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 A088023 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. 3
 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; text; internal format)
 OFFSET 1,2 COMMENTS Conjecture: a(n+1) >= a(n). Comments from Don Reble, 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. LINKS 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)

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Last modified April 1 22:48 EDT 2023. Contains 361717 sequences. (Running on oeis4.)