\[\frac{\mathop{\mathrm{i}^{n}\mathrm{erfc}}\left(z\right)}{\mathop{\mathrm{i}^{%
n-1}\mathrm{erfc}}\left(z\right)}=\cfrac{1/2}{z+\cfrac{(n+1)/2}{z+\cfrac{(n+2)%
/2}{z+}}}\cdots,\]