924 Hexachords

Take the 924 hexachords (12C6). The sum of their interval vectors
is (10C4 * 12) for 1,2,3,4,5 intervals and (10C4 * 6) for tritones.
(The proof is in a previous post).

Take Pascal's triangle, and reduce for values involved in obtaining
12C6. (This is really a triangle->diamond)

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
7 21 35 35 21 7
28 56 70 56 28
84 126 126 84
210 252 210
462 462
924

Take the first row that has a common denominator. (7 21 35 35 21 7)
Pull out the 7 factor. (1 3 5 5 3 1). Now to obtain 924 we fold
this over five times and get 7*((2*1)+(3*10)+(5*20)) By the same
procedure get 7 *((1*1)+(3*3)+(5*4)) for (10C4).

Since 11 is not in the 7 limit, it is nice to get rid of it. Now
although 924=7*11*3*4, the sums of each vector value column
is 2520 2520 2520 2520 1260. Divide this out by 12 12 12 12 12 6,
and you obtain exactly how many times each interval occurs, and
at which transposition, (where tritones get split in two of course)

Therefore, for the 924 hexachords, each interval occurs exactly
(10C4) times, which is (7*5*3*2), the primes of 7-limit tuning.

I should have posted this before my 80,50,35 coincidence post,
but I am not sure it gives me that much more credibility!

Now before dividing by 12 (6 for tritones), you have 7 *(1+9+20)
or 7+63+140 (intervals) and 7*(2+30+100) = 14+210+700 (everything).
Could this tie into tuning? With my system the intervals 1,2,3,4,5,6
are based on 7-limit tuning as 5/4 X 6/5 X 7/5 to obtain
21/20, 35/32, 6/5, 5/4, 21/16, 7/5 as the expression starting from
1/1. Of course there are other expressions, like the obvious 4/3
for interval 5.

Crazy idea?

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> Take the 924 hexachords (12C6). The sum of their interval vectors
> is (10C4 * 12) for 1,2,3,4,5 intervals and (10C4 * 6) for tritones.
> (The proof is in a previous post).
>
> Take Pascal's triangle, and reduce for values involved in obtaining
> 12C6. (This is really a triangle->diamond)
>
> 1
> 1 1
> 1 2 1
> 1 3 3 1
> 1 4 6 4 1
> 1 5 10 10 5 1
> 1 6 15 20 15 6 1
> 7 21 35 35 21 7
> 28 56 70 56 28
> 84 126 126 84
> 210 252 210
> 462 462
> 924
>
> Take the first row that has a common denominator. (7 21 35 35 21 7)
> Pull out the 7 factor. (1 3 5 5 3 1). Now to obtain 924 we fold
> this over five times and get 7*((2*1)+(3*10)+(5*20)) By the same
> procedure get 7 *((1*1)+(3*3)+(5*4)) for (10C4).
>
> Since 11 is not in the 7 limit, it is nice to get rid of it. Now
> although 924=7*11*3*4, the sums of each vector value column
> is 2520 2520 2520 2520 1260. Divide this out by 12 12 12 12 12 6,
> and you obtain exactly how many times each interval occurs, and
> at which transposition, (where tritones get split in two of course)
>
> Therefore, for the 924 hexachords, each interval occurs exactly
> (10C4) times, which is (7*5*3*2), the primes of 7-limit tuning.
>
> I should have posted this before my 80,50,35 coincidence post,
> but I am not sure it gives me that much more credibility!
>
> Now before dividing by 12 (6 for tritones), you have 7 *(1+9+20)
> or 7+63+140 (intervals) and 7*(2+30+100) = 14+210+700 (everything).
> Could this tie into tuning? With my system the intervals 1,2,3,4,5,6
> are based on 7-limit tuning as 5/4 X 6/5 X 7/5 to obtain
> 21/20, 35/32, 6/5, 5/4, 21/16, 7/5 as the expression starting from
> 1/1. Of course there are other expressions, like the obvious 4/3
> for interval 5.
>
> Crazy idea?

Here's the full grid for the 924 hexachords, where each cell is
divided by 12 (6 for tritones):

0 0 0 0 0 0
5 4 4 0 5 80
50 55 48 81 50 120
100 90 112 54 100 10
50 60 36 72 50 0
5 0 10 0 5 0
0 1 0 3 0 0
210 210 210 210 210 210

The columns are intervals 1-6. The rows are the frequency in which
they occur (at each transpostion, since I divided it out).

>