Functions orthogonal in the Hermitian sense. A new application of basic numbers

To find a particular set of functions Hn(u) satisfying the Hermitian relation Im,n ≡ ∫∞ -∞ e^-1/2x^2 Hm(ix)Hn(-ix)dx = 0 in which the exponential factor is exp (-x2/2) as also in (14) we may put z = e^iax, where a is an arbitrary positive constant and assume that Hn(ix) is a polynomial of the nth degree in z with real coefficients.