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Table[Length[Select[IntegerPartitions[n], Length[#]-Length[Union[#]]==k&]], {n, 0, 15}, {k, 0, n}] (* _augmented version, _Gus Wiseman_, Jan 23 2019 *)
1
1 0
1 1 0
2 0 1 0
2 2 0 1 0
3 2 1 0 1 0
4 2 3 1 0 1 0
5 4 2 2 1 0 1 0
6 6 3 3 2 1 0 1 0
8 7 5 4 2 2 1 0 1 0
10 8 10 3 5 2 2 1 0 1 0
12 13 8 9 4 4 2 2 1 0 1 0
15 15 14 10 8 5 4 2 2 1 0 1 0
18 21 15 16 8 9 4 4 2 2 1 0 1 0
22 25 23 17 17 7 10 4 4 2 2 1 0 1 0
27 30 32 21 19 16 8 9 4 4 2 2 1 0 1 0
From Gus Wiseman, Jan 23 2019: (Start)
It is possible to augment the triangle to cover the n = 0 and k = n cases, giving:
1
1 0
1 1 0
2 0 1 0
2 2 0 1 0
3 2 1 0 1 0
4 2 3 1 0 1 0
5 4 2 2 1 0 1 0
6 6 3 3 2 1 0 1 0
8 7 5 4 2 2 1 0 1 0
10 8 10 3 5 2 2 1 0 1 0
12 13 8 9 4 4 2 2 1 0 1 0
15 15 14 10 8 5 4 2 2 1 0 1 0
18 21 15 16 8 9 4 4 2 2 1 0 1 0
22 25 23 17 17 7 10 4 4 2 2 1 0 1 0
27 30 32 21 19 16 8 9 4 4 2 2 1 0 1 0
Row seven {5, 4, 2, 2, 1, 0, 1, 0} counts the following integer partitions (empty columns not shown).
(7) (322) (2221) (22111) (211111) (1111111)
(43) (331) (4111) (31111)
(52) (511)
(61) (3211)
(421)
(End)
Table[Length[Select[IntegerPartitions[n], Length[#]-Length[Union[#]]==k&]], {n, 0, 15}, {k, 0, n}] (* Gus Wiseman, Jan 23 2019 *)
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b[n_, i_] := b[n, i] = Expand[If[n == 0, 1, If[i < 1, 0, Sum[x^If[j == 0, 0, j-1]*b[n - i*j, i - 1], {j, 0, n/i}]]]]; T[n_] := Function [p, Table[ Coefficient[p, x, i], {i, 0, n - 1}]][b[n, n]]; Table[T[n], {n, 1, 16}] // Flatten (* Jean-François Alcover, Jan 23 2016, after Alois P. Heinz *)
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