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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).
3
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.
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
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 28 2014
STATUS
approved