Let u = A000201 = (lower Wythoff sequence) and v = A001950 = (upper Wythoff sequence. Conjecture: This sequence is the sequence p given by p(1) = 0 and p(u(k)) = p(k); p(v(k)) = 1p(k). [Clark Kimberling, Apr 15 2011]
[base 2] 0.111010010001100... = 0.9105334708635617... [Joerg Arndt, May 13 2011]
From Michel Dekking, Nov 28 2019: (Start)
This sequence is a morphic sequence.
Let the morphism sigma be given by
sigma(1) = 12, sigma(2) = 4, sigma(3) = 1, sigma(4) = 43,
and let the lettertoletter map lambda be given by
lambda(1) = 0, lambda(2) = 1, lambda(3) = 0, lambda(4) = 1.
Then a(n) = lambda(x(n)), where x(0)x(1)... = 1244343... is the fixed point of sigma starting with 1.
See footnote number 4 in the paper by Emmanuel Ferrand. (End)
From Michel Dekking, Nov 29 2019: (Start)
Proof of Kimberling's conjecture, by virtue of the four symbol morphism sigma in the previous comment.
We first show that the fixed point x = 1244343143112... of sigma has the remarkable property that the letters 1 and 4 exclusively occur at positions u(k), k=1,2,..., and the letters 2 and 3 exclusively occur at positions v(k) k=1,2,....
To see this, let alpha be the Fibonacci morphism on the alphabet {a,b}:
alpha(a) = ab, alpha(b) = a.
It is well known that the lower Wythoff sequence u gives the positions of a in the fixed point abaababaab... of alpha, and that the upper Wythoff sequence v gives the positions of b in this infinite Fibonacci word.
Let pi be the projection from {1,2,3,4} to {a,b} given by
pi(1) = pi(4) = a, pi(2) = pi(3) = b.
One easily checks that (composition of morphisms)
pi sigma = alpha pi.
This implies the remarkable property stated above.
What remains to be proved is that lambda(x(u(k)) = a(k) and lambda(x(v(k)) = a(k), where lambda is the lettertoletter map in the previous comment. We tackle this problem by an extensive analysis of the return words of the letter 1 in the sequence x= 1244343143112...
These are the words occurring in x which start with 1 and have no other occurrences of 1 in them.
One easily finds that they are given by
A:=1, B:=12, C:=124, D:=143, E:=1243, F:=12443, G:=1244343.
These words partition x. Moreover, sigma induces a morphism rho on the alphabet {A,B,C,D,E,F,G} given by
rho(A) = B, rho(B) =C, rho(C) = F, rho(D) = EA,
rho(E) = FA, rho(F) = GA, rho(G) = GDA.
CLAIM 1: Let mu be the lettertoletter map given by
mu(A)=mu(B) = 1, mu(C)=mu(D)=mu(E) = 2, mu(F) = 3, mu(G) = 4,
then mu(R) = (s(n+1)s(n)), where R = GDAEABFAB... is the unique fixed point of rho and s = 1,5,7,8,10,... is the sequence of positions of 1 in (x(u(k)).
Proof of CLAIM 1: The 1's in x occur exclusively at positions u(k), so if we want the differences s(n+1)s(n), we can strip the 2's and 3's from the return words of 1, and record how long it takes to the next 1. This is the length of the stripped words A~=1, B~=1, C~=14, ... which is given by mu.
CLAIM 2: Let delta be the 'decoration' morphism given by
delta(A) = 1, delta(B) = 2, delta(C) = 3, delta(D) = 21,
delta(E) = 31, delta(F) = 41, delta(G) = 421,
then delta(R) = (t(n+1t(n)), where rho(R) = R, and t is the sequence of positions of 1 or 3 in the sequence x = 1244343143112....
Proof of CLAIM 2: We have to record the differences in the occurrences of 1 and 3 in the return words of 1. These are given by delta. For example: F = 12443, where 1 and 3 occur at position 1 and 5; and the next 1 will occur at the beginning of any of the seven words A,...,G.
If we combine CLAIM 1 and CLAIM 2 with lambda(1) = lambda(3) = 0, we obtain the first half of Kimberling's conjecture, simply because delta = mu rho, and rho(R) = R.
The second half of the conjecture is obtained in a similar way. (End)
From Michel Dekking, Mar 10 2020: (Start)
The fact that this sequence is a morphic sequence can be easily deduced from the morphism tau given in A007895:
tau((j,0)) = (j,0) (j+1,1),
tau((j,1)) = (j,0).
The sum of digits of a number n in Zeckendorf expansion is given by projecting the nth letter in the fixed point of tau starting with (0,0) on its first coordinate: (j,i)>j for i=0,1.
If we consider all the j's modulo 2, we obtain a morphism sigma' on 4 letters:
sigma'((0,0)) = (0,0) (1,1),
sigma'((0,1)) = (0,0),
sigma'((1,0)) = (1,0)(0,1),
sigma'((1,1)) = (1,0).
The change of alphabet (0,0)>1, (0,1)>3, (1,0)>4, (1,1)>2, turns sigma' into sigma given in my Nov 28, 2019 comment. The projection map turns into the map 1>0, 2>1, 3>0, 4>1, as in my Nov 28 2019 comment.
(End)
Another proof that this sequence is a morphic sequence has been given in the monograph by Allouche and Shallit. They demonstrate on page 239 that (a(n)) is a letter to letter projection of a fixed point of a morphism on an alphabet of six letters. However, it can been seen easily that two pairs of letters can be merged, yielding a morphism on an alphabet of four letters, which after a change of alphabet equals the morphism from my previous comments.  Michel Dekking, Jun 25 2020
