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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 15:05:24

%S 1,0,0,2,1,2,7,5,7,17,14,18,39,32,42,76,71,88,157,143,182,293,292,357,

%T 562,558,692,1023,1060,1286,1854,1932,2347,3246,3464,4153,5639,6030,

%U 7207,9526,10324,12240,15912,17311,20444,26104,28585,33567,42326,46469

%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) + A241839(n) = A000041(n) for n >= 0.

%e a(8) counts these 7 partitions: 5111, 422, 3221, 3211, 22211, 221111, 2111111.

%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, A241739, A241740, A241743.

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Apr 28 2014