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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).
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%I #5 May 06 2014 15:05:38

%S 1,1,1,1,2,3,3,5,9,11,14,22,29,36,51,66,83,107,139,170,216,273,340,

%T 415,520,635,778,952,1177,1414,1724,2094,2527,3038,3691,4411,5286,

%U 6345,7586,9008,10778,12796,15163,17979,21288,25059,29608,34861,40927,48035

%N 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).

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

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

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

%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[2, p] < s[0, p]], {n, 0, z}] (* A241740 *)

%t Table[Count[f[n], p_ /; s[2, p] == s[0, p]], {n, 0, z}] (* A241741 *)

%t Table[Count[f[n], p_ /; s[2, p] > s[0, p]], {n, 0, z}] (* A241742 *)

%Y Cf. A241737, A241740, A241742, A241743.

%K nonn,easy

%O 0,5

%A _Clark Kimberling_, Apr 28 2014