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 A095076 Parity of 1-fibits 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 (list; graph; refs; listen; history; text; internal format)
 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)) = 1-p(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 letter-to-letter 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 letter-to-letter 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 letter-to-letter 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+1-t(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 Joerg Arndt, Matters Computational (The Fxtbook), section 38.11.1, pp. 754-756 E. Ferrand, An analogue of the Thue-Morse sequence, The Electronic Journal of Combinatorics, Volume 14 (2007), R30. Leonard Rozendaal, Pisano word, tesselation, plane-filling fractal, Preprint, 2017. MATHEMATICA r=(1+5^(1/2))/2; u[n_] := Floor[n*r];  (* A000201 *) a = 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

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Last modified December 9 19:51 EST 2019. Contains 329879 sequences. (Running on oeis4.)