A241415 Number of partitions p of n such that the number of numbers having multiplicity 1 in p is not a part and the number of numbers having multiplicity > 1 is a part.

0, 0, 1, 1, 1, 1, 2, 2, 3, 4, 7, 9, 16, 18, 31, 37, 56, 66, 92, 110, 153, 174, 231, 275, 357, 423, 542, 642, 825, 990, 1228, 1483, 1869, 2221, 2757, 3325, 4055, 4853, 5926, 7033, 8519, 10128, 12110, 14353, 17142, 20168, 23938, 28215, 33243, 39019, 45968

Offset: 0

Author

Clark Kimberling, Apr 23 2014

Keywords

• nonn
• easy

Examples

			a(6) counts these 2 partitions:  2211, 111111.
		

Crossrefs

Cf. A241413, A241414, A241416, A241417, A239737, A000041.

Programs

Mathematica

z = 30; f[n_] := f[n] = IntegerPartitions[n]; u[p_] := Length[DeleteDuplicates[Select[p, Count[p, #] ==  &]]]; e[q_] := Length[DeleteDuplicates[Select[q, Count[q, #] > 1 &]]]
Table[Count[f[n], p_ /; MemberQ[p, u[p]]], {n, 0, z}]  (* A241413 *)
Table[Count[f[n], p_ /; MemberQ[p, u[p]] && MemberQ[p, e[p]]], {n, 0, z}]  (* A241414 *)
Table[Count[f[n], p_ /; ! MemberQ[p, u[p]] && MemberQ[p, e[p]] ], {n, 0, z}] (* A241415 *)
Table[Count[f[n], p_ /; MemberQ[p, u[p]] && ! MemberQ[p, e[p]] ], {n, 0, z}] (* A241416 *)
Table[Count[f[n], p_ /; ! MemberQ[p, u[p]] && ! MemberQ[p, e[p]] ], {n, 0, z}] (* A241417 *)
Table[Count[f[n], p_ /; MemberQ[p, u[p]] || MemberQ[p, e[p]] ], {n, 0, z}] (* A239737 *)

Formula

a(n) + A241415(n) + A241416(n) = A239737(n) for n >= 0.