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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).
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%I #5 May 06 2014 15:06:04

%S 1,0,1,0,2,2,3,6,7,10,17,18,27,36,44,61,76,93,124,151,193,241,297,369,

%T 462,558,707,850,1044,1281,1561,1884,2323,2761,3367,4050,4857,5826,

%U 7024,8307,9982,11840,14058,16684,19785,23265,27585,32379,38125,44760

%N 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).

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

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

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

%t z = 40; f[n_] := f[n] = IntegerPartitions[n]; s[k_, p_] := Count[Mod[DeleteDuplicates[p], 3], k];

%t Table[Count[f[n], p_ /; s[0, p] < s[2, p]], {n, 0, z}] (* A241743 *)

%t Table[Count[f[n], p_ /; s[0, p] == s[1, p]], {n, 0, z}] (* A241744 *)

%t Table[Count[f[n], p_ /; s[0, p] > s[1, p]], {n, 0, z}] (* A241745 *)

%Y Cf. A241737, A241740, A241743, A241745.

%K nonn,easy

%O 1,5

%A _Clark Kimberling_, Apr 28 2014