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

0,5

Comments

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

Links

Table of n, a(n) for n=0..49.

Formula

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

Example

a(8) counts these 9 partitions:  71, 62, 53, 44, 41111, 332, 3221, 32111, 11111111.

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 *)

Crossrefs

Cf. A241737, A241740, A241742, A241743.

Sequence in context: A296725 A296588 A065460 * A175147 A001180 A228778

Adjacent sequences: A241738 A241739 A241740 * A241742 A241743 A241744

Keyword

nonn,easy

Author

Clark Kimberling, Apr 28 2014

Status

approved