A243574 Irregular triangular array of denominators of the positive rational numbers ordered as in Comments.

1, 2, 3, 1, 3, 5, 2, 5, 8, 4, 3, 1, 4, 8, 7, 13, 7, 3, 5, 2, 7, 13, 7, 12, 21, 11, 11, 5, 5, 8, 4, 3, 1, 5, 11, 11, 21, 12, 9, 19, 18, 34, 19, 10, 18, 9, 4, 8, 7, 13, 7, 3, 5, 2, 9, 18, 10, 19, 34, 18, 19, 9, 16, 31, 17, 31, 55, 29, 30, 14, 17, 29, 15, 14, 6

Offset: 1

Author

Clark Kimberling, Jun 07 2014

Keywords

• nonn
• easy
• tabf
• frac

Comments

Decree that row 1 is (1) and row 2 is (1/2). For n >=3, row n consists of numbers in increasing order generated as follows: 1/(x + 1) for each x in row n-1 together with x + 1 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 rational number occurs exactly once.

Examples

			First 6 rows of the array of rationals:
1/1
1/2
2/3 ... 2/1
1/3 ... 3/5 ... 3/2
2/5 ... 5/8 ... 3/4 ... 5/3 ... 3/1
1/4 ... 3/8 ... 4/7 ... 8/13 .. 5/7 .. 4/3 .. 8/5 .. 5/2
The denominators, by rows:  1,2,3,1,3,5,2,5,8,4,3,1,4,8,7,13,7,3,5,2,...
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..6000

Crossrefs

Cf. A242308, A242359, A000045.

Programs

Mathematica

z = 18; g[1] = {1}; f1[x_] := 1/x; f2[x_] := 1/(x + 1); 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];
Denominator[v]; (* A243574 *)
Numerator[v];   (* A242308 *)