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
- 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?
- What is the closed-form of the second sequence's continued fraction $B$ (or at least its decimal expansion up to 30 digits)?
- And if the search radius for $(a,b,c,d,e,f)$ is increased, can we find more solutions?