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A240218 Number of partitions p of n such that median(p) <= mean(p).

%I A240218 #11 Apr 15 2014 13:54:07
%S A240218 1,2,3,5,6,11,13,19,26,38,45,70,82,112,154,203,244,336,402,541,700,
%T A240218 878,1052,1386,1708,2095,2624,3328,3971,5071,6027,7377,9013,10783,
%U A240218 13220,16597,19615,23277,27939,34043,39982,48546,56854,68240,82828,97099,113268
%N A240218 Number of partitions p of n such that median(p) <= mean(p).
%F A240218 a(n) + A240220(n) = A000041(n) for n >= 1.
%F A240218 a(n) = A240217(n) + A240219(n) for n >= 1.
%e A240218 a(6) counts these 11 partitions:  6, 51, 42, 411, 33, 321, 3111, 222, 2211, 21111, 111111.
%t A240218 z = 60; f[n_] := f[n] = IntegerPartitions[n];
%t A240218 Table[Count[f[n], p_ /; Median[p] < Mean[p]], {n, 1, z}]  (* A240217 *)
%t A240218 Table[Count[f[n], p_ /; Median[p] <= Mean[p]], {n, 1, z}] (* A240218 *)
%t A240218 Table[Count[f[n], p_ /; Median[p] == Mean[p]], {n, 1, z}] (* A240219 *)
%t A240218 Table[Count[f[n], p_ /; Median[p] > Mean[p]], {n, 1, z}]  (* A240220 *)
%t A240218 Table[Count[f[n], p_ /; Median[p] >= Mean[p]], {n, 1, z}] (* A240221 *)
%Y A240218 Cf. A240217, A240219, A240220, A240221, A000041.
%K A240218 nonn,easy
%O A240218 1,2
%A A240218 _Clark Kimberling_, Apr 04 2014