For a fixed integer $n>2$, consider the values $$\Gamma\left(\frac1n\right),\Gamma\left(\frac2n\right),\ldots,\Gamma\left(\frac{n-1}{n}\right)$$ of the gamma function. Some of these are dependent in the following sense. The relation $$\Gamma\left(\frac{k}{n}\right)\Gamma\left(\frac{n-k}{n}\right)=\frac{\pi}{\sin(k\pi/n)}$$ is given by the reflection formula for $\Gamma$. This can be viewed as a nontrivial vanishing linear combination of $\log\Gamma(k/n)$, $\log\Gamma(1-k/n)$ and "something elementary" with integer coefficients. (This explains the "$n/2$" in the title.)
A less trivial example is $$\frac{\Gamma(1/14)\Gamma(9/14)\Gamma(11/14)}{\Gamma(3/14)\Gamma(5/14)\Gamma(13/14)}=2$$ appearing in my comment to this answer, effectively completing it. This relation follows from the duplication formula for $\Gamma$ and the basic relation $\Gamma(a+1)=a\Gamma(a)$, giving a relation between $\Gamma(1/14)$, $\Gamma(3/14)$ and $\Gamma(5/14)$ (via the reflection formula).
Now consider all relations on $\{\Gamma(r):r\in\mathbb{Q}_+\}$ generated by $\Gamma(r+1)=r\Gamma(r)$, the reflection formula, and the multiplication theorem for $\Gamma$ (not just the duplication formula). What is the number $a_n$ of independent values among $\Gamma(k/n)$, $0<k<n[/2]$ w.r.t. these relations?
So far it seems $a_3=1$, $a_4=1$, $a_5=2$, $a_6=2$; the above example could help for $a_{14}={?}$.
The problem can be stated purely in terms of rational numbers; the presence of $\Gamma$ is just a motivation. (Not sure if it's asked here already, or if $\{a_n\}$ is already in the OEIS.)