cp's OEIS Frontend

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.

%I A088023 #8 Feb 06 2022 06:32:11
%S A088023 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 A088023 16,17,17,17,18,19,20,20,21,22,22,22,23,24,25,25,25,26,26,26,26,27,28,
%U A088023 28,29,30,30,30,30,31,32,32,32,33,33,33
%N 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.
%C A088023 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 A088023 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 A088023 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 A088023 nonn
%O A088023 1,2
%A A088023 _Gary W. Adamson_, Sep 19 2003
%E A088023 Edited by _John W. Layman_, Oct 10 2003