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Number of partitions p of n such that (number of numbers in p of form 3k+1) > (number of numbers in p of form 3k+2).
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%I #5 May 06 2014 15:04:42

%S 0,1,1,1,3,3,3,8,9,10,20,24,27,49,58,69,109,132,153,234,279,331,469,

%T 565,662,918,1093,1290,1723,2056,2411,3165,3751,4411,5656,6700,7839,

%U 9932,11707,13699,17084,20099,23441,28939,33914,39498,48236,56392,65481

%N Number of partitions p of n such that (number of numbers in p of form 3k+1) > (number of numbers in p of form 3k+2).

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

%F a(n) + A241737(n) + A241838(n) = A000041(n) for n >= 0.

%e a(8) counts these 9 partitions: 71, 611, 44, 431, 4211, 3311, 311111, 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[1, p] < s[2, p]], {n, 0, z}] (* A241737 *)

%t Table[Count[f[n], p_ /; s[1, p] == s[2, p]], {n, 0, z}] (* A241738 *)

%t Table[Count[f[n], p_ /; s[1, p] > s[2, p]], {n, 0, z}] (* A241739 *)

%Y Cf. A241737, A241738, A241740, A241743.

%K nonn,easy

%O 0,5

%A _Clark Kimberling_, Apr 28 2014