# 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. – 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
• 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