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

0,6

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

Links

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

Formula

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

Example

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

Maple

b:= proc(n, i, d) option remember; `if`(min(i, n)<d+1, 0,
      `if`(irem(n, i)=0 and i=d+1, 1, b(n, i-1, d)+
      add(b(n-i*j, i-1, d+1), j=1..n/i)))
    end:
a:= 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]<d+1, 0, If[Mod[n, i]==0 && i == d+1, 1, b[n, i-1, d] + Sum[b[n-i*j, i-1, d+1], {j, 1, n/i}]]]; a[n_] := b[n, n, 0]; Table[a[n], {n, 0, 60}] (* Jean-François Alcover, Nov 17 2015, after Alois P. Heinz *)

Crossrefs

Cf. A239948, A239949, A239951, A239952.

Sequence in context: A029936 A114093 A345141 * A077768 A143038 A029040

Adjacent sequences: A239947 A239948 A239949 * A239951 A239952 A239953

Keyword

nonn,easy

Author

Clark Kimberling, Mar 30 2014

Status

approved