\[{H^{(1)}_{1/3}}\left(\zeta\right)=e^{-\pi i/3}{H^{(1)}_{-1/3}}\left(\zeta%
\right)=e^{-\pi i/6}\sqrt{3/z}\left(\operatorname{Ai}\left(-z\right)-i%
\operatorname{Bi}\left(-z\right)\right),\]