A241413 Number of partitions p of n such that the number of numbers having multiplicity 1 in p is a part of p.

0, 1, 0, 1, 1, 4, 5, 8, 10, 17, 21, 29, 38, 59, 68, 100, 124, 170, 214, 288, 351, 470, 576, 743, 921, 1176, 1430, 1816, 2214, 2753, 3364, 4176, 5015, 6215, 7478, 9120, 10966, 13351, 15916, 19301, 22982, 27618, 32846, 39354, 46515, 55570, 65598, 77842, 91730

Offset: 0

Author

Clark Kimberling, Apr 23 2014

Keywords

• nonn
• easy

Examples

			a(6) counts these 5 partitions:  42, 411, 321, 3111, 21111; e.g., 411 is counted because 1 part of 411 has multiplicity 1, and 1 is a part of 411.
		

Crossrefs

Cf. A241414, A241415, A241416, A241417, A239737.

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 *)