Re: 351 scales

hi Robert,

thanks for showing that -- it wasn't as hairy as I imagined.

> Now let's try n = 19

[note - this n (meaning the size of the available pitch-class
universe) isn't the same as Robert's other n (which meant "n out of 12"),
which might have been confusing in the post]

> 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

Here's what seems odd: in the 19-tone universe, 9-chords are the
complements of 10-chords, so you should have as many of one as you do of
the other. You say you can read the number of each from the appropriate
row of Pascal's triangle, but the 9th and 10th terms are different here.
Do you need to repeat the value 4862? --Jon

Clearly the 20th, not 19th, row of Pascal's triangle is the appropriate one.

Hi Jon,

> > 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

> Here's what seems odd: in the 19-tone universe, 9-chords are the
> complements of 10-chords, so you should have as many of one as you do of
> the other. You say you can read the number of each from the appropriate
> row of Pascal's triangle, but the 9th and 10th terms are different here.
> Do you need to repeat the value 4862? --Jon

Yes you do
Looking again at
http://www-chem430.scs.uiuc.edu/nmr-resources/pascal/pascal1.htm

row 19 is
1 19 171 969 3876 11628 27132 50388 75582 92378 92378 75582 50388 27132 11628 3876 969 171 19 1

Results listed are all these numbers divided by 19, except for the first and last, which remain as 1 (because of the 18*(x^19+y^19) term).

So the 4862 is 92378/19.

So I left out one of the 4862s, sorry.

> thanks for showing that -- it wasn't as hairy as I imagined.

Glad to be of help!

Robert

Cool! I just got Maple to expand the generating function for 19-tet
including inversions. You get this:

1+x+9*x^2+30*x^3+120*x^4+324*x^5+756*x^6+1368*x^7+2052*x^8+2494*x^9
+2494*x^10+2052*x^11+1368*x^12+756*x^13+324*x^14+120*x^15+30*x^16
+9*x^17+x^18+x^19

which agrees (phew) with my results that used a brute-force computer
search, posted earlier here: http://www.egroups.com/message/tuning/14154

And for 31-tet, which someone asked about earlier too, you get these
results:

cardinality number of chords
0, 31 1
1, 30 1
2, 29 15
3, 28 80
4, 27 560
5, 26 2793
6, 25 12103
7, 24 42640
8, 23 127920
9, 22 325845
10, 21 716859
11, 20 1367184
12, 19 2278640
13, 18 3329165
14, 17 4280355
15, 16 4850640

In a stunning coincidence, the last number also happens to be the phone
number of an old girlfriend!! Did someone just turn the Improbability
Drive on?

Lastly, the number of scales of 26 notes in 53-tet is: 9183681736376

that's 9 quadrillion, 183 trillion, 681 million, 736 thousand and 376. I
doubt we'll be seeing a 53-tet sequel to Slominsky's compendium of
scales...

I wrote (too quickly):

> Lastly, the number of scales of 26 notes in 53-tet is: 9183681736376
>
> that's 9 quadrillion, 183 trillion, 681 million, 736 thousand and
> 376.

oops, I forgot the "billions" column... adjust accordingly (9
trillion etc). And if anyone is relieved by the adjustment, there's a
story about the physicist Feynman lecturing on the length of time
remaining before the sun swells up and engulfs the earth. A nervous
student puts up his hand and says "E-e-excuse me, *how* many years
did you say?". Feynman says "Ten billion" at which the student
breathes a huge sigh of relief, and says "Thank God! I thought you
said ten million!"

--Jon

Hi Paul,

I've looked up the paper R.C. Read "Combinatorial problems in the theory of
music", Discrete Mathematics, 167/168. He has some other nice results
there.

He enumerates the numbers of notes for each scale according to symmetry.

His numbers agree with the numbers in the original post

http://www.egroups.com/message/tuning/16442

He lists the scales by symmetry, so the whole tone scale has symmetry 6
(all six of its modes are identical).

His numbers are: (number of notes, then numbers of scales in brackets)
2-fold symmetry: 2 (1), 4 (2), 6 (3) 8 (2) 10 (1)
3-fold symmetry: 3 (1), 6 (1), 9 (1).
4-fold symmetry: 4 (1), 8 (1)
6-fold symmetry: 6 (1).

He says of course that the numbers can be found by exhaustive enumeration
of the possibilities, but finds them using "Mobius inversion over the
lattice of subsets of the cyclic group C12."

Unfortunately, I don't yet have the background to follow his explanation
there. But would be interested to look it up some time and find out, and
then investigate how one can find the same results for other numbers of
notes.

Or, I wonder if anyone else on TL has the necessary group theoretic /
number theoretic background and knows how to do this already?

Robert