Wallis' product exploration

[Allen]

What I find mind blowing about it is that it's a combinatoric function, not of just triangles, but how many different ways triangles can tile a nearly-circular polygon.

I tried yesterday to find a printed/published formula for \( \pi \) in terms of catalan numbers, but could not. combining terms with Albert's observation above, we get

\(\pi = \lim\limits_{n\to\infty} \frac {2^{4n+2} } { (2n+1) (n+1)^2 C_n^2} \)

interesting scaling polynomial in the bottom

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

Out of curiosity, I think we can replace the \( (n+1)^2\) with a Triangluar number \( T_n\) and further drop a factor of 4 from the numerator:

\( \pi = \lim\limits_{n\to\infty} \frac {n^2 2^{4n} } { (2n+1) T_n^2 C_n^2} \)

in effort to reduce some polynomial terms, but I can't tell if that's a step forward or a step back.. Ideally, one could represent \( \pi \) as an integer divided by a hand full of (discrete) combinatoric functions.