Permutation(?): + Integers Increasingly Arranged

[Leroy Quet]

[Posted also to sci.math.]
 
Define an integer sequence as follows:

a(1) = 1; 

And place the positive integers, once per integer, into the sequence so 
that:

(m+1) is positioned in the sequence such that there are exactly (m-1) 
higher-valued terms between it and the term equal to m.
And the (m+1) is always less that the m whenever this is possible.

As to hopefully eliminate ambiguity, here is the arrangement of the first
16 integers: (figured by-hand)

1, 2, 13, 3, 6, *, 4, 11, *, 9, 5, * 7, *, *, *, *, *, *, 8,
 *, 10, 16, 12, *, 14, *, *, *, *, *, *, *, *, *, *, *, *, *, 15,...


Of course, the *'s are yet to be determined and are each > 16.


Because of the "higher-valued" part of the definition, the sequence will 
never get 'stuck', ie. unlike if we simply counted the number of terms 
in-between, where the sequence might get into a region where the nearest 
unused position is too far away.

But I must ask, is this sequence a permutation of the positive integers?
In other words, will every position eventually contain an integer?

If so, here is the inverse permutation's first few terms:

1, 2, 4, 7, 11, 5, 13, 20, 10,...

(Maybe the inverse sequence would be easier to study, actually.)

thanks,
Leroy Quet