Integers Arranged: Q & Puzzle

[Leroy Quet]

I am posting this to sci.math and rec.puzzles:
The reason for me posting it here is that I wonder about the sequence 
defined by:
n_th term is the number of solutions for n.

----------------
This seems like something that MUST have been thought about before.

For a positive integer, n, take all of the distinct integers 1 to n.
Place them along an integer number-line such that:
(Or you can imagine placing n marbles, labeled 1 to n, into a row of
boxes, one marble per box, such that...)

1 is placed anywhere along the line.
(m+1) is placed m steps (integer-positions or boxes) exactly from m,
for 1 <= m <= n-1.
n is placed as to be n steps from 1.
No two placed-integers share a position
(ie. no more than one marble per box).
(But some positions may be without any placed-integer,
ie. some boxes may be empty.)

Two examples:
n = 4  (* is an empty box.)

2,1,3,*,*,4

n = 11

11,9,7,*,*,*,*,*,6,8,10,1,2,5,3,*,*,4


First, the question: How many solutions are there for any given n?

Some n have no solution, such as n=2 or n = 5(?).

Obviously, for n >= 2, the number of solutions is even (because of
symmetry).

Now, I suspect that for some n's (which can be arbitrarily large),
that there is an algorithm to generate a solution.
But perhaps there is not any solution for any n > than some value.

So, the puzzle:

Either form an algorithm that generates a solution for some n > than
any arbitrarily large N.

Or simply try to find a solution for the largest n that you can.

Or, if the algorithm works for a given set of n's (such as n = odds),
then find a solution for the highest n not in that set (such as the
highest even, given my example).

Another question/puzzle: 
What about the most "compact" solution, ie. the solution where the
difference in the distance between the leftmost placed-integer
(leftmost filled box) and the rightmost placed-integer (rightmost
filled box) is minimized?
How would we prove that a certain solution is minimized without simply
looking at all possible solutions?

Thanks,
Leroy Quet