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a(1)=1. Then, if the first n terms contain all the numbers 1 to n, a(n+1)=2*a(n), otherwise a(n+1) is the smallest number not already in the sequence.
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%I #5 Mar 31 2012 10:29:23

%S 1,2,4,3,6,5,10,7,8,9,18,11,12,13,14,15,16,17,34,19,20,21,22,23,24,25,

%T 26,27,28,29,30,31,32,33,66,35,36,37,38,39,40,41,42,43,44,45,46,47,48,

%U 49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,130

%N a(1)=1. Then, if the first n terms contain all the numbers 1 to n, a(n+1)=2*a(n), otherwise a(n+1) is the smallest number not already in the sequence.

%e There are no gaps with just one term, so we double to get a(2)=2. Still no gaps, so we double again to get a(3)=4. Now 3 has not appeared yet, so a(4)=3. The sequence is compact again, so we double to get a(5)=6.

%K nonn,easy

%O 1,2

%A _Greg Knowles_, Jun 08 2011

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