A240305 Number of partitions p of n such that (maximal multiplicity of the parts of p) < (number of distinct parts of p).

1, 0, 0, 1, 1, 2, 3, 5, 7, 12, 14, 21, 29, 38, 50, 70, 90, 117, 156, 196, 253, 324, 411, 514, 650, 809, 1015, 1259, 1555, 1917, 2365, 2898, 3536, 4318, 5248, 6365, 7691, 9297, 11180, 13446, 16115, 19296, 23019, 27474, 32653, 38838, 46002, 54511, 64371, 76012

Offset: 0

Author

Clark Kimberling, Apr 05 2014

Keywords

• nonn
• easy

Examples

			a(6) counts these 3 partitions:  51, 42, 321.
		

Crossrefs

Cf. A240306, A239964, A240308, 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) = A240306(n) - A239964(n) for n >= 0.

a(n) + A239964(n) + A240309(n) = A000041(n) for n >= 0.