A364648 Starting position of the first occurrence of the longest monochromatic arithmetic progression of difference n in the Fibonacci infinite word (A003849).

2, 3, 20, 16, 11, 20, 0, 143, 2, 11, 54, 8, 32, 2, 11, 7, 70, 3, 7, 0, 986, 10, 3, 7, 16, 11, 2, 87, 376, 2, 3, 2, 21, 87, 2, 3, 7, 16, 3, 7, 0, 20, 23, 11, 20, 8, 11, 2, 11, 20, 36, 11, 7, 0, 6764, 31, 3, 376, 84, 11, 54, 0, 20, 2, 3, 2, 42, 87, 2, 3, 54, 304

Offset

1,1

Comments

From Gandhar Joshi, Jan 25 2025: (Start)

F(n) is the n-th Fibonacci number.

Conjecture: for n>0,

1. a(F(2n))=F(4n)-1; a(F(2n+1))=F(2n+3)-2.

2. a(F(6n)/2)=F(6n+3)/2-1; a(F(6n-3)/2)=F(6n)/2-2. (End)

Links

Gandhar Joshi, Table of n, a(n) for n = 1..1973

Ibai Aedo, U. Grimm, Y. Nagai, and P. Staynova, Monochromatic arithmetic progressions in binary Thue-Morse-like words, Theor. Comput. Sci., 934 (2022), 65-80.

Gandhar Joshi and D. Rust, Monochromatic arithmetic progressions in the Fibonacci word, arXiv:2501.05830 [math.DS], 2025. See p.12.

Example

For the difference n = 3, the longest monochromatic progression has length A339949(3) = 5 and thus defined by f(i)=f(i+3)=f(i+6)=f(i+9)=f(i+12), where f(i) is the i-th term of the Fibonacci word (A003849); the smallest i for which that holds is i=20, so a(3) = 20.

Prog

(Walnut)
# In the following line, replace every n with the desired constant difference, and every q with the longest MAP length for difference n given by A339949(n).
def f_n_map "?msd_fib Ak (k<q) => F[i]=F[i+n*k] & Aj (j<i) => ~(Ak (k<q) => F[j]=F[j+n*k])";
# Gandhar Joshi, Jan 25 2025

Crossrefs

Cf. A003849, A339949 (length of the longest monochromatic arithmetic progression).

Sequence in context: A344546 A279719 A279672 * A089181 A028425 A224987

Adjacent sequences: A364645 A364646 A364647 * A364649 A364650 A364651

Keyword

nonn

Author

Gandhar Joshi, Jul 31 2023

Status

approved