A286897 Sum T(n,k) of the k-th last entries in all blocks of all set partitions of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

1, 5, 1, 23, 6, 1, 109, 33, 7, 1, 544, 182, 45, 8, 1, 2876, 1034, 284, 59, 9, 1, 16113, 6122, 1815, 420, 75, 10, 1, 95495, 37927, 11931, 2987, 595, 93, 11, 1, 597155, 246030, 81205, 21620, 4665, 814, 113, 12, 1, 3929243, 1669941, 573724, 160607, 36900, 6979, 1082, 135, 13, 1

Offset: 1

Author

Alois P. Heinz, May 15 2017

Keywords

• nonn
• tabl

Examples

			T(3,2) = 6 because the sum of the second last entries in all blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 2+1+1+2 = 6.
Triangle T(n,k) begins:
      1;
      5,     1;
     23,     6,     1;
    109,    33,     7,    1;
    544,   182,    45,    8,   1;
   2876,  1034,   284,   59,   9,  1;
  16113,  6122,  1815,  420,  75, 10,  1;
  95495, 37927, 11931, 2987, 595, 93, 11, 1;
  ...
		

Links

• Alois P. Heinz, Row n = 1..50, flattened
• Wikipedia, Partition of a set

Crossrefs

Column k=1 gives A278677(n-1).

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

Cf. A285595.

Programs

Maple

b:= proc(n, l) option remember; `if`(n=0, [1, 0],
      (p-> p+[0, n*p[1]*x^1])(b(n-1, [l[], 1]))+
       add((p-> p+[0, n*p[1]*x^(l[j]+1)])(b(n-1,
       sort(subsop(j=l[j]+1, l), `>`))), j=1..nops(l)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n, [])[2]):
seq(T(n), n=1..14);

Mathematica

b[0, ] = {1, 0}; b[n, l_] := b[n, l] = Function[p, p + {0, n*p[[1]]*x^1} ][b[n - 1, Append[l, 1]]] + Sum[Function[p, p + {0, n*p[[1]]*x^(l[[j]] + 1)}][b[n - 1, Reverse @ Sort[ReplacePart[l, j -> l[[j]] + 1]]]], {j, 1, Length[l]}];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, n}]][b[n, {}][[2]]];
Table[T[n], {n, 1, 14}] // Flatten (* Jean-François Alcover, May 26 2018, from Maple *)