Squares

Since you can find a squared ratio from two triangulars,
and in turn, two squares from each triangular, you can
find a squared ratio based on four squared ratios. (And
a triangular from four triangulars). Here are the first
few squares from squares:

4/3: 9/8 16/15 16/15 25/24
9/8: 25/24 36/35 36/35 49/48
16/15: 49/48 64/63 64/63 81/80
25/24: 81/80 100/99 100/99 121/120

etc.

Squares are all n/n-1 * n/n+1
Triangulars are all n/n-1 * n+1/n+2

Multiplying the triangular by n/n+1 and n+1/n
gives n^2/n^2-1 and n^2+2n+1/n^2+2n which is the squared
ratios triangulars are composed of.

Here are the first few triangulars from triangulars

3/2: 6/5 10/9 15/14 21/20
6/5: 15/14 21/20 28/27 36/35
10/9 28/27 36/35 45/44 55/54

which in turn can be composed of squares (I'll do one)

3/2: 9/8 16/15 16/15 25/24 25/24 36/35 36/35 49/48

Just another onion!

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> Since you can find a squared ratio from two triangulars,
> and in turn, two squares from each triangular, you can
> find a squared ratio based on four squared ratios. (And
> a triangular from four triangulars). Here are the first
> few squares from squares:
>
> 4/3: 9/8 16/15 16/15 25/24
> 9/8: 25/24 36/35 36/35 49/48
> 16/15: 49/48 64/63 64/63 81/80
> 25/24: 81/80 100/99 100/99 121/120

Which tetrachord division is (9/8) (16/15)^2 (25/24)? It doesn't seem
to be exactly diatonic.

One thing to note about this is that we can take a division of the
octave into terms consisting of only square numerators, for instance
2 = (9/8)(4/3)^2, and obtain scales which have superparticular steps
with only square numerators. This might be fun to try.

> 3/2: 6/5 10/9 15/14 21/20
> 6/5: 15/14 21/20 28/27 36/35
> 10/9 28/27 36/35 45/44 55/54

The same comment applies. Starting from 2 = (10/9)(6/5)(3/2) we can
decompose the octave into scales with only triangular numbers in the
numerator.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@> wrote:
> >
> > Since you can find a squared ratio from two triangulars,
> > and in turn, two squares from each triangular, you can
> > find a squared ratio based on four squared ratios. (And
> > a triangular from four triangulars). Here are the first
> > few squares from squares:
> >
> > 4/3: 9/8 16/15 16/15 25/24
> > 9/8: 25/24 36/35 36/35 49/48
> > 16/15: 49/48 64/63 64/63 81/80
> > 25/24: 81/80 100/99 100/99 121/120
>
> Which tetrachord division is (9/8) (16/15)^2 (25/24)? It doesn't
seem
> to be exactly diatonic.

I thought about that. Take 9/8, 16/15, 25/24 and 16/15 again.
That is like a normal whole step, diatonic half-step, chromatic
half-step and another diatonic half-step: C, D, Eb, E, F, which
is a typical up-hill passage in music

>
> One thing to note about this is that we can take a division of the
> octave into terms consisting of only square numerators, for
instance
> 2 = (9/8)(4/3)^2, and obtain scales which have superparticular steps
> with only square numerators. This might be fun to try.
>
> > 3/2: 6/5 10/9 15/14 21/20
> > 6/5: 15/14 21/20 28/27 36/35
> > 10/9 28/27 36/35 45/44 55/54
>
> The same comment applies. Starting from 2 = (10/9)(6/5)(3/2) we can
> decompose the octave into scales with only triangular numbers in the
> numerator.
>

Cool! Are there any other interesting ratios besides squared and
triangular superparticulars?

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:

> Cool! Are there any other interesting ratios besides squared and
> triangular superparticulars?

Given any starting point p/q, and any limit point a/b, you can
find succesive mediants (na+p)/(nb+q). Taking ratios gives
(na+a+p)(nb+q) / (nb+b+q)(na+p), so you have quadratic numerators and
denominators.

If you take n^2/(n^2-1), it can be written (n/(n-1))/((n+1)/n), then
subtracting numberators and denominators of (n+1)/n and n/(n-1) gives
1/1, and putting n=2 in gives 2/1 as the starting point. Similarly,
approaching 2/1 starting from 1/1 gives 1/1, 3/2, 5/3, 7/4 with
rations 3/2, 10/9, 21/20 etc. Starting from 3/1, we get 3/1, 5/2, 7/3,
9/4 with ratios 6/5, 15/14, 28/27 etc. Putting this together, the
whole triangula r denomimator thing is for approaching 2/1, the whole
square denominator thing for approaching 1/1.

What happens when we break out of the mold? 50/49 has a denominator
which is neither triangular nor square. Where does that fit in?
50/49 = (10/7)/(7/5), and subtracting numerators and denominators
gives 3/2. Hence, now we want to look at a family approaching 3/2.
Starting from 1/1, we have 1/1, 4/3, 7/5, 10/7, 13/9, with ratios of
4/3, 21/20, 50/49, 91/90 and so forth. The denominators are all of the
form (2n+1)(3n+4). Also interesting are the negative values of n,
which give 56/55, 99/98 etc.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:
The denominators are all of the
> form (2n+1)(3n+4).

Numerators, sorry.

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

We have the following from the integer sequences handbook:

http://www.research.att.com/~njas/sequences/A033569

-1,6,25,56,99,154,221,300 returns no entry; I don't know why one and
not the other. Nothing about music in the description.