A239950 Number of partitions of n such that (number of distinct parts) = least part.

0, 1, 1, 1, 1, 2, 2, 3, 4, 4, 6, 6, 9, 8, 14, 11, 19, 18, 25, 24, 37, 31, 50, 46, 61, 64, 86, 79, 112, 115, 136, 149, 190, 184, 239, 255, 293, 329, 382, 408, 489, 531, 595, 675, 772, 827, 952, 1066, 1176, 1320, 1468, 1627, 1827, 2030, 2219, 2493, 2769, 3053

Offset: 0

Author

Clark Kimberling, Mar 30 2014

Keywords

• nonn
• easy

Comments

Also for n>0 the number of partitions of n such that (number of distinct parts) = multiplicity of the greatest part (by conjugation of the partition table). - Joerg Arndt, Apr 28 2014

Examples

			a(8) counts these 4 partitions :  62, 422, 332, 11111111.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A239948, A239949, A239951, A239952.

Programs

Maple

b:= proc(n, i, d) option remember; `if`(min(i, n) b(n$2, 0):
seq(a(n), n=0..60);  # Alois P. Heinz, Apr 02 2014

Mathematica

z = 50; d[p_] := d[p] = Length[DeleteDuplicates[p]]; f[n_] := f[n] = IntegerPartitions[n];
Table[Count[f[n], p_ /; d[p] < Min[p]], {n, 0, z}]  (*A239948*)
Table[Count[f[n], p_ /; d[p] <= Min[p]], {n, 0, z}] (*A239949*)
Table[Count[f[n], p_ /; d[p] == Min[p]], {n, 0, z}] (*A239950*)
Table[Count[f[n], p_ /; d[p] > Min[p]], {n, 0, z}]  (*A239951*)
Table[Count[f[n], p_ /; d[p] >= Min[p]], {n, 0, z}] (*A239952*)
b[n_, i_, d_] := b[n, i, d] = If[Min[i, n]Jean-François Alcover, Nov 17 2015, after Alois P. Heinz *)

Formula

A239948(n) + a(n) + A239951(n) = A000041(n) for n >= 0.