A240205 Number of partitions p of n such that mean(p) = multiplicity(min(p)).

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, 0, 149, 119, 2, 34, 480, 0, 2, 215, 788, 0, 288, 0, 1147, 923, 2, 0, 2528, 1, 1585, 611, 3319, 0, 1150, 3963, 5366, 986, 2, 0, 20317

Offset: 0

Author

Clark Kimberling, Apr 03 2014

Keywords

• nonn
• easy

Comments

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

Examples

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

Crossrefs

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

Programs

Mathematica

z = 60; f[n_] := f[n] = IntegerPartitions[n];
t1 = Table[Count[f[n], p_ /; Mean[p] < Count[p, Min[p]]], {n, 0, z}]  (* A240203 *)
t2 = Table[Count[f[n], p_ /; Mean[p] <= Count[p, Min[p]]], {n, 0, z}] (* A240204 *)
t3 = Table[Count[f[n], p_ /; Mean[p] == Count[p, Min[p]]], {n, 0, z}] (* A240205 *)
t4 = Table[Count[f[n], p_ /; Mean[p] > Count[p, Min[p]]], {n, 0, z}] (* A240206 *)
t5 = Table[Count[f[n], p_ /; Mean[p] >= Count[p, Min[p]]], {n, 0, z}] (* A240079 *)

Formula

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

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