I-a. Some functions
As these will be used in the continued fraction evaluations below, recall the Riemann zeta function $\zeta(s),$ and Dirichlet beta function $\beta(s),$ \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}
and special cases of the Clausen function $\operatorname{Cl}_s(x),$
$$\operatorname{Cl}_2(x) = \sum_{n=1}^\infty\frac{\sin(n\,x)}{n^2}$$
\begin{align} K &=\operatorname{Cl}_2\left(\tfrac12\pi\right) = \beta(2) = \sum_{n=0}^\infty\frac{1}{(4n+1)^2}-\sum_{n=0}^\infty \frac{1}{(4n+3)^2} \\ \kappa &= \operatorname{Cl}_2\left(\tfrac13\pi\right) \,=\, \frac{3\sqrt{3}}{4} \left(\sum_{n=0}^\infty\frac{1}{(3n+1)^2}-\sum_{n=0}^\infty \frac{1}{(3n+2)^2} \right) \end{align}
with Catalan's constant $K$ and its cubic counterpart Gieseking's constant $\kappa$.
I-b. Zagier's 6 sporadic sequences
Inspired by Apery's result in proving the irrationality of $\zeta(3)$ using certain integer sequences, Zagier (via a computer) searched for sequences with recurrence relation and deg-$2$ coefficients in $n$ of form,
$$(n+1)^2\,u_{n+1} = (\color{blue}{an^2+an+b})u_k+\color{blue}{cn^2}\,u_{n-1}$$
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.
II. Degree 2
Recall Zagier's recurrence above and, using the coefficients on the RHS, define the polynomial functions,
\begin{align} p(n) &= \color{blue}{an^2+an+b}\\ q(n) &= \color{blue}{cn^2}\times n^2 = c\,n^4 \end{align}
where we affix the monomial $n^2$ so that $q(n)$ will have degree twice that of $p(n).$ Then define the continued fraction,
$$C_2(a,b,c)=\cfrac{1}{p(0) + \cfrac{q(1)}{p(1) + \cfrac{q(2)}{p(2)+ \cfrac{q(3)}{p(3)+\ddots } }}}$$
where we start with $q(1)$ to avoid $q(0)=0$.
Q: Is it true that,
\begin{align} C_2(11,3,1) &= \frac15\,\zeta(2)\\ C_2(-17,-6,-72) &=\color{green}{-\frac5{6\sqrt3}\operatorname{Cl}_2\left(\tfrac13\pi\right) = -\frac5{6\sqrt3}\kappa}\\ C_2(10,3,-9) &=\frac2{3\sqrt3}\operatorname{Cl}_2\left(\tfrac13\pi\right) = \frac2{3\sqrt3}\kappa\\ C_2(7,2,8) &= \frac14\,\zeta(2)\\ C_2(12,4,-32) &= \frac12\operatorname{Cl}_2\left(\tfrac12\pi\right) = \frac12\beta(2)=\frac12K\\ C_2(-9,-3,-27) &=\;\color{red}{??} \end{align}
where $K$ is Catalan's constant and $\kappa$ is Gieseking's constant, both of which not yet proven to be irrational.
Note: The first evaluation is valid since it was found by Apery, while the second (in $\color{green}{\text{green}}$) is courtesy of H. Cohen's answer (though it has slow convergence which is why I missed it).
III. Degree 3
In Cooper's paper, we find the recurrence relation with deg-$3$ coefficients in $n$,
$$(n+1)^3\,v_{n+1} = -(2n+1)(an^2+an+a-2b)v_n -(a^2+4c)n^3\,v_{n-1}$$
where Zagier's $(a,b,c)$ also apply. Again, using the coefficients on the RHS, define the polynomial functions,
\begin{align} r(n) &= -(2n+1)(an^2+an+a-2b)\\ s(n) &= -(a^2+4c)n^3\times n^3 = -(a^2+4c)\,n^6 \end{align}
where we affix the monomial $n^3$ so that $s(n)$ will have degree twice that of $r(n).$ Then define the continued fraction,
$$C_3(a,b,c)=\cfrac{1}{r(0) + \cfrac{s(1)}{r(1) + \cfrac{s(2)}{r(2)+ \cfrac{s(3)}{r(3)+\ddots } }}}$$
where again we start with $s(1)$ to avoid $s(0)=0$.
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}
and $d = a^2+4c =125, 1, 64, 81, 16, -27,$ respectively (and all powers of the smallest primes $2,3,5$).
Note: The second closed-form is valid since it was also found by Apery which he used (together with other methods) to prove the irrationality of $\zeta(3)$.
IV. Degree 4
Curiously, there is no known 3-term recurrence,
$$P_1(n) v_{n+1} = P_2(n) v_n + P_3(n) v_{n-1}$$
where $P_i$ are polynomials of deg-$4$. Why?
V. Degree 5
But Zudilin found a 3-term recurrence,
$$Q_1(n) v_{n+1} = Q_2(n) v_n + Q_3(n) v_{n-1}$$
where $Q_i$ are polynomials of deg-$5$ and used it in an analogous continued fraction for $\zeta(4)$,
$$\frac{\zeta(4)}{13}=\cfrac{1}{u(0) + \cfrac{v(1)}{u(1) + \cfrac{v(2)}{u(2)+ \cfrac{v(3)}{u(3)+\ddots } }}}$$
with $u(n), v(n)$ as polynomial functions where, as usual, the latter has degree twice that of the former. (To be discussed in the next post.)
VI. Questions
- Are all cfracs with proposed closed-forms correct? (I know two of them are.)
- What are the closed-forms of the others?