A284621 Positions of 0 in A284620.

1, 5, 11, 15, 21, 27, 31, 37, 41, 47, 53, 57, 63, 69, 73, 79, 83, 89, 95, 99, 105, 109, 115, 121, 125, 131, 137, 141, 147, 151, 157, 163, 167, 173, 179, 183, 189, 193, 199, 205, 209, 215, 219, 225, 231, 235, 241, 247, 251, 257, 261, 267, 273, 277, 283, 287

Offset: 1

Author

Clark Kimberling, May 02 2017

Keywords

• nonn
• easy

Comments

This sequence and A005843 and A130568 partition the positive integers into sequences with slopes t = 1+sqrt(5), u = 3+sqrt(5), v = 2, where 1/t + 1/u + 1/v = 1. The positions of 1 in A284620 are given by A005843, and of 2, by A130568.

From Michel Dekking, Mar 17 2020: (Start)

This sequence is a generalized Beatty sequence.

It was shown in the Comments of A284620 that A284620 is the letter-to-letter image of the fixed point x = ABCDABCDCD... of the morphism

mu: A->AB, B->CD, C->ABCD, D->CD,

with the letter-to-letter map lambda defined by

lambda: A->0, B->1, C->2, D->1.

Note that A284620(n)=0 iff x(n) = A, where x = ABCDABCDCD... is the fixed point of mu. The return words of A in x are ABCD and ABCDCD. Coding these two return words by their lengths, mu induces a morphism rho on the coded return words given by

rho(4) = 46, rho(4) = 466.

The difference sequence (a(n+1)-a(n)) equals the unique fixed point r = 4646646466... of rho.

The morphism g on the alphabet {a,b} given by

g(a) = ab, g(b) =abb

was introduced in A284620. We see that rho is just an alphabet change of the morphism g.

Let f given by f(b) = ba, f(a) = b be the Fibonacci morphism on the alphabet {b,a} with fixed point x_F = babbababba....

Let x_G = ababbababb... be the fixed point of g. It is well-known (see, e.g., Lemma 12 in "Morphic words..."), that x_G = a x_F.

In general the partial sums of x_F are equal to the generalized Beatty sequence V given by V(n) = p*floor(n*phi) +q*n+r, where p = a-b and q = 2*b-a. See Lemma 8 in the Allouche and Dekking paper. Here we obtain p = 2, q = 2. So a(n) = 2*floor((n-1)*phi) + 2*n - 1, for n>0.

(End)

Examples

			As a word, A284620 = 012101212101210121..., in which 0 is in positions 1,5,11,15,...
		

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, Moscow Journal of Combinatorics and Number Theory 8, 325-341 (2019).
• M. Dekking, Morphic words, Beatty sequences and integer images of the Fibonacci language, Theoretical Computer Science 809, 407-417 (2020).

Crossrefs

Cf. A005843, A130568, A284620, A001950.

Programs

Mathematica

s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 13]  (* A003849 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"00" -> "2"}]
st = ToCharacterCode[w1] - 48 (* A284620 *)
Flatten[Position[st, 0]]  (* A284621 *)
Flatten[Position[st, 1]]  (* A005843 *)
Flatten[Position[st, 2]]  (* A130568 *)

Python

from math import isqrt
def A284621(n): return (n-1+isqrt(5*(n-1)**2)&-2)+(n<<1)-1 # Chai Wah Wu, May 22 2025

Formula

a(n+1) = 2*A001950(n) + 1, n>0. - Michel Dekking, Mar 17 2020