More on Hendecachords in 22t-ET

All,

Thanks for indulging me in this a little more. This is largely for my
own reference and those few who might still be interested in this
subject.

Let's just focus on 11-sets in 22tET:

Subset C(22,11) 16159 Types

Singles 554 554 0

Doubles 13970 6985 6985

Triples 15 5 10

Quads 1560 390 1170

Quints 0 0 0

Sext's 60 10 50
Totals 16159 7944 8215

These indicate Z-relations. There are 32066 11-sets in 22-tET,
reducing for transposition. There is only one (whole tone set)
that doesn't divide by 22, it divides by 2. Now reducing for mirror-
image, we get 32066+252/2=16159. That is the "Types" listed above.
Now, Singles are all the sets with no Z-relations. Doubles, Triples,
Quads and Sext's are when that number has the same interval vector,
but the sets are of different Tn/TnI types.

Now, tonight, I calculated this based on necklaces formulas and some
math on periodic sequences. I'll just give the basics:

Total sets: 32066
Symmetric sets: 252 (pretty obvious)
Complementable sets: 94 (not so obvious)
Reverse complementable sets: 1024 (not so obvious)

Now if one adds all four and divide by 4, you obtain 8359, this would
be sets reduced for all but "strange" Z-relations.

(32066+252)/2 = 16159
(94+1024)/2 = 559

There are 20 sets (Formula on Sloanes') which are both symmetrical
and complementable.

I believe I have solved this. In the table above the first column
lists the plurality, then we have total Z-relations in that class,
and third, we have that number divided by plurality of class. The
last column is just what's left, not too meaningful.

So the answer is that there is 7944 interval vectors.

Now my total with Polya math is 8359 sets. So 415 are "strange"
Z-relations. Here's where they fall:

116+116+20 = 252
7684+7684+539=15907
_____________
7800,7800,559 16159

We can see that Z-related doubles are 7800, these are sets whose
complements are Z-related. There are 559 whose complements are mirror-
image, or perhaps forward image. (Numbers come out the same, because
mirror is cut)

These is kind of neat: There are 145 strange Z-relations, which are
not based on a complementary pair

So taking Doubles (6985), and 2+2 from Quads (780), and 3+3 from
Sext's (30), but just 1+1 from Triples (5), these add to 7800. so
this is

6985+6985
780+780
30+30
5+5
_________
7800+7800

So, the last part of Triples, (5) must go to the right side, in the
559 group, 5 of them. This leaves 554 which are pure.

Also shows that all even Z-related sets make up the 7800 columns.

So the 415 reduction for strange Z-relations are the orphaned triple
piece (5) and collapsing further for the quads (390 piece) and
sextets (10+10=20 collapse) = 415 QED

Not a proof, just a derivation:) But the pieces snap in place like a
crisp, jigsaw puzzle.

PGH

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<phjelmstad@...> wrote:
>
> All,
>
> Thanks for indulging me in this a little more. This is largely for
my
> own reference and those few who might still be interested in this
> subject.
>
> Let's just focus on 11-sets in 22tET:
>
> Subset C(22,11) 16159 Types
>
> Singles 554 554 0
>
> Doubles 13970 6985 6985
>
> Triples 15 5 10
>
> Quads 1560 390 1170
>
> Quints 0 0 0
>
> Sext's 60 10 50
> Totals 16159 7944 8215
>
> These indicate Z-relations. There are 32066 11-sets in 22-tET,
> reducing for transposition. There is only one (whole tone set)
> that doesn't divide by 22, it divides by 2. Now reducing for mirror-
> image, we get 32066+252/2=16159. That is the "Types" listed above.
> Now, Singles are all the sets with no Z-relations. Doubles,
Triples,
> Quads and Sext's are when that number has the same interval vector,
> but the sets are of different Tn/TnI types.
>
> Now, tonight, I calculated this based on necklaces formulas and
some
> math on periodic sequences. I'll just give the basics:
>
> Total sets: 32066
> Symmetric sets: 252 (pretty obvious)
> Complementable sets: 94 (not so obvious)
> Reverse complementable sets: 1024 (not so obvious)
>
> Now if one adds all four and divide by 4, you obtain 8359, this
would
> be sets reduced for all but "strange" Z-relations.
>
> (32066+252)/2 = 16159
> (94+1024)/2 = 559
>
> There are 20 sets (Formula on Sloanes') which are both symmetrical
> and complementable.
>
> I believe I have solved this. In the table above the first column
> lists the plurality, then we have total Z-relations in that class,
> and third, we have that number divided by plurality of class. The
> last column is just what's left, not too meaningful.
>
> So the answer is that there is 7944 interval vectors.
>
> Now my total with Polya math is 8359 sets. So 415 are "strange"
> Z-relations. Here's where they fall:
>
> 116+116+20 = 252
> 7684+7684+539=15907
> _____________
> 7800,7800,559 16159
>
> We can see that Z-related doubles are 7800, these are sets whose
> complements are Z-related. There are 559 whose complements are
mirror-
> image, or perhaps forward image. (Numbers come out the same,
because
> mirror is cut)
>
> These is kind of neat: There are 145 strange Z-relations, which are
> not based on a complementary pair
>
> So taking Doubles (6985), and 2+2 from Quads (780), and 3+3 from
> Sext's (30), but just 1+1 from Triples (5), these add to 7800. so
> this is
>
> 6985+6985
> 780+780
> 30+30
> 5+5
> _________
> 7800+7800
>
> So, the last part of Triples, (5) must go to the right side, in the
> 559 group, 5 of them. This leaves 554 which are pure.
>
> Also shows that all even Z-related sets make up the 7800 columns.
>
> So the 415 reduction for strange Z-relations are the orphaned triple
> piece (5) and collapsing further for the quads (390 piece) and
> sextets (10+10=20 collapse) = 415 QED
>
> Not a proof, just a derivation:) But the pieces snap in place like
a
> crisp, jigsaw puzzle.
>
> PGH

Forgot this piece:

20+502+502=1024 (RevComp)
20+37+37=94 (Comp)

20 is common to both, being symmetrical. Obviously 502+37=539.
1098 is the whole RH column. 539 after reducing for mirror image.

PGH