Reference for, and/or proof of, $\prod_{n=1}^\infty(\frac{4n+1}{4n-1})^{4n}(\frac{2n^2-2n+1}{2n^2+2n+1})^n=\sqrt2\cosh(\pi/2)e^{-2G/\pi}$ Context:
I have derived some infinite products that I think are not well known. This is the easiest of them:
$$\prod_{n=1}^{\infty}\left(\frac{4n+1}{4n-1} \right)^{4n}\left(\frac{2n^2-2n+1}{2n^2+2n+1} \right)^n=\sqrt{2}\cosh(\pi/2)\,e^{-\frac{2G}{\pi}},\tag{1}$$
where $G$ is Catalan's constant. After some searching I can't find any reference to it.
Take a look at WolframAlpha.
Question 1:
Do you know any reference to this or something like that?

Question 2:
Could you find a solution to $(1)$?
 A: My goal being to obtain the partial product and then use asymptotics
$$P_1=\prod_{n=1}^p (4n+1)^{4n}=\frac{e^{\frac{1}{24}-\frac{C}{\pi }} \,16^{p (p+1)}\, \Gamma
   \left(\frac{1}{4}\right)}{\sqrt{A}\,\, \Gamma \left(p+\frac{5}{4}\right)}\,\exp\left(4 \zeta ^{(1,0)}\left(-1,p+\frac{5}{4}\right) \right)$$
$$P_2=\prod_{n=1}^p (4n-1)^{4n}=\frac{e^{\frac{1}{24}+\frac{C}{\pi }} \,16^{p (p+1)}\, \Gamma
   \left(p+\frac{3}{4}\right)}{\sqrt{A}\,\, \Gamma \left(\frac{3}{4}\right)}\,\exp\left(4 \zeta ^{(1,0)}\left(-1,p+\frac{3}{4}\right) \right)$$
$$\color{blue}{\frac{P_1}{P_2}=\sqrt{\pi }\, e^{-\frac{2 C}{\pi }}\frac{2^{2 p+1}}{\Gamma \left(2 p+\frac{3}{2}\right)}\,\times} $$ $$\color{blue}{\exp\left(4\left(\zeta ^{(1,0)}\left(-1,p+\frac{5}{4}\right)-\zeta
   ^{(1,0)}\left(-1,p+\frac{3}{4}\right)\right)\right)}$$
For large values of $p$
$$4\left(\zeta ^{(1,0)}\left(-1,p+\frac{5}{4}\right)-\zeta
   ^{(1,0)}\left(-1,p+\frac{3}{4}\right)\right)=$$ $$2 p \log (p)+(\log (p)+1)+\frac{3}{16 p}+O\left(\frac{1}{p^2}\right)$$
$$\color{blue}{\frac{P_1}{P_2}\sim\sqrt{\pi }\, e^{1-\frac{2 C}{\pi }}\frac{2^{2 p+1}}{\Gamma \left(2 p+\frac{3}{2}\right)}\,p^{2p+1}}$$
This was the easy part.
Now, writing
$$\frac{2 n^2-2 n+1}{2 n^2+2 n+1}=\frac {(n-a_1)(n-a_2) } {(n+a_1)(n+a_2) }$$ where
$$a_1=\frac {1-i}2 \qquad \text{and}\qquad a_2=\frac {1+i}2$$
$$Q_a=\prod_{n=1}^p (n+a)^n=e^{A_a}$$
$$A_a=a \left(\zeta
   ^{(1,0)}(0,a+1)-\zeta ^{(1,0)}(0,a+p+1)\right)+$$
$$\left(\zeta ^{(1,0)}(-1,a+p+1)-\zeta ^{(1,0)}(-1,a+1)\right)$$ and using the tedious expansion of $A_a$ for large $p$ the expansion of the second product is
$$\color{blue}{\prod_{n=1}^p \left(\frac{2 n^2-2 n+1}{2 n^2+2 n+1}\right)^n = e^{-(2 p+1)} \sqrt{2 (1+\cosh (\pi ))}+O\left(\frac{1}{p^2}\right)}$$
All the above make that, at the limit, the infinite product is
$$\color{red}{\prod_{n=1}^{\infty}\left(\frac{4n+1}{4n-1} \right)^{4n}\left(\frac{2n^2-2n+1}{2n^2+2n+1} \right)^n=e^{-\frac{2 C}{\pi }} \sqrt{1+\cosh (\pi )}}$$ which is your formula.
In terms of asymptotics, it would be
$$\sqrt{2}\, e^{-\frac{2 C}{\pi }} \cosh \left(\frac{\pi }{2}\right) \exp\left( -\frac{3}{8 p}+\frac{18}{97 p^2}-\frac{13}{188 p^3}+O\left(\frac{1}{p^4}\right)\right)$$
For $p=9$, the relative error is less than $0.01$% and for $p=18$ it reduces to $0.001$% .
A: Being:
$$P_{k}(a)=\prod_{n=1}^{k}\left( \frac{1}{e}(\frac{n+a}{n-1+a})^{n} \right),$$
with arbitrary $a$. The partial product is equivalent:
$$P_{k}(a)=\frac{(k+a)^{k}}{e^{k}a(a+1)\cdot \cdot \cdot (k-1+a)}=\frac{(k+a)^{k+1}\Gamma{(a)}}{e^{k}\Gamma{(k+1+a)}}\sim\frac{e^{a}\Gamma{(a)}}{\sqrt{2\pi}}(k+a)^{1/2-a} .$$
Where we have used Stirling's formula in the last step.
Putting $a=1/2$, then we get for first:
$$\prod_{n=1}^{\infty}\left(\frac{1}{e}(\frac{2n+1}{2n-1})^{n}\right)=\sqrt{\frac{e}{2}}.$$
Extending this to the quadratic case $a=(1+it)/2$ with $t \in {\rm I\!R}$ then $\lim |P_{k}(a)|^{2}$ exists and
since:
$$|\Gamma(a)|^2=\frac{\pi}{\cosh{(\pi t/2)}},$$
we have:
$$\prod_{n=1}^{\infty}\left(\frac{1}{e^{2}}(\frac{4n^2+4n+t^2+1}{4n^2-4n+t^2+1})^n \right)=\frac{e}{2\cosh({\pi t/2})}$$
this last product evaluated at $t=1$ combined with prove that: $\sqrt{2}=e^{1-{2K\over \pi}}\prod\limits_{n=1}^{\infty}\left({4n-1\over 4n+1}\right)^{4n}e^2$ gives the result.
