0,1

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}.

a(8) = 68 = floor(17*2.1297 + 12*1.2641 + 11*1.5428) = floor(68.3463).

Cf. A086880, A000996, A000997, A000998.

Adjacent sequences: A088019 A088020 A088021 this_sequence A088023 A088024 A088025

Sequence in context: A058713 A025259 A130030 this_sequence A016732 A077948 A077971

nonn

Paul D. Hanna (pauldhanna(AT)juno.com), Sep 19 2003