A n/(n-2) comma formula

I was discussing sums of reciprocals of tetrahedronal numbers on
tuning, and that led me to a formula for commas of the form n/(n-2)
with a prime limit small compared to the size of the denominator
(which is the point.) I've pointed out how to get an infinity of these
for superparticular commas; this extends that. This particular formula is

(2n+3)(4n+3)^2 / (2n+1)(2n+5)^2

When n=0, we get 27/25 (5 limit), for n=1, 245/243 (7 limit), for n=2,
847/845 (13 limit) and so forth.

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

This particular formula is
>
> (2n+3)(4n+3)^2 / (2n+1)(2n+5)^2

This should be

(2n+3)(4n+3)^2 / (2n+1)(4n+5)^2

If we substitute n-1 for n, it's a little more symmetrical

(2n+1)(4n-1)^2 / (2n-1)(4n+1)^2

This can be rewritten as follows

((2n+1)/(2n-1)) / ((4n+1)/(4n-1))^2

The squared term is a new feature. We already had considered such
things as

n^2/(n^2-1) = (n/(n-1)) / ((n+1)/n)

(n(n+1)/2) / (n(n+1)/2 -1) = (n/(n-1)) / ((n+2)/(n+1))