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A110963
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Fractalization of Kimberling's paraphrases sequence beginning with 1.
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5
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1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 2, 1, 4, 1, 1, 1, 5, 3, 3, 2, 6, 2, 2, 1, 7, 4, 4, 1, 8, 1, 1, 1, 9, 5, 5, 3, 10, 3, 3, 2, 11, 6, 6, 2, 12, 2, 2, 1, 13, 7, 7, 4, 14, 4, 4, 1, 15, 8, 8, 1, 16, 1, 1, 1, 17, 9, 9, 5, 18, 5, 5, 3, 19, 10, 10, 3, 20, 3, 3, 2, 21, 11, 11, 6, 22, 6, 6, 2, 23, 12, 12, 2, 24, 2, 2, 1, 25, 13
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OFFSET
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1,5
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COMMENTS
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Self-descriptive sequence: terms at even indices are the sequence itself, terms at odd indices (the skeleton of this sequence) are the terms of Kimberling's paraphrases sequence (A003602) beginning with 1.
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LINKS
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Antti Karttunen, Table of n, a(n) for n = 1..65537
Clark Kimberling, Fractal sequences.
Index entries for sequences related to binary expansion of n
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FORMULA
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For even n, a(n) = a(n/2), for odd n, a(n) = A003602((1+n)/2). - Antti Karttunen, Apr 03 2022
For n >= 0, (Start)
a(4n+2) = a(4n+3) = A003602(1+n).
a(8n+1) = A005408(n) = 2*n + 1.
a(4n+1) = a(8n+2) = a(8n+3) = 1+n.
a(n) = A110962(n-1) + 1.
(End)
a(n) = A353367(4*n). - Antti Karttunen, Apr 20 2022
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PROG
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(PARI)
A003602(n) = (1+(n>>valuation(n, 2)))/2;
A110963(n) = if(n%2, A003602((1+n)/2), A110963(n/2)); \\ Antti Karttunen, Apr 03 2022
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CROSSREFS
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One more than A110962 (but note the different starting offsets).
Cf. A000265, A003602, A005408, A110812, A110779, A110766, A351565 [= 2*a(n) - 1].
Cf. A353366 (Dirichlet inverse), A353367 (sum with it).
Sequence in context: A288003 A304382 A304717 * A292622 A292869 A106348
Adjacent sequences: A110960 A110961 A110962 * A110964 A110965 A110966
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KEYWORD
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base,easy,nonn
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AUTHOR
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Alexandre Wajnberg, Sep 26 2005
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EXTENSIONS
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Entry edited, starting offset corrected (from 0 to 1), and the offsets in formulas changed accordingly, and more terms added by Antti Karttunen, Apr 03 2022
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STATUS
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approved
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