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$\zeta(3)$ 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?
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    $\begingroup$ I like the second part of your question. I wouldn't be surprised to learn that the functional equation of zeta is involved here. $\endgroup$ May 12, 2019 at 11:00
  • $\begingroup$ The first example of polynomial you give fulfills $p(1-n)=-p(n)$. $\endgroup$ May 12, 2019 at 11:05
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    $\begingroup$ 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 :-) $\endgroup$ May 12, 2019 at 14:40
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    $\begingroup$ 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 $\zeta(-2)=0$. $\endgroup$ May 12, 2019 at 16:24
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    $\begingroup$ Can this be translated into an Apery-like sequence ? $\endgroup$
    – F. C.
    Aug 29, 2020 at 20:36

2 Answers 2

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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

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    $\begingroup$ Excellent! Thank you for this thorough answer. $\endgroup$
    – Wolfgang
    Jun 1 at 16:43
  • $\begingroup$ 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...). $\endgroup$
    – Wolfgang
    Jun 21 at 7:09
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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. :)

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