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 A005713 Define strings S(0)=0, S(1)=11, S(n) = S(n-1)S(n-2); iterate. 4
 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 (list; graph; refs; listen; history; text; internal format)
 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 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}; s = {1, 1}; s[n_] := s[n] = Join[s[n-1], s[n-2]]; s (* Jean-François Alcover, May 15 2013 *) nxt[{a_, b_}]:={b, Join[a, b]}; Drop[Nest[nxt, {{0}, {1, 1}}, 10][], 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  [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: A267813 A323000 A181183 * A188031 A305387 A085241 Adjacent sequences:  A005710 A005711 A005712 * A005714 A005715 A005716 KEYWORD nonn,easy,nice AUTHOR EXTENSIONS Corrected by Michael Somos STATUS approved

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Last modified August 4 09:50 EDT 2021. Contains 346446 sequences. (Running on oeis4.)