A239960 Number of partitions of n such that (number of distinct parts) = number of 1s.

1, 1, 0, 0, 1, 1, 2, 1, 4, 2, 6, 6, 8, 10, 16, 15, 22, 32, 31, 47, 54, 72, 81, 111, 123, 166, 189, 244, 274, 366, 411, 509, 614, 736, 872, 1056, 1256, 1479, 1785, 2099, 2479, 2942, 3498, 4028, 4870, 5600, 6655, 7712, 9127, 10512, 12431, 14327, 16776, 19401

Offset: 0

Author

Clark Kimberling, Mar 30 2014

Keywords

• nonn
• easy

Examples

			a(8) counts these 4 partitions :  611, 3311, 32111, 22211.
		

Links

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

Crossrefs

Cf. A239959, A239961.

Programs

Maple

b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
      b(n, i-1)+`if`(i=1, 0, add(b(n-1-i*j, i-1), j=1..(n-1)/i))))
    end:
a:= n-> `if`(n=0, 1, b(n-1$2)):
seq(a(n), n=0..70);  # Alois P. Heinz, Apr 03 2014

Mathematica

z = 54; d[p_] := d[p] = Length[DeleteDuplicates[p]]; Table[Count[IntegerPartitions[n], p_ /; d[p] == Count[p, 1]], {n, 0, z}]
(* Second program: *)
b[n_, i_] := b[n, i] = If[n == 0, 1, If[i < 1, 0, b[n, i - 1] + If[i == 1, 0, Sum[b[n - 1 - i*j, i - 1], {j, 1, (n - 1)/i}]]]]; a[n_] := If[n == 0, 1, b[n - 1, n - 1]]; Table[a[n], {n, 0, 70}] (* Jean-François Alcover, Aug 29 2016, after Alois P. Heinz *)