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$f(2,0)=\frac{\pi}{2}$ is very popular result, but there are a lot of them, not involving $\pi$, for example $$f(4,2)=\prod\limits_{n=1}^{\infty}\left[\frac{(4n-2)^2}{(4n-1)(4n-3)}\right]=\sqrt{2}$$ $$f(6,3)=\prod\limits_{n=1}^{\infty}\left[\frac{(6n-3)^2}{(6n-2)(6n-4)}\right]=\frac{2}{\sqrt{3}}$$ $$f(6,4)=\prod\limits_{n=1}^{\infty}\left[\frac{(6n-4)^2}{(6n-3)(6n-5)}\right]=\frac{3^{1/2}}{2^{1/3}}$$ How to find values of $f(m,k)$ for all $m>1$ and $(m-1)>k\geqslant0$?

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Hint. One may recall that $$ \prod_{n=1}^{N} \left(1 + \frac{z}{n+a}\right)=\frac{\Gamma(a+1)\Gamma(a+N+z+1)}{\Gamma(a+N+1)\Gamma(a+z+1)} \tag1 $$ where $\Gamma$ is the Euler gamma function, one may then write $$ \prod\limits_{n=1}^{N}\left[\frac{(mn-k)^2}{(mn-k+1)(mn-k-1)}\right]=\left[\prod\limits_{n=1}^{N}\left(1 + \frac{1/m}{n-k/m}\right)\left(1 - \frac{1/m}{n-k/m}\right)\right]^{-1} \tag2 $$ and apply $(1)$ to $(2)$, then letting $N \to \infty$.

Edit. The preceding steps give

$$ \prod\limits_{n=1}^{\infty}\left[\frac{(mn-k)^2}{(mn-k+1)(mn-k-1)}\right]=\frac{\Gamma \left(1-\frac{k+1}{m}\right) \Gamma \left(1-\frac{k-1}{m}\right)}{\Gamma \left(1-\frac{k}{m}\right)^2} $$

from which one deduces many particular cases.

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