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%I #9 Mar 06 2015 18:58:17
%S 0,2,4,5,8,9,10,12,16,17,18,20,22,24,28,32,33,34,35,38,40,44,48,49,52,
%T 56,58,64,65,66,67,68,72,73,76,80,81,84,88,92,96,97,100,104,106,112,
%U 113,118,124,128,129,130,131,134,136,140,144,145,148,152,154,160
%N a(0)=0; for n>0, choose a(n) to be the smallest number > a(n-1) such that the condition a(n) > Sum_{k=0..n} (number of 1's in binary expansion of n a(k)) holds
%C See A244510 for a decimal version.
%H Lars Blomberg, <a href="/A248893/b248893.txt">Table of n, a(n) for n = 0..10000</a>
%e a(7)=12 because the sum of 1's in the binary expansions of a(0)..a(6) is 9, and 12 is the smallest number greater than a(6)=10 and also greater than 9 + sum of 1's in the binary expansion of 12 (which is 2).
%Y Cf. A000120, A244510
%K nonn,base
%O 0,2
%A _Lars Blomberg_, Mar 06 2015
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