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Made cfrac notation consistent with literature, and added update.
Tito Piezas III
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On 12 cfracs: for Catalan's $K$, Gieseking's $\kappa$, and $\pi^2$, $\pi^3$, plus three for $\zeta(3)$ using Zagier's "six sporadic sequences"

I. Some functions

As these will be used in the continued fraction evaluations below, recall the Riemann zeta function $\zeta(s),$ and Dirichlet beta function $\beta(s),$

$$\beta(s) = \sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n-1)^s}$$

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} \operatorname{Cl}_2\left(\tfrac12\pi\right) &= K = \beta(2) \\ \operatorname{Cl}_2\left(\tfrac13\pi\right) &= \kappa \end{align}

with Catalan's constant $K$ and its cubic counterpart Gieseking's constant $\kappa$.


II. 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 of form,

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

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.


III. Continued fractions

Given a 3-term recurrence relation of form,

$$F_1(n)\,u_{n+1} = F_2(n)\,u_n + F_3(n)\,u_{n-1}$$

where $F_i(n)$ are polynomials of degree $k$. Define two polynomial functions using the rules,

\begin{align} p(n) &= F_1(n-1)\, F_3(n)\\ q(n) &= F_2(n) \end{align}

which implies $p(n)$ has degree twice that of $q(n)$. Define the continued fraction,

$$C =\cfrac{1}{q(0) + \cfrac{p(1)}{q(1) + \cfrac{p(2)}{q(2)+ \cfrac{p(3)}{q(3)+\ddots } }}}$$

More compactly,

$$C(m) = \frac1{q(0) + \large{\underset{n=1}{\overset{m}{\mathrm K}} ~ \frac{p(n)}{q(n)}}}$$

or in Mathematica notation,

$$C(m) = \frac1{q(0) + \text{ContinuedFractionK}[p(n),\;q(n),\, \text{{n, 1, m}}]}$$

It seems $C$ may have a nice closed-form based on the properties of the recurrence relation. Examples below.


IV. Degree 2

Recall Zagier's recurrence,

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

Define $p(n)$ and $q(n)$ according to the rules in the previous section,

\begin{align} p(n) &= \color{blue}{n^2}\times \color{blue}{cn^2} = cn^4\\ q(n) &= \color{blue}{an^2+an+b} \end{align}

Then define the cfrac,

$$C_2(a,b,c) = \frac1{q(0) + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{p(n)}{q(n)}}}$$

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. (Update: May 22, 2023) It turns out $C_2(-9,-3,-27)$ has six limits, one of which is divergent. See this MO post.


V. Degree 3

In Cooper's paper, we find the recurrence relation with deg-$3$ coefficients in $n$,

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

and Zagier's $(a,b,c).$ Using the same rules, let,

\begin{align} r(n) &= n^3\times(-a^2-4c)n^3 = -(a^2+4c)n^6\\ s(n) &= -(2n+1)(an^2+an+a-2b) \end{align}

Define the cfrac,

$$C_3(a,b,c) = \frac1{s(0) + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{r(n)}{s(n)}}}$$

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}

where $d = a^2+4c =125, 1, 64, 81, 16, -27,$ 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 $\zeta(3)$.


VI. Degree 4 & 5

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? But Zudilin found,

$$Q_1(n) v_{n+1} = Q_2(n) v_n + Q_3(n) v_{n-1}$$

where $Q_i$ are polynomials of deg-$5$ and used it in an analogous continued fraction for $\zeta(4).$ (To be discussed in the next post.)


VII. 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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