Up and down and up again!

$\begin{aligned} \prod_{n=1}^{\infty}{\large\left(4+(-1)^n\right)}^{^{\displaystyle\frac{1}{n^2}}}&=\prod_{n=1}^{\infty}{\huge 5}^{^{\displaystyle\frac{1}{(2n)^2}}}\cdot\prod_{n=1}^{\infty}{\huge 3}^{^{\displaystyle\frac{1}{(2n-1)^2}}}\\ &={\huge 5}^{^{\displaystyle\sum_{n=1}^{\infty}\frac{1}{(2n)^2}}}\cdot{\huge 3}^{^{\displaystyle\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}}}\\ &={\huge 5}^{^{\displaystyle\sum_{n=1}^{\infty}\frac{1}{(2n)^2}}}\cdot{\huge 3}^{^{\displaystyle\frac{\pi^2}{6}-\sum_{n=1}^{\infty}\frac{1}{(2n)^2}}}\\ &={\huge 5}^{^{\displaystyle\dfrac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}}}\cdot{\huge 3}^{^{\displaystyle\frac{\pi^2}{6}-\dfrac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}}}\\ &={\huge 5}^{^{\dfrac{\pi^2}{24}}}\cdot{\huge 3}^{^{\dfrac{\pi^2}{8}}}\\ &={\huge 135}^{^{\dfrac{\pi^2}{24}}} \end{aligned}$ So, $\large A+24=135+24=\boxed{159}$