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FORMULA
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B(n) := sum_{k>=0} k^n/(k!)^3 = A000996(n)*B(0) + A000997(n)*B(1) + A000998(n)*B(2) where B(0)=2.129702548983..., B(1)=1.264181150389..., B(2)=1.542838638501...; observe that these shift 3 places left under binomial transform: A000996={1, 0, 0, 1, 1, 1, 2, 6, 17, 44, 112, 304, 918, ...}, A000997={0, 1, 0, 0, 1, 2, 3, 5, 12, 36, 110, 326, 963, ...}, A000998={0, 0, 1, 0, 0, 1, 3, 6, 11, 24, 69, 227, 753, ...}; here A000998 is offset with 5 leading terms: {0, 0, 1, 0, 0}.
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