In a previous [MO post](https://mathoverflow.net/a/447123/12905), H. Cohen suggested [Gorodetsky's 2021 paper](https://arxiv.org/abs/2102.11839) which discussed $6+6+3=15$ "*sporadic sequences*". The first 6 are Zagier's sporadic sequences, the second 6 are by Almkvist-Zudilin, while the last 3 is credited to Cooper.


The first 12 were discussed in the previous post, so the last 3 will be discussed here **PLUS** a $16\text{th}$ sequence which seems to have been missed.




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**I. Recurrence relations**


In [Cooper's 2012 paper](https://www.researchgate.net/publication/257642843_Sporadic_sequences_modular_forms_and_new_series_for_1p), we find the 3-term recurrences,


$$(n+1)^3 u\_{n+1} = (2n+1)(13n^2+13n+4)u\_n + 3n(9n^2-1)u\_{n-1}$$
$$(n+1)^3 v\_{n+1} = 2(2n+1)(3n^2+3n+1)v\_n + 4n(16n^2-1)v\_{n-1}$$
$$(n+1)^3 w\_{n+1} = 2(2n+1)(7n^2+7n+3)w\_n -12n(16n^2-1)w\_{n-1}$$


These are for Cooper's sequences $s\_7,\, s\_{10},\, s\_{18},$ respectively. However, in [Zudilin's 2002 paper](https://arxiv.org/abs/math/0201024), there is **another** sequence also with a 3-term recurrence but with polynomial coefficients of deg-$5$,


$$(n+1)^5 x\_{n+1} = 3(2n + 1)(3n^2 + 3n + 1)(15n^2 + 15n + 4)x\_n +3n^3(9n^2-1)x\_{n-1}$$


**P.S.** By coincidence, it seems H. Cohen also found this recurrence (in 1980) as Zudilin mentions in page 9 of his paper.




---


**II. Continued fraction**


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 $m$. Define two polynomial functions using the same rules in the previous [MO post](https://mathoverflow.net/a/447123/12905),


\begin{align}
p\_k(n) &= F\_1(n-1)\, F\_3(n)\\
q\_k(n) &= F\_2(n)
\end{align}


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


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


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




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**III. The $s\_7$ sequence**


Using the first recurrence,


$$(n+1)^3 u\_{n+1} = (2n+1)(13n^2+13n+4)u\_n + \color{blue}{3n(9n^2-1)}u\_{n-1}$$


which is for $s\_7$. Applying the rules,


\begin{align}
p\_1(n) &= n^3 \times \color{blue}{3n(9n^2-1)}\\
q\_1(n) &= (2n+1)(13n^2+13n+4)
\end{align}


Then,


$$C(p\_1,\,q\_1) = \frac1{4 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{p\_1(n)}{q\_1(n)}}} \overset{\color{red}?}= \frac{\zeta(2)}7 \quad$$




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**IV. The $s\_{10}$ sequence**


Using the second recurrence,


$$(n+1)^3 v\_{n+1} = 2(2n+1)(3n^2+3n+1)v\_n + 4n(16n^2-1)v\_{n-1}$$


which is for $s\_{10}$. Let,


\begin{align}
p\_2(n) &= n^3\times 4n(16n^2-1)\\
q\_2(n) &= 2(2n+1)(3n^2+3n+1)\\
\end{align}


Then,


$$C(p\_2,\,q\_2) = \frac1{2 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{p\_2(n)}{q\_2(n)}}} \overset{\color{red}?}= \frac{\zeta(2)}5 \quad$$




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**V. The $s\_{18}$ sequence**


Using the third recurrence,


$$(n+1)^3 w\_{n+1} = 2(2n+1)(7n^2+7n+3)w\_n + 12n(-16n^2+1)w\_{n-1}$$


which is for $s\_{18}$. Let,


\begin{align}
p\_3(n) &= n^3\times12n({-16}n^2+1)\\
q\_3(n) &= 2(2n+1)(7n^2+7n+3)\\
\end{align}


Given [*Gieseking's constant*](https://mathworld.wolfram.com/GiesekingsConstant.html) $\kappa \approx 1.01494$, then,


$$C(p\_3,\,q\_3) = \frac1{6 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{p\_3(n)}{q\_3(n)}}} \overset{\color{red}?}= \frac2{3\sqrt3}\kappa$$




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**VI. Zudilin's sequence and $\zeta(4)$**


For general zeta $\zeta(m)$ and $m\geq2,$ the expected $q$-polynomial is deg-$m$,


$$\zeta(m) = \frac1{1 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-n^{2m}}{n^m+(n+1)^m}}} $$


But the recurrence found by Zudilin is ***deg-5***,


$$(n+1)^5 x\_{n+1} = 3(2n + 1)(3n^2 + 3n + 1)(15n^2 + 15n + 4)x\_n +3n^3(9n^2-1)x\_{n-1}$$


Define,


\begin{align}
p\_{4}(n) &= n^5\times3n^3(9n^2-1)\\
q\_{4}(n) &= 3(2n + 1)(3n^2 + 3n + 1)(15n^2 + 15n + 4)\\
\end{align}


Then,


$$C(p\_4,\,q\_4) = \frac1{12 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{p\_4(n)}{q\_4(n)}}} = \frac{\zeta(4)}{13}$$


which has been proven by Zudilin. The recurrence satisfies a sequence ([A220119](https://oeis.org/A220119)) with closed-form,


\begin{align}Z\_n 
&= \sum\_{k=0}^n\left(\sum\_{j=0}^n\binom{n}{j}^2\binom{n}{k}^2\binom{n+j}n\binom{n+k}n\binom{j+k}n\right)\\
&=1, 12, 804, 88680, 12386340,\dots
\end{align}


found in the first page of the [Krattenthaler-Rivoal 2009 paper](https://arxiv.org/abs/0907.2597).




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**VII. Questions**


1. Are the evaluations using $s\_{7},\,s\_{10},\,s\_{18}$ correct?
2. Is it possible to find a $3$-term recurrence relation (**analogous** to the $16$ known so far) with polynomial coefficients of deg-$n$, but $n\neq 0,1,2,3,5$? (*Surely there must be one for $n=4$?*)



On the continued fractions using Cooper's sequences $s_7,\, s_{10},\, s_{18}$ and the Zudilin-Cohen sequence
*(This answers Question 2.)*


Thanks to [Cohen's 2022 paper](https://arxiv.org/abs/2212.01095), turns out *there* is a deg-$4$ and one can find polynomials $Q\_k(n)$ for general deg-$k$ such that,


$$(n+1)^k s\_{n+1} = Q\_k (n)\, s\_n - n^k s\_{n-1}\qquad\tag{eq.1}$$


is a 3-term recurrence relation. This can then be used to create *continued fractions with closed-forms.* Recall the general cfrac for zeta $\zeta(k)$,


$$\zeta(k) = \frac1{1 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{2k}}{n^k+(n+1)^k}}} \qquad$$


We employ "denominators" with similar forms, including deg-$4$, the first being Apery's,


\begin{align}
Q\_3(n) &= n^3 + (n + 1)^3 + 4(2n + 1)^3\\
Q\_4(n) &= n^4 + (n + 1)^4 + 2n^2 + 2(n + 1)^2\\
Q\_5(n) &= n^5 + (n + 1)^5 + 6n^3 + 6(n + 1)^3\\
Q\_6(n) &= n^6 + (n + 1)^6 + 3 n^4 + 3(n + 1)^4 -(2n + 1)^2
\end{align}


then we get the nice cfracs,


\begin{align}
F\_3 &= \frac1{5 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{6}}{Q\_3(n)}}} = \frac{\zeta(3)}6\\[8pt]
F\_4 &= \frac{\color{red}{-1}}{3 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{8}}{Q\_4(n)}}} = \zeta(4)+4\,\zeta(2)-8 \\[8pt]
F\_5 &= \frac1{7 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{10}}{Q\_5(n)}}} = \zeta(5)+3\,\zeta(3)-9/2\\[8pt]
F\_6 &= \frac{\color{red}{-1}}{3 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{12}}{Q\_6(n)}}} = \zeta(6)+4\,\zeta(4)+16\,\zeta(2)-32
\end{align}


and so on. *Note*: These converge slightly faster (with Apery's the fastest) because to the general $n^k + (n + 1)^k$ expression, more terms have been added.




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**P.S.** For the recurrence, one may also choose exponents $\alpha,\beta,$


$$(n+1)^\alpha s\_{n+1} = Q\_k (n)\, s\_n - n^\beta s\_{n-1}\qquad$$


such that $\alpha \leq \beta$ and $\alpha+\beta = 2k.$ **Eq.1** was just the case $\alpha=\beta=k.$


