A240308 Number of partitions p of n such that (maximal multiplicity of the parts of p) >= (number of distinct parts of p).

0, 1, 2, 2, 4, 5, 8, 10, 15, 18, 28, 35, 48, 63, 85, 106, 141, 180, 229, 294, 374, 468, 591, 741, 925, 1149, 1421, 1751, 2163, 2648, 3239, 3944, 4813, 5825, 7062, 8518, 10286, 12340, 14835, 17739, 21223, 25287, 30155, 35787, 42522, 50296, 59556, 70243, 82902

Offset: 0

Author

Clark Kimberling, Apr 05 2014

Keywords

• nonn
• easy

Examples

			a(6) counts these 8 partitions:  6, 411, 33, 3111, 222, 2211, 21111, 111111.
		

Crossrefs

Cf. A240305, A240306, A239964, A240309, A000041.

Programs

Mathematica

z = 60; f[n_] := f[n] = IntegerPartitions[n]; m[p_] := Max[Map[Length, Split[p]]]  (* maximal multiplicity *); d[p_] := d[p] = Length[DeleteDuplicates[p]] (* number of distinct terms *)
t1 = Table[Count[f[n], p_ /; m[p] < d[p]], {n, 0, z}]  (* A240305 *)
t2 = Table[Count[f[n], p_ /; m[p] <= d[p]], {n, 0, z}] (* A240306 *)
t3 = Table[Count[f[n], p_ /; m[p] == d[p]], {n, 0, z}] (* A239964 *)
t4 = Table[Count[f[n], p_ /; m[p] >= d[p]], {n, 0, z}] (* A240308 *)
t5 = Table[Count[f[n], p_ /; m[p] > d[p]], {n, 0, z}]  (* A240309 *)

Formula

a(n) = A239964(n) + A240308(n) for n >= 0.

a(n) + A240305(n) = A000041(n) for n >= 0.