$\zeta(3)$ 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?
  • $\begingroup$ I like the second part of your question. I wouldn't be surprised to learn that the functional equation of zeta is involved here. $\endgroup$ – Sylvain JULIEN May 12 '19 at 11:00
  • $\begingroup$ The first example of polynomial you give fulfills $p(1-n)=-p(n)$. $\endgroup$ – Sylvain JULIEN May 12 '19 at 11:05
  • $\begingroup$ The same holds for the two other polynomials. $\endgroup$ – Sylvain JULIEN May 12 '19 at 11:14
  • $\begingroup$ @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... $\endgroup$ – Wolfgang May 12 '19 at 14:15
  • 1
    $\begingroup$ 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 :-) $\endgroup$ – Sylvain JULIEN May 12 '19 at 14:40

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