Re: 351 possible

Hi Paul

> 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)

352 is the sum of all the coefficients of
(
(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

which I didn't work out as it would take a while, though not nearly so long as
enumerating all the scales by hand.

(x+1) is probably easier to use than (x+y) in calcs - you don't need to remember
the powers of the y as in e.g. x^3*y^9, the two indices have to add up to 12.

So expand

(
(x+1)^12
+ ((x^2+1)^6
+ 2*((x^3+1)^4
+ 2*((x^4+1)^3
+ 2*((x^6+1)^2
+ 4*(x^12+1)
)/12

and the coefficient of x^n is the number of scales of n notes in 12-tet, and sum of all
the coeffs is 352 (counting scale of 0 notes as a scale!).

When doing it by hand, you can use Pascal's triangle to expand (1+a)^n

(1+a)^n = 1 + (1,n)a+ (2,n) a^2 + ...

where (k,n) = kth term in nth row of Pascal's triangle.

E.g.
(1+x)^5 = 1 + 5x + 10 x^2 + 10 x^3 + 5 x^4 + x^5

You need to find Polya's formula anew for each number of notes.

For scales with prime numbers of notes

Polya's formula:
( (x+1)^n + (n-1) *(x^n+1) )/n

the (x+1)^n is the term for the identity (single permutation consisting of n 1-cycles)
and the (n-1) *(x^n+1) is the term for all the other
transpositions - n-1 permutations, each consisting of one n-cycle.

Again, expand the formula, and find the coefficients.

For 15 tet
Polya's formula:

((x+1)^15 + 2*(x^3+1)^5 + 3*(x^5+1)^3 + 9*(x^15+1))/15

and for 12-tet with inversions

(
(1+x)^12
+ 7*(1+x^2)^6
+ 2*(1+x^4)^3
+ 2*(1+x^3)^4
+ 6*(1+x^5)*(1+x)^2
+ 2*(1+x^6)^2
+ 4*(1+x^12)
)/24

Again, expand the formula, and look at the nth coefficient to find the number of scales of n notes.

To apply to scales with other numbers of notes, you need to look at all the transpositions, see what cycles they
generate, and so make the cycle index polynomial. Then substitute (1+x^n) for xn to get Polya's formula.

E.g. in 22-tet, transposing by any amount except the identity, and 11 notes, gives a 22-cycle (take any note, and keep
transposing it by the same amount, and you go through all 22 notes). The transposition by 11 notes puts the notes into
11 2-cycles. Identity yields 22 1-cycles.

22-cycle -> x22. There are 20 of these

11 2-cycles -> x2^11

22 1-cycles -> x1^22

So the cycle index polynomial is
(x1^22+x2^11+20*x22)/22
and to get Polya's formula, replace xn by (1+x^n)

result:
(
(1+x)^22
+(1+x^2)^11
+20*(1+x^22)
)/22

You could use Mathematica or the like to expand the polynomials instead of doing it by hand.

Robert