A280511 Index sequence of the block-fractal sequence A001468.

2, 2, 5, 5, 5, 5, 5, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 89, 89, 89, 89, 89, 89, 89, 89, 89, 89, 89, 89

Offset: 1

Author

Clark Kimberling, Jan 06 2017

Keywords

• nonn
• easy

Comments

The index sequence (a(n)) of a 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(0), s(1), ..., s(n)). Following are definitions of block-fractal, reverse block-fractal, complementary block-fractal, and reverse complementary block-fractal, as pertain to any sequence s = (s(n)): s is block-fractal if every finite block s* of consecutive terms in s occurs more than once in s, and reverse block-fractal if reversal(s*) occurs in s; a zero-one sequence s is complement block-fractal if 1-s* occurs in s for every finite block S* of consecutive terms in s, and reverse complement block-fractal if reverse(1-s*) occurs in s.

Clearly each of the 4 containment conditions holds for all blocks s* if it holds for every initial block in s. Moreover, in all 4 cases, such a sequence s* occurs infinitely many times in s. This proper containment of infinitely many identical copies is comparable to proper containment of similar images in geometric fractals, hence the use of the word "fractal" for sequences.

The standard term for "block-fractal sequence" in the combinatorics on words literature is "recurrent sequence". The standard term for "reverse block-fractal" is "mirror-invariant". - Jeffrey Shallit, May 28 2023

Examples

			A001468 = (1,2,1,2,2,1,2,1,2,2,1,2,2,...) = (s(0), s(1), ... ).
(initial block #1) = (1) first repeats at s(2), so that a(1) = 2;
(initial block #2) = (1,2) first repeats at s(2), so that a(2) = 2;
(initial block #3) = (1,2,1) first repeats at s(5), so that a(3) = 5.
		

Links

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

Crossrefs

Cf. A000045, A001468.

Programs

Mathematica

r = GoldenRatio; seq = Table[Floor[(n + 1) r] - Floor[n r], {n, 0, 300}] (*A001468*)
seq = StringJoin[Map[ToString, seq]]
u = -1 + Most[Flatten[Rest[Reap[NestWhile[# + 1 &, 1,       Sow[First[Last[StringPosition[seq, StringTake[seq, #], 2]]]] >
1 &]]]]] (* A280511, Peter J. C. Moses, Jan 05 2017 *)

Formula

Concatenate F(2n+1) copies of F(2n+1), for n >= 1, where F = A000045, the Fibonacci numbers.