Background
Consider the Legendre duplication formula:
$$ \begin{aligned} \prod_{i=1}^{k-1}Γ\left(n+\frac{i}{k}\right)=(2π)^{\frac{k-1}{2}}k^{\frac{1}{2}-nk}Γ(nk) \end{aligned} $$
I chose different k values to calculate the Γ product table, but I discovered something strange in the calculation.
When written in factorial form, it seems to have some connection with Ramanujan's reciprocal formula of π
$$ \begin{aligned} Γ\left(n+\frac{1}{2}\right)Γ\left(n+\frac{1}{2}\right)Γ\left(n+\frac{1}{2}\right)=8π^{3/2}\frac{\color{red}{(2 n)!^3}}{2^{6n}×(2n)^3} \end{aligned} $$
$$ \begin{aligned} \frac{1}{π}=\frac{1}{2}\sum_{k=0}^∞(-1)^k\left(\frac{\color{red}{(2 k)!}}{(k!)^2}\right)^{\color{red}{3}}\frac{(4 k+1)}{2^{6k}} \end{aligned} $$
$$ \begin{aligned} Γ \left(n+\frac{1}{2}\right) Γ \left(n+\frac{1}{3}\right) Γ \left(n+\frac{2}{3}\right)=\frac{2\color{blue}{\sqrt{3}}π^{3/2}}{3}\frac{\color{red}{(2 n)!(3 n)!}}{n^2×2^{2n}×3^{3n}} \end{aligned} $$
$$ \begin{aligned} \frac{1}{π}=\frac{\color{blue}{\sqrt{3}}}{36}\sum_{k=0}^∞\frac{\color{red}{(2k)!(3k)!}}{(k!)^5}\frac{(51 k+7)}{(-2^6×3^3)^{k}} \end{aligned} $$
$$ \begin{aligned} Γ \left(n+\frac{1}{2}\right) Γ \left(n+\frac{1}{4}\right) Γ \left(n+\frac{3}{4}\right)=\color{blue}{\sqrt{2}}π^{3/2}\frac{\color{red}{(4 n)!}}{n× 4^{4n}} \end{aligned} $$
$$ \begin{aligned} \frac{1}{π}=\frac{2\color{blue}{\sqrt{2}}}{9801}\sum_{k=0}^∞\frac{\color{red}{(4k)!}}{(k!)^4}\frac{(58⋅455 k+1103)}{396^{4k}} \end{aligned} $$
$$ \begin{aligned} Γ\left(n+\frac{1}{2}\right) Γ\left(n+\frac{1}{6}\right) Γ\left(n+\frac{5}{6}\right)=2 π ^{3/2} \frac{\color{red}{(6 n)!}}{1728^n (3 n)!} \end{aligned} $$
$$ \begin{aligned} \frac{1}{π}=\frac{\sqrt{10005}}{66733350}\sum_{k=0}^∞\frac{\color{red}{(6 k)!}}{(3k)!}\frac{(163⋅3344418k+13591409)}{(-2)^{15k+6} (20010^k⋅k!)^3} \end{aligned} $$
Question
Is there any connection between the two, or is it just a coincidence?
If it is not a coincidence, does the following equation predict the existence of Ramanujan's equation for $π^2$?
$$ \begin{aligned} Γ\left(n+\frac{1}{2}\right) Γ\left(n+\frac{1}{5}\right) Γ\left(n+\frac{2}{5}\right) Γ\left(n+\frac{3}{5}\right) Γ\left(n+\frac{4}{5}\right) &=π^{5/2} 2^{3-2 n} 5^{\frac{1}{2}-5 n} Γ(2 n) Γ(5 n)\\ Γ\left(n+\frac{1}{2}\right) Γ\left(n+\frac{1}{8}\right) Γ\left(n+\frac{3}{8}\right) Γ\left(n+\frac{5}{8}\right) Γ\left(n+\frac{7}{8}\right) &=π^{5/2} 2^{\frac{7}{2}-18 n} \frac{Γ(2 n) Γ(8 n)}{Γ(4 n)}\\ Γ\left(n+\frac{1}{2}\right) Γ\left(n+\frac{1}{12}\right) Γ\left(n+\frac{5}{12}\right) Γ\left(n+\frac{7}{12}\right) Γ\left(n+\frac{11}{12}\right) &=π^{5/2} 2^{3-14 n} 3^{-6 n} \frac{Γ(2 n)^2 Γ(12 n)}{Γ(4 n) Γ(6 n)} \end{aligned} $$
