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1

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 efficientMathematicacode to find anrational $A,B,C,$ andalternativepolynomial $v_n$ with integer coefficients?quintic

Is there any special reason for the quadratic in the expression for $\zeta(5)$? – J. M.'s ennui – 2016-03-09T02:40:52.437

@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.) – Tito Piezas III – 2016-03-09T02:46:18.810