A241744 Number of partitions p of n such that (number of numbers in p of form 3k) = (number of numbers in p of form 3k+1).

1, 0, 1, 0, 2, 2, 3, 6, 7, 10, 17, 18, 27, 36, 44, 61, 76, 93, 124, 151, 193, 241, 297, 369, 462, 558, 707, 850, 1044, 1281, 1561, 1884, 2323, 2761, 3367, 4050, 4857, 5826, 7024, 8307, 9982, 11840, 14058, 16684, 19785, 23265, 27585, 32379, 38125, 44760

Offset

1,5

Comments

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

Links

Table of n, a(n) for n=1..50.

Formula

a(n) + A241744(n) + A241845(n) = A000041(n) for n >= 0.

Example

a(8) counts these 7 partitions:  8, 611, 3311, 3221, 32111, 311111, 2222.

Mathematica

z = 40; f[n_] := f[n] = IntegerPartitions[n]; s[k_, p_] := Count[Mod[DeleteDuplicates[p], 3], k];
Table[Count[f[n], p_ /; s[0, p] < s[2, p]], {n, 0, z}]  (* A241743 *)
Table[Count[f[n], p_ /; s[0, p] == s[1, p]], {n, 0, z}] (* A241744 *)
Table[Count[f[n], p_ /; s[0, p] > s[1, p]], {n, 0, z}]  (* A241745 *)

Crossrefs

Cf. A241737, A241740, A241743, A241745.

Sequence in context: A358088 A091770 A032058 * A308514 A039866 A106369

Adjacent sequences: A241741 A241742 A241743 * A241745 A241746 A241747

Keyword

nonn,easy

Author

Clark Kimberling, Apr 28 2014

Status

approved