A120613 -id:A120613 - cp's OEIS frontend

A110007 a(n) = n-F(F(F(n))) where F(x)=A120613(x)=floor(phi*floor(x/phi)) and phi=(1+sqrt(5))/2.

1, 2, 3, 4, 4, 5, 4, 5, 5, 4, 5, 4, 4, 5, 4, 5, 5, 4, 5, 4, 5, 5, 4, 5, 4, 4, 5, 4, 5, 5, 4, 5, 4, 4, 5, 4, 5, 5, 4, 5, 4, 5, 5, 4, 5, 4, 4, 5, 4, 5, 5, 4, 5, 4, 5, 5, 4, 5, 4, 4, 5, 4, 5, 5, 4, 5, 4, 4, 5, 4, 5, 5, 4, 5, 4, 5, 5, 4, 5, 4, 4, 5, 4, 5, 5, 4, 5, 4, 4, 5, 4, 5, 5, 4, 5, 4, 5, 5, 4, 5, 4, 4, 5, 4, 5

Offset: 1

Benoit Cloitre, Sep 02 2005

To build the sequence start from the infinite Fibonacci word: b(k)=floor(k/phi)-floor((k-1)/phi) for k>=1 giving 0,1,0,1,1,0,1,0,1,1,0,1,1,0,1,0,1,1,..... Then replace each 0 by the block {4,5,4} and each 1 by the block {5,5,4,5,4}. Append the initial string {1,2,3,4}.

Benoit Cloitre, On properties of irrational numbers related to the floor function, in preparation, 2005.

Cf. A005614 (infinite Fibonacci binary word), A120613.

Cf. sequences for a(n) = n-F^k(n): A003842 (k=1), A110006 (k=2), A110010 (k=4), A110011 (k=5).

Join[{1,2,3,4},Flatten[Table[Floor[k/GoldenRatio]-Floor[(k-1)/ GoldenRatio],{k,30}]/.{0->{4,5,4},1->{5,5,4,5,4}}]] (* Harvey P. Dale, Dec 12 2017 *)

F(x)=floor((1+sqrt(5))/2*floor((-1+sqrt(5))/2*x))
a(n)=n-F(F(F(n)))

A110010 a(n) = n-F(F(F(F(n)))) where F(x)=A120613(x)=floor(phi*floor(x/phi)) and phi=(1+sqrt(5))/2.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 6, 7, 6, 6, 7, 6, 5, 6, 6, 7, 6, 6, 7, 6, 7, 6, 6, 7, 6, 5, 6, 6, 7, 6, 6, 7, 6, 5, 6, 6, 7, 6, 6, 7, 6, 7, 6, 6, 7, 6, 5, 6, 6, 7, 6, 6, 7, 6, 7, 6, 6, 7, 6, 5, 6, 6, 7, 6, 6, 7, 6, 5, 6, 6, 7, 6, 6, 7, 6, 7, 6, 6, 7, 6, 5, 6, 6, 7, 6, 6, 7, 6, 5, 6, 6, 7, 6, 6, 7, 6, 7, 6, 6, 7, 6, 5, 6, 6, 7

Offset: 1

Benoit Cloitre, Sep 02 2005

To built the sequence start from the infinite Fibonacci word b(k)=floor(k/phi)-floor((k-1)/phi) for k>=1 giving 0,1,0,1,1,0,1,0,1,1,0,1,1,0,1,0,1,1,..... Then replace each 0 by the block {5,6,6} and each 1 by the block {7, 6, 6, 7, 6}. Append the initial string {1,2,3,4}.

Benoit Cloitre, On properties of irrational numbers related to the floor function, in preparation, 2005.

Cf. A005614 (infinite Fibonacci binary word), A120613.

Cf. sequences for a(n) = n-F^k(n): A003842 (k=1), A110006 (k=2), A110007 (k=3), A110011 (k=5).

F(x)=floor((1+sqrt(5))/2*floor((-1+sqrt(5))/2*x)); a(n)=n-F(F(F(F(n))))

A110011 a(n) = n-F(F(F(F(F(n))))) = n-F^5(n) where F(x)=A120613(x)=floor(phi*floor(x/phi)) and phi=(1+sqrt(5))/2.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 8, 7, 8, 8, 7, 8, 7, 8, 8, 7, 8, 8, 9, 8, 7, 8, 8, 7, 8, 7, 8, 8, 7, 8, 8, 7, 8, 7, 8, 8, 7, 8, 8, 9, 8, 7, 8, 8, 7, 8, 7, 8, 8, 7, 8, 8, 9, 8, 7, 8, 8, 7, 8, 7, 8, 8, 7, 8, 8, 7, 8, 7, 8, 8, 7, 8, 8, 9, 8, 7, 8, 8, 7, 8, 7, 8, 8, 7, 8, 8, 7, 8, 7, 8, 8, 7, 8, 8, 9, 8, 7, 8, 8, 7, 8, 7, 8

Offset: 1

Benoit Cloitre, Sep 02 2005

To built the sequence start from the infinite Fibonacci word b(k)=floor(k/phi)-floor((k-1)/phi) for k>=2 giving 1,0,1,1,0,1,0,1,1,0,1,1,0,1,0,1,1,..... Then replace each 0 by the block {9,8,7,8,8,7,8,7,8,8,7,8,8} and each 1 by the block {9,8,7,8,8,7,8,7,8,8,7,8,8,7,8,7,8,8,7,8,8}. Append the initial string {1,2,3,4,5,6,7,8,8,7,8,8,7,8,7,8,8,7,8,8}.

Benoit Cloitre, On properties of irrational numbers related to the floor function, in preparation, 2005.

Cf. A005614 (infinite Fibonacci binary word), A120613.

Cf. sequences for a(n) = n-F^k(n): A003842 (k=1), A110006 (k=2), A110007 (k=3), A110010 (k=4).

F(x)=floor((1+sqrt(5))/2*floor((-1+sqrt(5))/2*x)); a(n)=n-F(F(F(F(F(n)))))

A003842 The infinite Fibonacci word: start with 1, repeatedly apply the morphism 1->12, 2->1, take limit; or, start with S(0)=2, S(1)=1, and for n>1 define S(n)=S(n-1)S(n-2), then the sequence is S(oo).

Original entry on oeis.org

1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1

Offset: 0

N. J. A. Sloane

Or, fixed point of the morphism 1->12, 2->1, starting from a(1) = 2.
A Sturmian word, as are all versions of this sequence. This means that if one slides a window of length n along the sequence, one sees exactly n+1 different subwords (see A213975). For a proof, see for example Chap. 2 of Lothaire (2002).
The limiting mean of the first n terms is 3 - phi, where phi is the golden ratio (A001622); the limiting variance is 2 - phi. - Clark Kimberling, Mar 12 2014
The Wikipedia article on L-system Example 1 is "Algae" given by the axiom: A and rules: A -> AB, B -> A. The sequence G(n) = G(n-1)G(n-2) yields this sequence when A -> 1, B -> 2. - Michael Somos, Jan 12 2015
In the limit #1's : #2's = phi : 1. - Frank M Jackson, Mar 12 2018

			Over the alphabet {a,b} this is the sequence 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, ...

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. Berstel and J. Karhumaki, Combinatorics on words - a tutorial, Bull. EATCS, #79 (2003), pp. 178-228.
E. Bombieri and J. Taylor, Which distribution of matter diffracts? An initial investigation, in International Workshop on Aperiodic Crystals (Les Houches, 1986), J. de Physique, Colloq. C3, 47 (1986), C3-19 to C3-28.
Aldo de Luca and Stefano Varricchio, Finiteness and regularity in semigroups and formal languages. Monographs in Theoretical Computer Science. An EATCS Series. Springer-Verlag, Berlin, 1999. x+240 pp. ISBN: 3-540-63771-0 MR1696498 (2000g:68001). See p. 25.
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.
G. Melançon, Factorizing infinite words using Maple, MapleTech journal, vol. 4, no. 1, 1997, pp. 34-42, esp. p. 36.

T. D. Noe, Table of n, a(n) for n=0..10945 (20 iterations)
J.-P. Allouche and M. Mendes France, Automata and Automatic Sequences, in: Axel F. and Gratias D. (eds), Beyond Quasicrystals. Centre de Physique des Houches, vol 3. Springer, Berlin, Heidelberg, pp. 293-367, 1995; DOI https://doi.org/10.1007/978-3-662-03130-8_11.
J.-P. Allouche and M. Mendes France, Automata and Automatic Sequences, in: Axel F. and Gratias D. (eds), Beyond Quasicrystals. Centre de Physique des Houches, vol 3. Springer, Berlin, Heidelberg, pp. 293-367, 1995; DOI https://doi.org/10.1007/978-3-662-03130-8_11. [Local copy]
Scott Balchin and Dan Rust, Computations for Symbolic Substitutions, Journal of Integer Sequences, Vol. 20 (2017), Article 17.4.1.
Jean Berstel, Home Page
Julien Cassaigne, On extremal properties of the Fibonacci word, RAIRO-Theor. Inf. Appl. 42 (2008) 701-715.
J. Endrullis, D. Hendriks and J. W. Klop, Degrees of streams.
S. Ferenczi, Complexity of sequences and dynamical systems, Discrete Math., 206 (1999), 145-154.
J. Grytczuk, Infinite semi-similar words, Discrete Math. 161 (1996), 133-141.
A. Hof, O. Knill and B. Simon, Singular continuous spectrum for palindromic Schrödinger operators, Commun. Math. Phys. 174 (1995), 149-159.
Clark Kimberling, A Self-Generating Set and the Golden Mean, J. Integer Sequences, 3 (2000), #00.2.8.
Clark Kimberling, Intriguing infinite words composed of zeros and ones, Elemente der Mathematik (2021).
M. Lothaire, Algebraic Combinatorics on Words, Cambridge, 2002, see p. 41, etc.
G. Melançon, Lyndon factorization of sturmian words, Discr. Math., 210 (2000), 137-149.
F. Mignosi, A. Restivo and M. Sciortino, Words and forbidden factors, WORDS (Rouen, 1999). Theoret. Comput. Sci. 273 (2002), no. 1-2, 99--117. MR1872445 (2002m:68096) - From _N. J. A. Sloane_, Jul 10 2012
F. Mignosi and L. Q. Zamboni, On the number of Arnoux-Rauzy words, Acta arith., 101 (2002), no. 2, 121-129.
T. D. Noe, The first 1652 subwords of A003849, including leading zeros.
Giuseppe Pirillo, Fibonacci numbers and words, Discrete Math. 173 (1997), no. 1-3, 197--207. MR1468849 (98g:68135).
Patrice Séébold, Look and Say Fibonacci, RAIRO-Theor. Inf. Appl. 42 (2008) 729-746.
N. J. A. Sloane, The first 10946 terms, concatenated
N. J. A. Sloane, Families of Essentially Identical Sequences, Mar 24 2021 (Includes this sequence)
P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 22-31.
Eric Weisstein's World of Mathematics, Golden Ratio
Wikipedia, L-system Example 1: Algae
Index entries for sequences that are fixed points of mappings

A003849 is another common version of this sequence.
See also A014675, A005614, A001468, A106750, A213975, A213976, A214317, A214318, A182028, A120613.
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

a003842 n = a003842_list !! n
a003842_list = tail $ concat fws where
   fws = [2] : [1] : (zipWith (++) fws $ tail fws)
-- Reinhard Zumkeller, Oct 26 2013

Nest[ Flatten[ # /. {1 -> {1, 2}, 2 -> {1}}] &, {1}, 10] (* Robert G. Wilson v, Mar 04 2005 *)
Table[n + 1 - Floor[((1 + Sqrt[5])/2)*Floor[2*(n + 1)/(1 + Sqrt[5])]], {n, 1, 50}] (* G. C. Greubel, May 18 2017 *)
SubstitutionSystem[{1->{1,2},2->{1}},{1},{10}][[1]] (* Harvey P. Dale, Nov 19 2022 *)

for(n=1,50, print1(n+1 - floor(((1+sqrt(5))/2)*floor(2*(n+1)/(1+sqrt(5)))), ", ")) \\ G. C. Greubel, May 18 2017

def A003842(length):
    a = [1]
    while len(a)Nicholas Stefan Georgescu, Jun 14 2022

def aupto(nn):
    S, Fnm2, Fnm1 = [1, 2], 1, 2
    while len(S) < nn+1:
        S += S[:min(Fnm2, nn+1-len(S))]
        Fnm2, Fnm1 = Fnm1, Fnm1+Fnm2
    return S
print(aupto(104)) # Michael S. Branicky, Jun 06 2022

from math import isqrt
def A003842(n): return n+2-((m:=(n+2+isqrt(5*(n+2)**2)>>1)-n-2)+isqrt(5*m**2)>>1) # Chai Wah Wu, Aug 26 2022

Define strings S(0)=2, S(1)=1, S(n)=S(n-1)S(n-2); iterate. Sequence is S(infinity).

a(n) = n + 2 - A120613(n+1). - Benoit Cloitre, Jul 28 2005 [Corrected by N. J. A. Sloane, Jun 30 2018]

Entry revised by N. J. A. Sloane, Jul 03 2012

A120614 a(n) = g(n+1) - g(n) where g(k) = floor(phi*floor(k/phi)) and phi = (1+sqrt(5))/2.

Original entry on oeis.org

1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0

Offset: 1

Benoit Cloitre, Jun 17 2006

nonn

From Michel Dekking, Oct 29 2018: (Start)
Here is a proof that (a(n)) is fixed point of the morphism 0->102, 1->102, 2->02.
Let alpha:=phi-1. Then alpha*phi = 1. So
      g(k) = floor(phi*floor(k*alpha)).
Write k*alpha = floor(k*alpha) + {k*alpha}, i.e., {k*alpha} is the fractional part of k*alpha. Then
      g(k) = floor(phi*(k*alpha-{k*alpha})) = k + floor(-phi*{k*alpha}).
Thus
      a(n) = n+1 +floor(-phi*{(n+1)*alpha})-n -floor(-phi*{n*alpha}).
It follows that
      a(n)  = 1 - floor(phi*{(n+1)*alpha})  + floor(phi*{n*alpha}).
The difference -floor(phi*{(n+1)*alpha}) + floor(phi*{n*alpha}) is equal to -1, 0 or 1, since floor(phi*{n*alpha}) is equal to 0 or 1.
In fact, phi*{n*alpha} can only take values between 0 and 1.619, and floor(phi*{n*alpha}) = 0 if and only if
      {n*alpha} < 1/phi = alpha.
This is the same (putting rho:=1-alpha) as requiring
      {n*alpha+rho} < 1-alpha.
Via the rotation description of Sturmian sequences (see, e.g., Lothaire), one sees that this sequence is the inhomogeneous Sturmian sequence s(alpha, rho), but with offset 1, and with 0 and 1 exchanged. Since rho+alpha=1, it follows that s(alpha, rho) with offset 2 equals  s(1-alpha, 1-alpha), the classical Fibonacci sequence xF:=A003849, fixed point of 0->01, 1->0. We have found that
      a(n+1)=0 iff xF(n)=0, xF(n+1)=1,
      a(n+1)=1 iff xF(n)=0, xF(n+1)=0,
      a(n+1)=2 iff xF(n)=1, xF(n+1)=0.
This means that (a(n+1)) equals the 3-symbol Fibonacci sequence A270788 on the alphabet {0,2,1}. Then Proposition 5 in "Morphisms, Symbolic Sequences, and Their Standard Forms" yields that (a(n)) is fixed point of the morphism 0->102, 1->102, 2->02. (End)

Muniru A Asiru, Table of n, a(n) for n = 1..1000
F. Michel Dekking, Morphisms, Symbolic Sequences, and Their Standard Forms, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.
M. Lothaire, Algebraic Combinatorics on Words, Cambridge, 2002.

Cf. A003849, A120613, A120615, A270788.

[Floor((1+Sqrt(5))*Floor(2*(k+1)/(1+Sqrt(5)))/2) -
Floor((1+Sqrt(5))*Floor(2*k/(1+Sqrt(5)))/2): k in [1..100]]; // G. C. Greubel, Oct 23 2018

g:=k->floor((1+sqrt(5))/2*floor(k/((1+sqrt(5))/2))): seq(g(n+1)-g(n),n=1..110); # Muniru A Asiru, Oct 21 2018

#[[2]]-#[[1]]&/@Partition[Table[Floor[GoldenRatio*Floor[n/GoldenRatio]],{n,0,110}],2,1] (* Harvey P. Dale, Dec 14 2012 *)

{phi=(1+sqrt(5))/2; g(k)=floor(phi*floor(k/phi))};
vector(100, n, g(n+1)-g(n)) \\ G. C. Greubel, Oct 23 2018

from math import isqrt
def A120614(n): return ((m:=(n+1+isqrt(5*(n+1)**2)>>1)-n-1)+isqrt(5*m**2)>>1)-((k:=(n+isqrt(5*n**2)>>1)-n)+isqrt(5*k**2)>>1) # Chai Wah Wu, May 22 2025

a(floor(k*phi)+k+1)=0; a(floor(k*phi)+k+2)=2, if n is not in {floor(k*phi)+k+1}U{floor(k*phi)+k+2}_{k>=1} a(n)=1.
(a(n)) is a fixed point of the morphism 02-->10202 and 102-->10210202. [Corrected by Michel Dekking, Oct 29 2018]
Fixed point of the morphism 0->102, 1->102, 2->02. - Michel Dekking, Oct 21 2018

Initial 0 removed from data by Michel Dekking, Oct 22 2018

cp's OEIS Frontend

A110006 a(n) = n-F(F(n)) where F(x)=A120613(x)=floor(phi*floor(x/phi)) and phi=(1+sqrt(5))/2.

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A110007 a(n) = n-F(F(F(n))) where F(x)=A120613(x)=floor(phi*floor(x/phi)) and phi=(1+sqrt(5))/2.

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A110010 a(n) = n-F(F(F(F(n)))) where F(x)=A120613(x)=floor(phi*floor(x/phi)) and phi=(1+sqrt(5))/2.

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A110011 a(n) = n-F(F(F(F(F(n))))) = n-F^5(n) where F(x)=A120613(x)=floor(phi*floor(x/phi)) and phi=(1+sqrt(5))/2.

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A003842 The infinite Fibonacci word: start with 1, repeatedly apply the morphism 1->12, 2->1, take limit; or, start with S(0)=2, S(1)=1, and for n>1 define S(n)=S(n-1)S(n-2), then the sequence is S(oo).

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A120614 a(n) = g(n+1) - g(n) where g(k) = floor(phi*floor(k/phi)) and phi = (1+sqrt(5))/2.

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A120615 a(n) = Sum_{k=0..n} floor(phi*floor(k/phi)) where phi = (1+sqrt(5))/2.

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