Recurrence Relation Involving the gamma Function I'm having some doubts about my approach to the following problem. I am given that the function $k(z)$ is defined such that,
$$k(z)=\Gamma\left(\frac{1}{2}+z\right)\Gamma\left(\frac{1}{2}-z\right)\cos{\pi z}$$
I'm required to find the recurrence relation linking $k(z+1)$ and $k(z)$ and to then evaluate $k(z)$ for one specific integer value and thus find $k(z)$ for any real, integer value. My attempt was as follows.
Note that $\Gamma(s+1)=s\Gamma(s)$ and so
\begin{align*}
k(z+1)&=\Gamma\left(\frac{1}{2}+z+1\right)\Gamma\left(\frac{1}{2}-z+1\right)\cos{(\pi z + \pi)} \\
&=\left(\frac{1}{2}+z\right)\Gamma\left(\frac{1}{2}+z\right)\left(\frac{1}{2}-z\right)\Gamma\left(\frac{1}{2}-z\right)\cos{(\pi z + \pi)}
\end{align*}
Then since $\cos{(\pi z + \pi)}=-\cos{\pi z}$ we have that 
$$k(z+1)=\left(z^2-\frac{1}{4}\right)k(z)$$
If we then consider the case $z=0$, we have 
$$k(0)=\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)=\pi$$
From this we see that 
\begin{align*}
&k(1)=-\frac{1}{4}\pi\\
&k(2)=-\frac{3}{16}\pi\\
&k(3)=-\frac{45}{64}\pi\\
&\vdots
\end{align*}
I can't spot any pattern here other than the $4^z$ in the denominator which is making me think i've done something wrong, maybe in my use of $\Gamma(s+1)=s\Gamma(s)$? Any advice would be really appreciated.

You made a sign error.
\begin{align}
\frac{k(n+1)}{k(n)}
&=\frac{\Gamma\left(\frac{1}{2}+n+1\right)\Gamma\left(\frac{1}{2}-n\color{red}{-1}\right)(-\cos(\pi n))}{\Gamma\left(\frac{1}{2}+n\right)\Gamma\left(\frac{1}{2}-n\right)\cos(\pi n)}\\
&=\frac{\left(\frac{1}{2}+n\right)\Gamma\left(\frac{1}{2}+n\right)\frac{\Gamma\left(\frac{1}{2}-n\right)}{\left(\frac{1}{2}-n-1\right)}(-1)}{\Gamma\left(\frac{1}{2}+n\right)\Gamma\left(\frac{1}{2}-n\right)}\\
&=-\frac{\frac{1}{2}+n}{-\frac{1}{2}-n}\\
&=1
\end{align}
Hence $k(n)=k(0)=\pi$.
