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

%I #4 Apr 12 2014 16:23:26

%S 0,0,1,1,1,1,3,3,4,5,6,8,11,14,18,24,27,35,41,52,65,79,96,122,145,177,

%T 213,256,307,368,434,524,623,741,876,1049,1230,1458,1714,2019,2362,

%U 2778,3234,3792,4412,5148,5980,6970,8069,9372,10846,12559,14491,16754

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

%F a(n) = A240209(n) - A240208(n) for n >= 0.

%F a(n) + A240209(n) + A240210(n) = A000041(n) for n >= 0.

%e a(6) counts these 3 partitions: 222, 2211, 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. A240208, A240209, A240210, A240211, A000041.

%K nonn,easy

%O 0,7

%A _Clark Kimberling_, Apr 03 2014