Re: 351 possible scales

> Unfortunately there is no simple mathematical answer to this question that
> I know of... The total number of possible sets in 12tet is 12^2 or 4096,

Here is how it's done:

To find the answer, I looked up the sequence 1,6,19,43,66,80 in Sloane's On-Line Encyclopedia of Integer Sequences

http://www.research.att.com/~njas/sequences/index.html

Result:

Sequence: 1,1,6,19,43,66,80,66,43,19,6,1,1
Name: Musical scales consisting of n notes.
References R. C. Read, Combinatorial problems in theory of music, Discrete Math.
167 (1997), 543-551.
Formula: Coefft. of x^n in cycle index Z(C_12; 1+x).

However that formula is a bit cryptic. What does it mean?

It uses Polya's counting theorem.

First, we need the idea of a cycle. If you transpose a mode up by 3 notes in the
12 tone scale, then repeat that transposition, after four repeats, each note gets
back to where it started. Each note belongs to a cycle,and the length of the cycle
is the number of repeats needed to get the note back to where it started.

So in this example, each note belongs to a cycle of four notes, and there are three such
cycles altogether. In fact, these cycles are the notes of the three diminished seventh chords.

Transposition by 8 notes generates the same three cycles.

Similarly, for transposition up by 2 notes, you have two cycles, each of six notes,
corresponding to the two possible whole tone scales. Ditto for 10 notes.

For transposition by 1, 5, 7 and 11 notes, you have a single cycle of 12 notes.

For transposition by 0 notes, you have 12 cycles each of one note (each note just stays
where it is).

For transposition by 6 notes you have 6 cycles each of 2 notes, and for transposition by 4 notes, you have 4 cycles each
of 3 notes (the four possible tritones).

Now to make all this into an equation.

The equation for each permutation consists of x1^a*x2^b*....
where a is the number of cycles of length 1, b the number of cycles of length 2, and so on.

For instance, for the transposition by 4 notes, you have the four tritones, each of which is a cycle of 3 notes, so the
term that this contributes is x3^4. The other tritone ransposition by 8 notes adds an identical term.

For some pictures of these permutations, see
http://www.arch.gatech.edu/~economou/research/m&d98/

Look under Cycle index | computation

This shows the first seven of the twelve rotations of the dodecagon (rotation by 12-n has same
picture as rotation by n). It also has the reflections, which we
ignore for this problem (but would include if counting inversions as equivalent).

You then add together all the terms, and divide by the total number of permutations in the group:

(x1^12 + x2^6 + 2*x3^4 + 2*x4^3 +2*x6^2 + 4*x12)/12

This is called the cycle index polynomial.

Then, to apply Polya's counting theorem, you replace x1 by
(t1+t2+...+tn), x2 by (t1^2+t2^2+...+tn^2), etc, where n is the number of colourings -
in this case, there are just two colourings for each note - included, or not included.
But if you wanted to have, say, notes that are muted as a third poss, then you could
use n=3, and so on.

Since there are just two possibilities, let's call them x and y. (in the ref. above, they
were called x and 1, which is also okay).

So equation becomes:
(
(x+y)^12
+ ((x^2+y^2)^6
+ 2*((x^3+y^3)^4
+ 2*((x^4+y^4)^3
+ 2*((x^6+y^6)^2
+ 4*(x^12+y^12)
)/12

Then Polya's counting theorem says that the number of distinct ways of
colouring n notes is the coefficient of x^n*y^(12-n) in this equation.

The odd numbers are particularly easy to find since most of the
terms are for all powers even, and so can be ignored.

We only need to expand:
(((x+y)^12 + 2*(x^3+y^3)^4)/12

Let's look for the term x^3*y^9.
Using row 12 of Pascal's triangle:
1 12 66 220 495 792 924 792 495 220 66 12 1
(from e.g. http://www-chem430.scs.uiuc.edu/nmr-resources/pascal/pascal1.htm)
then it's coeff in (x+y)^12 is 220
It's coeff in 2*(x^3+y^3)^4 is 8
So we want 228/12 = 19.

Now try n = 5

then it's coeff in (x+y)^12 is 792
It's coeff in 2*(x^3+y^3)^4 is 0
So we want 792/12 = 66.

Now let's try n = 19
This is easy because it is prime, so all the entries, except the identity are made up of 19-cycles.
Cycle index formula:
(x^19+18*x19)/19

Polya's formula:
(x+y)^19 + 18*(x^19+y^19))/19

So, except for the 0 and 19 note cases, we only need to look at nth term in (x+y)^19
Using row 19 of Pascal's triangle:

1, 1, 9, 51, 204, 612, 1428, 2652, 3978, 4862, 3978, 2652, 1428, 612, 204, 51, 9, 1, 1

For Thai equal 7 tone scale,
1 7 21 35 35 21 7 1
->
1, 1, 3, 5, 5, 3, 1.

For example, the five three note scales are (showing the steps, e.g. 0 2 5 is an example of 2 3):
1 1, 1 2, 1 3, (1 4 is same as 1 3)
2 2, 2 3
3 2 (same as 2 3).

Let's work out one more example

For 15-tet:

Cycle index formula:
transposition by 1, 2, 4, 7, 8, 11, 13, 14 all yield single cycle of 15 notes, i.e. x15.

Identity gives x1^15 (as usual).

Transposition by 3, 6, 9 and 12 yield x5^3 (each is made up of three 5-cycles).

Transposition by 5, 10 yield x3^5 (each made of five 3-cycles).

So cycle index formula is
(x1^15 + 2*x3^5 + 3*x5^3 + 9*x15)/15

Polya's formula
((x+y)^15 + 2*(x^3+y^3)^5 + 3*(x^5+y^5)^3 + 9*(x^15+y^15))/15

So for example, to find the number of 6 note scales in 15-tet,
need coeff. of x^6, so only need first two terms.

Coeff,. of x^6 in ((x+y)^15
Using 15th row of Pascal's triangle
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
we see it is 5005

Coeff. of x^6 in 2*(x^3+y^3)^5, using
1 5 10 10 5 1
is 20

So answer is 5025/15 = 335.

To treat inversions as identical, need to add extra terms for these.

In 12-tet there are 12 possible inversions - 6 reflecting in a note, and also 6 in gap
between two notes.

E.g.
1 2 3 4 5 6 7 8 9 10 11 12
->
12 11 10 9 8 7 6 5 4 3 2 1
(inverting in gap between 7 and 8 - all notes change)
or
->
1 12 11 10 9 8 7 6 5 4 3 2
(1 and 7 remain unchanged)

First type has 6 cycles each of length 2, e.g. 1 -> 12 -> 1.
term in the cycle index equation before the divide by 24:
6*x2^6 (12 such inversions).

Second type has 5 cycles of length 2, and two cycles of length 1.
You multiply the results, so the contribution to the cycle index equation is
6*x2^5 * x1^2

So equation now is
(x1^12 + 7*x2^6 + 2*x4^3 + 2*x3^4 + 6*x^5*x1^2 +2*x6^2 + 4*x12)/24

Robert

Thanks a bunch, Robert, though I got lost as to where you moved from 12-tET
to another tuning and then back again, and didn't see an explicit
calculation yielding 351 (or 352).