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A191598 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.

%I A191598 #9 Nov 13 2024 17:00:41
%S A191598 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 A191598 26,27,28,29,30,31,32,33,66,35,36,37,38,39,40,41,42,43,44,45,46,47,48,
%U A191598 49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,130
%N A191598 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.
%H A191598 Michael S. Branicky, <a href="/A191598/b191598.txt">Table of n, a(n) for n = 1..10000</a>
%e A191598 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.
%o A191598 (Python)
%o A191598 from itertools import count, islice
%o A191598 def agen():  # generator of terms
%o A191598     an, aset, m = 1, {1}, 2
%o A191598     for n in count(1):
%o A191598         yield an
%o A191598         an = 2*an if m > n else m
%o A191598         aset.add(an)
%o A191598         while m in aset: m += 1
%o A191598 print(list(islice(agen(), 67))) # _Michael S. Branicky_, Nov 13 2024
%K A191598 nonn,easy
%O A191598 1,2
%A A191598 _Greg Knowles_, Jun 08 2011