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I. Two recurrence relations

The first one was also discussed in this MO post. We have the similar,

\begin{align} (n+1)^5 u_{n+1} &= (2n + 1)(9n^2 + 9n + 3)(15n^2 + 15n + 4)u_n +3n^3(9n^2-1)u_{n-1}\\ (n+1)^5 v_{n+1} &= (2n + 1)(9n^2 + 9n + 3)(n^2 + n + 2)v_n -9n^3(9n^2-1)v_{n-1} \end{align}

for sequences involving binomials,

\begin{align} u(n) &= 1, 12, 804, 88680\dots\\ v(n) &= 1, 6, 45, 360, 2970\dots \end{align}

with $u(0) = v(0) = 1,$ and are A220119 and A004988, respectively. The first recurrence is by Zudilin-Cohen, while I found the second by computer search so is conjectural (though verified up to my computer's limit of 256 terms). The general form used for the search is,

$$(n+1)^5 u_{n+1} = (2n + 1)(a n^2 + a n + b)(c n^2 + c n + d)u_n + e f n^3(e^2n^2 - 1)u_{n-1}$$

for positive integers $a,b,c,d,e$ and a very limited search radius of,

$$(a < 20,\; b < 15,\; c < 20,\; d < 15,\; e < 10)$$

with $-5<f<5$.


II. Continued fractions

The interest in such 3-term recurrence relations (as discussed in the linked post) is they can be used to create continued fractions with nice closed-forms. The first one is,

\begin{align} A &= \frac1{12 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{3n^8(9n^2-1)}{3(2n + 1)(3n^2 + 3n + 1)(15n^2 + 15n + 4)}}} = \frac{\zeta(4)}{13}\\ B &= \frac1{6 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-9n^8(9n^2-1)}{3(2n + 1)(3n^2 + 3n + 1)(n^2 + n + 2)}}} = 0.17718095881044959\dots \end{align}

Note the "numerator" is deg-$10$. My computer can evaluate $B$ for only about 15 decimal places so its closed-from, if any, is yet unknown.


III. Questions

  1. The second sequence $v(n) = (-9)^n\binom{-2/3}{n} = 1, 6, 45, 360, 2970\dots$ has a much simpler recurrence given in the OEIS link. But it is known a sequence can have many recurrence relations. So is the alternative one for $v(n)$ valid as well?
  2. What is the closed-form of the second sequence's continued fraction $B$ (or at least its decimal expansion up to 30 digits)?
  3. And if the search radius for $(a,b,c,d,e,f)$ is increased, can we find more solutions?
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    $\begingroup$ Regarding 3: no way to know without searching more :) I think it would be interesting to increase the search radius. Regarding 1: yes, the alternative recurrence is valid. Notice that $v(n+1)=v(n) 3(3n+2)/(n+1)$ and by iterating $v(n+1) = v(n-1) 3^2 (3n+2)(3n-1)/((n+1)n)$. It also implies $v(n)=v(n-1)3(3n-1)/n$. This means your recurrence $(n+1)^5 v(n+1)=3(2n+1)(3n^2+3n+1)(n^2+n+2)v(n)n-9n^3(9n^2-1)v(n-1)$ simplifies to $v(n-1) \cdot P(n) = 0$ where $P$ is some rational function that can easily verified to be identically $0$. Regarding 2: I have no idea. $\endgroup$ May 24, 2023 at 14:01
  • $\begingroup$ @OfirGorodetsky: I hope someone will calculate $B$ to enough decimal digits. Like the three Cooper-type sequences, two of the cfracs involved $\zeta(2)$ but the third involve the Gieseking constant. I have a feeling $B$ is a linear combination of special constants. Hopefully. $\endgroup$ May 24, 2023 at 14:06
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    $\begingroup$ $$B=0.17718095881045019823324895123752890422554125603949180664059032531594162340064300076996956935170071463703973788350537026151515637489943$$ As I mentioned, I have a powerful program which can compute thousands of digits of cfracs in less than a second, so you can send a whole list to me if you want. $\endgroup$ May 24, 2023 at 21:22
  • $\begingroup$ Added: and contrary to many "private" programs, if you know how to install Pari/GP from a GIT repository, anybody can use it. $\endgroup$ May 24, 2023 at 21:34
  • $\begingroup$ @HenriCohen: Thank you for the generous offer. I do have a list. Kindly see these four new cfracs based on Ramanujan-type "sporadic" sequences in this post. $\endgroup$ May 28, 2023 at 17:37

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