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

0,13

Links

Table of n, a(n) for n=0..65.

Formula

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

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

Example

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

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*)

Crossrefs

Cf. A241086, A241091, A241093, A000009.

Sequence in context: A239499 A241139 A294242 * A055656 A078571 A362970

Adjacent sequences: A241089 A241090 A241091 * A241093 A241094 A241095

Keyword

nonn,easy

Author

Clark Kimberling, Apr 18 2014

Status

approved