A239737 Number of partitions p of n such that the number of numbers having multiplicity 1 in p is a part or the number of numbers having multiplicity > 1 is a part.

0, 1, 1, 2, 2, 5, 7, 10, 13, 21, 28, 38, 54, 77, 99, 137, 180, 236, 306, 398, 504, 644, 807, 1018, 1278, 1599, 1972, 2458, 3039, 3743, 4592, 5659, 6884, 8436, 10235, 12445, 15021, 18204, 21842, 26334, 31501, 37746, 44956, 53707, 63657, 75738, 89536, 106057

Offset: 0

Author

Clark Kimberling, Apr 23 2014

Keywords

• nonn
• easy

Examples

			a(6) counts these 7 partitions:  42, 411, 321, 3111, 2211, 21111, 111111.
		

Crossrefs

Cf. A241413, A241414, A241415, A241416, A241417, 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) + A241417(n) = A000041(n) for n >= 0.