This question is a refinement of my related MSE question which was asked over 2 years ago and no answers have yet been posted.

Consider the following formulas for the Dirichlet eta function $\eta(s)$ and the Dirichlet beta function $\beta(s)$ which I believe are globally convergent and are related to formula (3) in this MSE question.

$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^{K+1}}\sum\limits_{n=0}^K\frac{(-1)^n}{(n+1)^s}\sum\limits_{k=0}^{K-n} \binom{K+1}{K-n-k}\right)\tag{1}$$

$$\beta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^{K+1}}\sum\limits_{n=0}^K \frac{(-1)^n}{(2 n+1)^s} \sum\limits_{k=0}^{K-n} \binom{K+1}{K-n-k}\right)\tag{2}$$

Mathematica indicates the formulas above for $\eta(s)$ and $\beta(s)$ can be represented by difference roots at odd and even positive integers respectfully. 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.

For example, these difference root representations lead to the following recursive formulas for $\eta(5)$ and Catalan's constant $\beta(2)$ where formula (3) below is exactly equivalent to formula (1) above for $\eta(s)$ evaluated at $s=5$ and formula (4) below is exactly equivalent to formula (2) above for $\beta(s)$ evaluated at $s=2$.

$$\eta(5)=\underset{K\to\infty}{\text{lim}}\left(2^{-K-1} y_{\eta(5)}(K,K+1)\right)\tag{3a}$$

where $y_{\eta(5)}(K,n)$ is defined recursively as

$$y_{\eta(5)}(K,n)=$$ $$\left\{\begin{array}{cc} 0 & n=0 \\ 2^{K+1}-1 & n=1 \\ \frac{1}{32} \left(K+31\ 2^{K+1}-30\right) & n=2 \\ \frac{(n-2)^5 (K-n+3)\, y_{\eta(5)}(K,n-3)+(K (5 (n-3) n ((n-3) n+5)+31)+n (n ((n-6) n ((n-5) n+30)+300)-262)+94)\, y_{\eta(5)}(K,n-2)+(n-1) \left(-K (n-1)^4+n (n (n ((n-2) n+8)-12)+8)-2\right)\, y_{\eta(5)}(K,n-1)}{(n-1) n^5} & n>2 \\ \end{array}\right.\tag{3b}.$$

$$C=\beta(2)=\underset{K\to\infty}{\text{lim}}\left((-1)^K\, 2^{-K-1}\, y_{\beta(2)}(K,K+1)\right)\tag{4a}$$

where $y_{\beta(2)}(K,n)$ is defined recursively as

$$y_{\beta(2)}(K,n)=$$ $$\left\{\begin{array}{cc} 0 & n=0 \\ \frac{1}{(2 K+1)^2} & n=1 \\ -\frac{K (4 K (K+2)+13)+1}{\left(1-4 K^2\right)^2} & n=2 \\ \frac{(K-n+3) (2 K-2 n+7)^2\, y_{\beta(2)}(K,n-3)+\left(93 n+4 \left(n^3-2 (K+4) n^2+K (K+11) n-K (3 K+17)\right)-97\right)\, y_{\beta(2)}(K,n-2)-\left(-4 n^3+12 (K+2) n^2-4 K (3 K+14) n-61 n+K (4 K (K+8)+77)+59\right)\, y_{\beta(2)}(K,n-1)}{(2 K-2 n+3)^2 (n-1)} & n>2 \\ \end{array}\right.\tag{4b}.$$

I've read Apéry's irrationality proof of $\zeta(3)$ is based on a recursive formula and I was wondering if recursive formulas for $\eta(2 n+1)$ and $\beta(2 n)$ such as the examples above are useful in any way.

Question: Do recursive formulas for $\eta(2 n+1)$ and $\beta(2 n)$ such as formulas (3) and (4) above for $\eta(5)$ and $\beta(2)$ provide any insight into the rationality of $\eta(2 n+1)$ (and hence $\zeta(2 n+1)$) and/or $\beta(2 n)$ for $n\in\mathbb{N}$?

I'll note that I believe formulas (1) and (2) above are exactly equivalent to

$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=0}^K \frac{1}{2^{n+1}} \sum\limits_{k=0}^n \frac{(-1)^k \binom{n}{k}}{(k+1)^s}\right)\tag{5}$$

$$\beta(s)=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=0}^K \frac{1}{2^{n+1}} \sum\limits_{k=0}^n \frac{(-1)^k \binom{n}{k}}{(2 k+1)^s}\right)\tag{6}$$

and can also be evaluated as follows

$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^{K+1}} \sum\limits_{n=0}^K \frac{(-1)^n}{(n+1)^s}\, \binom{K+1}{K-n} \, _2F_1(1,n-K;n+2;-1)\right)\tag{7}$$

$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^{K+1}} \sum\limits_{n=0}^K \frac{(-1)^n}{(n+1)^s}\, P_{K-n}^{(n+1,-K-1)}(3)\right)\tag{8}$$

$$\beta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^{K+1}}\sum\limits_{n=0}^K \frac{(-1)^n}{(2 n+1)^s} \binom{K+1}{K-n}\,_2F_1(1,n-K;n+2;-1)\right)\tag{9}$$

$$\beta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^{K+1}} \sum\limits_{n=0}^K \frac{(-1)^n}{(2 n+1)^s}\, P_{K-n}^{(n+1,-K-1)}(3)\right)\tag{10}$$

where $_2F_1(a,b;c;z)$ is the Gauss hypergeometric function and $P_n^{(a,b)}(x)$ is the Jacobi polynomial.