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%I #13 Oct 12 2018 20:26:53
%S 1,2,5,3,9,7,13,4,17,12,21,10,25,18,29,6,33,23,37,16,41,28,45,14,49,
%T 34,53,24,57,39,61,8,65,44,69,31,73,50,77,22,81,55,85,38,89,60,93,19,
%U 97,66,101,46,105,71,109,32,113,76,117,52,121,82,125,11,129,87,133
%N Let k be the greatest odd divisor of n and let S be the sequence of positive integers not in the sequence so far in increasing order. Then a(n) = S(k).
%C In other words, a(n) = A000265(n)-th positive integer unused so far.
%C A permutation of the positive integers.
%C a(n) = 2n - 1 for odd n, a(n) < 2n - 1 otherwise.
%H Ivan Neretin, <a href="/A318578/b318578.txt">Table of n, a(n) for n = 1..10000</a>
%e For n=6, the highest odd divisor of n is k = 3. The sequence up to that point is 1, 2, 5, 3, 9. The numbers which are not yet in the sequence (in increasing order) are S = 4, 6, 7, 8, 10, ... and the 3rd of these is 7, which is therefore a(6).
%t Fold[Append[#1, Complement[Range[Max[#1] + (r = #2/2^IntegerExponent[#2, 2])], #1][[r]]] &, {1}, Range[2, 67]]
%Y Cf. A119435, A101319, A000265.
%K nonn
%O 1,2
%A _Ivan Neretin_, Aug 29 2018
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