A240312 Number of partitions p of n such that (maximal multiplicity of the parts of p) = (maximal part of p).

1, 1, 0, 0, 2, 1, 2, 0, 2, 3, 5, 5, 9, 7, 11, 11, 18, 15, 28, 27, 41, 43, 62, 64, 91, 96, 127, 140, 184, 200, 260, 287, 365, 410, 511, 573, 717, 803, 985, 1120, 1359, 1538, 1859, 2106, 2522, 2870, 3407, 3872, 4586, 5207, 6128, 6976, 8167, 9284, 10844, 12321

Offset: 0

Author

Clark Kimberling, Apr 05 2014

Keywords

• nonn
• easy

Examples

			a(6) counts these 2 partitions:  3111, 2211.
		

Crossrefs

Cf. A240310, A240311, A240313, A240314, A000041.

Programs

Mathematica

z = 60; f[n_] := f[n] = IntegerPartitions[n]; m[p_] := Max[Map[Length, Split[p]]]  (* maximal multiplicity *)
Table[Count[f[n], p_ /; m[p] < Max[p]], {n, 0, z}]  (* A240310 *)
Table[Count[f[n], p_ /; m[p] <= Max[p]], {n, 0, z}] (* A240311 *)
Table[Count[f[n], p_ /; m[p] == Max[p]], {n, 0, z}] (* A240312 *)
Table[Count[f[n], p_ /; m[p] >= Max[p]], {n, 0, z}] (* A240313 *)
Table[Count[f[n], p_ /; m[p] > Max[p]], {n, 0, z}]  (* A240314 *)

Formula

a(n) = A240311(n) - A240310(n) for n >= 0.

a(n) + A240310(n) + A240314(n) = A000041(n) for n >= 0.