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A269838 a(1)=1, then a(n) is the smallest unused multiple of a(A062050(n-1)).

%I A269838 #6 Mar 29 2016 06:05:02
%S A269838 1,2,3,4,5,6,9,8,7,10,12,16,15,18,27,24,11,14,21,20,25,30,36,32,28,40,
%T A269838 48,64,45,54,81,72,13,22,33,44,35,42,63,56,49,50,60,80,75,90,108,96,
%U A269838 55,70,84,100,125,120,144,128,112,160,192,256,135,162,243,216,17
%N A269838 a(1)=1, then a(n) is the smallest unused multiple of a(A062050(n-1)).
%C A269838 A permutation of positive integers.
%C A269838 Alternative construction: start with the infinite sequence of all 1s, at step n look up the value of the term at n'th place (k), and if it has been used before, find the smallest unused multiple of k (m*k), find all occurrences of k in the sequence (there are going to be infinitely many), and change every second of them to m*k (see Example).
%H A269838 Ivan Neretin, <a href="/A269838/b269838.txt">Table of n, a(n) for n = 1..10000</a>
%e A269838 Start with
%e A269838 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
%e A269838 At step 1, do nothing.
%e A269838 At step 2, change every second 1 to 2.
%e A269838 2: 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2
%e A269838 At step 3, change every other remaining 1 to 3.
%e A269838 3: 1 2 3 2 1 2 3 2 1 2 3 2 1 2 3 2
%e A269838 At step 4, change every other 2 to 4.
%e A269838 4: 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
%e A269838 And so on.
%e A269838 5: 1 2 3 4 5 2 3 4 1 2 3 4 5 2 3 4
%e A269838 6: 1 2 3 4 5 6 3 4 1 2 3 4 5 6 3 4
%e A269838 7: 1 2 3 4 5 6 9 4 1 2 3 4 5 6 9 4
%e A269838 8: 1 2 3 4 5 6 9 8 1 2 3 4 5 6 9 8
%e A269838 9: 1 2 3 4 5 6 9 8 7 2 3 4 5 6 9 8
%e A269838 ...
%t A269838 Fold[Append[#1, Min@Complement[Range[Max@#1 + 1]*#1[[#2 - 2^Floor@Log2[#2 - 1]]], #1]] &, {1}, Range[2, 65]] (* _Ivan Neretin_, Mar 06 2016 *)
%K A269838 nonn
%O A269838 1,2
%A A269838 _Ivan Neretin_, Mar 06 2016