A239958 Number of partitions p of n such that (number of distinct parts of p) >= max(p) - min(p).

1, 1, 2, 3, 5, 6, 9, 11, 16, 18, 25, 30, 39, 47, 59, 69, 89, 105, 126, 153, 184, 215, 259, 307, 362, 426, 501, 583, 687, 800, 923, 1080, 1252, 1439, 1666, 1917, 2202, 2533, 2900, 3311, 3792, 4326, 4915, 5605, 6366, 7205, 8180, 9259, 10458, 11815, 13322

Offset: 0

Author

Clark Kimberling, Mar 30 2014

Keywords

• nonn
• easy

Examples

			a(6) counts all of the 15 partitions of 7 except these 4:  61, 52, 511, 1111111.
		

Crossrefs

Cf. A239954, A239955, A239956, A034296.

Programs

Mathematica

z = 60; d[p_] := d[p] = Length[DeleteDuplicates[p]]; f[p_] := f[p] = Max[p] - Min[p]; g[n_] := g[n] = IntegerPartitions[n];
Table[Count[g[n], p_ /; d[p] < f[p]], {n, 0, z}]  (*A239954*)
Table[Count[g[n], p_ /; d[p] <= f[p]], {n, 0, z}] (*A239955*)
Table[Count[g[n], p_ /; d[p] == f[p]], {n, 0, z}] (*A239956*)
Table[Count[g[n], p_ /; d[p] > f[p]], {n, 0, z}]  (*A034296*)
Table[Count[g[n], p_ /; d[p] >= f[p]], {n, 0, z}] (*A239958*)
ndpQ[p_]:=Module[{prt=Union[p]},Length[prt]>=(Max[prt]-Min[prt])]; Table[Length[Select[ IntegerPartitions[ n],ndpQ]],{n,0,50}] (* Harvey P. Dale, Dec 31 2023 *)

Formula

a(n) + A239954(n) = A000041(n) for n >= 0.