I've noticed the Dirichlet eta function $\eta(s)$ and the Dirichlet beta function $\beta(s)$ can be represented by difference roots at odd and even positive integers respectfully. For example, consider the formulas below for a few positive integers.
$$\eta(3)=\underset{K\to\infty}{\text{lim}}\left(2^{-K-1} \text{DifferenceRoot}\left[\{y,n\},\left\{\left(-3 K n^2-9 K n-7 K-n^4-5 n^3-15 n^2-25 n-16\right) y(n+1)+(n+2) \left(K n^2+4 K n+4 K-n^3-7 n^2-19 n-19\right) y(n+2)+\left(-(n+1)^3\right) (K-n) y(n)+(n+2) (n+3)^3 y(n+3)=0,y(0)=0,y(1)=2^{K+1}-1,y(2)=\frac{1}{8} \left(K-2^{K+1}+2\right)+2^{K+1}-1\right\}\right][K+1]\right)\tag{1}$$
$$\eta(5)=\underset{K\to\infty}{\text{lim}}\left(2^{-K-1} \text{DifferenceRoot}\left[\{y,n\},\left\{\left(-5 K n^4-30 K n^3-70 K n^2-75 K n-31 K-n^6-7 n^5-30 n^4-90 n^3-165 n^2-161 n-64\right) y(n+1)+(n+2) \left(K n^4+8 K n^3+24 K n^2+32 K n+16 K-n^5-13 n^4-74 n^3-222 n^2-341 n-211\right) y(n+2)+\left(-(n+1)^5\right) (K-n) y(n)+(n+2) (n+3)^5 y(n+3)=0,y(0)=0,y(1)=2^{K+1}-1,y(2)=\frac{1}{32} \left(K-2^{K+1}+2\right)+2^{K+1}-1\right\}\right][K+1]\right)\tag{2}$$
$$\beta(2)=\underset{K\to\infty}{\text{lim}}\left((-1)^K 2^{-K-1} \text{DifferenceRoot}\left[\{y,n\},\left\{\left(4 (2 K-1) n^2-4 (K-1) K n+8 K-4 n^3-9 n-2\right) y(n+1)+\left(12 (K-1) n^2+4 K (4-3 K) n+K (4 (K-1) K+17)-4 n^3-25 n-16\right) y(n+2)+(n+2) (2 K-2 n-3)^2 y(n+3)-(K-n) (2 K-2 n+1)^2 y(n)=0,y(0)=0,y(1)=\frac{1}{(2 K+1)^2},y(2)=-\frac{K (4 K (K+2)+13)+1}{\left(1-4 K^2\right)^2}\right\}\right][K+1]\right)\tag{3}$$
$$\beta(4)=\underset{K\to\infty}{\text{lim}}\left((-1)^K 2^{-K-1} \text{DifferenceRoot}\left[\{y,n\},\left\{\left(8 \left(2 K \left(4 K^2+6 K+15\right)-5\right) n^2+64 K n^4-8 (4 K (3 K+1)+11) n^3-8 K (K (2 K (K+6)+27)-9) n+16 K (2 K (K (K+2)-1)+1)-16 n^5-17 n-2\right) y(n+1)+\left(16 (5 K-6) n^4-8 (20 (K-2) K+43) n^3+8 (K (4 K (5 K-12)+105)-75) n^2+8 K (K (2 (12-5 K) K-81)+124) n+K (8 K (K (2 (K-2) K+19)-49)+417)-16 n^5-497 n-160\right) y(n+2)+(n+2) (2 K-2 n-3)^4 y(n+3)-(K-n) (2 K-2 n+1)^4 y(n)=0,y(0)=0,y(1)=\frac{1}{(2 K+1)^4},y(2)=\frac{1}{(2 K+1)^4}-\frac{K+2}{(1-2 K)^4}\right\}\right][K+1]\right)\tag{4}$$
Question: Do the difference root representations provide any insight into the rationality of $\eta(2 n+1)$ (and hence $\zeta(2 n+1)$) and/or $\beta(2 n)$ (e.g. Catalan's constant $G=\beta(2)$) for $n\in\mathbb{N}$?
Mathematica explains difference root mathematical sequence (also known as holonomic sequence and P-recursive sequence) as follows:
The holonomic sequence $h(k)$ defined by a DifferenceRoot function satisfies a holonomic difference equation $p_n(k)\ h(k+n)+p_{n-1}(k)\ h(k+n-1)+...+p_0(k)\ h(k)=0$ with polynomial coefficients $p_i(k)$ and initial values $h(n-1)=h_{n-1},...,h'(1)=h_1,h(0)=h_0$.
Also see Holonomic function and P-recursive equation.
Here's the information contained in formula (1) above in slightly different form. Note DifferenceRoot is followed by the holonomic difference equation and then three initial values $y(0)$, $y(1)$, and $y(2)$.
$\left( \begin{array}{c} \text{DifferenceRoot} \\ \left( \begin{array}{c} \left(-3 K n^2-9 K n-7 K-n^4-5 n^3-15 n^2-25 n-16\right) y(n+1)+(n+2) \left(K n^2+4 K n+4 K-n^3-7 n^2-19 n-19\right) y(n+2)+\left(-(n+1)^3\right) (K-n) y(n)+(n+2) (n+3)^3 y(n+3)=0 \\ y(0)=0 \\ y(1)=2^{K+1}-1 \\ y(2)=\frac{1}{8} \left(K-2^{K+1}+2\right)+2^{K+1}-1 \\ \end{array} \right) \\ \end{array} \right)$
Formulas (1) to (4) above are based on the following formulas for $\eta(s)$ and $\beta(s)$ which I believe are globally convergent and are related to formula (3) in this question. Formulas (1) and (2) above are equivalent to formula (5a)/(5b) below evaluated at $s=3$ and $s=5$ respectively. Formulas (3) and (4) above are equivalent to formulas (6a)/(6b) below evaluated at $s=2$ and $s=4$ respectively.
$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^{K+1}}\sum\limits_{n=0}^K\frac{(-1)^n}{(n+1)^s}\sum\limits_{k=0}^{K-n} \binom{K+1}{K-n-k}\right)\tag{5a}$$
$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^{K+1}}\sum\limits_{n=0}^K\frac{(-1)^n}{(n+1)^s} \binom{K+1}{K-n}\,_2F_1(1,n-K;n+2;-1)\right)\tag{5b}$$
$$\beta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^{K+1}}\sum\limits_{n=0}^K \frac{(-1)^n}{(2 n+1)^s} \sum\limits_{k=0}^{K-n} \binom{K+1}{K-n-k}\right)\tag{6a}$$
$$\beta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^{K+1}}\sum\limits_{n=0}^K \frac{(-1)^n}{(2 n+1)^s} \binom{K+1}{K-n}\,_2F_1(1,n-K;n+2;-1)\right)\tag{6b}$$
In an attempt to provide further clarification, here's a Mathematica implementation of Catalan's constant based on (3) above
$C=\beta(2)=\underset{K\to\infty}{\text{lim}}\left(\text{Block}\left[\left\{\text{y0}=0,\text{y1}=\frac{1}{(2 K+1)^2},\text{y2}=-\frac{K (4 K (K+2)+13)+1}{\left(1-4 K^2\right)^2},\text{y3},n=0\right\},\text{While}\left[n+3<K+2, \text{y3}=-\frac{\text{y0} (K-n) \left(-(2 K-2 n+1)^2\right)+\text{y1} \left(4 (2 K-1) n^2-4 (K-1) K n+8 K-4 n^3-9 n-2\right)+\text{y2} \left(12 (K-1) n^2+4 K (4-3 K) n+K (4 (K-1) K+17)-4 n^3-25 n-16\right)}{(n+2) (2 K-2 n-3)^2}; \text{y0}=\text{y1};\text{y1}=\text{y2};\text{y2}=\text{y3};n\text{++}\right];(-1)^K 2^{-K-1} \left(\left\{\begin{array}{cc} \text{y1} & K=0 \\ \text{y2} & K=1 \\ \text{y3} & K>1 \\ \end{array}\right.\right)\right]\right)\tag{7}$
where I believe the implementation in (7) above is exactly equivalent to formulas (6a) and (6b) above evaluated at $s=2$ and when (6a), (6b), and (7) above are all three evaluated at the same non-negative integer value of $K$.
I believe the Mathematica implementation defined in (7) above is exactly equivalent to
$$C=\beta(2)=\underset{K\to\infty}{\text{lim}}\left((-1)^K\, 2^{-K-1}\, y(K,K+1)\right)\tag{8}$$
where $y(K,n)$ is defined recursively as
$$y(K,n)=$$ $$\left\{\begin{array}{cc} 0 & n=0 \\ \frac{1}{(2 K+1)^2} & n=1 \\ -\frac{K (4 K (K+2)+13)+1}{\left(1-4 K^2\right)^2} & n=2 \\ \frac{(K-n+3) y(K,n-3) (2 K-2 n+7)^2+\left(93 n+4 \left(n^3-2 (K+4) n^2+K (K+11) n-K (3 K+17)\right)-97\right) y(K,n-2)-\left(-4 n^3+12 (K+2) n^2-4 K (3 K+14) n-61 n+K (4 K (K+8)+77)+59\right) y(K,n-1)}{(2 K-2 n+3)^2 (n-1)} & n>2 \\ \end{array}\right.\tag{9}.$$