\[a_{k}(\nu)=\frac{(4\nu^{2}-1^{2})(4\nu^{2}-3^{2})\cdots(4\nu^{2}-(2k-1)^{2})}{%
k!8^{k}}=\frac{{\left(\frac{1}{2}-\nu\right)_{k}}{\left(\frac{1}{2}+\nu\right)%
_{k}}}{(-2)^{k}k!},\]