A339949 - OEIS

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

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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

Jeffrey Shallit, Table of n, a(n) for n = 1..231

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

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.