A240079 Number of partitions p of n such that mean(p) >= multiplicity(min(p)).

0, 1, 1, 2, 3, 4, 6, 9, 13, 17, 24, 31, 43, 56, 73, 98, 126, 157, 212, 263, 329, 431, 531, 649, 850, 1033, 1255, 1575, 1961, 2357, 2917, 3497, 4328, 5272, 6281, 7493, 9360, 11079, 13091, 15650, 19226, 22596, 26802, 31423, 37930, 45259, 52829, 61570, 74181

Offset: 0

Author

Clark Kimberling, Apr 03 2014

Keywords

• nonn
• easy

Examples

			a(6) counts these 5 partitions:  6, 51, 42, 33, 321.
		

Crossrefs

Cf. A240203, A240204, A240205, A240206, 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) = A240206(n) - A240205(n) for n >= 0.

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