How to Tribonacci

The basis I got for the Tribonacci process seemed a little
mysterious, but the mystery is solved. The transformation matrix is
actually the transpose of the Tribonacci one, meaning the usual
operation is on vals, and I was looking at a transposed one on
[1,t^2-t,t].

The transformation matrix one gets from multiplication by t sends
C->a, B->a+c, C->b+c, and is

[0 0 1]
[1 0 1]
[0 1 1]

If we take the diatonic scale Dan started with, we have intervals of
9/8, 10/9 and 16/15, with a matrix

[ 4 -1 -1]
[ 1 -2 1]
[-3 2 0]

The inverse of this is

[2 2 3]
[3 3 5]
[4 5 7]

If we set our "a" value so that a*(2+2*(t^2-t)+3*t) = 1, and then use
the next two rows to calculate the corresponding approximations of 3
and 5, we get

log_2(3) ~ a*(3 + 3*(t^2-t)+5*t) 2.1 cents sharp
log_2(5) ~ a*(4 + 5*(t^2-t)+7*t) -3.5 cents flat

If we now transform the above val matrix by the Tribonacci
transformation, we get

[2 2 3] [0 0 1] [2 3 7]
[3 3 5] [1 0 1] = [3 5 11]
[4 5 7] [0 1 1] [5 7 16]

If we do the same calculation with this new val matrix we get the
same approximation to 3 and 5 as before; instead of a Golden Meantone
we are getting a Tribonacci 3 *and* 5. The sequence in question is
2 2 3 7 12 22 41 75 ... and the fifth is indeed the same fifth I
started out by yowling about, but we need the Tribonacci Third also
to define this thing.

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

> I'd like to be wrong though, so here's a couple for you to try:
>
> 2, 1, 4, ...
> 1, 4, 2, ...

I need to get corresponding Tribonacci sequences for the 3 and the 5,
as well as the 2, so I extend your first example to

2 1 4 7 12 23 ...
3 2 6 11 19 36 ...
4 3 9 16 28 53 ...

If t is the Tribonacci constant, this gives me an octave multiplier of
d = 2+(t^2-t)+4t = 2+3t+t^2, a 3 of (3 + 2(t^2-t)+6t)/d, 17 cents
flat, and a 5 of (4+3(t^2-t)+9t)/d, 14 cents flat.

Do you think you could do the other one, or is this not clear?

--- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:
> Gene,
>
> If you lattice out the results you gave in 2D they create a 2, 3, 2,
> ... scale. What I'm looking for are results that would lattice out
to
> either a 2, 1, 4, ... or a 1, 4, 2, ... scale.

If you start with a block with steps of size 27/25, 9/8, and 10/9,
which corresponds to the first three steps I gave, and approximate
using the approximate 3 and 5 I gave, you get 27/25:9/8:10/9
approximated by 1:(t^2-t):t. The 27/25 comes in at about 110 cents,
9/8 at about 170 cents, and 10/9 at about 202.5 cents (so they are
effectively reversed.) The JI block is

1-10/9-6/5-4/3-3/2-5/3-9/5-(2),

which has 2 27/25, 1 9/8, and 4 10/9. Since the 10/9 is largest, this
is actually a 2, 1, 4 scale, which is what you wanted.

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

> If you take a Tribonacci like 1, 1, 4, ..., it's something that
> shouldn't be based on 3s and 5s. Or rather there's nothing saying it
> should (and from the looks of it I think it would probably be 7s and
> 11s). Anyway, using your method, if I'm understanding correctly,
> deciding the primes (or 2D planes) is arbitrary. Is this correct or
am
> I mistaken?

You have three generators, including the 2, and so the most obvious
thing is to approximate 2,3 and 5. What would make more sense? I
don't see why you would want to do 2,7 and 11 instead, as you seem to
be suggesting.

> What would be ideal would be something akin to the 1D generator
> achieved by way of adjacent fractions in a Fibonacci series that is
> not tied into an arbitrary function--well calling that generator
some
> rational approximation is arbitrary, but that comes after the fact
and
> not before it.

We start off with certain scale steps, and nothing prevents us from
using these as generators instead--it's a change of basis, but no
more. You could, for instance, have approximate values for 9/8, 16/15
and 1125/1024 as your three generators, just as you could use 9/8 and
4/3 instead of 3/2 and 2 for the Golden case.

> Also, like the 1D Fibonacci case, I would think that successive
scales
> in a given series should create the same 2D "generators". Take the
> prototypical 2, 2, 3, ... example--what's the Fokker block for 2, 3,
> 7, ...? And how's that and the ones that follow related to the 3s
and
> 5s of the 2, 2, 3, ...?

This is determined by your transpose basis, which tells you how your
intervals break up into smaller ones.

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

> <<What would make more sense?>>

> Again, what would make more sense would be something akin to the 1D
> generator achieved by way of adjacent fractions in a Fibonacci
series
> that is not tied into an arbitrary function.

I'll take a shot at it, but meanwhile down below I show why I think
this is already very much like the Golden Meantone.

> <<I don't see why you would want to do 2,7 and 11 instead, as you
seem
> to be suggesting.>>

> You could see the 1, 1, 4, ... as a PB based on 2, 7, and 11 with
UVs
> of 353/343 and 121/112. Why? Say we think of this scale as having a
> two stepsize cardinality at 1, 5, ..., the 1D generator would have
> nothing (or nothing "most obvious") to do with a 3 or a 5. A 1D
chain
> of 7s would work if you've got to approximate it by a most likely
> rational.

If you like that better, I suppose you could do it that way.

> <<This is determined by your transpose basis, which tells you how
your
> intervals break up into smaller ones.>>
>
> Okay, what exactly does the near 3/2 and near 5/4 of the 2, 2,
3, ...
> have to do with the 2, 3, 7, ... (please use examples, I'm slow)?

The near 3/2 and near 5/4 has to do with more than just 2,3,7... The
recurrence is really one on a 3D vector, just as what you call "the
1D case" (which actually has two dimensions) involves a recurrence on
a 2D vector. If you wanted to work with some other vectors, they
could have the same 2,3,7... even so.

In other words, the Golden Meantone really is

7 12 19 31 50 ...
11 19 30 49 79 ...

We don't pick the closest approximation to log_2(3), even when a
better one becomes available, we stick with the one we get by taking
the ratio of the above.

We can write this as a sequence of mediants:

med(11/7, 19/12) = 30/19, med(19/12, 30/19) = 49/31 ...

The corresponding mediants (which might be generalized mediants) are
easily found for other recurrences:

med(3/2,3/2,5/3) = 11/7 med(4/2,5/2,7/3) = 16/7; then
med(3/2,5/3,11/7) = 19/12 and med(5/2,7/2,16/7) = 28/12

and so forth. This is your 223 Tribonacci example. Just as we find the
golden generators using [1, phi] we use [1,t^2-t,t] to get a weighted
mediant, which tells us the limit of the sequence. The two situations
seem quite adequately analogous to me.