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A096270
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Fixed point of the morphism 0->01, 1->011.
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35
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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
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OFFSET
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0,1
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COMMENTS
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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.
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
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REFERENCES
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J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
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LINKS
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FORMULA
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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
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
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MATHEMATICA
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Nest[ Function[l, {Flatten[(l /. {0 -> {0, 1}, 1 -> {0, 1, 1}})]}], {0}, 6] (* Robert G. Wilson v, Feb 04 2005 *)
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PROG
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(PARI) a(n)=-1+floor(n*(1+sqrt(5))/2)-floor((n-1)*(1+sqrt(5))/2) \\ Benoit Cloitre, Apr 01 2006
(Python)
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
(Magma) [-1+Floor(n*(1+Sqrt(5))/2)-Floor((n-1)*(1+Sqrt(5))/2): n in [1..100]]; // Wesley Ivan Hurt, Aug 29 2022
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CROSSREFS
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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
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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