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

1,2

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.

Links

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

Example

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

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 *)

Crossrefs

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

Sequence in context: A111792 A113197 A113199 * A378194 A233559 A285332

Adjacent sequences: A243569 A243570 A243571 * A243573 A243574 A243575

Keyword

nonn,easy,tabf

Author

Clark Kimberling, Jun 07 2014

Status

approved