$\zeta(3)$ 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 reason?

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