I. Some functions

As we will use these in the continued fraction (cfrac) evaluations below, recall the Riemann zeta function and the Dirichlet beta function,

\begin{align} \zeta(s) &= \sum_{n=1}^\infty\frac{1}{n^s}\\ \beta(s) &= \sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n-1)^s} \end{align}

II. Zagier's 6 sporadic sequences

Inspired by Apery's result in proving the irrationality of $\zeta(3)$ using certain integer sequences, Zagier (using a computer) searched for sequences with recurrence relation and deg-2 coefficients of form,

$$(n+1)^2u_{n+1} = (\color{blue}{an^2+an+b})u_k+\color{blue}c\,n^2u_{n-1}$$

that produced only integer values. Only six $(a,b,c)$ were found, namely,

$$(11,3,1),\quad (7,2,8) ,\quad (12,4,-32)$$ $$(-17,-6,-72),\quad (10,3,-9), \quad (-9,-3,-27)$$

It seems we can use ALL these coefficients to produce nice cfracs.

III. Level 2

Define the polynomial function,

$$s_n = \color{blue}{an^2+an+b}$$

based on the recurrence above and the continued fraction,

$$C_2(a,b,c)=\cfrac{1}{s_0 + \cfrac{1^4\, \color{blue}c}{s_1 + \cfrac{2^4\, \color{blue}c}{s_2+ \cfrac{3^4\,\color{blue}c}{s_3+\ddots } }}}$$

Q: Is it true that,

\begin{align} C_2(11,3,1) &= \frac15\,\zeta(2)\\ C_2(7,2,8) &= \frac14\,\zeta(2)\\ C_2(12,4,-32) &= \frac12\,{\beta(2)}=\frac12K \end{align}

Note that $K$ is Catalan's constant which is not yet proven to be irrational. The first one is valid since it was found by Apery. We will use the remaining $(a,b,c)$ for level-3.

IV. Level 3

In Cooper's paper, we find the recurrence relation with cubic coefficients,

$$(n+1)^3 v_{n+1} = \color{blue}{-(2n+1)(an^2+an+a-2b)}v_n \color{blue}{- (a^2+4c)}n^3v_{n-1}$$

where Zagier's $(a,b,c)$ also apply. Define the polynomial function,

$$t_n = \color{blue}{-(2n+1)(an^2+an+a-2b)}$$

and the continued fraction with constant $\color{blue}{d = -(a^2+4c)}$,

$$C_3(a,b,c)=\cfrac{1}{t_0 + \cfrac{1^6\, \color{blue}d}{t_1 + \cfrac{2^6\, \color{blue}d}{t_2+ \cfrac{3^6\,\color{blue}d}{t_3+\ddots } }}}$$

Q: Is it true that,

\begin{align} C_3(-17,-6,-72) &= \frac16\,\zeta(3)\\ C_3(10,3,-9) &= -\frac{7}{24}\,\zeta(3)\\ C_3(-9,-3,-27) &= \frac{128}{243\sqrt3}\,\beta(3) = \frac{4\pi^3}{243\sqrt3} \end{align}

The first one is valid since it was also found by Apery.

V. Level 4

Zudilin also found an analogous continued fraction for $\zeta(4)$,

$$C_4(a_1, a_2,\dots a_n) =\cfrac{1}{p_0 + \cfrac{1^8\, q_1}{p_1 + \cfrac{2^8\, q_2}{p_2+ \cfrac{3^8\,q_3}{p_3+\ddots } }}}$$

where $p_i, q_i$ are polynomial functions. Makes you wonder if there is for $\beta(4) = \frac1{768}\left[\psi_3(\frac14)-8\pi^4\right]$ as well.

VI. Questions

1. So are all 6 cfracs evaluated correctly? (I know two of them are.)
2. If we use the level-2 triplets $(a,b,c)$ into the level-3 cfrac $C_3(a,b,c)$, would they have closed-forms? (And vice versa, namely the level-3 triplets into the level-2 cfrac.)