$\zeta(3)$ 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$$ p(n) =n^3+(n-1)^3=(2n-1)(n^2-n+1)=1,9,35,91,\dots \qquad $$](https://tpiezas.wordpress.com/2012/05/04/continued-fractions-for-zeta2-and-zeta3/) (this one generalizes in the obvious way to the odd zeta values $\zeta(5),\zeta(7),...$) or, as shown by Apéry, [$k=6$ and
$$ p(n) = (2n-1)(17n^2-17n+5)= 5,117,535,1463,\dots 
 . $$](https://mathoverflow.net/questions/265688/about-a-ramanujan-sato-formula-of-level-10-a-recurrence-and-zeta5) **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?
> 
> 
> 



Representations of $\zeta(3)$ as continued fractions involving cubic polynomials
See **NOTE** below.


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](http://www.ramanujanmachine.com/). These findings for $\zeta(3)$ and several other constants were published in [Arxiv](https://arxiv.org/pdf/1907.00205v4.pdf) (Table 5. pg. 16) in May, 2020 and also in [Science Journal Nature](https://www.nature.com/articles/s41586-021-03229-4) (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).


This [youtube video shows at 26:10](https://youtu.be/h0FW7l7z-C4?t=1578) $\zeta(3)$ Continued Fractions for $k=8/7$ (Wolfgang's) and $k=12/7$.


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](https://arxiv.org/ftp/arxiv/papers/2111/2111.04468.pdf) 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.




---


**NOTE**.


As Wolfgang has pointed out in his answer, there was a "Séminaire de Théorie des Nombres" held in Bordeaux University on March 21th, 1980. In the [Exposé N°23 by Christian Batut and Michael Olivier](https://gdz.sub.uni-goettingen.de/download/pdf/PPN320141322_0009/LOG_0026.pdf) "Sur l'accéléracion de la convergence de certaines fractions continues" (in French), the $\zeta(3)$ CF with $k=8/7$ is found [on page 23-20 3.2.4](https://gdz.sub.uni-goettingen.de/id/PPN320141322_0009?tify=%7B%22pages%22:%5B246%5D,%22panX%22:0.482,%22panY%22:0.73,%22view%22:%22export%22,%22zoom%22:0.349%7D). In this presentation, Apéry's equivalent $\zeta(3)$ CF with $k=6$ is also found (23-19 3.2.2), together with some CFs for Catalan's and other constants. Other $\zeta(3)$ CFs like $k=5/2$ or $k=12/7$ are not shown.


To preserve the spirit of the original response, I will leave this answer at this point.




---


**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>


