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a(1) = 1; a(n) = (largest odd divisor of a(n-1))th smallest positive integer not yet in the sequence.
3

%I #14 Sep 06 2018 14:16:22

%S 1,2,3,6,7,12,8,4,5,14,17,28,18,21,35,50,40,15,30,31,51,72,23,43,66,

%T 56,20,16,9,27,54,55,87,120,38,45,79,115,153,192,13,37,73,112,26,41,

%U 81,126,105,152,52,42,59,104,44,36,33,78,88,46,67,124,80,24,19,64,10,32,11,58

%N a(1) = 1; a(n) = (largest odd divisor of a(n-1))th smallest positive integer not yet in the sequence.

%C It seems likely, but not certain, that this sequence is a permutation of the positive integers, which it is if and only if there are an infinite number of powers of 2 in the sequence.

%H Ivan Neretin, <a href="/A101319/b101319.txt">Table of n, a(n) for n = 1..10000</a>

%e a(6) = 12 and the highest odd divisor of 12 is 3. Among the first 6 terms of the sequence is not 4, 5, 8, 9, ... and the 3rd of these is 8, which is therefore a(7).

%t Nest[Append[#, Complement[Range[Max[#] + (r = #[[-1]]/2^IntegerExponent[#[[-1]], 2])], #][[r]]] &, {1}, 69] (* _Ivan Neretin_, Sep 03 2018 *)

%K nonn

%O 1,2

%A _Leroy Quet_, Dec 23 2004

%E More terms from _Hans Havermann_, Dec 24 2004