TITLE: Representations of $\zeta(3)$ as continued fractions involving cubic polynomials
QUESTION [29 upvotes]: $\zeta(3)$ has at least two well-known representations of the form $$\zeta(3)=\cfrac{k}{p(1) - \cfrac{1^6}{p(2)- \cfrac{2^6}{ p(3)- \cfrac{3^6}{p(4)-\ddots } }}},$$
where $k\in\mathbb Q$ and $p$ is a cubic polynomial with integer coefficients. Indeed, we can take $k=1$ and$$ p(n) =n^3+(n-1)^3=(2n-1)(n^2-n+1)=1,9,35,91,\dots \qquad $$ (this one generalizes in the obvious way to the  odd zeta values $\zeta(5),\zeta(7),...$) or, as shown by Apéry, $k=6$ and
$$  p(n) = (2n-1)(17n^2-17n+5)= 5,117,535,1463,\dots 
 . $$ Numerically, I have found that $k=\dfrac87$ and $p(n) = (2n-1)(3n^2-3n+1)$ also works. (Is that known? Maybe Ramanujan obtained that as some by-product?)
The question:  


Are there other values of $k$ where such a polynomial exists? 
Must all those polynomials have a zero at $\dfrac12$ for some deeper reason?

REPLY [17 votes]: This MO inquiry is over 3 yrs old now.
By the date the question about the $\zeta(3)$ CF with $k=8/7$ was made (Feb, 2019), it can be answered in the negative nowadays, since it was '(re)-discovered' (and tagged as a new conjectured CF for Apéry's constant) by a team from Technion - Institute of Technology (Israel) using a highly specialized CAS named The Ramanujan Machine. See here. These findings for $\zeta(3)$ and several other constants were published in Arxiv (Table 5. pg. 16) in May, 2020 and also in Science Journal Nature (Feb, 2021). See Ref. below.
So, I think it deserves to be called Wolfgang's $\zeta(3)$ CF. It provides about 1.5 decimal digits per iteration.
Are there other values of k where such a polynomial exists?.
Answer is yes.
In addition to known $k=1,\,8/7,\,6$, ($k=6$ CF is equivalent to Apéry's recursion employed to prove $\zeta(3)$'s irrationality), $k=5/2$ is currently also known (June, 2019) and $k=12/7$ is conjectured (2020).
Must all those polynomials have a zero at 1/2 for some deeper reason?
$k=5/2$ CF does not have a zero at 1/2, but $k=1,6,8/7,12/7$ do (all have companion numerator sequence polynomials $q(n)=-n^6$). There are some conjectures in this paper to look for candidates to prove (or improve) the irrationality (measure) of some constants based on the type of factors and roots that $q(n)$ must have.

Ref: Raayoni, G., Gottlieb, S., Manor, Y. et al. Generating conjectures on fundamental constants with the Ramanujan Machine. Nature 590, 67–73 (2021). https://doi.org/10.1038/s41586-021-03229-4