%I #27 Sep 21 2018 22:40:11

%S 0,0,1,0,1,2,0,1,3,3,0,1,4,6,4,0,1,6,10,10,5,0,1,8,19,20,15,6,0,1,12,

%T 33,46,35,21,7,0,1,16,63,100,94,56,28,8,0,1,24,111,220,242,172,84,36,

%U 9,0,1,32,201,488,633,514,290,120,45,10,0,1,48,369,1104,1643,1518,984,460,165,55,11

%N Square array A(n,k), n,k>=0, read by antidiagonals: A(n,k) is the total element sum of the k-fold f transform applied to the length n sequence of 1's. And f returns a sorted result after multiplying the elements in its input sequence with 1, 2, 3,... in descending size order.

%H Alois P. Heinz, <a href="/A166278/b166278.txt">Antidiagonals n = 0..140, flattened</a>

%e A(3,4) = 33, because f([1,1,1]) = [1,2,3], (f^2)([1,1,1]) = [3,3,4], (f^3)([1,1,1]) = [4,6,9], (f^4)([1,1,1]) = [9,12,12], and 9+12+12 = 33.

%e Square array A(n,k) begins:

%e 0, 0, 0, 0, 0, 0, ...

%e 1, 1, 1, 1, 1, 1, ...

%e 2, 3, 4, 6, 8, 12, ...

%e 3, 6, 10, 19, 33, 63, ...

%e 4, 10, 20, 46, 100, 220, ...

%e 5, 15, 35, 94, 242, 633, ...

%p f:= l-> sort([seq(sort(l, `>`)[i]*i, i=1..nops(l))]):

%p A:= (n, k)-> add(i, i=(f@@k)([1$n])):

%p seq(seq(A(n, d-n), n=0..d), d=0..15);

%t f[L_List] := f[L] = Sort[Reverse[Sort[L]]*Range[Length[L]]];

%t A[0, _] = 0; A[n_, 0] := n; A[n_, k_] := Total[Nest[f, Range[n], k-1]];

%t Table[A[n, k-n], {k, 0, 15}, {n, 0, k}] // Flatten (* _Jean-François Alcover_, Jun 07 2016 *)

%Y Columns k=0-3 give: A001477, A000217, A000292, A070893.

%Y Rows n=0-2 give: A000004, A000012, A029744(k+2).

%Y Main diagonal gives A261490.

%K easy,nonn,tabl

%O 0,6

%A _Alois P. Heinz_, Oct 10 2009