Re: Set-theory quirk in 29-tet

Paul Hj. wrote, quoting me:

> > 173173 nonachords
> > 346346 decachords
> >
> > ("set-class": equivalence class of pc collections, under transposition
> > and inversion)
> >
> > My guess is that's it's more of a coincidence, rather than something
> > you can explain with a quirk of number theory in Z_29. Anyone know
> > any better?
>
> Yes one runs across quite a bit of this sort of thing. It took me a > while to figure out the formulas for symmetric sets, (It varies, > depending on the evenness and oddness of n and k, among other things) So > a prime ET like 29 with k=10 is Combin(29,10)/29 and with k=9 is > Combin(29,9)/29, which is exactly half, the number of symmetric sets is > 2002 and 1001 respectively, obviously one is half the other, so the > number of set types is just ((Combin(29,10)/29)+2002)/2 and > ((Combin(29,9)/29)+1001)/2.

I don't think this is quite right: you're _adding_ the symmetrical sets to the count of _all_ sets C_29,10 (which already includes the symmetrical sets). But the general principle of separating them is right. So another aspect of the "coincidence" here is that the number of symmetrical types is a multiple of 1001, as is the number of non-symmetrical types (you can tell because the overall numbers are of the form abcabc, which is always a multiple of 1001).

Best --Jon

--- In tuning-math@yahoogroups.com, Jon Wild <wild@m...> wrote:
>
>
> > Yes one runs across quite a bit of this sort of thing. It took me
a
> > while to figure out the formulas for symmetric sets, (It varies,
> > depending on the evenness and oddness of n and k, among other
things) So
> > a prime ET like 29 with k=10 is Combin(29,10)/29 and with k=9 is
> > Combin(29,9)/29, which is exactly half, the number of symmetric
sets is
> > 2002 and 1001 respectively, obviously one is half the other, so
the
> > number of set types is just ((Combin(29,10)/29)+2002)/2 and
> > ((Combin(29,9)/29)+1001)/2.
>
> I don't think this is quite right: you're _adding_ the symmetrical
sets to
> the count of _all_ sets C_29,10 (which already includes the
symmetrical
> sets). But the general principle of separating them is right. So
another
> aspect of the "coincidence" here is that the number of symmetrical
types
> is a multiple of 1001, as is the number of non-symmetrical types
(you can
> tell because the overall numbers are of the form abcabc, which is
always a
> multiple of 1001).

It's an average - you average the total with the symmetric sets to
trim off half the asymmetric sets.

1001 is C(14,Floor(9/2)= C(14,4)
2002 is C(14,5)

Paul Hj