A242359 Irregular triangular array of denominators of the positive rational numbers ordered as in Comments.

1, 1, 1, 2, 1, 2, 3, 1, 2, 3, 3, 4, 1, 2, 3, 3, 4, 4, 5, 5, 1, 2, 3, 3, 4, 4, 5, 5, 5, 5, 7, 7, 6, 1, 2, 3, 3, 4, 4, 5, 5, 5, 5, 7, 7, 6, 6, 7, 7, 9, 8, 10, 9, 7, 1, 2, 3, 3, 4, 4, 5, 5, 5, 5, 7, 7, 6, 6, 7, 7, 9, 8, 10, 9, 7, 7, 9, 10, 8, 9, 11, 12, 11, 13

Offset

1,4

Comments

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

Links

Clark Kimberling, Table of n, a(n) for n = 1..2500

Example

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

Mathematica

z = 12; g[1] = {1}; f1[x_] := x + 1; f2[x_] := 1/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]]]; t = Table[Reverse[g[n]], {n, 1, z}]
Denominator[Flatten[t]] (* A242359 *)
Numerator[Flatten[t]]   (* A242360 *)

Crossrefs

Cf. A226080, A242360, A000045.

Sequence in context: A369320 A115452 A039676 * A368485 A113126 A348043

Adjacent sequences: A242356 A242357 A242358 * A242360 A242361 A242362

Keyword

nonn,easy,tabf,frac

Author

Clark Kimberling, Jun 07 2014

Status

approved