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Many people have tried and failed to extend Apery's Irrationality proof of $\zeta(3)$ to Catalan's constant, by looking for a fast converging series for Catalan's constant analogous to the one for $\zeta(3)$ that Apery utilized:

$${\displaystyle {\begin{aligned}\zeta (3)&={\frac {5}{2}}\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{{\binom {2k}{k}}k^{3}}}\end{aligned}}}.$$

see this question: Why can we not establish the irrationality of Catalan's constant the same way as $\zeta(3)$?

which unfortunately didn't receive an answer.

My question is different and specifically relates to Beukers-like irrationality proof for $\zeta(2)$/$\zeta(3)$ as most clearly articulated recently by F. M. S. Lima in Beukers-like irrationality proofs for $\zeta(2)$ and $\zeta(3)$

For analogous Beukers-like irrationality proof applied in the case Catalan's Constant, does a proof fail in the initial lemma's or in evaluating and applying the unit square integral?

Lets rewrite the Lemma's applying to $\zeta(2)$ given by F. M. S. Lima and apply them to an analogous unit square integral for Catalan's Constant, $G$, that is "Lemma N" in the paper becomes "Lemma N_G" here:

Lemma 1G A unit square integral for Catalan's Constant $$\int_0^1 \int_0^1 \frac{1}{1+(x y)^2} \,dx\,dy= G$$

Lemma 2G ($I_{2r,2r}$) For all odd integers $r>0$ $$\int_0^1 \int_0^1 \frac{x^{2r}y^{2r}}{1+(x y)^2} \,dx\,dy= G-\sum_{m=1}^{2r}\frac{(-1)^{m-1}}{(2m-1)^2}$$

Lemma 3G ($I_{2r,2s}$) Let r and s be non-negative odd integers, $r\ne s$. Then $$\int_0^1 \int_0^1 \frac{x^{2r}y^{2s}}{1+(x y)^2} \,dx\,dy=\frac{\widetilde{h_s}-\widetilde{h_r}}{2(r-s)}$$ where $\widetilde{h_n}=\sum_{m=1}^n \frac{(-1)^{m-1}}{(2m-1)}$, an alternating analog of the Harmonic Number.

Lemma 4G ($I_{2r,2r}$ as a linear form). For all odd integers $r>0$ $$I_{2r,2r}=G-\frac{z_{2r}}{(d_{2r})^2} $$ for some $z_{2r} \in {\mathbb{N}}^*$. Where $d_{r}=lcm(1^2,3^2,5^2,...,r^2)$

Lemma 5G ($I_{2r,2s}$ is a positive rational). For all odd $r,s \in {\mathbb{N}},\, r \ne s,$ $$I_{2r,2s}=\frac{z_{2r,2s}}{(d_{2r})^2} $$ for some $z_{2r,2s}\in {\mathbb{N}^*}$

Lemma 6G and Lemma 7G These are written in terms of $n$ and therefore both Lemma's can be written in terms of $2n$

If the analogous lemma's can be all proved correct for odd integer $r,s>0$ then presumably a proof in regards to $G$ must fail in the main part of the proof, what Lima labels Theorem 1, i.e. in the process of evaluating the new unit square integral and applying the result; which for me is the hardest part of this proof.

Added One difficulty is defining the even powered polynomials to multiply together - the resultant terms will have more than one factor of 2 so the lemmas above need to be modified.

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  • $\begingroup$ $(+1)$. Elegant question. If $d_{r}=lcm(1^2,3^2,5^2,...,r^2)$ then what will be $d_{2r}$? Here we are taking lcm of squares of odd numbers then how the last term can be even? Also what will be the asymptotic equality of $d_{2r}$ as $r\to\infty$? $\endgroup$
    – Max
    Aug 10, 2023 at 18:41

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Thank you for citing my arXiv paper on the irrationality of zeta(2) and zeta(3). I've also tried to develop a similar proof for Catalan's constant, but the last steps need some magic modifications, so maybe this is not the proper way to deal with the irrationality of this constant. A more promising way is the search for rapidly-converging Apery-like series, as those in my other arXiv paper https://arxiv.org/abs/1207.3139

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  • $\begingroup$ $(+1)$. Nice answer! If $d_{r}=lcm(1^2,3^2,5^2,...,r^2)$ then what will be $d_{2r}$? Here we are taking lcm of squares of odd numbers then how the last term can be even? Also what will be the asymptotic equality of $d_{2r}$ as $r\to\infty$? $\endgroup$
    – Max
    Aug 10, 2023 at 18:40

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