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%I #25 May 19 2021 16:04:23
%S 0,1,2,3,3,4,6,7,4,5,7,8,11,12,14,15,5,6,8,9,12,13,15,16,20,21,23,24,
%T 27,28,30,31,6,7,9,10,13,14,16,17,21,22,24,25,28,29,31,32,37,38,40,41,
%U 44,45,47,48,52,53,55
%N Starting with an empty list, n is inserted after the a(n)th element such that the binary representations of the list's elements are always sorted lexicographically.
%C The last occurrence of any positive n in this sequence is a(2^(n - 1)).
%C As the list in question expands, its initial terms converge toward A131577.
%C The last item of the list is always zero or an element of A075427.
%H T. D. Noe, <a href="/A188215/b188215.txt">Table of n, a(n) for n = 0..1023</a>
%F a(2^n + b) = n + b + 1 for b = 0 or 1.
%F a(2^n - b) = 2^n - b for b = 1 or 2.
%e For example, an a(n) of 3 means that n should be inserted after the 3rd element of the list to keep the elements lexicographically ordered.
%e [] (Initial empty list)
%e [0] (Zero inserted at the beginning: a(0) = 0)
%e [0, 1] (One inserted after element 1: a(1) = 1)
%e [0, 1, 10] (Two inserted after element 2: a(2) = 2)
%e [0, 1, 10, 11] (Three inserted after element 3: a(3) = 3)
%e [0, 1, 10, 100, 11] (Four inserted after element 3: a(4) = 3)
%t lst = {}; Table[s = IntegerString[n, 2]; lst = Sort[Append[lst, s]]; Position[lst, s][[1, 1]] - 1, {n, 0, 63}] (* _T. D. Noe_, Apr 19 2011 *)
%o (Python)
%o l = []
%o for i in range(17):
%o b = bin(i)[2:]
%o l.append(b)
%o l.sort()
%o print(l.index(b))
%Y Cf. A264596.
%K nonn,base
%O 0,3
%A _Grant Garcia_, Mar 24 2011
%E Program added by _Grant Garcia_, Mar 30 2011
%E Edited by _Grant Garcia_, Apr 13 2011