A241738 Number of partitions p of n such that (number of numbers in p of form 3k+1) = (number of numbers in p of form 3k+2).

1, 0, 0, 2, 1, 2, 7, 5, 7, 17, 14, 18, 39, 32, 42, 76, 71, 88, 157, 143, 182, 293, 292, 357, 562, 558, 692, 1023, 1060, 1286, 1854, 1932, 2347, 3246, 3464, 4153, 5639, 6030, 7207, 9526, 10324, 12240, 15912, 17311, 20444, 26104, 28585, 33567, 42326, 46469

Offset: 0

Author

Clark Kimberling, Apr 28 2014

Keywords

• nonn
• easy

Comments

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

Examples

			a(8) counts these 7 partitions:  5111, 422, 3221, 3211, 22211, 221111, 2111111.
		

Crossrefs

Cf. A241737, A241739, A241740, A241743.

Programs

Mathematica

z = 40; f[n_] := f[n] = IntegerPartitions[n]; s[k_, p_] := Count[Mod[DeleteDuplicates[p], 3], k];
Table[Count[f[n], p_ /; s[1, p] < s[2, p]], {n, 0, z}]  (* A241737 *)
Table[Count[f[n], p_ /; s[1, p] == s[2, p]], {n, 0, z}] (* A241738 *)
Table[Count[f[n], p_ /; s[1, p] > s[2, p]], {n, 0, z}]  (* A241739 *)

Formula

a(n) + A241737(n) + A241839(n) = A000041(n) for n >= 0.