A339949 a(n) is the greatest runlength in all n-sections of the infinite Fibonacci word A014675.

2, 3, 5, 6, 7, 3, 2, 12, 4, 4, 4, 4, 18, 2, 3, 6, 20, 5, 3, 2, 30, 4, 3, 4, 4, 9, 2, 3, 9, 4, 4, 3, 4, 47, 2, 3, 5, 10, 6, 3, 2, 15, 4, 4, 4, 4, 13, 2, 3, 7, 8, 5, 3, 2, 77, 4, 3, 5, 6, 8, 3, 2, 10, 4, 4, 3, 4, 24, 2, 3, 6, 78, 6, 3, 2, 22, 4, 3, 4, 4, 11, 2

Offset

1,1

Comments

Equivalently a(n) is the greatest runlength in all n-sections of the infinite Fibonacci word A003849.

From Jeffrey Shallit, Mar 23 2021: (Start)

We know that the Fibonacci word has exactly n+1 distinct factors of length n.

So to verify a(n) we simply verify there is a monochromatic arithmetic progression of length a(n) and difference n by examining all factors of length (n*a(n) - n + 1) (and we know when we've seen all of them). Next we verify there is no monochromatic AP of length a(n)+1 and difference n by examining all factors of length n*a(n) + 1.

Again, we know when we've seen all of them. (End)

Links

Gandhar Joshi, Table of n, a(n) for n = 1..10000 (terms 1..232 from Jeffrey Shallit).

Dmitry Badziahin and Jeffrey Shallit, Badly approximable numbers, Kronecker's theorem, and diversity of Sturmian characteristic sequences, arXiv:2006.15842 [math.NT], 2020.

Gandhar Joshi and Dan Rust, Monochromatic arithmetic progressions in the Fibonacci word, arXiv:2501.05830 [math.NT], 2025. See p. 9.

Formula

From Gandhar Joshi, Jan 14 2025: (Start)

phi = the golden ratio. g(n) = min {n*phi mod 1, 1 - (n*phi mod 1)}.

If g(n) <= (phi)^(-2), a(n) = ceiling{((phi)^(-1))/g(n)};

otherwise, a(n) = 2*ceiling{((phi)^(-1)-g(n))/g(2n)}. (End)

Example

For n  >= 1,  r =  0..n, k >= 0, let A014675(n*k+r) denote the k-th term of the r-th n-section of A014675; i.e.,
(A014675(k)) = 212212122122121221212212212122122121221212212212122121...
  has runlengths 1,1,2,1,1,1,2,1,2,1,...; a(1) = 2.
(A014675(2k)) = 22112211222122212221122112221222122211221122112221222...
  has runlengths 2,2,2,2,3,1,3,1,3,2,...
(A014675(2k+1)) = 122212221122112211222122211221122112221222122211221...
   has runlengths 1,3,1,3,2,2,2,2,2,3,...; a(2) = 3.
(A014675(3k)) = 22111222211122221122222112222211222211122221112222111...
  has runlengths 2,3,4,3,4,2,5,2,5,2,4,3,4,3,...
(A014675(3k+1)) = 112222111222211122221112222111222211222221122221112...
  has runlengths 2,4,3,4,3,4,3,4,3,4,,5,2,4,3,...
(A014675(3k+2)) = 222211222221122221112222111222211122221112222112222...
  has runlengths 4,2,5,2,4,3,4,3,4,3,4,3,4,2,...; a(3) = 5.

Mathematica

r = (1 + Sqrt[5])/2; z = 4000;
f[n_] := Floor[(n + 2) r] - Floor[(n+1) r];  (* A014675 *)
t = Table[Max[Map[Length, Union[Split[Table [f[n m], {n, 0, Floor[z/m]}]]]]], {m, 1, 20}, {n, 1, m}];
Map[Max, t] (* A339949 *)

Prog

(PARI)
phi = quadgen(5);
g(n) = min(frac(n * phi), 1 - frac(n * phi));
a(n) = if (g(n) <= (1 / phi)^2, ceil((1 / phi) / g(n)), 2*ceil(((1 / phi) - g(n)) / g(2 * n))); \\ Gandhar Joshi, Jan 14 2025

Crossrefs

Cf. A001622, A003849, A014675, A339950.

Sequence in context: A084735 A002734 A367417 * A160100 A371257 A247891

Adjacent sequences: A339946 A339947 A339948 * A339950 A339951 A339952

Keyword

nonn,easy

Author

Clark Kimberling, Dec 26 2020

Extensions

a(61) corrected by Jeffrey Shallit, Mar 23 2021

Status

approved