A241093 Number of partitions p of n into distinct parts such that max(p) > 1 + 2*(number of parts of p).

0, 0, 0, 0, 1, 1, 1, 2, 3, 4, 5, 7, 8, 11, 13, 17, 21, 26, 31, 38, 45, 54, 65, 77, 92, 108, 128, 149, 175, 203, 237, 274, 318, 366, 424, 486, 559, 640, 733, 836, 953, 1084, 1232, 1398, 1583, 1792, 2025, 2286, 2576, 2902, 3262, 3666, 4111, 4610, 5160, 5774

Offset

0,8

Links

Table of n, a(n) for n=0..55.

Formula

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

Example

a(12) counts these 8 partitions: {12}, {11,1}, {10,2}, {9,3}, {9,2,1}, {8,4}, {8,3,1}, {7,5}.

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, A241092, A000009.

Sequence in context: A354816 A199120 A118083 * A116470 A115649 A191168

Adjacent sequences: A241090 A241091 A241092 * A241094 A241095 A241096

Keyword

nonn,easy

Author

Clark Kimberling, Apr 18 2014

Status

approved