cp's OEIS Frontend

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)