A284624 Positions of 1 in A284749.

2, 5, 9, 12, 16, 20, 23, 27, 30, 34, 38, 41, 45, 49, 52, 56, 59, 63, 67, 70, 74, 77, 81, 85, 88, 92, 96, 99, 103, 106, 110, 114, 117, 121, 125, 128, 132, 135, 139, 143, 146, 150, 153, 157, 161, 164, 168, 172, 175, 179, 182, 186, 190, 193, 197, 200, 204, 208

Offset: 1

Author

Clark Kimberling, May 02 2017

Keywords

• nonn
• easy

Comments

This sequence and A214971 and A284625 partition the positive integers into sequences with slopes t = (5+sqrt(5))/2, u = (5+sqrt(5))/2, v = sqrt(5), where 1/t + 1/u + 1/v = 1. The positions of 0 in A284749 are given by A214971, and of 2, by A284625.

Examples

			As a word, A284749 = 012012201201220122..., in which 1 is in positions 2,5,9,12,...
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..10000
• J.-P. Allouche, F. M. Dekking, Generalized Beatty sequences and complementary triples, arXiv:1809.03424v3 [math.NT], 2018-2019.

Crossrefs

Cf. A214971, A284625, A284749.

Programs

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 conjectured *)
Flatten[Position[st, 1]]  (* A284624 *)
Flatten[Position[st, 2]]  (* A284625 *)

Python

from math import isqrt
def A284624(n): return (n<<1)+(n-1+isqrt(5*(n-1)**2)>>1) # Chai Wah Wu, May 22 2025

Formula

Conjectured: a(n) = 1 + A214971(n).

Proof of this conjecture: it is easy to see that the infinite Fibonacci word A003849 is a concatenation of the words 01 and 001. So if we replace all 001 by 2, only the 01 remain, i.e., every 0 is directly followed by 1. - Michel Dekking, Aug 19 2018

a(n) = floor((n-1)*phi) + 2*(n-1) + 2 (Theorem 31 in Allouche and Dekking). - Michel Dekking, Oct 08 2018