A243572 Irregular triangular array generated as in Comments; contains every positive integer exactly once.

1, 2, 3, 4, 6, 9, 5, 7, 10, 12, 18, 27, 8, 11, 13, 15, 19, 21, 28, 30, 36, 54, 81, 14, 16, 20, 22, 24, 29, 31, 33, 37, 39, 45, 55, 57, 63, 82, 84, 90, 108, 162, 243, 17, 23, 25, 32, 34, 38, 40, 42, 46, 48, 56, 58, 60, 64, 66, 72, 83, 85, 87, 91, 93, 99, 109

Offset: 1

Author

Clark Kimberling, Jun 07 2014

Keywords

• nonn
• easy
• tabf

Comments

Decree that row 1 is (1), row 2 is (2, 3), and row 3 is (4, 6, 9). Let r(n) = A001590(n+2), so that r(r) = r(n-1) + r(n-2) + r(n-3) with r(1) =1, r(2) = 2, r(3) = 3. Row n of the array, for n >= 4, consists of the numbers, in increasing order, defined as follows: all 3*x from x in row n-1, together with all 1 + 3*x from x in row n-2, together with all 2 + 3*x from x in row n-3. Thus, the number of numbers in row n is r(n), a tribonacci number. Every positive integer occurs exactly once in the array, so that the resulting sequence is a permutation of the positive integers.

Examples

			First 5 rows of the array:
1
2 ... 3
4 ... 6 ... 9
5 ... 7 ... 10 .. 12 .. 18 .. 27
8 ... 11 .. 13 .. 15 .. 19 .. 21 .. 28 .. 30 .. 36 .. 54 .. 81
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1500

Crossrefs

Cf. A232561, A243571, A243573, A232561, A002590.

Programs

Mathematica

z = 10; g[1] = {1}; f1[x_] := x + 1; f2[x_] := 3 x; h[1] = g[1];
b[n_] := b[n] = DeleteDuplicates[Union[f1[g[n - 1]], f2[g[n - 1]]]];
h[n_] := h[n] = Union[h[n - 1], g[n - 1]];
g[n_] := g[n] = Complement [b[n], Intersection[b[n], h[n]]]
u = Table[g[n], {n, 1, z}]; v = Flatten[u]  (* A243572 *)