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.

Hint: $\quad\Gamma(1+z)~\Gamma(1-z)=\dfrac{\pi z}{\sin\pi z}\qquad$ and $\qquad\Gamma\bigg(\dfrac12+z\bigg)~\Gamma\bigg(\dfrac12-z\bigg)=\dfrac\pi{\cos\pi z}$
See Euler's reflection formula, Legendre's duplication formula, and Gauss' multiplication theorem.
