Given the Riemann zeta function $\zeta(n)$.

I. $x=\zeta(3)$

Using Euler's continued fraction formula, we can form $\zeta(3)$'s cfrac as,

$$Ax+B = \cfrac{1}{v_1 - \cfrac{1^6}{v_2 - \cfrac{2^6}{v_3 - \cfrac{3^6}{v_4 -\ddots}}}}\tag1$$

A solution to $(1)$ is $A,B = 1,0,$ where,

$$v_n := (n-1)^3+n^3 = (2n - 1)(n^2 - n + 1)$$

starting with $n=1$. However, Apéry also found $A,B = \tfrac{1}{6},0,$ and,

$$v_n := 34n^3 - 51n^2 + 27n - 5 = (2n - 1)(17n^2 - 17n + 5)$$

and proved that the accelerated rate of convergence was such that $x=\zeta(3)$ could not be rational.

II. $x=\zeta(5)$

$$Ax^2+Bx+C = \cfrac{1}{v_1 - \cfrac{1^{10}}{v_2 - \cfrac{2^{10}}{v_3 - \cfrac{3^{10}}{v_4 -\ddots}}}}\tag2$$

A solution to $(2)$ is $A,B,C=0,1,0,$ where,

$$v_n := (n-1)^5+n^5 = (2 n-1) (n^4-2 n^3+4 n^2-3 n+1) $$

Question: For $\zeta(5)$, what would be an efficient Mathematica code to find an alternative rational $A,B,C,$ and quintic polynomial $v_n$ with integer coefficients?

  • $\begingroup$ Is there any special reason for the quadratic in the expression for $\zeta(5)$? $\endgroup$ Mar 9, 2016 at 2:40
  • $\begingroup$ @J.M. I've tried a crude and limited search with $A=0$, but there was no hit. So perhaps I was missing another term. (Or maybe my bounds for the quintic's coefficients were too small.) $\endgroup$ Mar 9, 2016 at 2:46

1 Answer 1


My guess would be to search first for polynomials of shape $$(2n-1)(a(n^2-n)^2+b(n^2-n)+c),$$ as it might be a good assumption from here that the zeros are symmetric w.r.t. $n=\frac12$. But of course that is pure speculation. I have done some experiments with GP Pari for positive odd $a,b,c$ but not too thoroughly. Needless to say, I didn't yet find anything...

Your solution to $(2)$ corresponds to $(a,b,c)=(1,3,1)$.

  • $\begingroup$ This response would be more appropriate as a comment $\endgroup$
    – Max Coplan
    Feb 14, 2019 at 21:09
  • 3
    $\begingroup$ @MaxCoplan Well, granted that it's not about Mathematica code, but it does provide (hopefully) an idea for a more efficient search. Apart from that, it may be a bit long for a comment anyway. $\endgroup$
    – Wolfgang
    Feb 14, 2019 at 21:20

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