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A280511 Index sequence of the block-fractal sequence A001468.

%I A280511 #11 May 28 2023 20:45:32
%S A280511 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,
%T A280511 34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,
%U A280511 34,34,34,34,34,34,89,89,89,89,89,89,89,89,89,89,89,89
%N A280511 Index sequence of the block-fractal sequence A001468.
%C A280511 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.
%C A280511 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.
%C A280511 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
%H A280511 Clark Kimberling, <a href="/A280511/b280511.txt">Table of n, a(n) for n = 1..1000</a>
%F A280511 Concatenate F(2n+1) copies of F(2n+1), for n >= 1, where F = A000045, the Fibonacci numbers.
%e A280511 A001468 = (1,2,1,2,2,1,2,1,2,2,1,2,2,...) = (s(0), s(1), ... ).
%e A280511 (initial block #1) = (1) first repeats at s(2), so that a(1) = 2;
%e A280511 (initial block #2) = (1,2) first repeats at s(2), so that a(2) = 2;
%e A280511 (initial block #3) = (1,2,1) first repeats at s(5), so that a(3) = 5.
%t A280511 r = GoldenRatio; seq = Table[Floor[(n + 1) r] - Floor[n r], {n, 0, 300}] (*A001468*)
%t A280511 seq = StringJoin[Map[ToString, seq]]
%t A280511 u = -1 + Most[Flatten[Rest[Reap[NestWhile[# + 1 &, 1,       Sow[First[Last[StringPosition[seq, StringTake[seq, #], 2]]]] >
%t A280511 1 &]]]]] (* A280511, _Peter J. C. Moses_, Jan 05 2017 *)
%Y A280511 Cf. A000045, A001468.
%K A280511 nonn,easy
%O A280511 1,1
%A A280511 _Clark Kimberling_, Jan 06 2017