A non-Tribonacci example

Let's see what can be done with the following recurrence:

6 1 2 7 3 9 10 12 19 22 31 41 53 72 94 ...
9 2 3 11 5 14 16 19 30 35 49 65 84 114 149 ...
14 2 5 16 7 21 23 28 44 51 72 95 123 167 218 ...

The characteristic polynomial is x^3-x-1, and in terms of the power
basis [1, x, x^2] the matrix corresponding to x is

[0 0 1]
m = [1 0 1]
[0 1 0]

if tm denotes the transpose of m, we want a transpose basis, in which
multiplication by x acts like tm, not m. This means we want to find a
matrix n such that n^(-1) m n = tm; and since the first basis element
in both cases may as well be 1, we want the first row of n to be
[1 0 0]. This determines n, and solving for it gives us

[1 0 0]
n = [0 0 1]
[0 1 0]

This means the transpose basis, corresponding to [1,t^2-t,t] in the
case of Tribonacci, will be [1,x^2,x]. From the above we now have
that the approximation to log_2(3) will be given by

(9+28x^2+3*x)/(6+x^2+2*x) = (356-7*x+20x^2)/241

and to log_2(5) by

(14+2*x^2+5*x)/(6+x^2+2*x) = (565+18*x-17*x^2)/241

This works out to a fifth flat by -0.75 cents and a third flat by
-2.85 cents.

--- In tuning-math@y..., genewardsmith@j... wrote:
> Let's see what can be done with the following recurrence:
>
> 6 1 2 7 3 9 10 12 19 22 31 41 53 72 94 ...
> 9 2 3 11 5 14 16 19 30 35 49 65 84 114 149 ...
> 14 2 5 16 7 21 23 28 44 51 72 95 123 167 218 ...
>
> The characteristic polynomial is x^3-x-1, and in terms of the power
> basis [1, x, x^2] the matrix corresponding to x is
>
> [0 0 1]
> m = [1 0 1]
> [0 1 0]
>
> if tm denotes the transpose of m, we want a transpose basis, in
which
> multiplication by x acts like tm, not m. This means we want to find
a
> matrix n such that n^(-1) m n = tm; and since the first basis
element
> in both cases may as well be 1, we want the first row of n to be
> [1 0 0]. This determines n, and solving for it gives us
>
> [1 0 0]
> n = [0 0 1]
> [0 1 0]
>
> This means the transpose basis, corresponding to [1,t^2-t,t] in the
> case of Tribonacci, will be [1,x^2,x]. From the above we now have
> that the approximation to log_2(3) will be given by
>
> (9+28x^2+3*x)/(6+x^2+2*x) = (356-7*x+20x^2)/241
>
> and to log_2(5) by
>
> (14+2*x^2+5*x)/(6+x^2+2*x) = (565+18*x-17*x^2)/241
>
> This works out to a fifth flat by -0.75 cents and a third flat by
> -2.85 cents.

Kornerup eat your heart out! (Not really jumping into this right now,
just observing from a distance . . .)

--- In tuning-math@y..., genewardsmith@j... wrote:
> Let's see what can be done with the following recurrence:
>
> 6 1 2 7 3 9 10 12 19 22 31 41 53 72 94 ...
> 9 2 3 11 5 14 16 19 30 35 49 65 84 114 149 ...
> 14 2 5 16 7 21 23 28 44 51 72 95 123 167 218 ...

Let's revisit this, and add rows for 7 and 11:

6 1 2 7 3 9 10 12 19 22 31 41 ...
9 2 3 11 5 14 16 19 30 35 49 65 ...
14 2 5 16 7 21 23 28 44 51 72 95 ...
16 3 6 19 9 25 28 34 53 62 87 115 ...
21 4 6 25 10 31 35 41 66 76 107 142 ...

The rows are no longer linearly independent, and we can find the
dependency by inverting the matrix of the first three columns, getting
the matrix of step sizes, and multiplying this by the first three
elements of our new rows. The matrix is

[ 4 -1 -1]
[-3 2 0]
[10 2 3]

Multiplying [16 3 6] by this gives [-5 2 3], and 2^(-5)*3^2*5^2/7 =
225/224, which tells us that this is a kernel element for this system.
Similarly, multiplying by [21 4 6] gives us [12 -1 -3], and this
divided by 11 is 4096/4125, which is therefore also in the kernel. If
we take (4125/4096)/(225/224) = 385/384, we find another kernel
element. Calculating the approximations to 7 and 11 shows that 7 is
1/2 cent sharp and 11 2.9 cents flat. We therefore have 3,5,7, and 11
all quite well approximated.