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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?
GH from MO
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Wolfgang
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    I like the second part of your question. I wouldn't be surprised to learn that the functional equation of zeta is involved here. – Sylvain JULIEN May 12 '19 at 11:00
  • The first example of polynomial you give fulfills $p(1-n)=-p(n)$. – Sylvain JULIEN May 12 '19 at 11:05
  • The same holds for the two other polynomials. – Sylvain JULIEN May 12 '19 at 11:14
  • @SylvainJULIEN Putting $x:=n-\frac12$, we get indeed odd polynomials in $x$: $$ k=1\ \text{ with } p=34 x^3 + \frac32 x\k=\frac87\ \text{ with }p=6 x^3 + \frac12 x\ k=6\ \text{ with }p=2x^3 + \frac32 x.$$ But going from there to the functional equation of zeta (which also contains a gamma function) seems a bit too far-fetched... – Wolfgang May 12 '19 at 14:15
  • I just see the same symmetry group (namely ${Id,s\mapsto 1-s}$) in the polynomials of degree 2 you listed and the functional equation of $\zeta$ (more exactly of its completed L-function), nothing less, nothing more. As you ask for a deeper reason for $1/2$ being a root of the relevant polynomials, and as we deal with the Riemann zeta function, I suggest a possible one which may be or not the right one. – Sylvain JULIEN May 12 '19 at 14:22
  • Sure. I could have said "a bit too optimistic" instead of "a bit too far-fetched". :) – Wolfgang May 12 '19 at 14:25
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    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 :-) – Sylvain JULIEN May 12 '19 at 14:40
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    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 $\zeta(-2)=0$. – Sylvain JULIEN May 12 '19 at 16:24
  • Replace $k$ with $\tau(k)$ in the above comment, where $\tau(v):=1-v$ (I unpurposedly deleted a crucial part of my sentence trying to fix the Latex). – Sylvain JULIEN May 12 '19 at 16:30
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    Can this be translated into an Apery-like sequence ? – F. C. Aug 29 '20 at 20:36
  • Not exactly of the form, but a remarkably simple looking one from MSE: https://math.stackexchange.com/questions/141066/a-new-continued-fraction-for-ap%c3%a9rys-constant-zeta3?rq=1. I copy-paste it here for ease of access: $\zeta(3) = \cfrac{6}{v_1 + \cfrac{1^3}{1 + \cfrac{1^3}{v_2 + \cfrac{2^3}{1 + \cfrac{2^3}{v_3 +\ddots}}}}}\tag{2}$

    where the $v_n$ are given by the cubic function

    $v_n = 4(2n-1)^3 = 4, 108, 500, 1372, \dots$

     – D.R. Jan 08 '24 at 05:42
  • @D.R. Yes, this is somewhat related to what Michael Somos calls "bisection", but I don't think this gives additional insights here. :( – Wolfgang Jan 08 '24 at 11:48

2 Answers2

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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

Wolfgang
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Jorge Zuniga
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    Excellent! Thank you for this thorough answer. – Wolfgang Jun 01 '22 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...). – Wolfgang Jun 21 '22 at 07:09
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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. :)

Wolfgang
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