\[{H^{(2)}_{1/3}}\left(\zeta\right)=e^{\pi i/3}{H^{(2)}_{-1/3}}\left(\zeta\right%
)=e^{\pi i/6}\sqrt{3/z}\left(\operatorname{Ai}\left(-z\right)+i\operatorname{%
Bi}\left(-z\right)\right),\]