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Number of partitions p of n such that median(p) < multiplicity(min(p)).
5

%I #5 Apr 12 2014 16:24:03

%S 0,0,1,1,2,3,6,7,10,14,19,26,37,48,65,87,115,150,194,250,322,407,518,

%T 653,823,1029,1287,1598,1984,2449,3021,3706,4540,5540,6752,8197,9933,

%U 12004,14482,17421,20924,25070,29992,35797,42661,50735,60254,71421,84536

%N Number of partitions p of n such that median(p) < multiplicity(min(p)).

%F a(n) = A240213(n) - A240214(n) for n >= 0.

%F a(n) + A240214(n) + A240216(n) = A000041(n) for n >= 0.

%e a(8) counts these 10 partitions: 611, 5111, 4211, 41111, 32111, 311111, 2222, 221111, 2111111, 11111111.

%t z = 40; f[n_] := f[n] = IntegerPartitions[n]; t1 = Table[Count[f[n], p_ /; Median[p] < Count[p, Min[p]]], {n, 0, z}] (* A240212 *)

%t t2 = Table[Count[f[n], p_ /; Median[p] <= Count[p, Min[p]]], {n, 0, z}] (* A240213 *)

%t t3 = Table[Count[f[n], p_ /; Median[p] == Count[p, Min[p]]], {n, 0, z}] (* A240214 *)

%t t4 = Table[Count[f[n], p_ /; Median[p] > Count[p, Min[p]]], {n, 0, z}] (* A240215 *)

%t t5 = Table[Count[f[n], p_ /; Median[p] >= Count[p, Min[p]]], {n, 0, z}] (* A240216 *)

%Y Cf. A240213, A240214, A240215, A240216, A000041.

%K nonn,easy

%O 0,5

%A _Clark Kimberling_, Apr 04 2014