6
$\begingroup$

In a previous MO post, H. Cohen suggested Gorodetsky's 2021 paper 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.


I. Recurrence relations

In Cooper's 2012 paper, 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, 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,

\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.


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,

$$C(p_1,\,q_1) = \frac1{4 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{p_1(n)}{q_1(n)}}} \overset{\color{red}?}= \frac{\zeta(2)}7 \quad$$


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,

$$C(p_2,\,q_2) = \frac1{2 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{p_2(n)}{q_2(n)}}} \overset{\color{red}?}= \frac{\zeta(2)}5 \quad$$


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 $\kappa \approx 1.01494$, then,

$$C(p_3,\,q_3) = \frac1{6 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{p_3(n)}{q_3(n)}}} \overset{\color{red}?}= \frac2{3\sqrt3}\kappa$$


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) 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.


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 $n=4$?)
$\endgroup$
0

1 Answer 1

1
$\begingroup$

(This answers Question 2.)

Thanks to Cohen's 2022 paper, 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.


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 $\alpha=\beta=k.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.