Double Diads

I have calculated the distinct energy states (interval vectors)
between every combination of diads in 12-et (stacked necklaces) as
being 57.

In order to calculate the interval vectors between every
pair of hexachords will require checking C{81,2}*7=22,680
combinations. (80+79+78+77...3+2+1 times 7 offsets, 0 thru 6)

Checking for diads only takes C{7,2} * 7 combinations = 147. They
reduce to 57. Below is the interval vector times the multiplicity
Note that the vector measures (0,1,2,3,4,5,6). With two stacked
necklaces, it is now neccessary to measure "0" interval as well
as 1 thru 6)

0,0,0,0,0,2,2 1
0,0,0,0,1,2,1 3
0,0,0,0,2,0,2 1
0,0,0,0,2,2,0 1
0,0,0,1,0,3,0 1
0,0,0,1,1,1,1 5
0,0,0,1,2,1,0 3
0,0,0,2,0,0,2 1
0,0,0,2,0,2,0 1
0,0,0,2,2,0,0 1
0,0,0,4,0,0,0 1
0,0,1,0,1,1,0 1
0,0,1,0,1,2,0 1
0,0,1,0,2,0,1 3
0,0,1,1,0,1,1 5
0,0,1,1,1,1,0 5
0,0,1,2,1,0,0 5
0,0,2,0,0,0,2 1
0,0,2,0,0,2,0 1
0,0,2,0,2,0,0 4
0,0,2,1,0,1,0 2
0,0,3,0,0,0,1 1
0,1,0,0,2,1,0 2
0,1,0,1,0,2,0 2
0,1,0,1,1,0,1 5
0,1,0,1,2,0,0 2
0,1,0,2,0,1,0 6
0,1,1,0,0,1,1 5
0,1,1,0,1,1,0 10
0,1,1,1,0,0,1 3
0,1,1,1,1,0,0 1
0,1,2,0,0,1,0 2
0,1,2,1,0,0,0 1
0,2,0,0,0,0,2 1
0,2,0,0,0,2,0 4
0,2,0,0,1,0,1 2
0,2,0,1,0,1,0 2
0,2,1,0,1,0,0 2
0,3,0,1,0,0,0 1
1,0,0,0,3,0,0 1
1,0,0,1,1,1,0 3
1,0,0,2,0,0,1 4
1,0,1,0,0,2,0 2
1,0,1,0,1,0,1 7
1,0,1,1,0,1,0 3
1,0,2,0,1,0,0 2
1,1,0,0,0,1,1 7
1,1,0,0,1,1,0 3
1,1,0,1,1,0,0 3
1,1,1,1,0,0,0 3
1,2,1,0,0,0,0 2
2,0,0,0,0,0,2 2
2,0,0,0,0,2,0 1
2,0,0,0,2,0,0 1
2,0,0,2,0,0,0 1
2,0,2,0,0,0,0 1
2,2,0,0,0,0,0 1