\[\operatorname{Bi}\left(z\right)=\frac{1}{2^{1/3}\Gamma\left(\tfrac{2}{3}\right%
)}z^{-1/4}M_{0,-1/3}\left(2\zeta\right)+\frac{3}{2^{5/3}\Gamma\left(\tfrac{1}{%
3}\right)}z^{-1/4}M_{0,1/3}\left(2\zeta\right),\]