A280514 - OEIS

A280514

Index sequence of the reverse block-fractal sequence A003849.

1, 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

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OFFSET

1,2

COMMENTS

The sequence is the concatenation of blocks, the n-th of which, for n >=0, consists of the integers from F(n+1) down to F(2) = 1, where F = A000045, the Fibonacci numbers. See A280511 for the definition of reverse block-fractal sequence. The index sequence (a(n)) of a reverse block-fractal sequence (s(n)) is defined (at A280513) by a(n) = least k > 0 such that (s(k), s(k+1), ..., s(k+n)) = (s(n), s(n-1), ..., s(0)).

Apparently (up to offset) a duplicate of A246105. - R. J. Mathar, Jan 10 2017

Let W be the Fibonacci word A003849. Then a(n) is the least k such that the reversal of the first n-block in W occurs in W beginning at the k-th term. Since (a(n)) is unbounded, the reversal of every block in W occurs infinitely many times in W. - Clark Kimberling, Dec 19 2020

LINKS

Clark Kimberling, Table of n, a(n) for n = 1..1000

EXAMPLE

A003849 = (0,1,0,0,1,0,1,0,0,1,0,0,1,...) = (s(1), s(2), ... ).

(init. block #1) = (1); reversal (0) first occurs at s(1), so a(1) = 1;

(init. block #2) = (0,1); rev. (1,0) first occurs at s(2), so a(2) = 2;

(init. block #3) = (0,1,0); rev. (0,1,0) first occurs at s(1), so a(3) = 1;

(init. block #4) = (0,1,0,0); rev. (0,0,1,0) first occurs at s(3), so a(4) = 3.

MATHEMATICA

r = GoldenRatio; t = Table[Floor[(n + 2) #] - Floor[(n + 1) #], {n, 0, 220}] &[

2 - GoldenRatio] (* A003849 *)

u = StringJoin[Map[ToString, t]]