I stumbled upon something rather interesting today while noodling around. Given that this general formula for all $\zeta(x)$ produces a continued fraction, does it therefore stand that all $\zeta(x)$ are irrational?
$\zeta(x)$= $1+
\cfrac{1^{2x}}{
2^x-\cfrac{2^{2x}}{K_2 + K_3-\cfrac{3^{2x}}{K_3 + K_4-\cfrac{4^{2x}}{K_4 + K_5-\ddots}}}}$
Where $K_n$ is the denominator of the $n^{th}$ Riemann (e.g. $\zeta(3)$, $K_n$:{1,8,27,64 $\dots$}, otherwise noted as $n^x$). $$-$$ Notice, that $K_n$ + $K_{n+1}$ in Apery's $\zeta(3)$ is nothing more than https://oeis.org/A005898 from 35 onwards, or https://oeis.org/A001844 for $\zeta(2)$