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A384001 Irregular triangle T(n,j,k), j = 1..A024718(n), k = 1..n, where row 1 = {(0), (1)}, and row n = union of n-tuples whose sum s < n, and the n-tuples formed by appending s to the (n-1)-tuples in row n-1.

%I A384001 #9 Jul 16 2025 19:58:34
%S A384001 0,1,0,0,0,1,0,2,1,0,1,1,0,0,0,0,0,1,0,0,2,0,0,3,0,1,0,0,1,1,0,1,2,0,
%T A384001 2,0,0,2,1,1,0,0,1,0,1,1,0,2,1,1,0,1,1,1,2,0,0,0,0,0,0,0,0,0,1,0,0,0,
%U A384001 2,0,0,0,3,0,0,0,4,0,0,1,0,0,0,1,1,0,0,1,2
%N A384001 Irregular triangle T(n,j,k), j = 1..A024718(n), k = 1..n, where row 1 = {(0), (1)}, and row n = union of n-tuples whose sum s < n, and the n-tuples formed by appending s to the (n-1)-tuples in row n-1.
%C A384001 Terms in row n are sorted lexicographically.
%C A384001 Row n is created by finding n-tuples w with elements from 0..n-1, taking only those w whose sums are less than n.
%C A384001 For example, row n = 3 contains 3-tuples w that have elements from 0..2, i.e., {(0,0,0), (0,0,1), (0,0,2), (0,1,0), (0,1,1), (0,2,0), (1,0,0), (1,0,1), (1,1,0), (2,0,0)}.
%C A384001 Let s be the sum of w. Then we take all elements w of row n-1 and append n-s to w to obtain certain 3-tuples with elements from 0..n whose sum s = n.
%C A384001 Continuing the example, row 2 = {(0,0), (0,1), (0,2), (1,0), (1,1)}, which, adding n-s to the right end gives {(0,0,3), (0,1,2), (0,2,1), (1,0,2), (1,1,1)}.
%C A384001 Let p_i be the i-th smallest prime divisor of N = A384000(n) (where i is not necessarily the i-th prime). Then, the terms m in row N of A162306 are of the form m = Product_{i..n} p_i^T(n,j,n-k+1). For instance, for N = 6, we have row 6 of A162306 = {1, 2, 3, 4, 6}, which is {2^0*3^0, 2^1*3^0, 2^2*3^0, 2^0*3^1, 2^1*3^1} = {1, 2, 4, 3, 6}, sorted.
%F A384001 Length of row n = n*A024718(n).
%e A384001 Table begins:
%e A384001   1:   (0), (1);
%e A384001   2:   (0, 0), (0, 1), (0, 2), (1, 0), (1, 1);
%e A384001   3:   (0,0,0), (0,0,1), (0,0,2), (0,0,3), (0,1,0),
%e A384001        (0,1,1), (0,1,2), (0,2,0), (0,2,1), (1,0,0),
%e A384001        (1,0,1), (1,0,2), (1,1,0), (1,1,1), (2,0,0)
%e A384001   etc.
%e A384001 Row 2 arranged as a rank 2 table, concatenating T(2,j,k), k = 1..2:
%e A384001 00   10   20
%e A384001 01   11
%e A384001 .
%e A384001 Row 3 arranged as a rank 3 table, concatenating T(3,j,k), k = 1..3:
%e A384001 000  001  002  003     100  101  102    200
%e A384001 010  011  012          110  111
%e A384001 020  021
%t A384001 nn = 4; w[0] = {{0}};
%t A384001 Do[If[n == 1, Set[w[1], {{0}, {1}}],
%t A384001   Set[w[n], Union@ Join[Select[Tuples[Range[0, n - 1], n], Total[#] < n &],
%t A384001     Map[Append[#, n - Total[#]] &, w[n - 1] ] ] ] ], {n, nn}];
%t A384001 Flatten@ Array[w, nn]
%Y A384001 Cf. A001700, A024718, A384000, A384002.
%K A384001 nonn,tabf,easy
%O A384001 1,8
%A A384001 _Michael De Vlieger_, May 21 2025