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

1,2

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.

Links

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

Example

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,...

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 *)

Crossrefs

Cf. A242308, A242359, A000045.

Sequence in context: A194760 A035517 A099471 * A121775 A127951 A208814

Adjacent sequences: A243571 A243572 A243573 * A243575 A243576 A243577

Keyword

nonn,easy,tabf,frac

Author

Clark Kimberling, Jun 07 2014

Status

approved