A082389 -id:A082389 - cp's OEIS frontend

A014675 The infinite Fibonacci word (start with 1, apply 1->2, 2->21, take limit).

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

Offset: 0

N. J. A. Sloane

The limiting mean and variance of the first n terms are both equal to the golden ratio (A001622). - Clark Kimberling, Mar 12 2014
Let F = A000045 (Fibonacci numbers).  For n >= 3, the first F(n)-2 terms of A014675 form a palindrome; see A001911.  If k is not one of the numbers F(n)-2, then the first k terms of A014675 do not form a palindrome. - Clark Kimberling, Jul 14 2014
First differences of A000201. - Tom Edgar, Apr 23 2015 [Editor's note: except for the offset: as for A022342, below. - M. F. Hasler, Oct 13 2017]
Also first differences of A022342 (which starts at offset 1): a(n)=A022342(n+2)-A022342(n+1), n >= 0. Equal to A001468 without its first term: a(n) = A001468(n+1), n >= 0. - M. F. Hasler, Oct 13 2017
The word is a concatenation of three runs: 1, 2, and 22. The limiting proportions of these are respectively 1/2, 1 - phi/2, and (phi - 1)/2, where phi = golden ratio.  The mean runlength is (phi + 1)/2. - Clark Kimberling, Dec 26 2010

D. Gault and M. Clint, "Curiouser and curiouser" said Alice. Further reflections on an interesting recursive function, Internat. J. Computer Math., 26 (1988), 35-43. See Table 2.
D. E. Knuth, The Art of Computer Programming, Vol. 4A, Section 7, p. 36.
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)
M. Bunder and K. Tognetti, On the self matching properties of [j tau], Discrete Math., 241 (2001), 139-151.
D. Gault and M. Clint, "Curiouser and curiouser said Alice. Further reflections on an interesting recursive function, Intern. J. Computer. Math., 26 (1988), 35-43. (Annotated scanned copy)
J. Grytczuk, Infinite semi-similar words, Discrete Math. 161 (1996), 133-141.
G. Melançon, Lyndon factorization of sturmian words, Discr. Math., 210 (2000), 137-149.
N. J. A. Sloane, Families of Essentially Identical Sequences, Mar 24 2021 (Includes this sequence)
Index entries for sequences that are fixed points of mappings

This is the {2,1} version. The standard form is A003849 (alphabet {0,1}). See also A005614 (alphabet {1,0}), A003842 (alphabet {1,2} instead of {2,1}).
Cf. A082389, A008351, A000045, A001622, A001911.
Equals A001468 except for initial term.
Differs from A025143 in many entries starting at entry 8.
First differences of A000201 and of A022342.
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

Digits := 50: t := evalf( (1+sqrt(5))/2); A014675 := n->floor((n+2)*t)-floor((n+1)*t);

Nest[ Flatten[ # /. {1 -> 2, 2 -> {2, 1}}] &, {1}, 11] (* Robert G. Wilson v *)
SubstitutionSystem[{1->{2},2->{2,1}},{1},{11}][[1]] (* Harvey P. Dale, Jan 01 2023 *)

first(n)=my(v=[1],u); while(#vCharles R Greathouse IV, Jun 21 2017

apply( {A014675(n,r=quadgen(5)-1)=(n+2)\r-(n+1)\r}, [0..99]) \\ M. F. Hasler, Apr 07 2021, improved on suggestion from Kevin Ryde, Apr 23 2021

from math import isqrt
def A014675(n): return (n+2+isqrt(m:=5*(n+2)**2)>>1)-(n+1+isqrt(m-10*n-15)>>1) # Chai Wah Wu, Aug 10 2022

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

a(n) = floor((n+2)*phi) - floor((n+1)*phi) = A000201(n+2) - A000201(n+1), phi = (1 + sqrt(5))/2.

Corrected by N. J. A. Sloane, Nov 07 2001

A096270 Fixed point of the morphism 0->01, 1->011.

Original entry on oeis.org

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, 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, 1, 1, 0, 1, 0

Offset: 0

N. J. A. Sloane, Jun 22 2004

This is another version of the Fibonacci word A005614.
(With offset 1) for k>0, a(ceiling(k*phi^2))=0 and a(floor(k*phi^2))=1 where phi=(1+sqrt(5))/2 is the Golden ratio. - Benoit Cloitre, Apr 01 2006
(With offset 1) for n>1 a(A000045(n)) = (1-(-1)^n)/2.
Equals the Fibonacci word A005614 with an initial zero.
Also the Sturmian word of slope phi (cf. A144595). - N. J. A. Sloane, Jan 13 2009
More precisely: (a(n)) is the inhomogeneous Sturmian word of slope phi-1 and intercept 0: a(n) = floor((n+1)*(phi-1)) - floor(n*(phi-1)), n >= 0. - Michel Dekking, May 21 2018
The ratio of number of 1's to number of 0's tends to the golden ratio (1+sqrt(5))/2 = 1.618... - Zak Seidov, Feb 15 2012

J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.

A.H.M. Smeets, Table of n, a(n) for n = 0..20000
Scott Balchin and Dan Rust, Computations for Symbolic Substitutions, Journal of Integer Sequences, Vol. 20 (2017), Article 17.4.1.
J-P. Borel and François Laubie, Construction de mots de Christoffel, Comptes rendus de l'Académie des sciences. Série 1, Mathématique 313.8 (1991): 483-485.
N. J. A. Sloane, Families of Essentially Identical Sequences, Mar 24 2021 (Includes this sequence)
Richard Southwell and Jianwei Huang, Complex Networks from Simple Rewrite Systems, arXiv preprint arXiv:1205.0596 [cs.SI], 2012.
Index entries for sequences that are fixed points of mappings

Cf. A003849, A096268, A001519. See A005614, A114986 for other versions.

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

[-1+Floor(n*(1+Sqrt(5))/2)-Floor((n-1)*(1+Sqrt(5))/2): n in [1..100]]; // Wesley Ivan Hurt, Aug 29 2022

Nest[ Function[l, {Flatten[(l /. {0 -> {0, 1}, 1 -> {0, 1, 1}})]}], {0}, 6] (* Robert G. Wilson v, Feb 04 2005 *)

a(n)=-1+floor(n*(1+sqrt(5))/2)-floor((n-1)*(1+sqrt(5))/2) \\ Benoit Cloitre, Apr 01 2006

from math import isqrt
def A096270(n): return (n+1+isqrt(5*(n+1)**2)>>1)-(n+isqrt(5*n**2)>>1)>>1 # Chai Wah Wu, Aug 29 2022

Conjecture: a(n) is given recursively by a(1)=0 and, for n>1, by a(n)=1 if n=F(2k+1) and a(n)=a(n-F(2k+1)) otherwise, where F(2k+1) is the largest odd-indexed Fibonacci number smaller than or equal to n. (This has been confirmed for more than nine million terms.) The odd-indexed bisection of the Fibonacci numbers (A001519) is {1, 2, 5, 13, 34, 89, ...}. So by the conjecture, we would expect that a(30) = a(30-13) = a(17) = a(17-13) = a(4) = a(4-2) = a(2) = 1, which is in fact correct. - John W. Layman, Jun 29 2004
From Michel Dekking, Apr 13 2016: (Start)
Proof of the above conjecture:
Let g be the morphism above: g(0)=01, g(1)=011. Then g^n(0) has length F(2n+1), and (a(n)) starts with g^n(0) for all n>0. Obviously g^n(0) ends in 1 for all n, proving the first part of the conjecture.
We extend the semigroup of words with letters 0 and 1 to the free group, adding the inverses 0*:=0^{-1} and 1*:=1^{-1}. Easy observation: for any word w one has g(w1)= g(w0)1. We claim that for all n>1 one has g^n(0)=u(n)v(n)v(n)0*1, where u(n)=g(u(n-1))0 and v(n)=0*g(v(n-1))0. The recursion starts with u(2)=0, v(2)=10. Indeed: g^2(0)=01011=u(2)v(2)v(2)0*1. Induction step:
  g^{n+1}(0)=g(g^n(0))= g(u(n)v(n)v(n)0*1)= g(u(n)v(n)v(n))1= g(u(n))00*g(v(n))00*g(v(n))00*1=u(n+1)v(n+1)v(n+1)0*1.
Since v(n) has length F(2n-1), which is the largest odd-indexed Fibonacci number smaller than or equal to m for all m between F(2n-1) and F(2n+1), the claim proves the second part of the conjecture. (End)
(With offset 1) a(n) = -1 + floor(n*phi) - floor((n-1)*phi) where phi=(1+sqrt(5))/2 so a(n) = -1 + A082389(n). - Benoit Cloitre, Apr 01 2006

More terms from John W. Layman, Jun 29 2004

A082844 Start with 3,2 and apply the rule a(a(1)+a(2)+...+a(n)) = a(n), fill in any undefined terms with a(t) = 2 if a(t-1) = 3 and a(t) = 3 if a(t-1) = 2.

Original entry on oeis.org

3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3

Offset: 1

Benoit Cloitre, Apr 15 2003; revised Jun 07 2003

a(1)=3, a(2)=2, a(a(1)+a(2)+...+a(n)) = a(n) and a(a(1)+a(2)+...+a(n)+1) = 5-a(n).
More generally, sequence a(n) = floor(r*(n+2))-floor(r*(n+1)), r = (1/2) *(z+sqrt(z^2+4)), z integer >=1, is defined with a(1), a(2) and a(a(1)+a(2)+...+a(n)+f(z)) = a(n); a(a(1)+a(2)+...+a(n)+f(z)+1) = (2z+1)-a(n) where f(1)=0, f(z)=z-2 for z>=2.
Conjecture: a(n) = A097509(n+1). - Benedict W. J. Irwin, Mar 13 2016. [See the discussion in A097509. - N. J. A. Sloane, Mar 09 2021]
Theorem: Referring to the solution to Problem B6 in the 81st William Lowell Putnam Mathematical Competition (see link), in the notation of the first solution, the sequence a(n) = c_{n+1} indexed from 1 equals the present sequence, A082844. - Manjul Bhargava, Kiran Kedlaya, and Lenny Ng, Sep 09 2021.

Manjul Bhargava, Kiran Kedlaya, and Lenny Ng, Solutions to the 81st William Lowell Putnam Mathematical Competition
Luke Schaeffer, Jeffrey Shallit, and Stefan Zorcic, Beatty Sequences for a Quadratic Irrational: Decidability and Applications, arXiv:2402.08331 [math.NT], 2024. See pp. 17-19.
N. J. A. Sloane, Families of Essentially Identical Sequences, Mar 24 2021 (Includes this sequence)

Cf. A082389, A082845, A097509.

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

[Floor((1+Sqrt(2))*(n+2))-Floor((1+Sqrt(2))*(n+1)) : n in [1..100]]; // Wesley Ivan Hurt, Mar 13 2016

A082844:=n->floor((1+sqrt(2))*(n+2))-floor((1+sqrt(2))*(n+1)): seq(A082844(n), n=1..100); # Wesley Ivan Hurt, Mar 13 2016

With[{r=1+Sqrt[2]},Table[Floor[r*(n+2)]-Floor[r*(n+1)],{n,110}]] (* Harvey P. Dale, Oct 10 2012 *)

from math import isqrt
def A082844(n): return 1+isqrt((n+2)**2<<1)-isqrt((n+1)**2<<1) # Chai Wah Wu, May 24 2025

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

A058065 Complement of A057843.

Original entry on oeis.org

0, 1, 3, 5, 6, 8, 9, 11, 13, 14, 16, 18, 19, 21, 22, 24, 26, 27, 29, 30, 32, 34, 35, 37, 39, 40, 42, 43, 45, 47, 48, 50, 52, 53, 55, 56, 58, 60, 61, 63, 64, 66, 68, 69, 71, 73, 74, 76, 77, 79, 81, 82, 84, 85, 87, 89, 90, 92, 94, 95, 97, 98, 100, 102, 103, 105, 107, 108, 110

Offset: 0

N. J. A. Sloane, Nov 24 2000

Jon Asier Bárcena-Petisco, Luis Martínez, María Merino, Juan Manuel Montoya, and Antonio Vera-López, Fibonacci-like partitions and their associated piecewise-defined permutations, arXiv:2503.19696 [math.CO], 2025. See p. 3.

Cf. A057843, A082389.

Table[Floor[n*GoldenRatio+Sqrt[5]],{n,0,100}]-2 (* Harvey P. Dale, Mar 01 2019 *)

a(n) = floor(tau*n+sqrt(5))-2 where tau is the Golden ratio = (1+sqrt(5))/2. - Benoit Cloitre, Apr 14 2003

A245977 Limit-reverse of the infinite Fibonacci word A014675 = (s(0),s(1),...) = (2,1,2,2,1,2,1,2, ...) using initial block (s(2),s(3)) = (2,2).

Original entry on oeis.org

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

Offset: 0

Clark Kimberling and Peter J. C. Moses, Aug 10 2014

nonn

Suppose, as in A245920, that S = (s(0),s(1),s(2),...) is an infinite sequence such that every finite block of consecutive terms occurs infinitely many times in S.  (It is assumed that A014675 is such a sequence.)  Let B = B(m,k) = (s(m-k),s(m-k+1),...,s(m)) be such a block, where m >= 0 and k >= 0.  Let m(1) be the least i > m such that (s(i-k), s(i-k+1),...,s(i)) = B(m,k), and put B(m(1),k+1) = (s(m(1)-k-1),s(m(1)-k),...,s(m(1))).  Let m(2) be the least i > m(1) such that (s(i-k-1),s(i-k),...,s(i)) = B(m(1),k+1), and put B(m(2),k+2) = (s(m(2)-k-2),s(m(2)-k-1),...,s(m(2))).  Continuing in this manner gives a sequence of blocks B(m(n),k+n).  Let B'(n) = reverse(B(m(n),k+n)), so that for n >= 1, B'(n) comes from B'(n-1) by suffixing a single term; thus the limit of B'(n) is defined; we call it the "limit-reverse of S with initial block B(m,k)", denoted by S*(m,k), or simply S*.
...
The sequence (m(i)), where m(0) = 0, is the "index sequence for limit-reversing S with initial block B(m,k)" or simply the index sequence for S*, as in A245921 and A245978.
Apparently a(n) = A082389(n+2) = A059426(n+1). - R. J. Mathar, Sep 01 2014

			S = the infinite Fibonacci word A014675, with B = (s(2), s(3)); that is, (m,k) = (2,3)
S = (2,1,2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,...)
B'(0) = (2,2)
B'(1) = (2,2,1)
B'(2) = (2,2,1,2)
B'(3) = (2,2,1,2,1)
B'(4) = (2,2,1,2,1,2)
B'(5) = (2,2,1,2,1,2,2)
S* = (2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,1,...), with index sequence (3,8,11,16,21,29,...)

Clark Kimberling, Table of n, a(n) for n = 0..300

Cf. A245978, A245979, A245920, A014675.

z = 100; seqPosition2[list_, seqtofind_] := Last[Last[Position[Partition[list, Length[#], 1], Flatten[{_, #, _}], 1, 2]]] &[seqtofind]; x = GoldenRatio; s = Differences[Table[Floor[n*x], {n, 1, z^2}]]; ans = Join[{s[[p[0] = pos = seqPosition2[s, #] - 1]]}, #] &[{s[[3]], s[[4]]}]; (* Initial block is (s(3),s(4)) [OR (s(2),s(3)) if using offset 0] *); cfs = Table[s = Drop[s, pos - 1]; ans = Join[{s[[p[n] = pos = seqPosition2[s, #] - 1]]}, #] &[ans], {n, z}]; rcf = Last[Map[Reverse, cfs]] (* A245977 *)

cp's OEIS Frontend

A014675 The infinite Fibonacci word (start with 1, apply 1->2, 2->21, take limit).

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

Python

Formula

Extensions

A096270 Fixed point of the morphism 0->01, 1->011.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Python

Formula

Extensions

A082844 Start with 3,2 and apply the rule a(a(1)+a(2)+...+a(n)) = a(n), fill in any undefined terms with a(t) = 2 if a(t-1) = 3 and a(t) = 3 if a(t-1) = 2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Python

Formula

A058065 Complement of A057843.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

A245977 Limit-reverse of the infinite Fibonacci word A014675 = (s(0),s(1),...) = (2,1,2,2,1,2,1,2, ...) using initial block (s(2),s(3)) = (2,2).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica