A241655 Number of partitions p of n such that 2*(number of even numbers in p) = (number of odd numbers in p).

0, 0, 1, 1, 3, 4, 6, 9, 12, 17, 21, 31, 37, 54, 66, 92, 114, 159, 198, 268, 335, 448, 563, 736, 921, 1190, 1485, 1892, 2340, 2953, 3636, 4534, 5542, 6861, 8333, 10226, 12347, 15052, 18079, 21907, 26168, 31537, 37526, 44987, 53307, 63653, 75156, 89369, 105204

Offset

0,5

Comments

Each number in p is counted once, regardless of its multiplicity.

Links

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

Formula

a(n) = A241654(n) - A241653(n) for n >= 0.

a(n) + A241651(n) + A241653(n) = A000041(n) for n >= 0.

Example

a(6) counts these 6 partitions:  6, 42, 411, 222, 2211, 21111.

Mathematica

z = 30; f[n_] := f[n] = IntegerPartitions[n]; s0[p_] := Count[Mod[DeleteDuplicates[p], 2], 0];
s1[p_] := Count[Mod[DeleteDuplicates[p], 2], 1];
Table[Count[f[n], p_ /; 2 s0[p] < s1[p]], {n, 0, z}]  (* A241651 *)
Table[Count[f[n], p_ /; 2 s0[p] <= s1[p]], {n, 0, z}] (* A241652 *)
Table[Count[f[n], p_ /; 2 s0[p] == s1[p]], {n, 0, z}] (* A241653 *)
Table[Count[f[n], p_ /; 2 s0[p] >= s1[p]], {n, 0, z}] (* A241654 *)
Table[Count[f[n], p_ /; 2 s0[p] > s1[p]], {n, 0, z}]  (* A241655 *)

Crossrefs

Cf. A241651, A241652, A241653, A241654.

Sequence in context: A097922 A103109 A241639 * A288346 A373416 A078743

Adjacent sequences: A241652 A241653 A241654 * A241656 A241657 A241658

Keyword

nonn,easy

Author

Clark Kimberling, Apr 27 2014

Status

approved