Ambitonal Chord Properties

Let us define a chord Q as a set of positive odd integers with a GCD of 1, where each successive component in the set is greater than the former. Let us also define the fixed-inversion of the chord Q as H(Q) = LCM(Q)/Q (but rearranged such that the least-to-greatest order within the set is preserved).

This fixed-inversion has the following properties:
1a.) H(Q) = LCM(Q)/Q
1b.) H(H(Q)) = Q (it is a dual function)
2a.) LCM(Q) = LCM(H(Q))

Now let us consider an ambitonal triad T = A:B:C. By definition, the largest odd integer in set T is equal to the largest odd integer in H(T). In other words, max(T) = max(H(T)). The largest integer in the fixed-inversion can also be understood to be the fixed-inverse of the smallest integer within the original set, by property 1a. Therefore max(H(T)) = LCM(T)/min(T). By property 2a, we can then derive the following equalities: max(T) = max(H(T)), and min(T) = min(H(T)).

Using the above equalities, we can derive another property:
3a.) LCM(Q) = min(Q)*max(Q), if and only if Q is an ambitonal chord.

Because we know that the maximum and minimum components must remain fixed from the original set to the fixed-inversion, we can then say that the triad T = [A, B, C] has a fixed-inverse H(T) = [A, D C]. From property 2a we can then say LCM(A, B, C) = LCM(A, D, C). Therefore LCM(LCM(A,C), B) = LCM(LCM(A,C), D). Using property 3a we can reduce this to LCM(T, B) = LCM(T, D). This only holds true if B=D, or if B and D are both divisible by T.

Therefore if LCM(A*C)/B is an integer, the triad A:B:C (and all of its octave inversions/transpositions) is an ambitonal triad.

-Ryan

--- In tuning-math@yahoogroups.com, "Ryan Avella" <domeofatonement@...> wrote:
>
> Let us define a chord Q as a set of positive odd integers with a GCD of 1, where each successive component in the set is greater than the former. Let us also define the fixed-inversion of the chord Q as H(Q) = LCM(Q)/Q (but rearranged such that the least-to-greatest order within the set is preserved).

To be terribly picky, sets don't have an order. But thanks for putting this together.

--- In tuning-math@yahoogroups.com, "Ryan Avella" <domeofatonement@...> wrote:
>
> Let us define a chord Q as a set of positive odd integers with a GCD of 1, where each successive component in the set is greater than the former. Let us also define the fixed-inversion of the chord Q as H(Q) = LCM(Q)/Q (but rearranged such that the least-to-greatest order within the set is preserved).
>
> This fixed-inversion has the following properties:
> 1a.) H(Q) = LCM(Q)/Q
> 1b.) H(H(Q)) = Q (it is a dual function)
> 2a.) LCM(Q) = LCM(H(Q))
>
> Now let us consider an ambitonal triad T = A:B:C. By definition, the largest odd integer in set T is equal to the largest odd integer in H(T). In other words, max(T) = max(H(T)). The largest integer in the fixed-inversion can also be understood to be the fixed-inverse of the smallest integer within the original set, by property 1a. Therefore max(H(T)) = LCM(T)/min(T). By property 2a, we can then derive the following equalities: max(T) = max(H(T)), and min(T) = min(H(T)).
>
> Using the above equalities, we can derive another property:
> 3a.) LCM(Q) = min(Q)*max(Q), if and only if Q is an ambitonal chord.
>
> Because we know that the maximum and minimum components must remain fixed from the original set to the fixed-inversion, we can then say that the triad T = [A, B, C] has a fixed-inverse H(T) = [A, D C]. From property 2a we can then say LCM(A, B, C) = LCM(A, D, C). Therefore LCM(LCM(A,C), B) = LCM(LCM(A,C), D). Using property 3a we can reduce this to LCM(T, B) = LCM(T, D). This only holds true if B=D, or if B and D are both divisible by T.
>
>
> Therefore if LCM(A*C)/B is an integer, the triad A:B:C (and all of its octave inversions/transpositions) is an ambitonal triad.

This is good. It's like the Fundamental Theorem of Ambitonal Chords.

You can use this to list all the ambitonal chords (not only triads) with a given LCM. In fact, it only depends on the pattern of prime factorization of the LCM, not the specific prime factors.

So we have:

p^2 (9,25,49,81...)
1:p:p^2

pq (15,21,33,35,39,51,55,57...)
1:p:pq <-> 1:q:pq
1:p:q:pq

p^3 (27,125...)
1:p:p^3 <-> 1:p^2:p^3
1:p:p^2:p^3

p^2q (45,63,75,99...)
(Not assuming anything about the ordering of p, q, p^2, and pq)
1:p:p^2q <-> 1:pq:p^2q
1:q:p^2q <-> 1:p^2:p^2q
1:p:q:p^2q <-> 1:p^2:pq:p^2q
1:p:p^2:p^2q <-> 1:q:pq:p^2q
1:q:p^2:p^2q
1:p:q:p^2:p^2q <-> 1:q:p^2:pq:p^2q
1:p:pq:p^2q
1:p:p^2:pq:p^2q <-> 1:p:q:pq:p^2q
1:p:q:p^2:p^2q:pq
p:q:p^2:pq (this one is an ASS)

pqr (105,165,195,231...)
(Assuming WLOG p<q<r)
A bunch of subsets of 1:p:q:r:pq:pr:qr:pqr containing both 1 and pqr
A bunch of subsets of p:q:r:pq:pr:qr containing both p and qr
q:r:pr <-> q:pq:pr
q:r:pq:pr

So you see that as you go to higher and higher odd LCMs, more interesting patterns keep appearing. Unfortunately, the numbers get way too high to sound good to human ears before most of that happens. The most interesting ambitonal JI chords mere humans can actually appreciate are the well-known ASSes:

3:5:9:15
3:7:9:21
3:9:11:33
3:9:13:39
(and possibly 3:5:15:25)

and after that maybe things like 5:7:15:21, but this is already starting to get complex and dissonant.

Keenan

> > Therefore if LCM(A*C)/B is an integer, the triad A:B:C (and all of its octave inversions/transpositions) is an ambitonal triad.
>
> This is good. It's like the Fundamental Theorem of Ambitonal Chords.
>
> You can use this to list all the ambitonal chords (not only triads) with a given LCM. In fact, it only depends on the pattern of prime factorization of the LCM, not the specific prime factors.

Mike has actually made me aware that we can replace LCM(A*C) with A*C. This is because GCD(A,C) = A*C/LCM(A,C).

As long as A*C is divisible by B, A:B:C will be ambitonal. So therefore, if we have a dyad A:C, we can find notes to saturate the chord (without raising the odd limit) that satisfy the inequality A < S < C, where S is the set of integers that evenly divide A*C. This works even if A:C is not in lowest terms. And as you already mentioned, all of these ambitonal triads are subsets of much larger Anomalous Saturated Suspensions.

> So you see that as you go to higher and higher odd LCMs, more interesting patterns keep appearing. Unfortunately, the numbers get way too high to sound good to human ears before most of that happens. The most interesting ambitonal JI chords mere humans can actually appreciate are the well-known ASSes:
>
> 3:5:9:15
> 3:7:9:21
> 3:9:11:33
> 3:9:13:39
> (and possibly 3:5:15:25)
>
> and after that maybe things like 5:7:15:21, but this is already starting to get complex and dissonant.
>
> Keenan
>

Yes, that is a bit unfortunate. However, it is much easier to memorize a simple rule of ambitonal chord construction than to memorize a set of usable chords or to randomly test chords. Now whenever i am away from my computer, I can find ambitonal chords on the back of a napkin. This is really useful to me because I wasn't aware of how to do this before.

Ryan

Yeah, this is really handy. Great work with all of this man. Now, to
find those elusive ztonal chords...

-Mike

On Mon, Feb 6, 2012 at 4:04 PM, Ryan Avella <domeofatonement@yahoo.com> wrote:
>
> > > Therefore if LCM(A*C)/B is an integer, the triad A:B:C (and all of its octave inversions/transpositions) is an ambitonal triad.
> >
> > This is good. It's like the Fundamental Theorem of Ambitonal Chords.
> >
> > You can use this to list all the ambitonal chords (not only triads) with a given LCM. In fact, it only depends on the pattern of prime factorization of the LCM, not the specific prime factors.
>
> Mike has actually made me aware that we can replace LCM(A*C) with A*C. This is because GCD(A,C) = A*C/LCM(A,C).
>
> As long as A*C is divisible by B, A:B:C will be ambitonal. So therefore, if we have a dyad A:C, we can find notes to saturate the chord (without raising the odd limit) that satisfy the inequality A < S < C, where S is the set of integers that evenly divide A*C. This works even if A:C is not in lowest terms. And as you already mentioned, all of these ambitonal triads are subsets of much larger Anomalous Saturated Suspensions.
>
>
> > So you see that as you go to higher and higher odd LCMs, more interesting patterns keep appearing. Unfortunately, the numbers get way too high to sound good to human ears before most of that happens. The most interesting ambitonal JI chords mere humans can actually appreciate are the well-known ASSes:
> >
> > 3:5:9:15
> > 3:7:9:21
> > 3:9:11:33
> > 3:9:13:39
> > (and possibly 3:5:15:25)
> >
> > and after that maybe things like 5:7:15:21, but this is already starting to get complex and dissonant.
> >
> > Keenan
> >
>
> Yes, that is a bit unfortunate. However, it is much easier to memorize a simple rule of ambitonal chord construction than to memorize a set of usable chords or to randomly test chords. Now whenever i am away from my computer, I can find ambitonal chords on the back of a napkin. This is really useful to me because I wasn't aware of how to do this before.
>
> Ryan

Sorry, it appears I made a few mistakes. Gene caught an obvious one, and I just noticed some which I didn't see before. The version with corrections is below. I am going to avoid indicating and annotating corrections simply because that might make it more confusing than it was to start out with.

Note: I replaced the function H(Q) with Q*, and H(T) with T*.

MODIFIED PROOF:

> Let us define a chord Q as a set of positive odd integers with a GCD of 1. Let us also define the fixed-inversion of the chord Q as Q* = LCM(Q)/Q.
>
> This fixed-inversion has the following properties:
> 1a.) Q* = LCM(Q)/Q
> 1b.) FHQ* = Q (it is a dual function)
> 2a.) LCM(Q) = LCM(Q*)
>
> Now let us consider an ambitonal triad T = [A, B, C]. By definition, the largest odd integer in set T is equal to the largest odd integer in T*. In other words, max(T) = max(T*). The largest integer in the fixed-inversion can also be understood to be the fixed-inverse of the smallest integer within the original set, by property 1a. Therefore max(T*) = LCM(T)/min(T). By property 2a, we can then derive the following equalities: max(T) = max(T*), and min(T) = min(T*).
>
> Using the above equalities, we can derive another property:
> 3a.) LCM(Q) = min(Q)*max(Q), if and only if Q is an ambitonal chord.
>
> Because we know that the maximum and minimum components must remain fixed from the original set to the fixed-inversion, we can then say that the triad T = [A, B, C] has a fixed-inverse T* = [A, D C]. From property 2a we can then say LCM(A, B, C) = LCM(A, D, C).
>
> Now, this next part is tricky, but we can rewrite the above as LCM(A, B, B, C) = LCM(A, D, D, C). Therefore LCM(LCM(A,B,C), B) = LCM(LCM(A,D,C), D). Using property 3a we can reduce this to LCM(AC, B) = LCM(AC, D). This only holds true if B=D, or if B and D can both evenly divide AC.
>
>
> Therefore if (AC)/B is an integer, the triad A:B:C (and all of its octave inversions/transpositions) is an ambitonal triad.
>
>
> Ryan
>

--- In tuning-math@yahoogroups.com, "Ryan Avella" <domeofatonement@...> wrote:
>
> Sorry, it appears I made a few mistakes. Gene caught an obvious one, and I just noticed some which I didn't see before. The version with corrections is below. I am going to avoid indicating and annotating corrections simply because that might make it more confusing than it was to start out with.

Okay, I finally got a simplified proof worth mentioning.

Lets say we have a triad A:B:C composed of odd integers in lowest terms, which
we will write as the set S = {A, B, C}. The inverse of this chord we will call
S*, where S* = LCM(A, B, C)/S.

Now, if S is an ambitonal triad, this means that the largest odd number (and
consequentially the smallest odd number) of both sets will be equal, by
definition. Therefore, S\S* is either a null set, or it is equal to the set {B}.

The former equality is trivially true for symmetrical chords.

The latter is true whenever LCM(A, B, C)/C = A, using the definition of the inverse. Then implementing some properties of least common multiples, we can rewrite this as GCD(AB, AC, BC) = B. This is true as long as AC is divisible by B, and AB != C.

Therefore, the triad A:B:C is ambitonal as long as AC is divisible by B, and as
long as AB != C.

Ryan

--- In tuning-math@yahoogroups.com, "Ryan Avella" <domeofatonement@...> wrote:

> Ryan

Nice!