003
a(n) = A003714(n) mod 2 = A014417(n) mod 2. - Philippe Deléham, Jan 04 2004
The first formula by Cloitre is just one of an infinite family of formulas. Using phi^2=1+phi, it follows that r=phi/(1+2*phi)=2-phi. Then from floor(-x)=-floor(x)-1 for non-integer x, it follows that a(n)=2-A014675(n)=2-(floor((n+2)* phi)-floor((n+1)*phi)). - Michel Dekking, Aug 27 2016
a(n) = 1 - A096270(n+1), i.e., A096270 is the complement of this sequence. - A.H.M. Smeets, Mar 31 2024

Extensions

Revised by N. J. A. Sloane, Jul 03 2012

A003622 The Wythoff compound sequence AA: a(n) = floor(n*phi^2) - 1, where phi = (1+sqrt(5))/2.

Original entry on oeis.org

1, 4, 6, 9, 12, 14, 17, 19, 22, 25, 27, 30, 33, 35, 38, 40, 43, 46, 48, 51, 53, 56, 59, 61, 64, 67, 69, 72, 74, 77, 80, 82, 85, 88, 90, 93, 95, 98, 101, 103, 106, 108, 111, 114, 116, 119, 122, 124, 127, 129, 132, 135, 137, 140, 142, 145, 148, 150, 153, 156, 158, 161, 163, 166
Offset: 1

Views

Author

N. J. A. Sloane, Mira Bernstein, Marc LeBrun

Keywords

Comments

Also, integers with "odd" Zeckendorf expansions (end with ...+F_2 = ...+1) (Fibonacci-odd numbers); first column of Wythoff array A035513; from a 3-way splitting of positive integers. [Edited by Peter Munn, Sep 16 2022]
Also, numbers k such that A005206(k) = A005206(k+1). Also k such that A022342(A005206(k)) = k+1 (for all other k's this is k). - Michele Dondi (bik.mido(AT)tiscalenet.it), Dec 30 2001
Also, positions of 1's in A139764, the smallest term in Zeckendorf representation of n. - John W. Layman, Aug 25 2011
From Amiram Eldar, Sep 03 2022: (Start)
Numbers with an odd number of trailing 1's in their dual Zeckendorf representation (A104326), i.e., numbers k such that A356749(k) is odd.
The asymptotic density of this sequence is 1 - 1/phi (A132338). (End)
{a(n)} is the unique monotonic sequence of positive integers such that {a(n)} and {b(n)}: b(n) = a(n) - n form a partition of the nonnegative integers. - Yifan Xie, Jan 25 2025

References

  • A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 62.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 307-308 of 2nd edition.
  • C. Kimberling, "Stolarsky interspersions", Ars Combinatoria 39 (1995) 129-138.
  • D. R. Morrison, "A Stolarsky array of Wythoff pairs", in A Collection of Manuscripts Related to the Fibonacci Sequence. Fibonacci Assoc., Santa Clara, CA, 1980, pp. 134-136.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 10.
  • N. J. A. Sloane and Simon Plouffe, Encyclopedia of Integer Sequences, Academic Press, 1995: this sequence appears twice, as both M3277 and M3278.

Links

Crossrefs

Positions of 1's in A003849.
Complement of A022342.
Cf. A066096, A139764, A035516, A026273, A104326, A132338, A356749.
The Wythoff compound sequences: Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864. The eight triples AAA, AAB, ..., BBB are A134859, A134860, A035337, A134862, A134861, A134863, A035338, A134864, resp.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

Programs

  • Haskell
    a003622 n = a003622_list !! (n-1)
a003622_list = filter ((elem 1) . a035516_row) [1..]
-- Reinhard Zumkeller, Mar 10 2013
  • Maple
    A003622 := proc(n)
    n+floor(n*(1+sqrt(5))/2)-1 ;
end proc: # R. J. Mathar, Jan 25 2015
# Maple code for the Wythoff compound sequences, from N. J. A. Sloane, Mar 30 2016
# The Wythoff compound sequences: Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864. The eight triples AAA, AAB, ..., BBB are A134859, A134860, A035337, A134862, A134861, A134863, A035338, A134864, resp.
# Assume files out1, out2 contain lists of the terms in the base sequences A and B from their b-files
read out1; read out2; b[0]:=b1: b[1]:=b2:
w2:=(i,j,n)->b[i][b[j][n]];
w3:=(i,j,k,n)->b[i][b[j][b[k][n]]];
for i from 0 to 1 do
lprint("name=",i);
lprint([seq(b[i][n],n=1..100)]):
od:
for i from 0 to 1 do for j from 0 to 1 do
lprint("name=",i,j);
lprint([seq(w2(i,j,n),n=1..100)]);
od: od:
for i from 0 to 1 do for j from 0 to 1 do for k from 0 to 1 do
lprint("name=",i,j,k);
lprint([seq(w3(i,j,k,n),n=1..100)]);
od: od: od:
  • Mathematica
    With[{c=GoldenRatio^2},Table[Floor[n c]-1,{n,70}]] (* Harvey P. Dale, Jun 11 2011 *)
Range[70]//Floor[#*GoldenRatio^2]-1& (* Waldemar Puszkarz, Oct 10 2017 *)
  • PARI
    a(n)=floor(n*(sqrt(5)+3)/2)-1
  • PARI
    a(n) = (sqrtint(n^2*5)+n*3)\2 - 1; \\ Michel Marcus, Sep 17 2022
  • Python
    from sympy import floor
from mpmath import phi
def a(n): return floor(n*phi**2) - 1 # Indranil Ghosh, Jun 09 2017
  • Python
    from math import isqrt
def A003622(n): return (n+isqrt(5*n**2)>>1)+n-1 # Chai Wah Wu, Aug 11 2022

Formula

a(n) = floor(n*phi) + n - 1. [Corrected by Jianing Song, Aug 18 2022]
a(n) = floor(floor(n*phi)*phi) = A000201(A000201(n)). [See the Mathematics Stack Exchange link for a proof of the equivalence of the definition. - Jianing Song, Aug 18 2022]
a(n) = 1 + A022342(1 + A022342(n)).
G.f.: 1 - (1-x)*Sum_{n>=1} x^a(n) = 1/1 + x/1 + x^2/1 + x^3/1 + x^5/1 + x^8/1 + ... + x^F(n)/1 + ... (continued fraction where F(n)=n-th Fibonacci number). - Paul D. Hanna, Aug 16 2002
a(n) = A001950(n) - 1. - Philippe Deléham, Apr 30 2004
a(n) = A022342(n) + n. - Philippe Deléham, May 03 2004
a(n) = a(n-1) + 2 + A005614(n-2); also a(n) = a(n-1) + 1 + A001468(n-1). - A.H.M. Smeets, Apr 26 2024

A005614 The binary complement of the infinite Fibonacci word A003849. Start with 1, apply 0->1, 1->10, iterate, take limit.

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane

Keywords

Comments

Previous name was: The infinite Fibonacci word (start with 1, apply 0->1, 1->10, iterate, take limit).
Characteristic function of A022342. - Philippe Deléham, May 03 2004
a(n) = number of 0's between successive 1's (see also A003589 and A007538). - Eric Angelini, Jul 06 2005
With offset 1 this is the characteristic sequence for Wythoff A-numbers A000201=[1,3,4,6,...].
Eric Angelini's comment made me think that if 1 is defined to be the number of 0's between successive 1's in a string of 0's and 1's, then this string is 101. Applying the same operation to the digits of 101 leads to 101101, the iteration leads to successive palindromes of lengths given by A001911, up to a(n). - Rémi Schulz, Jul 06 2010
For generalized Fibonacci words see A221150, A221151, A221152, ... - Peter Bala, Nov 11 2013
The limiting mean of the first n terms is phi - 1; the limiting variance is phi (A001622). - Clark Kimberling, Mar 12 2014
Apply the difference operator to every column of the Wythoff difference array, A080164, to get an array of Fibonacci numbers, F(h). Replace each F(h) with h, and apply the difference operator to every column. In the resulting array, every column is A005614. - Clark Kimberling, Mar 02 2015
Binary expansion of the rabbit constant A014565. - M. F. Hasler, Nov 10 2018

Examples

			The infinite word is 101101011011010110101101101011...

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
  • G. Melançon, Factorizing infinite words using Maple, MapleTech journal, vol. 4, no. 1, 1997, pp. 34-42, esp. p. 36.

Links

Crossrefs

Binary complement of A003849, which is the standard form of this sequence.
Two other essentially identical sequences are A096270, A114986.
Subwords: A178992, A171676.
Cf. A000045 (Fibonacci numbers), A001468, A001911, A005206 (partial sums), A014565, A014675, A022342, A036299, A044432, A221150, A221151, A221152.
Cf. A339051 (odd bisection), A339052 (even bisection).
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

Programs

  • Haskell
    a005614 n = a005614_list !! n
a005614_list = map (1 -) a003849_list
-- Reinhard Zumkeller, Apr 07 2012
  • Magma
    [Floor((n+1)*(-1+Sqrt(5))/2)-Floor(n*(-1+Sqrt(5))/2): n in [1..100]]; // Vincenzo Librandi, Jan 17 2019
  • Maple
    Digits := 50; u := evalf((1-sqrt(5))/2); A005614 := n->floor((n+1)*u)-floor(n*u);
  • Mathematica
    Nest[ Flatten[ # /. {0 -> {1}, 1 -> {1, 0}}] &, {1}, 10] (* Robert G. Wilson v, Jan 30 2005 *)
Flatten[Nest[{#, #[[1]]} &, {1, 0}, 9]] (* IWABUCHI Yu(u)ki, Oct 23 2013 *)
SubstitutionSystem[{0 -> {1}, 1 -> {1, 0}}, {1, 0}, 9] // Last (* Jean-François Alcover, Feb 06 2020 *)
  • PARI
    a(n,w1,s0,s1)=local(w2); for(i=2,n,w2=[ ]; for(k=1,length(w1),w2=concat(w2, if(w1[ k ],s1,s0))); w1=w2); w2
for(n=2,10,print(n" "a(n,[ 0 ],[ 1 ],[ 1,0 ]))) \\ Gives successive convergents to sequence
  • PARI
    /* for m>=1 compute exactly A183136(m+1)+1 terms of the sequence */
r=(1+sqrt(5))/2;v=[1,0];for(n=2,m,v=concat(v,vector(floor((n+1)/r),i,v[i]));a(n)=v[n];) /* Benoit Cloitre, Jan 16 2013 */
  • Python
    from math import isqrt
def A005614(n): return (n+isqrt(m:=5*(n+2)**2)>>1)-(n+1+isqrt(m-10*n-15)>>1) # Chai Wah Wu, Aug 17 2022

Formula

Define strings S(0)=1, S(1)=10, thereafter S(n)=S(n-1)S(n-2); iterate. Sequence is S(oo). The individual S(n)'s are given in A036299.
a(n) = floor((n+2)*u) - floor((n+1)*u), where u = (-1 + sqrt(5))/2.
Sum_{n>=0} a(n)/2^(n+1) = A014565. - R. J. Mathar, Jul 19 2013
From Peter Bala, Nov 11 2013: (Start)
If we read the present sequence as the digits of a decimal constant c = 0.101101011011010 ... then we have the series representation c = Sum_{n >= 1} 1/10^floor(n*phi). An alternative representation is c = Sum_{n >= 1} 1/10^floor(n/phi) - 10/9.
The constant 9*c has the simple continued fraction representation [0; 1, 10, 10, 100, 1000, ..., 10^Fibonacci(n), ...]. See A010100.
Using this result we can find the alternating series representation c = 1/9 - 9*Sum_{n >= 1} (-1)^(n+1)*(1 + 10^Fibonacci(3*n+1))/( (10^(Fibonacci(3*n - 1)) - 1)*(10^(Fibonacci(3*n + 2)) - 1) ). The series converges very rapidly: for example, the first 10 terms of the series give a value for c accurate to more than 5.7 million decimal places. Cf. A014565. (End)
a(n) = A005206(n+1) - A005206(n). a(2*n) = A339052(n); a(2*n+1) = A339051(n+1). - Peter Bala, Aug 09 2022

Extensions

Corrected by Clark Kimberling, Oct 04 2000
Name corrected by Michel Dekking, Apr 02 2019

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

Views

Author

Karamanos Konstantinos, N. J. A. Sloane

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).

Links

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.
Cf. A060572

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

A014577 The regular paper-folding sequence (or dragon curve sequence). Alphabet {1,0}.

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Eric W. Weisstein

Keywords

Comments

a(n) is the complement of the bit to the left of the least significant "1" in the binary expansion of n+1. E.g., n = 3, n+1 = 4 = 100_2, so a(3) = (complement of bit to left of 1) = 1. - Robert L. Brown, Nov 28 2001 [Adjusted to match offset by N. J. A. Sloane, Apr 15 2021]
To construct the sequence: start from 1,(..),0,(..),1,(..),0,(..),1,(..),0,(..),1,(..),0,... and fill undefined places with the sequence itself. - Benoit Cloitre, Jul 08 2007
This is the characteristic function of A091072 - 1. - Gary W. Adamson, Apr 11 2010
Turns (by 90 degrees) of the Heighway dragon which can be rendered as follows: [Init] Set n=0 and direction=0. [Draw] Draw a unit line (in the current direction). Turn left/right if a(n) is zero/nonzero respectively. [Next] Set n=n+1 and goto (draw). See fxtbook link below. - Joerg Arndt, Apr 15 2010
Sequence can be obtained by the L-system with rules L->L1R, R->L0R, 1->1, 0->0, starting with L, and then dropping all L's and R's (see example). - Joerg Arndt, Aug 28 2011
From Gary W. Adamson, Jun 20 2012: (Start)
One half of the infinite Farey tree can be mapped one-to-one onto A014577 since both sequences can be derived directly from the binary. First few terms are
1, 1, 0, 1, 1, 0, 0, 1, 1, 1, ...
1/2 2/3 1/3 3/4 3/5 2/5 1/4 4/5 5/7 5/8, ...
Infinite Farey tree fractions can be derived from the binary by appending a repeat of rightmost binary term to the right, then recording the number of runs to obtain the continued fraction representation. Example: 9 = 1001 which becomes 10011 which becomes [1,2,2] = 5/7. (End)
The sequence can be considered as a binomial transform operator for a target sequence S(n). Replace the first 1 in A014577 with the first term in S(n), then for successive "1" term in A014577, map the next higher term in S(n). If "0" in A014577, map the next lower term in S(n). Using the sequence S(n) = (1, 3, 5, 7, ...), we obtain (1), (3, 1), (3, 5, 3, 1), (3, 5, 7, 5, 3, 5, 3, 1), .... Then parse the terms into subsequences of 2^k terms, adding the terms in each string. We obtain (1, 4, 12, 32, 80, ...), the binomial transform of (1, 3, 5, 7, ...). The 8-bit string has one 1, three 5's, three 7's and one 1) as expected, or (1, 3, 3, 1) dot (1, 3, 5, 7). - Gary W. Adamson, Jun 24 2012
From Gary W. Adamson, May 29 2013: (Start)
The sequence can be generated directly from the lengths of continued fraction representations of fractions in one half of the Stern-Brocot tree (fractions between 0 and 1):
1/2
1/3 2/3
1/4 2/5 3/5 3/4
1/5 2/7 3/8 3/7 4/7 5/8 5/7 4/5
...
and their corresponding continued fraction representations are:
[2]
[3] [1,2]
[4] [2,2] [1,1,2] [1,3]
[5] [3,2] [2,1,2] [2,3] [1,1,3] [1,1,1,2] [1,2,2] [1,4]
...
Record the lengths by rows then reverse rows, getting:
1,
2, 1,
2, 3, 2, 1,
2, 3, 4, 3, 2, 3, 2, 1,
...
start with "1" and if the next term is greater than the current term, record a 1, otherwise 0; getting the present sequence, the Harter-Heighway dragon curve. (End)
The paper-folding word "110110011100100111011000..." can be created by concatenating the terms of a fixed point of the morphism or string substitution rule: 00 -> 1000, 01 -> 1001, 10 -> 1100 & 11 -> 1101, beginning with "11". - Robert G. Wilson v, Jun 11 2015
From Gary W. Adamson, Jun 04 2021: (Start)
Since the Heighway dragon is composed of right angles, it can be mapped with unit trajectories (Right = 1, Left = (-1), Up = i and Down = -i) on the complex plane where i = sqrt(-1). The initial (0th) iterate is chosen in this case to be the unit line from (0,0) to (-1,0). Then follow the directions below as indicated, resulting in a reflected variant of the dragon curve shown at the Eric Weisstein link. The conjectured system of complex plane trajectories is:
0 -1
1 -1, i
2 -1, i, 1, i
3 -1, i, 1, i, 1, -i, 1, i
4 -1, i, 1, i, 1, -i, 1, i, 1, -i, -1, -i, 1, -i, 1, i
...
The conjecture succeeds through the 4th iterate. It appears that to generate the (n+1)-th row, bring down the n-th row as the left half of row (n+1). For the right half of row (n+1), bring down the n-th row but change the signs of the first half of row n. For example, to get the complex plane instructions for the third iterate of the dragon curve, bring down (-1, i, 1, i) as the left half, and the right half is (1, -i, 1, i). (End)
From Gary W. Adamson, Jun 09 2021: (Start)
Partial sums of the iterate trajectories produce a sequence of complex addresses for unit segments. Partial sums of row 4 are: -1, (-1+i), i, 2i, (1+2i), (1+i), (2+i), (2+2i), (3+2i), (3+i), (2+i), 2, 3, (3-i), (4-i), 4. (zeros are omitted with terms of the form a+0i). The reflected variant of the dragon curve has the 0th iterate from (0,0) to (1,0), and the respective addresses simply change the signs of the real terms. (End)

Examples

			1 + x + x^3 + x^4 + x^7 + x^8 + x^9 + x^12 + x^15 + x^16 + x^17 + x^19 + ...
From _Joerg Arndt_, Aug 28 2011: (Start)
Generation via string substitution:
Start: L
Rules:
  L --> L1R
  R --> L0R
  0 --> 0
  1 --> 1
-------------
0:   (#=1)
  L
1:   (#=3)
  L1R
2:   (#=7)
  L1R1L0R
3:   (#=15)
  L1R1L0R1L1R0L0R
4:   (#=31)
  L1R1L0R1L1R0L0R1L1R1L0R0L1R0L0R
5:   (#=63)
  L1R1L0R1L1R0L0R1L1R1L0R0L1R0L0R1L1R1L0R1L1R0L0R0L1R1L0R0L1R0L0R
Drop all L and R to obtain 1101100111001001110110001100100
(End)

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, pp. 155, 182.
  • Chandler Davis and Donald E. Knuth, Number Representations and Dragon Curves -- I and II, Journal of Recreational Mathematics, volume 3, number 2, April 1970, pages 66-81, and number 3, July 1970, pages 133-149. Reprinted in Donald E. Knuth, Selected Papers on Fun and Games, CSLI Publications, 2010, pages 571-614.
  • Michel Dekking, Michel Mendes 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).
  • Martin Gardner, Mathematical Magic Show, New York: Vintage, pp. 207-209 and 215-220, 1978.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

Links

Crossrefs

Essentially the same: A014707, A014709, A014710, A034947, A038189, A082410, A089013, A099545, A112347, A121238, A317335, A317336.
Cf. A059125, A065339, A005811, A220466, A088748, A091072, A343173 (first differences), A343174 (partial sums).
The two bisections are A000035 and the sequence itself.
See A343181 for prefixes of length 2^k-1.

Programs

  • Magma
    [(1+KroneckerSymbol(-1,n))/2: n in [1..100]]; // Vincenzo Librandi, Aug 05 2015
  • Magma
    [Floor(1/2*(1+(-1)^(1/2*((n+1)/2^Valuation(n+1, 2)-1)))): n in [0..100]]; // Vincenzo Librandi, Aug 05 2015
  • Maple
    nmax:=98: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 0 to ceil(nmax/(p+2))+1 do a((2*n+1)*2^p-1) := (n+1) mod 2 od: od: seq(a(n), n=0..nmax); # Johannes W. Meijer, Jan 28 2013
a014577 := proc(n) local p,s1,s2,i;
if n=0 then return(1); fi;
s1:=convert(n,base,2); s2:=nops(s1);
for i from 1 to s2 do if s1[i]=1 then p:=i; break; fi; od:
if p <= s2-1 then 1-s1[p+1]; else 1; fi; end;
[seq(a014577(i),i=1..120)]; # N. J. A. Sloane, Apr 08 2021
# third Maple program:
a:= n-> 1-irem(iquo((n+1)/2^padic[ordp](n+1, 2), 2), 2):
seq(a(n), n=0..120);  # Alois P. Heinz, Apr 08 2021
  • Mathematica
    a[n_] := Boole[EvenQ[((n+1)/2^IntegerExponent[n+1, 2]-1)/2]]; Table[a[n], {n, 0, 98}] (* Jean-François Alcover, Feb 16 2012, after Gary W. Adamson, updated Nov 21 2014 *)
Table[1-(((Mod[#1,2^(#2+2)]/2^#2)&[n,IntegerExponent[n,2]])-1)/2,{n,1,100,1}] (* WolframAlpha compatible code; Robert L. Brown, Jan 06 2015 *)
MapThread[(a[x_/;IntegerQ[(x-#1)/4]]:= #2)&,{{1,3},{1,0}}];a[x_/;IntegerQ[x/2]]:=a[x/2];a/@ Range[100] (* Bradley Klee, Aug 04 2015 *)
(1 + JacobiSymbol[-1, Range[100]])/2 (* Paolo Xausa, May 22 2024 *)
Array[Boole[BitAnd[#, BitAnd[#, -#]*2] == 0] &, 100] (* Paolo Xausa, May 22 2024, after Joerg Arndt C++ code *)
  • PARI
    {a(n) = if( n%2, a(n\2), 1 - (n/2%2))} /* Michael Somos, Feb 05 2012 */
  • PARI
    a(n)=1/2*(1+(-1)^(1/2*((n+1)/2^valuation(n+1,2)-1))) \\ Ralf Stephan, Sep 02 2013
  • PARI
    a(n)=!bittest(n+1,valuation(n+1,2)+1); \\ Robert L. Brown, Jul 07 2025
  • Python
    def A014577(n):
    s = bin(n+1)[2:]
    m = len(s)
    i = s[::-1].find('1')
    return 1-int(s[m-i-2]) if m-i-2 >= 0 else 1 # Chai Wah Wu, Apr 08 2021

Formula

a(n) = (1+Jacobi(-1,n+1))/2 (cf. A034947). - N. J. A. Sloane, Jul 27 2012 [Adjusted to match offset by Peter Munn, Jul 01 2022]
Set a=1, b=0, S(0)=a, S(n+1) = S(n), a, F(S(n)), where F(x) reverses x and then interchanges a and b; sequence is limit S(infinity).
From Ralf Stephan, Jul 03 2003: (Start)
a(4*n) = 1, a(4*n+2) = 0, a(2*n+1) = a(n).
a(n) = 1 - A014707(n) = 2 - A014709(n) = A014710(n) - 1. (End)
Set a=1, b=0, S(0)=a, S(n+1)=S(n), a, M(S(n)), where M(S) is S but the bit in the middle position flipped. (Proof via isomorphism of both formulas to a modified string substitution.) - Benjamin Heiland, Dec 11 2011
a(n) = 1 if A005811(n+1) > A005811(n), otherwise a(n) = 0. - Gary W. Adamson, Jun 20 2012
a((2*n+1)*2^p-1) = (n+1) mod 2, p >= 0. - Johannes W. Meijer, Jan 28 2013
G.f. g(x) satisfies g(x) = x*g(x^2) + 1/(1-x^4). - Robert Israel, Jan 06 2015
a(n) = 1 - A065339(n+1) mod 2. - Peter Munn, Jun 29 2022
From A.H.M. Smeets, Mar 19 2023: (Start)
a(n) = 1 - A038189(n+1).
a(n) = A082410(n+2).
a(n) = 1 - A089013(n+1)
a(n) = (3 - A099545(n+1))/2.
a(n) = (A112347(n+1) + 1)/2.
a(n) = (A121238(n+1) + 1)/2.
a(n) = (A317335(n) + 2)/3.
a(n) = (A317336(n) + 10)/3. (End)
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/2. - Amiram Eldar, Sep 14 2024

Extensions

More terms from Ralf Stephan, Jul 03 2003

A020985 The Rudin-Shapiro or Golay-Rudin-Shapiro sequence (coefficients of the Shapiro polynomials).

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane

Keywords

Comments

Other names are the Rudin-Shapiro or Golay-Rudin-Shapiro infinite word.
The Shapiro polynomials are defined by P_0 = Q_0 = 1; for n>=0, P_{n+1} = P_n + x^(2^n)*Q_n, Q_{n+1} = P_n - x^(2^n)*Q_n. Then P_n = Sum_{m=0..2^n-1} a(m)*x^m, where the a(m) (the present sequence) do not depend on n. - N. J. A. Sloane, Aug 12 2016
Related to paper-folding sequences - see the Mendès France and Tenenbaum article.
a(A022155(n)) = -1; a(A203463(n)) = 1. - Reinhard Zumkeller, Jan 02 2012
a(n) = 1 if and only if the numbers of 1's and runs of 1's in binary representation of n have the same parity: A010060(n) = A268411(n); otherwise, when A010060(n) = 1 - A268411(n), a(n) = -1. - Vladimir Shevelev, Feb 10 2016. Typo corrected and comment edited by Antti Karttunen, Jul 11 2017
A word that is uniform primitive morphic, but not pure morphic. - N. J. A. Sloane, Jul 14 2018
Named after the Austrian-American mathematician Walter Rudin (1921-2010), the mathematician Harold S. Shapiro (1928-2021) and the Swiss mathematician and physicist Marcel Jules Edouard Golay (1902-1989). - Amiram Eldar, Jun 13 2021

References

  • Jean-Paul Allouche and Jeffrey Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 78 and many other pages.

Links

Crossrefs

Cf. A022155, A005943 (factor complexity), A014081.
Cf. A020987 (0-1 version), A020986 (partial sums), A203531 (run lengths), A033999, A380667 (first differences).
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
    a020985 n = a020985_list !! n
a020985_list = 1 : 1 : f (tail a020985_list) (-1) where
   f (x:xs) w = x : x*w : f xs (0 - w)
-- Reinhard Zumkeller, Jan 02 2012
  • Maple
    A020985 := proc(n) option remember; if n = 0 then 1 elif n mod 2 = 0 then A020985(n/2) else (-1)^((n-1)/2 )*A020985( (n-1)/2 ); fi; end;
  • Mathematica
    a[0] = 1; a[1] = 1; a[n_?EvenQ] := a[n] = a[n/2]; a[n_?OddQ] := a[n] = (-1)^((n-1)/2)* a[(n-1)/2]; a /@ Range[0, 80] (* Jean-François Alcover, Jul 05 2011 *)
a[n_] := 1 - 2 Mod[Length[FixedPointList[BitAnd[#, # - 1] &, BitAnd[n, Quotient[n, 2]]]], 2] (* Jan Mangaldan, Jul 23 2015 *)
Array[RudinShapiro, 81, 0] (* JungHwan Min, Dec 22 2016 *)
  • PARI
    A020985(n)=(-1)^A014081(n)  \\ M. F. Hasler, Jun 06 2012
  • Python
    def a014081(n): return sum([((n>>i)&3==3) for i in range(len(bin(n)[2:]) - 1)])
def a(n): return (-1)**a014081(n) # Indranil Ghosh, Jun 03 2017
  • Python
    def A020985(n): return -1 if (n&(n>>1)).bit_count()&1 else 1 # Chai Wah Wu, Feb 11 2023

Formula

a(0) = a(1) = 1; thereafter, a(2n) = a(n), a(2n+1) = a(n) * (-1)^n. [Brillhart and Carlitz, in proof of theorem 4]
a(0) = a(1) = 1, a(2n) = a(n), a(2n+1) = a(n)*(1-2*(n AND 1)), where AND is the bitwise AND operator. - Alex Ratushnyak, May 13 2012
Brillhart and Morton (1978) list many properties.
a(n) = (-1)^A014081(n) = (-1)^A020987(n) = 1-2*A020987(n). - M. F. Hasler, Jun 06 2012
Sum_{n >= 1} a(n-1)*(8*n^2+4*n+1)/(2*n*(2*n+1)*(4*n+1)) = 1; see Allouche and Sondow, 2015. - Jean-Paul Allouche and Jonathan Sondow, Mar 21 2015

A022342 Integers with "even" Zeckendorf expansions (do not end with ...+F_2 = ...+1) (the Fibonacci-even numbers); also, apart from first term, a(n) = Fibonacci successor to n-1.

Original entry on oeis.org

0, 2, 3, 5, 7, 8, 10, 11, 13, 15, 16, 18, 20, 21, 23, 24, 26, 28, 29, 31, 32, 34, 36, 37, 39, 41, 42, 44, 45, 47, 49, 50, 52, 54, 55, 57, 58, 60, 62, 63, 65, 66, 68, 70, 71, 73, 75, 76, 78, 79, 81, 83, 84, 86, 87, 89, 91, 92, 94, 96, 97, 99, 100, 102, 104, 105, 107
Offset: 1

Views

Author

Marc LeBrun

Keywords

Comments

The Zeckendorf expansion of n is obtained by repeatedly subtracting the largest Fibonacci number you can until nothing remains; for example, 100 = 89 + 8 + 3.
The Fibonacci successor to n is found by replacing each F_i in the Zeckendorf expansion by F_{i+1}; for example, the successor to 100 is 144 + 13 + 5 = 162.
If k appears, k + (rank of k) does not (10 is the 7th term in the sequence but 10 + 7 = 17 is not a term of the sequence). - Benoit Cloitre, Jun 18 2002
From Michele Dondi (bik.mido(AT)tiscalenet.it), Dec 30 2001: (Start)
a(n) = Sum_{k in A_n} F_{k+1}, where a(n)= Sum_{k in A_n} F_k is the (unique) expression of n as a sum of "noncontiguous" Fibonacci numbers (with index >= 2).
a(10^n) gives the first few digits of g = (sqrt(5)+1)/2.
The sequences given by b(n+1) = a(b(n)) obey the general recursion law of Fibonacci numbers. In particular the (sub)sequence (of a(-)) yielded by a starting value of 2=a(1), is the sequence of Fibonacci numbers >= 2. Starting points of all such subsequences are given by A035336.
a(n) = floor(phi*n+1/phi); phi = (sqrt(5)+1)/2. a(F_n)=F_{n+1} if F_n is the n-th Fibonacci number.
(End)
From Amiram Eldar, Sep 03 2022: (Start)
Numbers with an even number of trailing 1's in their dual Zeckendorf representation (A104326), i.e., numbers k such that A356749(k) is even.
The asymptotic density of this sequence is 1/phi (A094214). (End)

Examples

			The successors to 1, 2, 3, 4=3+1 are 2, 3, 5, 7=5+2.

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 307-308 of 2nd edition.
  • E. Zeckendorf, Représentation des nombres naturels par une somme des nombres de Fibonacci ou de nombres de Lucas, Bull. Soc. Roy. Sci. Liège 41, 179-182, 1972.

Links

Crossrefs

Positions of 0's in A003849.
Complement of A003622.
Cf. A000201, A005206, A035336, A066096, A001950, A062879, A035516, A026274.
Cf. A035612, A094214, A104326, A356749.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

Programs

  • Haskell
    a022342 n = a022342_list !! (n-1)
a022342_list = filter ((notElem 1) . a035516_row) [0..]
-- Reinhard Zumkeller, Mar 10 2013
  • Magma
    [Floor(n*(Sqrt(5)+1)/2)-1: n in [1..100]]; // Vincenzo Librandi, Feb 16 2015
  • Maple
    A022342 := proc(n)
      local g;
      g := (1+sqrt(5))/2 ;
    floor(n*g)-1 ;
end proc: # R. J. Mathar, Aug 04 2013
  • Mathematica
    With[{t=GoldenRatio^2},Table[Floor[n*t]-n-1,{n,70}]] (* Harvey P. Dale, Aug 08 2012 *)
  • PARI
    a(n)=floor(n*(sqrt(5)+1)/2)-1
  • PARI
    a(n)=(sqrtint(5*n^2)+n-2)\2 \\ Charles R Greathouse IV, Feb 27 2014
  • Python
    from math import isqrt
def A022342(n): return (n+isqrt(5*n**2)>>1)-1 # Chai Wah Wu, Aug 17 2022

Formula

a(n) = floor(n*phi^2) - n - 1 = floor(n*phi) - 1 = A000201(n) - 1, where phi is the golden ratio.
a(n) = A003622(n) - n. - Philippe Deléham, May 03 2004
a(n+1) = A022290(2*A003714(n)). - R. J. Mathar, Jan 31 2015
For n > 1: A035612(a(n)) > 1. - Reinhard Zumkeller, Feb 03 2015
a(n) = A000201(n) - 1. First differences are given in A014675 (or A001468, ignoring its first term). - M. F. Hasler, Oct 13 2017
a(n) = a(n-1) + 1 + A005614(n-2) for n > 1; also a(n) = a(n-1) + A014675(n-2) = a(n-1) + A001468(n-1). - A.H.M. Smeets, Apr 26 2024

Extensions

Name edited by Peter Munn, Dec 07 2021
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