After 6 years from this old MO post, I finally find in the literature polynomials of deg-$5$ for the continued fraction of $\zeta(5)$.
I. Recurrences involving $\zeta(5)$
In Cohen's 2022 paper, there are three polynomials of deg-$5$ with an irreducible quartic factor (compare to the reducible ones in the previous post). These can be placed in a 3-term recurrence relation analogous to Apery's,
\begin{align} (n+1)^5 \alpha_{n+1} &= (2n + 1)(n^4+2n^3+10n^2+9n+3)\alpha_n -n^5\alpha_{n-1}\\ (n+1)^5 \beta_{n+1} &= (2n + 1)(n^4+2n^3+10n^2+9n+7)\beta_n -n^5\beta_{n-1}\\ (n+1)^5 \gamma_{n+1} &= (2n + 1)(n^4+2n^3+20n^2+19n+13)\gamma_n -n^5\gamma_{n-1} \end{align}
But in contrast to all the "sporadic sequences" in previous posts which involve binomial numbers, I found these recurrences satisfy sequences that are just polynomial functions,
\begin{align} \alpha(n) &= \frac{n^3+1}{n+1} = 1, 3, 7, 13, 21, 31, 43,\dots\\ \beta(n) &= n^3-(n-1)^3 = 1, 7, 19, 37, 61, 91, 127,\dots\\ \gamma(n) &= \frac{n^5 - (n - 1)^5 + 3n^3 - 3(n - 1)^3}{4} = 1, 13, 67, 223, 571, 1231,\dots \end{align}
where $\alpha(1)=\beta(1)=\gamma(1)=1.$ Edit: Different choices of exponents may also work. For example,
$$(n+1)^{\color{blue}4}\, a_{n+1} = (2n + 1)(n^4+2n^3+10n^2+9n+3)a_n -n^{\color{blue}6}\,a_{n-1}$$
seems obeyed by a more complicated sequence (A001564) which involves factorials,
$$a(n) = \frac{n^3+1}{n+1}\,(n-1)! = 1, 3, 14, 78, 504, 3720,\dots\qquad$$
II. Recurrences involving $\zeta(7)$
There is also an Apery-like recurrence for deg-$7$,
$$(n+1)^7 \delta_{n+1} = (2n + 1)(n^6 + 3n^5 + 17n^4 + 29n^3 + 35n^2 + 21n + 5)\delta_n -n^7\delta_{n-1}$$
which is satisfied by the sequence,
$$\delta(n) = n^2+(n-1)^2 = 1, 5, 13, 25, 41, 61, 85,\dots$$
III. Continued fractions
Given Apery's recurrence,
$$(n+1)^3 s_{n+1} = (2n + 1)(17n^2 + 17n + 5) s_n -n^3 s_{n-1}$$
the cubic polynomial can be expressed as a nice sum of cubes,
\begin{align}P(n) &= (2n + 1)(17n^2 + 17n + 5)\\ &= n^3 + (n + 1)^3 + 4(2n + 1)^3 \end{align}
used in his cfrac,
$$F_0 = \frac1{5 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-n^6}{P(n)}}} = \frac{\zeta(3)}{6}\quad\quad$$
Likewise, Cohen's quintics can also be expressed as a sum of powers,
\begin{align} A(n) &= n^5 + (n + 1)^5 + 6\left(n+\tfrac13\right)^3 + 6\left(n+\tfrac23\right)^3\\[6pt] B(n) &= n^5 + (n + 1)^5 + 6n^3+6(n+1)^3\\[6pt] C(n) &= n^5 + (n + 1)^5 + 16\left(n+\lambda_a\right)^3+16\left(n+\lambda_b\right)^3 \end{align}
where $\lambda_{a,b}=\frac12\pm\sqrt{\frac16}$ and revealing a little "pattern": the expressions inside the cubes sum to $2n+1. $ Cohen then evaluated the cfracs in terms of $\zeta(3)$ and $\zeta(5)$,
\begin{align} \quad F_1 &= \frac1{3 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-n^{10}}{A(n)}}} = \zeta(5)-\zeta(3)+1/2\\[8pt] \quad F_2 &= \frac1{7 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-n^{10}}{B(n)}}} = \zeta(5)+3\zeta(3)-(4+1/2)\\[8pt] \quad F_3 &= \frac1{13 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-n^{10}}{C(n)}}} = \zeta(5)+\tfrac{11}{4}\zeta(3)-(4+17/64) \end{align}
So not purely $\zeta(5)$, but a pleasant result nonetheless. Likewise for deg-$7$,
$$D(n) = n^7 + (n + 1)^7 + 8n^5 + 8(n + 1)^5 -8(n + \psi_a)^3 -8(n + \psi_b)^3$$
where $\psi_{a,b}=\frac12\pm\sqrt{\frac1{12}}.$ Then,
$$F_4 = \frac1{5 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-n^{14}}{D(n)}}} = \zeta(7)-4\zeta(3)+4\qquad\qquad$$
IV. Questions
For deg-$5$, the recurrences have the general form,
$$(n+1)^5 s_{n+1} = (2n + 1)(an^4+bn^3+cn^2+dn+e)s_n -f\, n^5 s_{n-1}$$
I made a search that wouldn't stress my old computer. So with a limited radius,
$$(a =1 ,\; b < 4,\; c < 100,\; d < 100,\; e < 50,\; f = 1)$$
for positive integers and, excluding the polynomial that appears for all $\zeta(s)$, those three are the only ones in the range.
- So, the obvious question, if $a\neq 1$ (like Apery's) and with a higher search radius, are there more solutions?
- And if there are only finitely many, is the reason similar to why Zagier found only six sporadic sequences?
