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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 15:05:32

%S 0,0,0,1,1,1,3,4,4,7,10,12,17,24,30,40,53,70,90,118,152,194,244,316,

%T 396,497,626,784,960,1202,1483,1816,2230,2738,3312,4042,4908,5922,

%U 7141,8627,10327,12388,14832,17703,21075,25120,29795,35321,41822,49439,58286

%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) + A241741(n) + A241842(n) = A000041(n) for n >= 0.

%e a(8) counts these 4 partitions: 611, 431, 3311, 311111.

%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, A241741, A241741, A241743.

%K nonn,easy

%O 0,7

%A _Clark Kimberling_, Apr 28 2014