Prove that
$$\prod_{k=1}^{\infty} \prod_{n=1}^{\infty} \prod_{m=1}^{\infty}(k+n+m)^{\frac{(-1)^{k+m+n}}{k+m+n}}=\frac{\mathbb{A}^{\frac{3}{2}}}{\pi^{\frac{3}{4}} e^{\frac{1}{8}-(\frac{7}{12}+\gamma) \log 2+\frac{1}{2} \log ^{2} 2}}.$$
Here $\mathbb{A}$ is the Glaisher-Kinkelin constant and $\gamma$ is the Euler-Mascheroni constant. I found this problem on this site at number $256$, there was a piece of advice just below:
Currently I do not have a solution on this but the most straight forward idea is to actually try to find the number of ways $n$ can be written as a sum of three numbers and reduce the triple product into a single one.
The number of ways to represent $n$ as a sum of three natural terms is equal to $\frac{(n-1)(n-2)}{2}$, and then, obviously, take the logarithm. However, I get an unusually big logarithm term coefficient. May I be missing something?
For $|x|<1$, $$\sum_{k=1}^\infty\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{(-x)^{k+n+m}}{k+n+m}\log(k+n+m) = \sum_{n=1}^\infty\frac{(-x)^n(n^2-3n+2)}{2n}\log n\\ = -\frac{f_{-1}(-x)-3f_0(-x)+2f_1(-x)}{2}$$ Where $f_s(x)=\frac{\partial }{\partial s}\text{Li}_s(x)$, and $\text{Li}_s(x)$ is the polylogarithm. Taking the limit as $x\rightarrow 1^{-}$, we get $$\sum_{k=1}^\infty\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{(-1)^{k+n+m}}{k+n+m}\log(k+n+m)=\frac{\eta'(-1)-3\eta'(0)+2\eta'(1)}{2},$$ where $\eta(s)$ is the Dirichlet Eta function, and hence $$\prod_{k=1}^{\infty} \prod_{n=1}^{\infty} \prod_{m=1}^{\infty}(k+n+m)^{\frac{(-1)^{k+m+n}}{k+m+n}}=\exp\left(\frac{\eta'(-1)-3\eta'(0)+2\eta'(1)}{2}\right)=\frac{\mathbb{A}^{\frac{3}{2}}}{\pi^{\frac{3}{4}} e^{\frac{1}{8}-(\frac{7}{12}+\gamma) \log 2+\frac{1}{2} \log ^{2} 2}}$$