Pentachords in 12t-ET - Tuning Coincidence

Here are interval vector counts for the 66 pentachords in 12t-ET (I
have divided C{12,5} by 12 since these pentachords divide out cleanly)

3,3,4,0,3,16
20,20,16,27,20,40
30,30,36,27,30,10
12,12,8,9,12,0
1,1,2,3,1,0

Columns are for 1 through 6, semitones through tritones, and rows
are for values 0 through 4, (For example, Row 1 Column 1 is how
many times 0 occurs in the semitone slot)

Here they are weighted

0,0,0,0,0,0
20,20,16,27,20,40
60,60,72,54,60,20
36,36,24,27,36,0
4,4,8,12,4,0

This just means you multiply by the value in the slot, so
that all the occurrences of say, "4" in the semitone slot
are multiplied by 4.

Now take unweighted values and sort: You obtain
1,2,3,8,9,10,12,16,20,27,30,36

Weighted values gives
4,8,12,16,20,24,27,36,40,54,60,72

Reducing for powers of 2 (put in the same octave), with either list
you get

16,18,20,24,27,30

which is a 5-limit hexachord C,D,E,G,A,B. Now if you take
a circle of 6 open fifths based on (1,2,3,4,5,6) you get
G,A,B,D,E,F#. Measuring these intervals from "C" you get
7,3,11,2,4,6 which is the same as (1,2,3,4,5,6) Coincidence?
Seeing as I am measuring interval vector value counts in
the (1,2,3,4,5,6) interval slots. Of course with C,D,E,G,A,B
you have to measure from "F" to get (1,2,3,4,5,6) intervals...