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

%I #4 Apr 12 2014 16:23:33

%S 0,1,1,1,3,4,6,7,11,14,19,26,35,44,59,74,97,120,158,192,247,304,383,

%T 470,587,714,885,1074,1317,1593,1943,2334,2831,3396,4086,4883,5859,

%U 6966,8319,9870,11726,13864,16422,19345,22834,26830,31548,36969,43354,50651

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

%F a(n) = A240207(n) + A240209(n) for n >= 0.

%F a(n) + A240210(n) = A000041(n) for n >= 0.

%e a(6) counts these 6 partitions: 411, 3111, 222, 2211, 21111, 111111.

%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, A240209, A240210, A240211, A000041.

%K nonn,easy

%O 0,5

%A _Clark Kimberling_, Apr 03 2014