Hi there,

Just seen it for the case s>1, was obvious really

but I was looking for something too compliciated

For repeat r/s, and s > 1,

(r+1)/(s+1) occurs at c = d = s, e = s+1.

So for x = (a+n)/a, (r+1)/(s+1) occurs at

(sa+sn)/(sa+(s+1)n)

Testing several values in the tree confirms this

It doesn't depend on r, and if you fix s, then

the ratio you are looking for is always in

the same place.

e.g. trying s = 4,

for x = 11/10

5/4: 6/5 occurs at 44/45,

3/2: 7/5 at 44/45

7/4: 8/5 at 44/45

2/1: 9/5 at 44/45

9/4: 2/1 at 44/45

...

For ratios of form r/1, the formula reduces to

(a+n)/(a+2n)

e.g. for x = 11/10, the required ratio is always

found at 11/12

e.g. r = 2

2/1: 3/2 at 11/12

3/1: 2/1 at 11/12

4/1: 5/2 at 11/12

5/1: 3/1 at 11/12

etc.

My previous observation was that for r integer,

(r+1)/2 occurs at (r-1)(a+n)/((r-1)(a+2n))

- which reduces to this formula if you

cancel the (r-1)s, as you can since the tree

has the same ratio at all integer multiples

of the degree, e.g. if it occurs in

11th position in 12th row, it will also

occur at 22nd position in 24th row etc.

Anyway I've written up the observations now on

the page, and will try to prove it after the

release of FTS.

http://tunesmithy.co.uk/uo_non_oct.htm

Robert