# Representations of $\zeta(3)$ as continued fractions involving cubic polynomials

$$\zeta(3)$$ 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?
• I like the second part of your question. I wouldn't be surprised to learn that the functional equation of zeta is involved here. May 12, 2019 at 11:00
• The first example of polynomial you give fulfills $p(1-n)=-p(n)$. May 12, 2019 at 11:05
• No worries. I do believe in the existence of a deep harmony lying in the core of the mathematical realm, otherwise I wouldn't be here :-) May 12, 2019 at 14:40
• Note that you still get an equality replacing the different variables $v$ appearing in the continued fraction, namely $$\zeta(\tau(3))=\cfrac{k}{p(\tau(1)) - \cfrac{\tau(1)^6}{p(\tau(2))- \cfrac{\tau(2)^6}{ p(\tau(3))- \cfrac{\tau(3)^6}{p(\tau(4))-\ddots } }}},$$, I.e $\zeta(-2)=0$. May 12, 2019 at 16:24
• Can this be translated into an Apery-like sequence ? Aug 29, 2020 at 20:36

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. 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).

This youtube video shows at 26:10 $$\zeta(3)$$ 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 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 "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. 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

• Excellent! Thank you for this thorough answer. Jun 1 at 16:43
• Great that you have found a better link, which provides directly a pdf. I have just corrected some typos in the French, as I happen to live in France (though far from Bordeaux...). Jun 21 at 7:09

In a comment to a related question, an article from 1980 (!) is quoted, where this has been proved as a by-product on p. 20 in a quite elementary way. Thus a by-product indeed, but not by Ramanujan himself. :)