A246105 Least m > 0 for which (s(m),...,s(n+m-1)) = (s(n),...,s(0)), the reverse of the first n+1 terms of the infinite Fibonacci word A003849.

2, 1, 3, 2, 1, 5, 4, 3, 2, 1, 8, 7, 6, 5, 4, 3, 2, 1, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13

Offset

0,1

Links

Table of n, a(n) for n=0..73.

Formula

Concatenation of blocks (F(k), F(k - 1), ..., F(3), F(2)) beginning with k = 3, where F = A000045 (Fibonacci numbers).

Example

reverse(s(0),...,s(6)) = reverse(0,1,0,0,1,0,1) = (1,0,1,0,0,1,0)
first repeats in A003849 at (s(4),...,s(10)), so that a(6) = 4.

Mathematica

s = Flatten[Nest[{#, #[[1]]} &, {0, 1}, 12]]; b[m_, n_] := b[m, n] = Take[s, {m, n}]; t = -1 + Flatten[Table[Select[Range[2, 1600], b[#, n + # - 1] == Reverse[b[1, n]] &, 1], {n, 1, 120}]]

Crossrefs

Cf. A003849, A000045, A246104, A241422.

Sequence in context: A279394 A308509 A280514 * A211980 A171730 A131243

Adjacent sequences: A246102 A246103 A246104 * A246106 A246107 A246108

Keyword

nonn,easy

Author

Clark Kimberling, Aug 14 2014

Status

approved