Tribonacci

Hey Gene, any thoughts on this year-old tuning list post? I think Dan
Stearns thinks he understands the phenomena, but I didn't follow his
answers too closely . . .

--- In tuning@y..., "Paul Erlich" <PERLICH@A...> wrote:
For MOS scales we've seen the noble generators allow for a uniquely
regular expansion of resources, in that each MOS will have L = s*phi
and the steps in the old MOS change into the steps of the new MOS as
follows:

L(old) -> L(new) + s(new)
s(old) -> L(new)

so that the sizes (number of notes) of any three consecutive scales
in the "evolution" obey a generalized Fibonacci recursion:

siz(n-1) + siz(n) = siz(n+1);

and the ratios of adjacent sizes approaches phi. The most famous
example is the Kornerup system, which has the same scale sizes as
Yasser's proposed evolution

2, 5, 7, 12, 19, 31, 50, 81 . . .

but maintains the same (noble) generator throughout.

What if we allow three step sizes, and posit the following evolution
rules:

siz(n-2) + siz(n-1) + siz(n) = siz(n+1)? Then the ratio of adjacent
sizes approaches the solution of

x^3 - x^2 - x - 1 = 0

= 1.839286755...

One example is the so-called "Tribonacci Sequence":

1, 1, 2, 4, 7, 13, 24, 44, 81,...

while another may be more musically relevant and may be familiar to
followers of Kraehenbuehl & Schmidt:

2, 2, 3, 7, 12, 22, 41, . . .

although the next term would be 75 instead of K&S's 78 -- since K&S
started with 3-limit JI and forced "inflections" reflecting
successively higher prime limits to "deform" the scale at each stage.

Like the Fibonacci sequence, the Tribonacci sequence and its
relatives can be derived from a continued fraction representation,
but this time using generalized, "third-order" continued fractions
(see http://www.mathsoft.com/asolve/constant/pythag/dgm.html).

Can anyone demonstrate a rule for how the steps of one scale
transform into the steps of the next scale, and find the set of
scales which are to Kraehenbuehl & Schmidt as Kornerup's set of
scales is to Yasser? Bonus question: What is the object here
analogous to the generator in MOS scales, and in what way is this
generator built upon itself to create these "hyper-MOS" scales? Is
there more than one solution?
--- End forwarded message ---

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> Hey Gene, any thoughts on this year-old tuning list post? I think
Dan
> Stearns thinks he understands the phenomena, but I didn't follow
his
> answers too closely . . .

Despite its name, the linear recurrence I gave is for musical
purposes more like the Fibonacci recurrence than the Tribonacci
recurrence is. From the point of view of generators, we would be
taking a generalized mediant, not an ordinary one, using
M(p1/q1,p2/q2,p3/q3)=(p1+p2+p3)/(q1+q2+q3) to get the successive
terms of a Tribonacci generator. We can make the Tribonnaci sequence
Dan gave into a Tribonacci-mediant sequence by

1/2,1/2,2/3,4/7,7/12,13/22,24/41,44/75,81/138...

Note that 81/138=27/46; we don't reduce fractions for the mediants,
or in other words, the numerators and denominators are linear
recurrences. The result is a slightly sharp (about 2 cents)
Tribonnaci fifth. I don't see anything hyper-MOS, but I never was
clear what that meant. One *could* reduce fractions and see where
that leads, but I don't see why it leads anywhere beyond a paper for
the Fibonnaci Quarterly.

I wouldn't mess with ternary continued fractions if I were you. :)

--- In tuning-math@y..., genewardsmith@j... wrote:
> --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:
>
> > Hey Gene, any thoughts on this year-old tuning list post? I think
> Dan
> > Stearns thinks he understands the phenomena, but I didn't follow
> his
> > answers too closely . . .
>
> Despite its name, the linear recurrence I gave is for musical
> purposes more like the Fibonacci recurrence than the Tribonacci
> recurrence is.

Yes, I see the difference between a two-term recurrence and a three-
term recurrence.

> From the point of view of generators, we would be
> taking a generalized mediant, not an ordinary one, using
> M(p1/q1,p2/q2,p3/q3)=(p1+p2+p3)/(q1+q2+q3) to get the successive
> terms of a Tribonacci generator. We can make the Tribonnaci
sequence
> Dan gave

Hmm . . . where did Dan give this? I thought it was original to me in
the post I just forwarded.

> into a Tribonacci-mediant sequence by
>
> 1/2,1/2,2/3,4/7,7/12,13/22,24/41,44/75,81/138...
>
> Note that 81/138=27/46; we don't reduce fractions for the mediants,

Of course.

> or in other words, the numerators and denominators are linear
> recurrences. The result is a slightly sharp (about 2 cents)
> Tribonnaci fifth. I don't see anything hyper-MOS, but I never was
> clear what that meant.

Well, it's not the same thing as "hyper-MOS" has meant on this list,
or at least not in an obvious way. But I guess the thinking behind
the question should be clear . . . ?

Would Dan like to chime in here?

--- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> What I was interested in was this: If you take the first two terms
in
> a Fibonacci series to be the number of two stepsizes in given n-tet
> where the fourth term is the n-tet and the third term is the number
of
> notes in the scale or subset of the n-tet, then the third term is
> always the generator that renders the scale (and this of course is
> consistent wherever you might be in the series).

In the notation I have suggested I think you are saying that, for
instance, 5,7,12,19,31,50... leads to the temperaments/scales
7;7+5, 12;12+7, 19;19+12, 31;31+19 etc. This is Yasser's sequence for
the music of the future, I understand.

Some rotation of this
> will also always agree with an L-out-of-M where L is the third term
> and M is the fourth term.

Rotation?

> So with all that in mind, my question basically was, "how then does
a
> three term Tribonacci analogue shake down?" I found and posted some
of
> my own answers, but I'd definitely be interested to hear others' as
> well

In the same notation, I presume the sequence 2,2,3,7,12,22 ... leads
to 7;7+5, 12;12+10, 22;22+19, 41;41+34 ... Here the white keys are
one term of the denominator sequence, and the black keys are the sum
of the two previous terms. This hops all over the place as far as
generators go--4/22, 13/41, 11/75 etc.

--- In tuning-math@y..., genewardsmith@j... wrote:

> In the same notation, I presume the sequence 2,2,3,7,12,22 ...
leads
> to 7;7+5, 12;12+10, 22;22+19, 41;41+34 ... Here the white keys are
> one term of the denominator sequence, and the black keys are the
sum
> of the two previous terms. This hops all over the place as far as
> generators go--4/22, 13/41, 11/75 etc.

Well I think the idea was that, while the Fibonacci case corresponds
to a series of MOS scales where the generator is the limit of the
sequence, the Tribonacci case might instead correspond in some way to
some sort of "hyper-MOS" scale, generated not 1-dimensionally by a
single interval but rather (in some sense) perhaps 2-dimensionally by
a triad or something . . . ?

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:
> --- In tuning-math@y..., genewardsmith@j... wrote:

> > In the same notation, I presume the sequence 2,2,3,7,12,22 ...
> leads
> > to 7;7+5, 12;12+10, 22;22+19, 41;41+34 ... Here the white keys
are
> > one term of the denominator sequence, and the black keys are the
> sum
> > of the two previous terms. This hops all over the place as far as
> > generators go--4/22, 13/41, 11/75 etc.

> Well I think the idea was that, while the Fibonacci case
corresponds
> to a series of MOS scales where the generator is the limit of the
> sequence, the Tribonacci case might instead correspond in some way
to
> some sort of "hyper-MOS" scale, generated not 1-dimensionally by a
> single interval but rather (in some sense) perhaps 2-dimensionally
by
> a triad or something . . . ?

The characteristic polynomial, x^3-x^2-x-1, defines a Pisot number,
meaning a real algebraic integer greater than one all of whose
conjugates are less than one in absolute value (another example would
be the golden ratio.) In consequence, the ratios of successive terms
converge, and the generator should settle down to something definite.

--- In tuning-math@y..., genewardsmith@j... wrote:
> --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:
> > --- In tuning-math@y..., genewardsmith@j... wrote:
>
> > > In the same notation, I presume the sequence 2,2,3,7,12,22 ...
> > leads
> > > to 7;7+5, 12;12+10, 22;22+19, 41;41+34 ... Here the white keys
> are
> > > one term of the denominator sequence, and the black keys are
the
> > sum
> > > of the two previous terms. This hops all over the place as far
as
> > > generators go--4/22, 13/41, 11/75 etc.
>
> > Well I think the idea was that, while the Fibonacci case
> corresponds
> > to a series of MOS scales where the generator is the limit of the
> > sequence, the Tribonacci case might instead correspond in some
way
> to
> > some sort of "hyper-MOS" scale, generated not 1-dimensionally by
a
> > single interval but rather (in some sense) perhaps 2-
dimensionally
> by
> > a triad or something . . . ?
>
> The characteristic polynomial, x^3-x^2-x-1, defines a Pisot number,
> meaning a real algebraic integer greater than one all of whose
> conjugates are less than one in absolute value (another example
would
> be the golden ratio.) In consequence, the ratios of successive
terms
> converge, and the generator should settle down to something
definite.

The ratio of successive terms converges to the "Tribonacci constant",
yes, but I don't think you're catching my drift. Don't the scales
have three, rather than two, step sizes, therefore not being MOS
scales, or scales with a single generator at all? Dan, can you help
here?

--- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> What does seem to
> work is taking the M-out-of-N idea as an L-out-of-M-out-of-N idea.

It sounds like what you are describing is what I was calling
a "muddle" in some recent postings.

> So take the Tribonacci syntonic diatonic sequence again:
>
> 2, 2, 3, 7, 12, 22, 41, ...
>
> 12-out-of-22-out-of-41 gives 0 3 7 9 12 16 18 22.

You could get for instance the sequence of muddles 7;12+10;22+19,
12;22+19;41+34; 22;41+34;75+63..., though what you give looks like
it is out of 22, not 41.

If you let the
> Tribonacci constant be X and the three terms (or stepsizes) be A, B
> and C, then converting this so that C/A=X and (C+A)/B=X gives:
>
> 0 175 383 496 671 879 992 1200
> 0 208 321 496 704 817 1025 1200
> 0 113 288 496 609 817 992 1200
> 0 175 383 496 704 879 1087 1200
> 0 208 321 529 704 912 1025 1200
> 0 113 321 496 704 817 992 1200
> 0 208 383 591 704 879 1087 1200

The way I think of it a muddle is not just one scale, but the whole
set.

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> The ratio of successive terms converges to the "Tribonacci
constant",
> yes, but I don't think you're catching my drift. Don't the scales
> have three, rather than two, step sizes, therefore not being MOS
> scales, or scales with a single generator at all? Dan, can you help
> here?

What I gave were MOS by definition, but muddles are quite another
matter.

--- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> I'm not quite getting exactly what a "muddle" is yet, but I'll take
a
> look in the archives.

It was over on the tuning list, I'm afraid. By p;q+r with q and r
relatively prime I mean a scale of p notes with a generator
(a+b)/(q+r), where the numerators are defined by taking the
pentultimate semiconvergent for q/r. If q and r are not relatively
prime, then for example 10;12+10 would refer to the various paultone
scales, with 5+5;12+10 and 6+4;12+10 being his two main kinds.

By a "muddle" I mean the set of related scales defined by a
temperament of a temperament--for instance 7;19+12;41+31 is the set
of scales one gets by treating Canasta as if it were 31-equal and
using it to define diatonic scales. One gets things very close to
major, minor and Indian diatonic scales out of it, all related by key
relationships.

--- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> So take the Tribonacci syntonic diatonic sequence again:
>
> 2, 2, 3, 7, 12, 22, 41, ...
>
> 12-out-of-22-out-of-41 gives 0 3 7 9 12 16 18 22. If you let the
> Tribonacci constant be X and the three terms (or stepsizes) be A, B
> and C, then converting this so that C/A=X and (C+A)/B=X gives:
>
> 0 175 383 496 671 879 992 1200
> 0 208 321 496 704 817 1025 1200
> 0 113 288 496 609 817 992 1200
> 0 175 383 496 704 879 1087 1200
> 0 208 321 529 704 912 1025 1200
> 0 113 321 496 704 817 992 1200
> 0 208 383 591 704 879 1087 1200
>
> Unlike the 2D (triadic) idea, I believe this works for any arbitrary
> A,B,C Tribonacci series. However, unlike the two term series scales,
> these three term scales do not guarantee trivalence (if you allow
that
> Myhill is equals "bivalence").

Dan, this is great, thanks.

You still need to show me that the Tribonacci constant is the right
constant. What are some other scales besides the 12-tone one? If this
is truly like the Fibonacci model wrt the Golden Ratio, you should
see all the notes in the 12-tone scale preserved without alteration
in the 22-tone scale, etc.

--- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> 7-out-of-12-out-of-22:
>
> 0 175 383 496 671 879 992 1200
> 0 208 321 496 704 817 1025 1200
> 0 113 288 496 609 817 992 1200
> 0 175 383 496 704 879 1087 1200
> 0 208 321 529 704 912 1025 1200
> 0 113 321 496 704 817 992 1200
> 0 208 383 591 704 879 1087 1200

These don't seem to be 22-et intervals, though some are close.

> So here's the Tribonacci
>
> 2, 3, 7, 12, 22, 41, ...
>
> 12-out-of-22-out-of-41:

When you say "12 out of 22" do you mean 12;17+5 (circle of 12 fifths
of a 22 et) or something else (such as some version of 12;12+10?)

this is the
> junction in this story when I always expect a Gene, or a Robert
> Walker, or anybody else with the math know-how to ride on in and
save
> the day!

I have to figure out what you are doing first.

--- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> <<When you say "12 out of 22" do you mean 12;17+5 (circle of 12
fifths
> of a 22 et) or something else (such as some version of 12;12+10?)>>

> Hmm, I mean N/M multiplied by zero through M and rounded down to the
> nearest integer, and from that I take the nearest L-tet.

When you do that for 12/22, you get
0,0,1,1,2,2,3,3,4,4,5,6,6,7,7,8,8,9,9,10,10,11,12, which I doubt is
what you mean.

--- In tuning-math@y..., genewardsmith@j... wrote:
> --- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:
>
> > 7-out-of-12-out-of-22:
> >
> > 0 175 383 496 671 879 992 1200
> > 0 208 321 496 704 817 1025 1200
> > 0 113 288 496 609 817 992 1200
> > 0 175 383 496 704 879 1087 1200
> > 0 208 321 529 704 912 1025 1200
> > 0 113 321 496 704 817 992 1200
> > 0 208 383 591 704 879 1087 1200
>
> These don't seem to be 22-et intervals, though some are close.
>
> > So here's the Tribonacci
> >
> > 2, 3, 7, 12, 22, 41, ...
> >
> > 12-out-of-22-out-of-41:
>
> When you say "12 out of 22" do you mean 12;17+5 (circle of 12
fifths
> of a 22 et) or something else (such as some version of 12;12+10?)
>
> this is the
> > junction in this story when I always expect a Gene, or a Robert
> > Walker, or anybody else with the math know-how to ride on in and
> save
> > the day!
>
> I have to figure out what you are doing first.

Look at the original post again. It looks like there should be a
simple generating rule that converts the steps of one scale in the
series to the steps of the next scale in the series, kind of like
the "Fibonacci" MOS generating rule illustrated in the original
message. One would think that there would be three step sizes in
these scales.

--- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:
> Hi Paul,
>
> <<You still need to show me that the Tribonacci constant is the
right
> constant. What are some other scales besides the 12-tone one? If
this
> is truly like the Fibonacci model wrt the Golden Ratio, you should
see
> all the notes in the 12-tone scale preserved without alteration in
the
> 22-tone scale, etc.>>
>
> Right, the other one was actually the
>
> 2, 2, 3, 7, 12, 22, ...
>
> 7-out-of-12-out-of-22:
>
> 0 175 383 496 671 879 992 1200
> 0 208 321 496 704 817 1025 1200
> 0 113 288 496 609 817 992 1200
> 0 175 383 496 704 879 1087 1200
> 0 208 321 529 704 912 1025 1200
> 0 113 321 496 704 817 992 1200
> 0 208 383 591 704 879 1087 1200
>
> So here's the Tribonacci
>
> 2, 3, 7, 12, 22, 41, ...
>
> 12-out-of-22-out-of-41:
>
> 0 95 208 321 416 529 591 704 817 912 1025 1087 1200
> 0 113 226 321 434 496 609 722 817 930 992 1105 1200
> 0 113 208 321 383 496 609 704 817 879 992 1087 1200
> 0 95 208 270 383 496 591 704 766 879 974 1087 1200
> 0 113 175 288 401 496 609 671 784 879 992 1105 1200
> 0 62 175 288 383 496 557 671 766 879 992 1087 1200
> 0 113 226 321 434 496 609 704 817 930 1025 1138 1200
> 0 113 208 321 383 496 591 704 817 912 1025 1087 1200
> 0 95 208 270 383 478 591 704 799 912 974 1087 1200
> 0 113 175 288 383 496 609 704 817 879 992 1105 1200
> 0 62 175 270 383 496 591 704 766 879 992 1087 1200
> 0 113 208 321 434 529 643 704 817 930 1025 1138 1200

Aha!! So the Tribonacci constant really works, Dan! I don't think I
ever fully realized this before -- great work, Dan! Gene, take note!
And don't let Dan's "L-out-of-M-out-of-N" notation confuse you into
thinking that N-tET has anything to do with this -- think instead of
the analogy of MOS meantone scales with exactly the Kornerup golden
generator, and how each scale in the series can be generated from the
previous scale by the rule

s(old) -> L(new)
L(old) -> L(new) + s(new)

So the question is, what's the analogous rule for the Tribonacci
case, and is there anything that serves as analogous to the generator?

--- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> The method I use appears to do this. I looked a ton of different
> Tribonacci A,B,C,... scales,

How are these defined, and where did you get the idea to use the
Tribonacci constant??

> and while I remember a few cases where I
> had to slightly alter the method, these were rare and I think they
> were tied to some slight bug in the way I did this. After a while I
> just got sick of working on it--I do this "by hand", so mistakes
> brought on by tedium are inevitable too--and I decided that someone
> else with a math background would eventually have to come along and
> tidy things up.

Gene, this is your chance!

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> So the question is, what's the analogous rule for the Tribonacci
> case, and is there anything that serves as analogous to the
generator?

I don't see that this has anything to do with MOS. If I take Dan at
his word about the 7-12-22 business, it seems to me you just get
1/t = t^2-t-1 (where t is the Tribonacci constant) as a generator. On
the other hand, if you do as I suggested, and look at
1/2,1/2,1/3,3/7,5/12,9/22,17/41... you get that the denominators tend
to ((8*t^2-7*t+9)/22)*t^n and the numerators to ((3*t^2-4*t+2)/11)*t^n
with the result that the generator approaches a slightly flat (two
cents worth) Tribonnaci fourth of (t^2-4*t+11)*(1200/17) cents. If
you look at muddles, if both parts of the muddle grow with n then the
whole thing approaches a MOS. It looks like the thing to do is to
toss all this out the window and look at what Dan is doing on its own
terms.

--- In tuning-math@y..., genewardsmith@j... wrote:
> --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:
>
> > So the question is, what's the analogous rule for the Tribonacci
> > case, and is there anything that serves as analogous to the
> generator?
>
> I don't see that this has anything to do with MOS. If I take Dan at
> his word about the 7-12-22 business, it seems to me you just get
> 1/t = t^2-t-1 (where t is the Tribonacci constant) as a generator.

Huh? How do you see a single generator operating here? These are
three-step-size scales, while a single generator would produce two-
step-size scales.

> On
> the other hand, if you do as I suggested, and look at
> 1/2,1/2,1/3,3/7,5/12,9/22,17/41... you get that the denominators
tend
> to ((8*t^2-7*t+9)/22)*t^n and the numerators to ((3*t^2-4*t+2)/11)
*t^n
> with the result that the generator approaches a slightly flat (two
> cents worth) Tribonnaci fourth of (t^2-4*t+11)*(1200/17) cents.

Such a fourth cannot generate the scales Dan posted, nor can any
single interval.

> If
> you look at muddles, if both parts of the muddle grow with n then
the
> whole thing approaches a MOS.

Here we are most definitely not approaching an MOS.

> It looks like the thing to do is to
> toss all this out the window and look at what Dan is doing on its
own
> terms.

I believe I am asking the same questions Dan is, which are the same
as the ones in my original year-old tuning list post with which I
started this thread (though Dan has made some progress over that by
actually constructing the relevant scales!).

--- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:
> Gene,
>
> <<When you do that for 12/22,>>
>
> But I wrote N/M, so for a 7-out-of-12-out-of-22 that would be 22/12
or
> 0 1 3 5 7 9 11 12 14 16 18 20 22. Then the L-tet (7-tet here) inside
> of that would be 0 3 7 9 12 16 18 22.
>
> The idea is that this should work for any arbitrary A, B, C, ...
> Tribonacci series.

This is a scale out of the 7;7+5;12+10 muddle--why do you need to
bring in the Tribonacci constant at all?

--- In tuning-math@y..., genewardsmith@j... wrote:
> --- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:
> > Gene,
> >
> > <<When you do that for 12/22,>>
> >
> > But I wrote N/M, so for a 7-out-of-12-out-of-22 that would be
22/12
> or
> > 0 1 3 5 7 9 11 12 14 16 18 20 22. Then the L-tet (7-tet here)
inside
> > of that would be 0 3 7 9 12 16 18 22.
> >
> > The idea is that this should work for any arbitrary A, B, C, ...
> > Tribonacci series.
>
> This is a scale out of the 7;7+5;12+10 muddle--why do you need to
> bring in the Tribonacci constant at all?

Gene, take a deep breath. You're evidently missing something
important even though it's being repeated (this has happened before),
so you need to slow down and empty your mind of preconceptions, and
give your genius room to function.

Now take a look at the scales Dan posted (giving cents for all
rotations) which were _not_ subsets of ETs at all. Notice how one
scale is precisely a subset of the next scale, and each scale has
_three_ step sizes. If I'm understanding Dan correctly (and I believe
that I am), the relative sizes of the three step sizes are governed
by the Tribonacci ratio -- in every iteration. This is similar to how
the relative sizes of the two step sizes, in each of the proper MOSs
of the Kornerup golden generator, or any other generator which is a
noble (CF ends in all 1s) fraction of an octave, is exactly the
Golden ratio.

If one were to use a different constant from the Tribonacci ratio,
and again assuming I'm understanding Dan correctly, one would see
that the scale in one iteration would _not_ be precisely a subset of
the scale in the next iteration -- just as if, in the sequence of
meantone scales with 7, 12, 19, 31, 50, 81 . . . notes, if the ratio
of small step to large step were not the Golden Ratio in each scale,
then one scale would not be precisely a subset of the next scale in
the sequence.

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:
> --- In tuning-math@y..., genewardsmith@j... wrote:

> > I don't see that this has anything to do with MOS. If I take Dan
at
> > his word about the 7-12-22 business, it seems to me you just get
> > 1/t = t^2-t-1 (where t is the Tribonacci constant) as a generator.

> Huh? How do you see a single generator operating here? These are
> three-step-size scales, while a single generator would produce two-
> step-size scales.

Not for a muddle--we could have a 7;7+5;1/t muddle, which would have
three step sizes and which *would* be a Tribonacci limit, but it
doesn't seem to be what Dan is talking about, since the scales don't
correspond.

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> Gene, take a deep breath. You're evidently missing something
> important even though it's being repeated (this has happened before)
...

Sorry to be so frustrating, and thanks for the explanation. I may
need things repeated again some time. :)

--- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> where C/A=X and (A+C)/B=X
> (and X is the Tribonacci constant)

That's what I was looking for! Now, how did you come upon the idea to
do it this way? It certainly ends up working just wonderfully!