

A095076


Parity of 1fibits in Zeckendorf expansion A014417(n).


8



0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0
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OFFSET

0,1


COMMENTS

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)


LINKS

Table of n, a(n) for n=0..101.
Joerg Arndt, Matters Computational (The Fxtbook), section 38.11.1, pp. 754756
E. Ferrand, An analogue of the ThueMorse sequence, The Electronic Journal of Combinatorics, Volume 14 (2007), R30.
Leonard Rozendaal, Pisano word, tesselation, planefilling fractal, Preprint, 2017.
Index entries for characteristic functions


MATHEMATICA

r=(1+5^(1/2))/2; u[n_] := Floor[n*r]; (* A000201 *)
a[1] = 0; h = 128;
c = (u[#1] &) /@ Range[2h];
d = (Complement[Range[Max[#1]], #1] &)[c]; (* A001950 *)
Table[a[d[[n]]] = 1  a[n], {n, 1, h  1}];
Table[a[c[[n]]] = a[n], {n, 1, h}] (* A095076 conjectured *)
Flatten[Position[%, 0]] (* A189034 *)
Flatten[Position[%%, 1]] (* A189035 *)


PROG

(Python)
def ok(n): return 1 if n==0 else n*(2*n & n == 0)
print [bin(n)[2:].count("1")%2 for n in range(0, 1001) if ok(n)] # Indranil Ghosh, Jun 08 2017


CROSSREFS

a(n) = A010060(A003714(n)). a(n) = 1  A095111(n). Characteristic function of A020899. Run counts are given by A095276.
Cf. A189034, A189035 (positions of 0 and 1 if the conjecture is valid.
Sequence in context: A174206 A265333 A159637 * A285080 A167392 A190201
Adjacent sequences: A095073 A095074 A095075 * A095077 A095078 A095079


KEYWORD

nonn,changed


AUTHOR

Antti Karttunen, Jun 01 2004


STATUS

approved



