WhittakerM[\[Nu], \[Mu], z] == (z^(\[Mu] + 1/2) HypergeometricPFQ[{\[Mu] - \[Nu] + 1/2, Subscript[a, 2]}, {2 \[Mu] + 1, Subscript[a, 2]}, z])/E^(z/2)

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2007-05-02