

A280513


Index sequence of the reverse blockfractal sequence A001468.


3



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
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OFFSET

1,2


COMMENTS

The sequence is the concatenation of blocks, the nth 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 blockfractal sequence. The index sequence (a(n)) of a reverse blockfractal 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(n1), ..., s(1)).
Let W be the Fibonacci word A096270. Then a(n) = least k such that the reversal of the first nblock in W occurs in W beginning at the kth 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



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]]]) === {}) &]]]];


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



