A240452 Number of partitions p of n such that (sum of parts with multiplicity 1) >= (sum of all other parts).

1, 1, 1, 2, 3, 4, 6, 8, 13, 17, 22, 28, 43, 55, 71, 87, 124, 153, 202, 243, 332, 401, 511, 608, 828, 984, 1236, 1458, 1903, 2245, 2826, 3301, 4245, 4963, 6119, 7108, 9064, 10508, 12837, 14834, 18584, 21442, 26150, 30028, 37139, 42599, 51356, 58742, 72370

Offset

0,4

Links

Table of n, a(n) for n=0..48.

Formula

a(n) + A240449(n) = A000041(n) for n >= 0.

Example

a(6) counts these 6 partitions:  6, 51, 42, 411, 321, 3111.

Mathematica

z = 30; p[n_] := p[n] = IntegerPartitions[n]; f[p_] := f[p] = First[Transpose[p]];
ColumnForm[t = Table[Select[p[n], 2 Total[f[Select[#, Last[#] == 1 &] /. {} -> {{0, 0}}]] &[Tally[#]] < n &], {n, 0, z}]] (* shows the partitions *)
Map[Length, t]  (* A240448 *)
ColumnForm[t = Table[Select[p[n], 2 Total[f[Select[#, Last[#] == 1 &] /. {} -> {{0, 0}}]] &[Tally[#]] <= n &], {n, 0, z}]] (* shows the partitions *)
Map[Length, t]  (* A240449 *)
ColumnForm[t = Table[Select[p[n], 2 Total[f[Select[#, Last[#] == 1 &] /. {} -> {{0, 0}}]] &[Tally[#]] == n &], {n, 0, z}]] (* shows the partitions *)
Map[Length, t]  (* A240447 with alternating 0's *)
ColumnForm[t = Table[Select[p[n], 2 Total[f[Select[#, Last[#] == 1 &] /. {} -> {{0, 0}}]] &[Tally[#]] > n &], {n, 0, z}]] (* shows the partitions *)
Map[Length, t]  (* A240451 *)
ColumnForm[t = Table[Select[p[n], 2 Total[f[Select[#, Last[#] == 1 &] /. {} -> {{0, 0}}]] &[Tally[#]] >= n &], {n, 0, z}]] (* shows the partitions *)
Map[Length, t]  (* A240452 *)
(* Peter J. C. Moses, Apr 02 2014 *)

Crossrefs

Cf. A240448, A240447, A240449, A240451, A000041.

Sequence in context: A294342 A094372 A039880 * A263359 A246905 A000029

Adjacent sequences: A240449 A240450 A240451 * A240453 A240454 A240455

Keyword

nonn,easy

Author

Clark Kimberling, Apr 05 2014

Status

approved