%I #22 Aug 28 2020 16:46:00
%S 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,
%T 1,428,245,149,100,64,29,8,1,899,533,341,228,163,93,37,9,1,1875,1142,
%U 765,512,383,256,130,46,10,1,3890,2420,1683,1144,859,638,386,176,56,11,1
%N 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.
%H Alois P. Heinz, <a href="/A322428/b322428.txt">Rows n = 1..50, flattened</a>
%e 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.
%e Triangle T(n,k) begins:
%e 1;
%e 3, 1;
%e 8, 3, 1;
%e 19, 8, 4, 1;
%e 43, 20, 11, 5, 1;
%e 94, 48, 27, 16, 6, 1;
%e 202, 110, 64, 42, 22, 7, 1;
%e 428, 245, 149, 100, 64, 29, 8, 1;
%e 899, 533, 341, 228, 163, 93, 37, 9, 1;
%e 1875, 1142, 765, 512, 383, 256, 130, 46, 10, 1;
%e ...
%p b:= proc(n, l) option remember; `if`(n=0, add(l[-i]*x^i,
%p i=1..nops(l)), add(b(n-j, sort([l[], j])), j=1..n))
%p end:
%p T:= n-> (p-> seq(coeff(p, x, i), i=1..degree(p)))(b(n, [])):
%p seq(T(n), n=1..12);
%t 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 T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 1, Exponent[p, x]}]][ b[n, {}]];
%t Array[T, 12] // Flatten (* _Jean-François Alcover_, Dec 29 2018, after _Alois P. Heinz_ *)
%Y Column k=1 gives A102712.
%Y Row sums give A001787.
%Y T(n+1,1+ceiling(n/2)) gives A027306.
%Y Cf. A322427.
%K nonn,tabl
%O 1,2
%A _Alois P. Heinz_, Dec 07 2018