A280513 Index sequence of the reverse block-fractal sequence A001468.

1, 2, 1, 5, 4, 3, 2, 1, 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, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 89, 88, 87, 86, 85, 84, 83, 82, 81, 80, 79, 78, 77, 76, 75, 74

Offset: 1

Author

Clark Kimberling, Jan 06 2017

Keywords

• nonn
• easy

Comments

The sequence is the concatenation of blocks, the n-th of which, for n >= 1, consists of the integers from F(2n+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 here by a(n) = least k > 0 such that (s(k), s(k+1), ..., s(k+n)) = (s(n), s(n-1), ..., s(1)).

Let W be the Fibonacci word A096270. Then a(n) = 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 17 2020

Examples

			A001468 = (1,2,1,2,2,1,2,1,2,2,1,2,2,...) = (s(1), s(2), ... ).
(init. block #1) = (1); reversal (1) first occurs at s(1), so a(1) = 1;
(init. block #2) = (1,2); rev. (2,1) first occurs at s(2), so a(2) = 2;
(init. block #3) = (1,2,1); rev. (1,2,1) first occurs at s(1), so a(3) = 1;
(init. block #4) = (1,2,1,2); rev. (2,1,2,1) first occurs at s(5), so a(4) = 5.
		

Links

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

Crossrefs

Cf. A000045, A001468, A096270.

Programs

Mathematica

r = GoldenRatio; t = Table[Floor[(n + 1) r] - Floor[n*r], {n, 0, 420}]
u = StringJoin[Map[ToString, t]]; breverse[seq_] :=
Flatten[Last[Reap[NestWhile[# + 1 &, 1, (StringLength[
str = StringTake[seq, Min[StringLength[seq], #]]] == # && ! (Sow[
StringPosition[seq, StringReverse[str], 1][[1]][[1]]]) === {}) &]]]];
breverse[u]  (* Peter J. C. Moses, Jan 02 2017 *)