A241741 Number of partitions p of n such that (number of numbers in p of form 3k+2) = (number of numbers in p of form 3k).

1, 1, 1, 1, 2, 3, 3, 5, 9, 11, 14, 22, 29, 36, 51, 66, 83, 107, 139, 170, 216, 273, 340, 415, 520, 635, 778, 952, 1177, 1414, 1724, 2094, 2527, 3038, 3691, 4411, 5286, 6345, 7586, 9008, 10778, 12796, 15163, 17979, 21288, 25059, 29608, 34861, 40927, 48035

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 9 partitions:  71, 62, 53, 44, 41111, 332, 3221, 32111, 11111111.
		

Crossrefs

Cf. A241737, A241740, A241742, 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[2, p] < s[0, p]], {n, 0, z}]  (* A241740 *)
Table[Count[f[n], p_ /; s[2, p] == s[0, p]], {n, 0, z}] (* A241741 *)
Table[Count[f[n], p_ /; s[2, p] > s[0, p]], {n, 0, z}]  (* A241742 *)

Formula

a(n) + A241740(n) + A241842(n) = A000041(n) for n >= 0.