login
Number of partitions p of n such that median(p) <= mean(p).
5

%I #11 Apr 15 2014 13:54:07

%S 1,2,3,5,6,11,13,19,26,38,45,70,82,112,154,203,244,336,402,541,700,

%T 878,1052,1386,1708,2095,2624,3328,3971,5071,6027,7377,9013,10783,

%U 13220,16597,19615,23277,27939,34043,39982,48546,56854,68240,82828,97099,113268

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

%F a(n) + A240220(n) = A000041(n) for n >= 1.

%F a(n) = A240217(n) + A240219(n) for n >= 1.

%e a(6) counts these 11 partitions: 6, 51, 42, 411, 33, 321, 3111, 222, 2211, 21111, 111111.

%t z = 60; f[n_] := f[n] = IntegerPartitions[n];

%t Table[Count[f[n], p_ /; Median[p] < Mean[p]], {n, 1, z}] (* A240217 *)

%t Table[Count[f[n], p_ /; Median[p] <= Mean[p]], {n, 1, z}] (* A240218 *)

%t Table[Count[f[n], p_ /; Median[p] == Mean[p]], {n, 1, z}] (* A240219 *)

%t Table[Count[f[n], p_ /; Median[p] > Mean[p]], {n, 1, z}] (* A240220 *)

%t Table[Count[f[n], p_ /; Median[p] >= Mean[p]], {n, 1, z}] (* A240221 *)

%Y Cf. A240217, A240219, A240220, A240221, A000041.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Apr 04 2014