I've been looking into Apéry's irrationality proof of $\zeta (3)$, and one of the first questions I instantly had, was how did he derive the following continued fraction? $$\begin{equation*} \zeta (3)=\dfrac{6}{5+\overset{\infty }{\underset{n=1}{\mathbb{K}}}\dfrac{-n^{6}}{34n^{3}+51n^{2}+27n+5}}\end{equation*}$$

Furthermore, is it possible to get a similar continued fraction for $\zeta(5)$, $\zeta(7)$ or $G$?

A rapidly converging central binomial series was recently found for Catalan's constant:

$$G = \frac{1}{2} \sum_{n=0}^{\infty} (-1)^n \frac{(3n+2) 8^n}{(2n+1)^3 \binom{2n}{n}^3}$$ If I understand correctly, this gives us the first stage of an analogue of Apéry's proof for $G$. The next stage in his proof is to use a fast recursion formula that approaches $\zeta(3)$: $$n^3 u_n + (n-1)^3 u_{n-2} = (34n^3 - 51n^2 + 27n - 5) u_{n-1},\, n \geq 2$$ A similar recursive formula that approaches $G$ was found: $$(2n+1)^2 (2n+2)^2 p(n) u_{n+1}-q(n)u_n = (2n-1)^2 (2n)^2 p(n+1) u_{n-1}$$ where $$p(n) = 20n^2 - 8n + 1$$ $$q(n) = 3520n^6 + 5632n^5 +2064n^4-384n^3-156n^2+16n+7$$ Now as Zudilin mentions in his paper,'the analogy is far from proving the desired irrationality of $G$', but why exactly is this recursive formula not good enough to prove the irrationality of $G$?