A280513 - OEIS

%I #23 Dec 18 2020 04:23:43

%S 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,

%T 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,

%U 2,1,89,88,87,86,85,84,83,82,81,80,79,78,77,76,75,74

%N Index sequence of the reverse block-fractal sequence A001468.

%C 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)).

%C 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

%H Clark Kimberling, <a href="/A280513/b280513.txt">Table of n, a(n) for n = 1..1000</a>

%e A001468 = (1,2,1,2,2,1,2,1,2,2,1,2,2,...) = (s(1), s(2), ... ).

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

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

%e (init. block #3) = (1,2,1); rev. (1,2,1) first occurs at s(1), so a(3) = 1;

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

%t r = GoldenRatio; t = Table[Floor[(n + 1) r] - Floor[n*r], {n, 0, 420}]

%t u = StringJoin[Map[ToString, t]]; breverse[seq_] :=

%t Flatten[Last[Reap[NestWhile[# + 1 &, 1, (StringLength[

%t str = StringTake[seq, Min[StringLength[seq], #]]] == # && ! (Sow[

%t StringPosition[seq, StringReverse[str], 1][[1]][[1]]]) === {}) &]]]];

%t breverse[u]  (* _Peter J. C. Moses_, Jan 02 2017 *)

%Y Cf. A000045, A001468, A096270.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Jan 06 2017