A284749 {001->2}-transform of the infinite Fibonacci word A003849.

0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 2, 0, 1, 2

Offset

1,3

Comments

From Daniel Rust, Aug 18 2018: (Start)

This word is the fixed point of the morphism 0->2, 1->01, 2->2201 with the finite string 0, 1, 2, 0, 1 appended to the beginning. This morphism comes from taking the 'proper' version of the Fibonacci morphism 0->01, 1->1, given by 0->001, 1->01 (A189661 but with the rightmost 0 moved to the left of each image word), then replacing 001 with 2 and noting that the new symbol 2 should map to 00100101 = 2201 in order to be consistent.

The finite string appended to the beginning comes from the process of finding a proper version of the Fibonacci morphism using a return word encoding and taking conjugates which causes a shift of the respective fixed points.

(End)

This sequence is the unique fixed point of the morphism 0->01, 1->2, 2->0122. See the paragraph following Lemma 23 in the paper by Allouche and me. - Michel Dekking, Oct 05 2018

Links

Clark Kimberling, Table of n, a(n) for n = 1..10000

J.-P. Allouche and F. M. Dekking, Generalized Beatty sequences and complementary triples, arXiv:1809.03424v3 [math.NT], 2018-2019.

Example

As a word, A003849 = 01001010010010100..., and replacing each 001 by 2 gives 01201220120...

Mathematica

s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 13]  (* A003849 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"001" -> "2"}]
st = ToCharacterCode[w1] - 48 (* A284749 *)
Flatten[Position[st, 0]]  (* A214971 *)
Flatten[Position[st, 1]]  (* A284624 *)
Flatten[Position[st, 2]]  (* A284625 *)

Prog

(Python)
from math import isqrt
def A284624(n): return (n<<1)+(n-1+isqrt(5*(n-1)**2)>>1)
def A284625(n): return (n+isqrt(5*n**2)&-2)-n+2
def A284749(n):
    def bsearch(f, n):
        kmin, kmax = 0, 1
        while f(kmax) <= n:
            kmax <<= 1
        kmin = kmax>>1
        while True:
            kmid = kmax+kmin>>1
            if f(kmid) > n:
                kmax = kmid
            else:
                kmin = kmid
            if kmax-kmin <= 1:
                break
        return kmin
    for i, f in enumerate((A284624, A284625), 1):
        if f(bsearch(f, n))==n: return i
    return 0 # Chai Wah Wu, May 22 2025

Crossrefs

Cf. A003849, A214971, A284624, A284625.

Sequence in context: A248806 A118407 A101663 * A062169 A113680 A316723

Adjacent sequences: A284746 A284747 A284748 * A284750 A284751 A284752

Keyword

nonn,easy

Author

Clark Kimberling, May 02 2017

Status

approved