A241091 Number of partitions p of n into distinct parts such that max(p) <= 1 + 2*(number of parts of p).

0, 1, 1, 2, 1, 2, 3, 3, 3, 4, 5, 5, 7, 7, 9, 10, 11, 12, 15, 16, 19, 22, 24, 27, 30, 34, 37, 43, 47, 53, 59, 66, 72, 82, 88, 99, 109, 120, 131, 146, 160, 176, 194, 212, 233, 256, 279, 304, 334, 362, 396, 431, 471, 510, 558, 604, 659, 714, 776, 839, 913, 985

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, A241092, A241093.

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) + A241092(n) for n >= 0.