OFFSET
1,2
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
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..1000
EXAMPLE
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.
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 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jan 06 2017
STATUS
approved