A286232 Sum T(n,k) of the entries in the k-th last blocks of all set partitions of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

1, 5, 1, 19, 10, 1, 75, 57, 17, 1, 323, 285, 145, 26, 1, 1512, 1421, 975, 317, 37, 1, 7630, 7395, 5999, 2865, 616, 50, 1, 41245, 40726, 36183, 22411, 7315, 1094, 65, 1, 237573, 237759, 221689, 163488, 72581, 16630, 1812, 82, 1, 1451359, 1468162, 1405001, 1160764, 649723, 206249, 34425, 2840, 101, 1

Offset: 1

Author

Alois P. Heinz, May 04 2017

Keywords

• nonn
• tabl

Examples

			T(3,2) = 10 because the sum of the entries in the second last blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 0+3+4+1+2 = 10.
Triangle T(n,k) begins:
      1;
      5,     1;
     19,    10,     1;
     75,    57,    17,     1;
    323,   285,   145,    26,    1;
   1512,  1421,   975,   317,   37,    1;
   7630,  7395,  5999,  2865,  616,   50,  1;
  41245, 40726, 36183, 22411, 7315, 1094, 65, 1;
  ...
		

Links

• Alois P. Heinz, Rows n = 1..100, flattened
• Wikipedia, Partition of a set

Crossrefs

Column k=1 gives A285424.

Main diagonal and first lower diagonal give: A000012, A002522.

Row sums give A000110(n) * A000217(n) = A105488(n+3).

Cf. A285362, A286231, A286416.

Programs

Mathematica

app[P_, n_] := Module[{P0}, Table[P0 = Append[P, {}]; AppendTo[P0[[i]], n]; If[Last[P0] == {}, Most[P0], P0], {i, 1, Length[P]+1}]];
setPartitions[n_] := setPartitions[n] = If[n == 1, {{{1}}}, Flatten[app[#, n]& /@ setPartitions[n-1], 1]];
T[n_, k_] := Select[setPartitions[n], Length[#] >= k&][[All, -k]] // Flatten // Total;
Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Aug 21 2021 *)