A304825 Sum of binomial(Y(2,p), 2) over the partitions p of n, where Y(2,p) is the number of part sizes with multiplicity 2 or greater in p.

1, 1, 3, 4, 9, 12, 22, 30, 50, 68, 105, 142, 210, 281, 400, 531, 736, 967, 1311, 1707, 2274, 2935, 3851, 4930, 6389, 8116, 10402, 13121, 16658, 20872, 26275, 32719, 40880, 50613, 62807, 77343, 95389, 116874, 143331, 174789, 213251, 258903, 314367, 380079, 459462

Offset: 6

Author

Emily Anible, May 19 2018

Keywords

• nonn

Examples

			For a(8), we sum over the partitions of eight. For each partition p, we take binomial(Y(2,p),2): that is, the number of parts with multiplicity at least two choose 2.
8................B(0,2) = 0
7,1..............B(0,2) = 0
6,2..............B(0,2) = 0
6,1,1............B(1,2) = 0
5,3..............B(0,2) = 0
5,2,1............B(0,2) = 0
5,1,1,1..........B(1,2) = 0
4,4..............B(1,2) = 0
4,3,1............B(0,2) = 0
4,2,2............B(1,2) = 0
4,2,1,1..........B(1,2) = 0
4,1,1,1,1........B(1,2) = 0
3,3,2............B(1,2) = 0
3,3,1,1..........B(2,2) = 1
3,2,2,1..........B(1,2) = 0
3,2,1,1,1........B(1,2) = 0
3,1,1,1,1,1......B(1,2) = 0
2,2,2,2..........B(1,2) = 0
2,2,2,1,1........B(2,2) = 1
2,2,1,1,1,1......B(2,2) = 1
2,1,1,1,1,1,1....B(1,2) = 0
1,1,1,1,1,1,1,1..B(1,2) = 0
---------------------------
Total.....................3
		

Crossrefs

Cf. A024786, A302347.

Programs

Maple

b:= proc(n, i, p) option remember; `if`(n=0 or i=1,
      binomial(`if`(n>1, 1, 0)+p, 2), add(
      b(n-i*j, i-1, `if`(j>1, 1, 0)+p), j=0..n/i))
    end:
a:= n-> b(n$2, 0):
seq(a(n), n=6..60);  # Alois P. Heinz, May 19 2018

Mathematica

Array[Total[Binomial[Count[Split@#, _?(Length@# >= 2 &)], 2] & /@IntegerPartitions[#]] &, 50]
(* Second program: *)
b[n_, i_, p_] := b[n, i, p] = If[n == 0 || i == 1,
     Binomial[If[n > 1, 1, 0] + p, 2], Sum[
     b[n-i*j, i-1, If[j>1, 1, 0]+p], {j, 0, n/i}]];
a[n_] := b[n, n, 0];
a /@ Range[6, 60] (* Jean-François Alcover, May 30 2021, after Alois P. Heinz *)

Formula

a(n) = (A301313(n) - A024788(n))/4.

G.f.: q^6 /((1-q^2)*(1-q^4))*Product_{j>=1} 1/(1-q^j).