A005713 Define strings S(0)=0, S(1)=11, S(n) = S(n-1)S(n-2); iterate.

1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1

Offset

0,1

Comments

a(A035336(n)) = 0. - Reinhard Zumkeller, Dec 30 2011

a(n) = 1 - A123740(n). This can be seen as follows. Define words T(0)=0, T(1)=1, T(n) = T(n-1)T(n-2). Then T(infinity) is the binary complement of the infinite Fibonacci word A003849. Obviously S(n) is the [1->11] transform of T(n). The claim now follows from the observation (see Comments of A123740) that doubling the 0's in the infinite Fibonacci word A003849 gives A123740. - Michel Dekking, Oct 21 2018

From Michel Dekking, Oct 22 2018: (Start)

Here is a proof of Cloitre's (corrected) formula

a(n) = abs(A014677(n+1)).

Since abs(-1) = abs(1) = 1, one has to prove that A014677(k)=0 if and only if there is an n such that AB(n) = k (using that a(n) = 1 - A123740(n)). Now A014677 is the sequences of first differences of A001468, and the 0's in A014677 occur if and only if there occurs a block 22 in A001468, which is given by

A001468(n) = floor((n+1)*phi) - floor(n*phi), n >= 0.

But then

A001468(n) = A014675(n-1), n > 0.

The sequence A014675 is fixed point of the morphism 1->2, 2->21, which is alphabet equivalent to the morphism 1->12, 2->1, the classical Fibonacci morphism in standard form. This implies that the 22 blocks in A001468 occur at position n+1 in if and only if 3 occurs in the fixed point A270788 of the 3-symbol Fibonacci morphism at k, which happens if and only if there is an n such that AB(n)=k (see Formula of A270788). (End)

Links

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

R. K. Guy, Letter to N. J. A. Sloane, 1987

Jeffrey Shallit and Anatoly Zavyalov, Transduction of Automatic Sequences and Applications, arXiv:2303.15203 [cs.FL], 2023, see p. 31.

Formula

From Benoit Cloitre, Apr 21 2003: (Start)

For n > 1, a(n-1) = floor(phi*ceiling(n/phi)) - ceiling(phi*floor(n/phi)) where phi=(1+sqrt(5))/2.

For n >= 0, a(n) = abs(A014677(n+1)). (End)

Example

The infinite word is S(infinity) = 110111101101111011110110...

Mathematica

s[0] = {0}; s[1] = {1, 1}; s[n_] := s[n] = Join[s[n-1], s[n-2]]; s[10] (* Jean-François Alcover, May 15 2013 *)
nxt[{a_, b_}]:={b, Join[a, b]}; Drop[Nest[nxt, {{0}, {1, 1}}, 10][[1]], 3] (* Harvey P. Dale, Jan 31 2019 *)

Prog

(PARI) a(n, f1, f2)=local(f3); for(i=3, n, f3=concat(f2, f1); f1=f2; f2=f3); f2
(PARI) printp(a(10, [ 0 ], [ 1, 1 ])) \\ Would give S(10). Sequence is S(infinity).
(Haskell)
a005713 n = a005713_list !! n
a005713_list = 1 : 1 : concat (sibb [0] [1, 1]) where
   sibb xs ys = zs : sibb ys zs where zs = xs ++ ys
-- Reinhard Zumkeller, Dec 30 2011

Crossrefs

Cf. A005614, A003849.

Cf. A001468, A014675, A014677, A123740, A270788.

Sequence in context: A285668 A267813 A323000 * A188031 A305387 A085241

Adjacent sequences: A005710 A005711 A005712 * A005714 A005715 A005716

Keyword

nonn,easy,nice

Author

N. J. A. Sloane

Extensions

Corrected by Michael Somos

Status

approved