A Permutation Defining Itself

[Leroy Quet]

I just submitted this sequence (which I find interesting) to the EIS:

>%S A000001 1,2,1,4,1,3,1,8,1,5,1,9,1,6,1,16,1,7,1,14,1,10,1,21,1,11,1
>%N A000001 a(2n-1) = 1;
>a(2n) = a(n)th lowest positive integer not among the earlier terms of the 
>sequence.
>%C A000001 The sequence {a(2k)} forms a permutation of the integers >= 2.
>%e A000001 a(12) = the a(6)th (the 3rd) lowest positive integer not among 
>the first 11 terms of the sequence. Not among the first 11 terms are 6, 7, 
>9, 10,...
>The 3rd of these is 9, which is a(12).
>%O A000001 1
>%K A000001 ,easy,more,nonn,
>%A A000001 Leroy Quet (qq-quet at mindspring.com), Dec 17 2004


Is there a direct (nonrecursive) way to calculate the terms of the 
sequence?


By the way, the "Quet transform (involving permutations)" of the sequence 
yields
(if I calculated right)
1,2,1,2,3,4,1,1,3,4,...

Does this second sequence have any significance, can its terms be 
directly calculated?

thanks,
Leroy Quet