**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][1]* $\beta(s),$
\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][2]* $\operatorname{Cl}_s(x),$
$$\operatorname{Cl}_2(x) = \sum_{n=1}^\infty\frac{\sin(n\,x)}{n^2}$$
\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][3]* $\kappa$.
---
**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][4], 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][5].*)
---
**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?
[1]: https://mathworld.wolfram.com/DirichletBetaFunction.html
[2]: https://mathworld.wolfram.com/ClausenFunction.html
[3]: https://mathworld.wolfram.com/GiesekingsConstant.html
[4]: https://www.researchgate.net/publication/257642843_Sporadic_sequences_modular_forms_and_new_series_for_1p
[5]: https://mathoverflow.net/q/447166/12905