%I #29 Apr 08 2021 03:19:31
%S 1,7,14,20,27,35,41,48,54,62,69,75,82,90,96,103,109,117,124,130,137,
%T 143,151,158,164,171,179,185,192,198,206,213,219,226,234,240,247,253,
%U 260,268,274,281,287,295,302,308,315,323,329,336,342,350,357,363,370,376,384,391,397,404
%N Numbers k such that all k-sections of the infinite Fibonacci word A014675 have just two different run-lengths.
%C Equivalent definition: these are the numbers n such that all n-sections of the infinite Fibonacci word A003849 have just two run-lengths.
%C The distinct terms of the difference sequence of the first 40 terms are 6, 7, and 8.
%C Conjecture: a(n) = A189378(n-1)+1 for n >= 2. - _Don Reble_, Apr 06 2021.
%C "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
%e Let W = A014675, so that as a word, W = 21221212212212122121221221212212212122121221221...
%e 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.
%e 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
%t r = (1 + Sqrt[5])/2; z = 80000;
%t f[n_] := Floor[(n + 1) r] - Floor[n r]; (* A014675 *)
%t t = Table[Max[Map[Length,
%t Union[Split[Table [f[n d], {n, 0, Floor[z/d]}]]]]], {d, 1,
%t 400}, {n, 1, d}];
%t u = Map[Max, t]
%t Flatten[Position[u, 2]] (* A339950 *)
%Y Cf. A001622, A003849, A014675, A339949.
%Y See also A189377, A189378, A189379.
%K nonn
%O 1,2
%A _Clark Kimberling_, Dec 26 2020
%E More terms from _Don Reble_, Apr 13 2021
|