A243571 Irregular triangular array generated as in Comments; contains every positive integer exactly once.

1, 2, 3, 4, 5, 6, 8, 7, 9, 10, 12, 16, 11, 13, 14, 17, 18, 20, 24, 32, 15, 19, 21, 22, 25, 26, 28, 33, 34, 36, 40, 48, 64, 23, 27, 29, 30, 35, 37, 38, 41, 42, 44, 49, 50, 52, 56, 65, 66, 68, 72, 80, 96, 128, 31, 39, 43, 45, 46, 51, 53, 54, 57, 58, 60, 67, 69

Offset: 1

Author

Clark Kimberling, Jun 07 2014

Keywords

• nonn
• easy
• tabf

Comments

Decree that row 1 is (1) and row 2 is (2). For n >= 3, row n consists of numbers in increasing order generated as follows: 2*x for each x in row n-1 together with 1+2*x for each x in row n-2. It is easy to prove that row n consists of F(n) numbers, where F = A000045 (the Fibonacci numbers), and that every positive integer occurs exactly once. Row n has F(n-1) even numbers and F(n-2) odd numbers.

The least and greatest numbers in row n are A083329(n-1) and 2^(n-1), for n >= 1.

Examples

			First 6 rows of the array:
  1
  2
  3 ... 4
  5 ... 6 ... 8
  7 ... 9 ... 10 .. 12 .. 16
  11 .. 13 .. 14 .. 17 .. 18 .. 20 .. 24 .. 32
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1500
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Cf. A232559, A243572, A243573, A000045.

Cf. A052955 for the first element in each row.

Programs

Mathematica

z = 10; g[1] = {1}; f1[x_] := x + 1; f2[x_] := 2 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] (* A243571 *)