\[\frac{{\mathrm{d}}^{m}}{{\mathrm{d}z}^{m}}\left(e^{\frac{1}{4}z^{2}}U\left(a,z%
\right)\right)=(-1)^{m}{\left(\tfrac{1}{2}+a\right)_{m}}e^{\frac{1}{4}z^{2}}U%
\left(a+m,z\right),\]