Magic recurrence

Here's a recurrence example. If we take x^7-x^4-x-1 as a
characteristic polynomial, we have 1.24564 as the large root, which
can be used as a generator for magic. Corresponding linear recurrences
are of the form

C[n] = C[n-3] + C[n-6] + C[n-7]

An example would be

3, 8, 9, 10, 15, 16, 22, 26, 33, 41, 51, 64, 79, 99, 123, 153, 191,
167, 248, 289, 324, 467, 570, 639, 882, 1107, 1252, 1673, 2144, 2461...

If anyone can figure out an actual use for this stuff I would be
interested.

Gene Ward Smith wrote:

>If anyone can figure out an actual use for this stuff I would be
>interested.
> >
The original motivation was for difference tones to match, but the patterns would be so complex for what you're looking at I don't think that's relevant. Another approach would be to keep to three non-zero coefficients (two on the RHS) that are +/- powers of two. That means first order difference tones between octave equivalents will match.

One case would be quartics with RHS coefficients from {-8, -4, -2, -1, 1, 2, 4, 8}. If they're of the form

X[n] = A*X[n-alpha] + B*X[n-beta]

where 0 < alpha < beta <= 4, the choices of alpha and beta are 4C2 or 6. For each of A and B thre are 8 choices. So the total number of series will be 6*8**2 = 384. So it would be possible to check all of them, but I haven't. Some won't work.

Expand it to quintics, and allow +/-16, there'll be 1000 different options.

I've come up with a meta-slendric. It keeps the equivalences from JI

4 + 3 = 7
8-7 = 7-6
7-6 = 4-3
8-7 = 4-3

The equation is x**4 = 4-2x, and the root is 1.4390... or 0.1936... octaves.

The trouble is, the sequence X[n] = 4X[n-4] - 2X[n-3] doesn'tmake much sense, and I don't think it converges. I don't really know what to do about this. The problem is the -2, so it goes up and down. If, above, you made A and B positive it'd make more sense, and that seems to be what you're doing with your examples. But it would also lose some useful scales like this one.

You could re-write it x**4 = 2x+4, and take a negative root, but I don't think that'd solve the problem.

Graham

--- In tuning-math@yahoogroups.com, Graham Breed <graham@m...> wrote:

> The original motivation was for difference tones to match, but the
> patterns would be so complex for what you're looking at I don't think
> that's relevant. Another approach would be to keep to three non-zero
> coefficients (two on the RHS) that are +/- powers of two. That means
> first order difference tones between octave equivalents will match.

Why do we want the difference tones to match?

> I've come up with a meta-slendric.

I've got a nice recurrence for tritonic, the 7-limit temperament with
[5, -11, -12, -29, -33, 3] as a wedgie. What's nice about it is that
the big root is a PV number, meaning all of its conjugates are less
than one in absolute value, so that the ratio converges quickly. The
polynomial is x^4-2*x^3-2*x^2-x+1, so the recurrence goes

c[n] = 2 c[n-1] + 2 c[n-2] + c[n-3] - c[n-4]

It doesn't have your RHS property, but I could search for that.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

> Why do we want the difference tones to match?

I think I just figured this out. Cool!

I've found another meta-magic. To get equal beating for 8:9:10,

9-8 = 10-9
10+8 = 2*9
5 + 4 = 3**2
4x + 4 = x**10

This has a solution of 1.24548... or 0.3167...octaves (380.04 cents). The sequence is

X[n] = 4(X[n-10] + X[n-9])

It converges, but very slowly. Here are some chunks -- there may be gaps

256, 256, 256, 256, 256, 256, 256, 512, 1216, 1856,
2048, 2048, 2048, 2048, 2048, 2048, 3072, 6912, 12288, 15616
16384, 16384, 16384, 16384, 16384, 20480, 39936, 76800, 111616, 128000

I think both magic and meantone work for JI-like equivalence because each prime is a positive number of generators. So the RHS terms are always positive.

Graham