Permutation(?): + Integers Increasingly Arranged

[Leroy Quet]

>Yes, it is a permutation.  We need to show that every space gets 
>filled eventually.  By induction, it is enough to prove that the 
>left-most space gets filled eventually.  We define a related sequence:
>     When n is placed, let b(n) be the number of empty spaces to the left 
>of n.
>This sequence obeys a recurrence:
>     If b(n) >= n, b(n + 1) = b(n) - n;
>     If b(n) < n,  b(n + 1) = b(n) + n - 1.
>What we need to show is that this sequence will always eventually get 
>back to 0.

>......

>...., so the sequence will always eventually 
>get back to 0.
>
>  - David Wasserman
>


Thanks for considering the sequence!

Has anybody calculated the sequence and/or its inverse to any number of 
terms greater than I have? 

Here is the sequence's definition again, for those who have deleted the 
original email.


>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 than

[ie. to the left of]

>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.

And again, here is the inverse permutation's first few terms:

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


thanks,
Leroy Quet