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

%I #4 Apr 12 2014 16:24:36

%S 0,1,1,2,3,4,5,8,12,16,23,30,40,53,70,89,116,147,191,240,305,385,484,

%T 602,752,929,1149,1412,1734,2116,2583,3136,3809,4603,5558,6686,8044,

%U 9633,11533,13764,16414,19513,23182,27464,32514,38399,45304,53333,62737

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

%F a(n) = A240215(n) + A240215(n) for n >= 0.

%F a(n) + A240212(n) = A000041(n) for n >= 0.

%e a(6) counts these 12 partitions: 8, 71, 62, 53, 521, 44, 431, 422, 332, 3311, 321, 22211.

%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. A240212, A240213, A240214, A240215, A000041.

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Apr 04 2014