A287775 Positions of 0 in A287772; complement of A050140 (conjectured and proved).

2, 3, 6, 7, 9, 10, 13, 14, 17, 18, 20, 21, 24, 25, 27, 28, 31, 32, 35, 36, 38, 39, 42, 43, 46, 47, 49, 50, 53, 54, 56, 57, 60, 61, 64, 65, 67, 68, 71, 72, 74, 75, 78, 79, 82, 83, 85, 86, 89, 90, 93, 94, 96, 97, 100, 101, 103, 104, 107, 108, 111, 112, 114

Offset: 1

Author

Clark Kimberling, Jun 03 2017

Keywords

• nonn
• easy

Comments

-1 < n*r - a(n) < 1 for n >= 1, where r = (5 + sqrt(5))/4.

From Michel Dekking, Dec 28 2017: (Start)

Let (d(n)) be the sequence of first differences: d(n)=a(n+1)-a(n).

CLAIM: d(n) = A108103(n+1) for n=1,2,….

Proof: As a word A287772 = 100110010011001100100110010011... obtained by substituting 0->1, 1->00 in the Fibonacci word F=0100101001001010010100...

This implies that A287772 is a concatenation of 00’s separated by 1’s and 11’s. Moreover, a 0110 occurs iff 1001 occurs in F, and a 010 occurs iff 101 occurs in F. Note also that occurrence of a 00 in A287772 yields a d(n)=1 (and so every other letter in d is a 1), occurrence of a 010 yields a d(n)=2, and occurrence of a 0110 yields a d(n)=3. Since the 1001’s and 101’s occur in 1F according to F itself with 1 prepended (see A001468 and A282162), we must have d(n)=A108103(n+1). (End)

Links

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

Crossrefs

Cf. A050140, A287772, A108103.

Programs

Mathematica

s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 10] (* A003849 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"0" -> "1", "1" -> "00"}]
st = ToCharacterCode[w1] - 48    (* A287772 *)
Flatten[Position[st, 0]]  (* A287775 *)
Flatten[Position[st, 1]]  (* A050140 conjectured *)

Python

from math import isqrt
def A287775(n):
    def f(x): return n+(r:=isqrt(x**2//5))+((isqrt(5*(r+1)**2)+r+1&-2)-r-1<=x)
    m, k = n, f(n)
    while m != k: m, k = k, f(k)
    return m # Chai Wah Wu, May 05 2025