A241092 Number of partitions p of n into distinct parts such that max(p) = 1 + 2*(number of parts of p).

0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 7, 7, 8, 9, 10, 10, 12, 13, 15, 17, 19, 21, 25, 26, 29, 32, 35, 38, 42, 46, 51, 57, 62, 69, 76, 83, 90, 100, 107, 117, 127, 139, 150, 165, 178, 195, 212, 231, 250, 273, 294, 319, 346, 373, 402

Offset: 0

Author

Clark Kimberling, Apr 18 2014

Keywords

• nonn
• easy

Examples

			a(12) counts these 5 partitions:  741, 732, 651, 642, 6321, 543, 5421.
		

Crossrefs

Cf. A241086, A241091, A241093, A000009.

Programs

Mathematica

z = 30; f[n_] := f[n] = Select[IntegerPartitions[n], Max[Length /@ Split@#] == 1 &];
Table[Count[f[n], p_ /; Max[p] < 1 + 2*Length[p]], {n, 0, z}] (*A241086*)
Table[Count[f[n], p_ /; Max[p] <= 1 + 2*Length[p]], {n, 0, z}](*A241091*)
Table[Count[f[n], p_ /; Max[p] == 1 + 2*Length[p]], {n, 0, z}](*A241092*)
Table[Count[f[n], p_ /; Max[p] >= 1 + 2*Length[p]], {n, 0, z}](*A241089*)
Table[Count[f[n], p_ /; Max[p] > 1 + 2*Length[p]], {n, 0, z}] (*A241093*)

Formula

a(n) + A241086(n) + A241093(n) = A000009(n) for n >= 1.

a(n) = A241091(n) - A241086(n) for n >= 0.