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Number of partitions p of n such that median(p) >= multiplicity(max(p)).
5

%I #4 Apr 12 2014 16:23:58

%S 0,1,1,2,4,6,8,12,18,25,36,48,66,87,117,152,204,262,344,438,562,713,

%T 906,1133,1430,1781,2223,2754,3411,4197,5170,6318,7726,9402,11434,

%U 13834,16747,20179,24301,29166,34976,41805,49940,59469,70763,83986,99578,117784

%N Number of partitions p of n such that median(p) >= multiplicity(max(p)).

%F a(n) = A240209(n) + A240210(n) for n >= 0.

%F a(n) + A240207(n) = A000041(n) for n >= 0.

%e a(6) counts these 8 partitions: 6, 51, 42, 411, 33, 321, 3111, 21111.

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

%t t1 = Table[Count[f[n], p_ /; Median[p] < Count[p, Max[p]]], {n, 0, z}] (* A240207 *)

%t t2 = Table[Count[f[n], p_ /; Median[p] <= Count[p, Max[p]]], {n, 0, z}] (* A240208 *)

%t t3 = Table[Count[f[n], p_ /; Median[p] == Count[p, Max[p]]], {n, 0, z}] (* A240209 *)

%t t4 = Table[Count[f[n], p_ /; Median[p] > Count[p, Max[p]]], {n, 0, z}] (* A240210 *)

%t t5 = Table[Count[f[n], p_ /; Median[p] >= Count[p, Max[p]]], {n, 0, z}] (* A240211 *)

%Y Cf. A240207, A240208, A240209, A240210, A000041.

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Apr 03 2014