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Tito Piezas III
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Finding two cfracs for $\zeta(2)$, three for $\zeta(3)$, one for Catalan's constant $K$, and one for $\pi^3$ using Zagier's "six sporadic sequences"?

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(-17,-6,-72) &=\;\color{red}{??}\\ C_2(10,3,-9) &=\;\color{red}{??}\\ C_2(7,2,8) &= \frac14\,\zeta(2)\\ C_2(12,4,-32) &= \frac12\,{\beta(2)}=\frac12K\\ C_2(-9,-3,-27) &=\;\color{red}{??} \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.


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(11,3,1) &=\;\color{red}{??}\\ C_3(-17,-6,-72) &= \frac16\,\zeta(3)\\ C_3(10,3,-9) &= -\frac{7}{24}\,\zeta(3)\\ C_3(7,2,8) &=\;\color{red}{??}\\ C_3(12,4,-32) &= \frac{7}{32}\,\zeta(3)\\ C_3(-9,-3,-27) &= \frac{128}{243\sqrt3}\,\beta(3) = \frac{4\pi^3}{243\sqrt3} \end{align}

The second 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. Are all cfracs with proposed closed-forms correct? (I know two of them are.)
  2. What are the closed-forms of the others?
Tito Piezas III
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