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

%I #4 Apr 12 2014 16:23:10

%S 0,1,0,0,1,0,1,0,2,1,2,0,4,0,2,7,3,0,17,0,5,26,2,0,60,1,2,61,59,0,91,

%T 0,149,119,2,34,480,0,2,215,788,0,288,0,1147,923,2,0,2528,1,1585,611,

%U 3319,0,1150,3963,5366,986,2,0,20317

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

%C a(n) = 0 if and only if n = 0 or n is a prime.

%F a(n) = A240204(n) - A240203(n) for n >= 0.

%F a(n) + A240203(n) + A240206(n) = A000041(n) for n >= 0.

%e a(12) counts these 4 partitions: 9111, 6222, 422211, 332211.

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

%t t1 = Table[Count[f[n], p_ /; Mean[p] < Count[p, Min[p]]], {n, 0, z}] (* A240203 *)

%t t2 = Table[Count[f[n], p_ /; Mean[p] <= Count[p, Min[p]]], {n, 0, z}] (* A240204 *)

%t t3 = Table[Count[f[n], p_ /; Mean[p] == Count[p, Min[p]]], {n, 0, z}] (* A240205 *)

%t t4 = Table[Count[f[n], p_ /; Mean[p] > Count[p, Min[p]]], {n, 0, z}] (* A240206 *)

%t t5 = Table[Count[f[n], p_ /; Mean[p] >= Count[p, Min[p]]], {n, 0, z}] (* A240079 *)

%Y Cf. A240203, A240204, A240206, A240079, A000041.

%K nonn,easy

%O 0,9

%A _Clark Kimberling_, Apr 03 2014