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%I #11 Jul 09 2015 15:15:18
%S 1,2,3,4,6,7,8,12,13,14,16,20,24,26,27,28,32,36,40,48,50,52,54,55,56,
%T 64,68,72,80,88,96,100,102,104,108,109,110,112,120,128,136,140,144,
%U 160,168,176,192,196,200,204,206,208,216,217,218,220,222,224,240,248,256
%N a(1) = 1, a(n) = a(floor((2n-1)/3)) + a(floor(2n/3)) for n > 1.
%C In other words, a(1)=1 and then any term is a sum of two earliest possible previous terms (not necessarily distinct), given that each term must be used in summation no more than three times. So a(2)=1+1 (thus 1 gets used twice), a(3)=1+2 (thus 1 gets used for the third and final time, then 2 steps in), and so on. - _Ivan Neretin_, Jul 09 2015
%H Joseph Myers, <a href="/A205591/b205591.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="http://www.bmoc.maths.org/home/bmo2-2012.pdf">2011/12 British Mathematical Olympiad Round 2</a>, Problem 2.
%Y Cf. A205592, A205593, A205594, A205595, A205596.
%K easy,nonn
%O 1,2
%A _Joseph Myers_, Jan 29 2012
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