A322428 Sum T(n,k) of k-th largest parts of all compositions of n; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

1, 3, 1, 8, 3, 1, 19, 8, 4, 1, 43, 20, 11, 5, 1, 94, 48, 27, 16, 6, 1, 202, 110, 64, 42, 22, 7, 1, 428, 245, 149, 100, 64, 29, 8, 1, 899, 533, 341, 228, 163, 93, 37, 9, 1, 1875, 1142, 765, 512, 383, 256, 130, 46, 10, 1, 3890, 2420, 1683, 1144, 859, 638, 386, 176, 56, 11, 1

Offset: 1

Author

Alois P. Heinz, Dec 07 2018

Keywords

• nonn
• tabl

Examples

			The 4 compositions of 3 are: 111, 12, 21, 3.  The sums of k-th largest parts for k=1..3 give: 1+2+2+3 = 8, 1+1+1+0 = 3, 1+0+0+0 = 1.
Triangle T(n,k) begins:
     1;
     3,    1;
     8,    3,   1;
    19,    8,   4,   1;
    43,   20,  11,   5,   1;
    94,   48,  27,  16,   6,   1;
   202,  110,  64,  42,  22,   7,   1;
   428,  245, 149, 100,  64,  29,   8,  1;
   899,  533, 341, 228, 163,  93,  37,  9,  1;
  1875, 1142, 765, 512, 383, 256, 130, 46, 10, 1;
  ...
		

Links

• Alois P. Heinz, Rows n = 1..50, flattened

Crossrefs

Column k=1 gives A102712.

Row sums give A001787.

T(n+1,1+ceiling(n/2)) gives A027306.

Cf. A322427.

Programs

Maple

b:= proc(n, l) option remember; `if`(n=0, add(l[-i]*x^i,
      i=1..nops(l)), add(b(n-j, sort([l[], j])), j=1..n))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..degree(p)))(b(n, [])):
seq(T(n), n=1..12);

Mathematica

b[n_, l_] := b[n, l] = If[n == 0, Sum[l[[-i]] x^i, {i, 1, Length[l]}], Sum[b[n - j, Sort[Append[l, j]]], {j, 1, n}]];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 1, Exponent[p, x]}]][ b[n, {}]];
Array[T, 12] // Flatten (* Jean-François Alcover, Dec 29 2018, after Alois P. Heinz *)