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

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

> Gene, if there's a simple algorithm that gets us only these

> semiconvergents and not the others (simpler, that is, than

> calculating all the semiconvergents and then throwing some out),

let

> us know on tuning-math.

Here's another way to do it: the convergents to 2^(7/24) are

5/4,11/9,71/58,224/183... which tells us that before 71/58 we have

semiconvergents of the form (11n+5)/(9n+4). We can solve the linear

equation for n which finds the value for which this is exactly as far

from 2^(7/24) as is 11/9, but on the other side, which gives

n=2.926... . This tells us we can toss out the values for n=1 and n=2.

If you were going to code this this would be a pretty good system,

but I doubt it's any kind of improvement by hand except in the cases

where we have a very good convergent, leading to a long line of

semiconvergents.