A241638 Number of partitions p of n such that (number of even numbers in p) = (number of odd numbers in p).

1, 0, 0, 1, 1, 4, 3, 8, 6, 13, 11, 20, 17, 31, 34, 47, 56, 78, 103, 125, 167, 203, 281, 315, 433, 487, 673, 745, 989, 1101, 1472, 1623, 2116, 2386, 3052, 3430, 4347, 4948, 6168, 7104, 8673, 10068, 12210, 14234, 17047, 20007, 23671, 27869, 32739, 38609, 45010

Offset: 0

Author

Clark Kimberling, Apr 27 2014

Keywords

• nonn
• easy

Comments

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

Examples

			a(6) counts these 3 partitions:  411, 2211, 21111.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A241636, A241637, A241639, A241640.

Programs

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_ /; s0[p] < s1[p]], {n, 0, z}]  (* A241636 *)
Table[Count[f[n], p_ /; s0[p] <= s1[p]], {n, 0, z}] (* A241637 *)
Table[Count[f[n], p_ /; s0[p] == s1[p]], {n, 0, z}] (* A241638 *)
Table[Count[f[n], p_ /; s0[p] >= s1[p]], {n, 0, z}] (* A241639 *)
Table[Count[f[n], p_ /; s0[p] > s1[p]], {n, 0, z}]  (* A241640 *)

Formula

a(n) = A241637(n) - A241636(n) = A241639(n) - A241640(n) for n >= 0.

a(n) + A241636(n) + A241640(n) = A000041(n) for n >= 0.

a(n) = A242618(n,0). - Alois P. Heinz, May 19 2014