A339950 - OEIS

A339950

Numbers k such that all k-sections of the infinite Fibonacci word A014675 have just two different run-lengths.

1, 7, 14, 20, 27, 35, 41, 48, 54, 62, 69, 75, 82, 90, 96, 103, 109, 117, 124, 130, 137, 143, 151, 158, 164, 171, 179, 185, 192, 198, 206, 213, 219, 226, 234, 240, 247, 253, 260, 268, 274, 281, 287, 295, 302, 308, 315, 323, 329, 336, 342, 350, 357, 363, 370, 376, 384, 391, 397, 404

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OFFSET

1,2

COMMENTS

Equivalent definition: these are the numbers n such that all n-sections of the infinite Fibonacci word A003849 have just two run-lengths.

The distinct terms of the difference sequence of the first 40 terms are 6, 7, and 8.

Conjecture: a(n) = A189378(n-1)+1 for n >= 2. - Don Reble, Apr 06 2021.

"All n-sections" means all subsequences S(k) = (A014675(n*i+k); i = 0, 1, 2, ...), for k = 0, ..., n-1. "Run-lengths" means the numbers of consecutive equal terms in the sequence: see examples. - M. F. Hasler, Apr 07 2021

LINKS

Table of n, a(n) for n=1..60.

EXAMPLE

Let W = A014675, so that as a word, W = 21221212212212122121221221212212212122121221221...

The unique 1-section of W is W itself, which is a concatenation of runs 1, 2, and 22, so that a(1) = 2. The sequence A339949 shows that a(n) > 2 for n = 2,3,4,5,6. For n = 7, the n-section of W that starts with its first letter, 2, is 221221221221221221221221221221221221121..., in which the runs are 22, 1, 11, supporting the conjecture that a(2) = 7.

Some run-lengths may appear quite late. For example, when n = 68, the third run-length appears in the n-section S(k=0) only with the 2829th element, corresponding to the 192372-th element of the original sequence. - M. F. Hasler, Apr 07 2021

MATHEMATICA

r = (1 + Sqrt[5])/2; z = 80000;

f[n_] := Floor[(n + 1) r] - Floor[n r]; (* A014675 *)

t = Table[Max[Map[Length,

Union[Split[Table [f[n d], {n, 0, Floor[z/d]}]]]]], {d, 1,

400}, {n, 1, d}];

u = Map[Max, t]

Flatten[Position[u, 2]] (* A339950 *)

CROSSREFS

Cf. A001622, A003849, A014675, A339949.