Revision History for A280511
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Showing entries 1-10
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#11 by N. J. A. Sloane at Sun May 28 20:45:32 EDT 2023
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#10 by Jon E. Schoenfield at Sun May 28 19:48:59 EDT 2023
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#9 by Jon E. Schoenfield at Sun May 28 19:48:56 EDT 2023
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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
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proposed
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#8 by Jeffrey Shallit at Sun May 28 18:35:37 EDT 2023
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#7 by Jeffrey Shallit at Sun May 28 18:35:34 EDT 2023
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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
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approved
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#6 by N. J. A. Sloane at Sat Jan 07 11:58:22 EST 2017
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#5 by Jon E. Schoenfield at Sat Jan 07 01:53:58 EST 2017
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#4 by Jon E. Schoenfield at Sat Jan 07 01:53:55 EST 2017
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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.
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proposed
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#3 by Clark Kimberling at Fri Jan 06 17:43:02 EST 2017
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#2 by Clark Kimberling at Fri Jan 06 12:22:18 EST 2017
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| NAME
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allocatedIndex sequence of the forblock-fractal Clarksequence KimberlingA001468.
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| DATA
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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
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| OFFSET
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1,1
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| COMMENTS
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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.
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| LINKS
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Clark Kimberling, <a href="/A280511/b280511.txt">Table of n, a(n) for n = 1..1000</a>
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| FORMULA
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Concatenate F(2n+1) copies of F(2n+1), for n >= 1, where F = A000045, the Fibonacci numbers.
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| EXAMPLE
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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.
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| MATHEMATICA
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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 *)
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| CROSSREFS
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Cf. A000045, A001468.
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| KEYWORD
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allocated
nonn,easy
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| AUTHOR
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Clark Kimberling, Jan 06 2017
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| STATUS
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approved
editing
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