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%I #8 Feb 06 2022 06:32:11
%S 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,
%T 16,17,17,17,18,19,20,20,21,22,22,22,23,24,25,25,25,26,26,26,26,27,28,
%U 28,29,30,30,30,30,31,32,32,32,33,33,33
%N 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.
%C 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.
%F 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.)
%e 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.
%K nonn
%O 1,2
%A _Gary W. Adamson_, Sep 19 2003
%E Edited by _John W. Layman_, Oct 10 2003
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