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A088023
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Set a(1) = 1. Then take the list of defined initial terms, reverse their order, add 1, 2, 3...to the reversed list in succession and append this new list to the right of the previously defined terms. Repeat this process indefinitely.
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3
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1, 2, 3, 3, 4, 5, 5, 5, 6, 7, 8, 8, 8, 9, 9, 9, 10, 11, 12, 12, 13, 14, 14, 14, 14, 15, 16, 16, 16, 17, 17, 17, 18, 19, 20, 20, 21, 22, 22, 22, 23, 24, 25, 25, 25, 26, 26, 26, 26, 27, 28, 28, 29, 30, 30, 30, 30, 31, 32, 32, 32, 33, 33, 33
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Conjecture: a(n+1) >= a(n). Comments from Don Reble (djr(AT)nk.ca), Nov 13 2005: The conjecture is plainly true. In fact, a(n+1)-a(n) = 0 or 1. Also a(A091072(n)) = n; a(A091072(n)+1) = n+1.
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FORMULA
| a(n)=2a(n/2)-1 if a=2^k else a(n)=a(2^k-n+1)+n-2^(k-1) if 2^(k-1)<n<2^k. (Ed.)
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EXAMPLE
| The sequence begins 1, 2, then reverse 1, 2 = 2, 1 then add 1, 2 to the latter getting 3, 3. Then append 3, 3, to the right of 1, 2, getting 1, 2, 3, 3. Then repeating the instructions, 1, 2, 3, 3 is reversed then add 1, 2, 3, 4 to 3, 3, 2, 1, = 4, 5, 5, 5. Append the latter to 1, 2, 3, 3 getting 1, 2, 3, 3, 4, 5, 5, 5...; and so on.
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CROSSREFS
| Sequence in context: A195778 A165706 A073092 * A061451 A205542 A086155
Adjacent sequences: A088020 A088021 A088022 * A088024 A088025 A088026
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KEYWORD
| nonn
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AUTHOR
| Gary W. Adamson (qntmpkt(AT)yahoo.com), Sep 19 2003
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EXTENSIONS
| Edited by John W. Layman (layman(AT)math.vt.edu), Oct 10 2003
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