Suggestion for a sequence: weights on a circle

[Hugo Pfoertner]

Edwin Clark wrote:
> 
> On Wed, 14 Sep 2005, Hugo Pfoertner wrote:
> 
> > Brendan McKay wrote:
> > >
> > > A recent puzzle in New Scientist suggests the following.
> > >
> > > Suppose you have n objects, with weights 1,2,3,...,n.
> > > (That is, exactly one object with each of those weights.)
> > >
> > > How many ways are there to place these objects evenly spaced
> > > around the circumference of a disk so that the disk will
> > > exactly balance on the centre point?
> > >

Now knowing that perfectly balanced arrangements of n weights will not
be possible if n is a prime power, one could ask the more general
question: How many solutions are there for a given n that minimize the
remaining imbalance?

With a slightly modified program I calculated:
(the floating point number gives the distance of the center of gravity
from (0,0)

 Number of weights: 3
 0.577350    1    1  2  3
 0.577350    2    1  3  2

 Number of weights: 4
 0.353553    1    1  3  2  4
 0.353553    2    1  4  2  3

 Number of weights: 5
 0.089806    1    1  4  3  2  5
 0.089806    2    1  5  2  3  4

 Number of weights: 7
 0.010927    1    1  4  7  2  3  5  6
 0.010927    2    1  6  5  3  2  7  4

 Number of weights: 8
 0.016415    1    1  4  7  3  6  2  5  8
 0.016415    2    1  7  4  3  6  5  2  8
 0.016415    3    1  8  2  5  6  3  4  7
 0.016415    4    1  8  5  2  6  3  7  4

 Number of weights: 9
 0.003184    1    1  5  9  2  7  3  4  8  6
 0.003184    2    1  5  9  4  2  6  7  3  8
 0.003184    3    1  6  5  4  9  2  3  7  8
 0.003184    4    1  6  8  4  3  7  2  9  5
 0.003184    5    1  8  3  7  6  2  4  9  5
 0.003184    6    1  8  7  3  2  9  4  5  6

 Number of weights: 11
 0.000019    1    1  8  9  5  2  6 10  7  3  4 11
 0.000019    2    1 11  4  3  7 10  6  2  5  9  8

 Number of weights: 13
 0.000028    1    1  2  7 12 13  4  5  3  8  6 11  9 10
 0.000028    2    1  4 11  6  5 12  7  2  9  8  3 10 13
 0.000028    3    1  5  3  8 12 10  7  4  2  6 11  9 13
 0.000028    4    1 10  9 11  6  8  3  5  4 13 12  7  2
 0.000028    5    1 13  9 11  6  2  4  7 10 12  8  3  5
 0.000028    6    1 13 10  3  8  9  2  7 12  5  6 11  4

Merging this with the numbers of perfectly balanced solutions we can
make the following sequence, starting at n=3:

2 2 2 4 2 4 6 48 2 1464 6 1440 96

Is there a better idea to get the next terms by something more clever
than trying all possible permutations?

Hugo Pfoertner