A239949 Number of partitions of n such that (number of distinct parts) <= least part.

1, 1, 2, 2, 3, 3, 5, 5, 8, 8, 12, 12, 18, 17, 26, 25, 36, 36, 50, 50, 69, 69, 93, 95, 123, 129, 164, 171, 215, 229, 278, 300, 365, 387, 468, 507, 595, 652, 760, 830, 966, 1055, 1214, 1336, 1530, 1674, 1910, 2104, 2380, 2617, 2953, 3253, 3656, 4019, 4504

Offset: 0

Author

Clark Kimberling, Mar 30 2014

Keywords

• nonn
• easy

Examples

			a(8) counts these 8 partitions:  8, 62, 53, 44, 422, 332, 2222, 11111111.
		

Links

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

Crossrefs

Cf. A239948, A239950, A239951, A239952.

Programs

Maple

b:= proc(n, i, d) option remember; `if`(n=0, 1, `if`(i<=d, 0,
      add(b(n-i*j, i-1, d+`if`(j=0, 0, 1)), j=0..n/i)))
    end:
a:= n-> b(n$2, 0):
seq(a(n), n=0..80);  # 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[n==0, 1, If[i <= d, 0, Sum[b[n-i*j, i-1, d + If[j==0, 0, 1]], {j, 0, n/i}]]]; a[n_] := b[n, n, 0]; Table[a[n], {n, 0, 80}] (* Jean-François Alcover, Nov 17 2015, after Alois P. Heinz *)

Formula

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