Re: 351 possible

On Dec. 12, 2000, Robert Walker wrote (in part)
>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.

While I agree with him on the result for 12-tet with inversions, I
get ((x+1)^15 + 2*(x^3+1)^5 + 4*(x^5+1)^3 + 8*(x^15+1))/15 for
15-tet, since 3, 6, 9, and 12 each generate a permutation with three
5-cycles, 5 and 10 generate a permutation with five 3-cycles, 1, 2,
4, 7, 8, 11, 13 and 14 generate permutations with a single 15-cycle
and 0 is a permutation with fifteen 1-cycles.

I also disagree with him on the results for 22-tet, since transposing
by an even number like 2 generates two 11-cycles, not the single
22-cycle that he claims. This would make the result:
((1+x)^22 + 10*(1+x^22) + 10*(1+x^11)^2 + (1-x^2)^11)/22

I'm sorry for the delay in responding to this. I saw the post and
thought it looked wrong just before I left for the holidays. I
checked my answers over the holidays, but it took me some time after
getting back to catch up and make sure no one had already made the
correction.

David Bowen

Dave Bowen wrote:

> While I agree with him on the result for 12-tet with inversions, I
> get ((x+1)^15 + 2*(x^3+1)^5 + 4*(x^5+1)^3 + 8*(x^15+1))/15 for
> 15-tet, since 3, 6, 9, and 12 each generate a permutation with three
> 5-cycles, 5 and 10 generate a permutation with five 3-cycles, 1, 2,
> 4, 7, 8, 11, 13 and 14 generate permutations with a single 15-cycle
> and 0 is a permutation with fifteen 1-cycles.

Hi Dave,

Thanks for the correction:

15-tet:
15 1-cycles -> x1^15. One of these: 0.
5 3-cycles -> x3^5. Two of these: 5 10
3 5-cycles -> x5^3. Four of these: 3 6 9 12
15-cycle -> x15. Eight of these: 1 2 4 7 8 11 13 14

So the cycle index polynomial is
(x1^15+2*x3^5+4*x5^3+8*x15)/22

and to get Polya's formula, replace xn by (1+x^n)

result:
(
(1+x)^15
+2*(1+x^3)^5
+4*(1+x^5)^3
+8*(1+x^15)
)/22

Expand the formula, and look at the nth coefficient to find the number of
scales of n notes in 15-tet.

22-tet:

22 1-cycles -> x1^22. One of these: 0.
11 2-cycles -> x2^11 (transpose by 11)
2 11-cycles -> x11^2. Eleven of these (2 4 6 8 10 12 14 16 18 20)
1 22-cycle -> x22. Nine of these (3 5 7 9 13 15 17 19 21)

So the cycle index polynomial is
(x1^22+x2^11+11*x11^2+9*x22)/22

and to get Polya's formula, replace xn by (1+x^n)

result:
(
(1+x)^22
+(1+x^2)^11
+11*(1+x^11)
+9*(1+x^22)
)/22

(here you got 1 1 10 10 as the coefficients, should be 1 1 11 9)

Thanks!

Robert

Dave Bowen wrote:

> While I agree with him on the result for 12-tet with inversions, I
> get ((x+1)^15 + 2*(x^3+1)^5 + 4*(x^5+1)^3 + 8*(x^15+1))/15 for
> 15-tet, since 3, 6, 9, and 12 each generate a permutation with three
> 5-cycles, 5 and 10 generate a permutation with five 3-cycles, 1, 2,
> 4, 7, 8, 11, 13 and 14 generate permutations with a single 15-cycle
> and 0 is a permutation with fifteen 1-cycles.

Hi Dave,

Sorry, my mistake again, you got the 22 coeffs right
Also I had /22 for /15 for 15-tet

....
So the cycle index polynomial is
(x1^15+2*x3^5+4*x5^3+8*x15)/15

and to get Polya's formula, replace xn by (1+x^n)

result:
(
(1+x)^15
+2*(1+x^3)^5
+4*(1+x^5)^3
+8*(1+x^15)
)/15

Expand the formula, and look at the nth coefficient to find the number of
scales of n notes in 15-tet.

22-tet:

22 1-cycles -> x1^22. One of these: 0.
11 2-cycles -> x2^11 (transpose by 11)
2 11-cycles -> x11^2. Ten of these (2 4 6 8 10 12 14 16 18 20)
1 22-cycle -> x22. Ten of these (1 3 5 7 9 13 15 17 19 21)

So the cycle index polynomial is
(x1^22+x2^11+10*x11^2+10*x22)/22

and to get Polya's formula, replace xn by (1+x^n)

result:
(
(1+x)^22
+(1+x^2)^11
+10*(1+x^11)
+10*(1+x^22)
)/22

....

Robert