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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
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.
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.)
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.
CROSSREFS
Sequence in context: A242453 A241151 A073092 * A324477 A287292 A260717
KEYWORD
nonn
AUTHOR
Gary W. Adamson, Sep 19 2003
EXTENSIONS
Edited by John W. Layman, Oct 10 2003
STATUS
approved