A240448 Number of partitions p of n such that (sum of parts with multiplicity 1) < (sum of all other parts).

0, 0, 1, 1, 2, 3, 5, 7, 9, 13, 20, 28, 34, 46, 64, 89, 107, 144, 183, 247, 295, 391, 491, 647, 747, 974, 1200, 1552, 1815, 2320, 2778, 3541, 4104, 5180, 6191, 7775, 8913, 11129, 13178, 16351, 18754, 23141, 27024, 33233, 38036, 46535, 54202, 66012, 74903

Offset

0,5

Links

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

Formula

a(n) + A240452(n) = A000041(n) for n >= 0.

Example

a(6) counts these 5 partitions:  33, 222, 2211, 21111, 111111.

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. A240449, A240447, A240451, A240452, A000041.

Sequence in context: A035563 A240063 A216374 * A218614 A338360 A028378

Adjacent sequences: A240445 A240446 A240447 * A240449 A240450 A240451

Keyword

nonn,easy

Author

Clark Kimberling, Apr 05 2014

Status

approved