%I #10 Dec 21 2024 05:49:19
%S 1,2,4,3,6,7,5,4,8,9,11,10,7,8,6,5,10,11,13,12,15,16,14,13,9,10,12,11,
%T 8,9,7,6,12,13,15,14,17,18,16,15,19,20,22,21,18,19,17,16,11,12,14,13,
%U 16,17,15,14,10,11,13,12,9,10,8,7,14,15,17,16,19,20,18,17,21,22,24
%N a(1)=1, a(2)=2, then the terms from a(2^k + 1) to a(2^(k+1)) are the first 2^k terms in reverse order, increased by a(2^k).
%H Ivan Neretin, <a href="/A378784/b378784.txt">Table of n, a(n) for n = 1..8192</a>
%e The first four terms are 1, 2, 4, 3. Hence the next four terms are (1+3, 2+3, 4+3, 3+3) taken in reverse order, that is: 6, 7, 5, 4.
%t Nest[Join[#, Reverse[#] + #[[-1]]] &, {1, 2}, 6]
%Y Cf. A378785 (Fibonacci-based analog).
%K nonn
%O 1,2
%A _Ivan Neretin_, Dec 07 2024