A242308 Irregular triangular array of numerators of the positive rational numbers ordered as in Comments.

1, 1, 2, 2, 1, 3, 3, 2, 5, 3, 5, 3, 1, 3, 4, 8, 5, 4, 8, 5, 2, 5, 3, 7, 13, 7, 8, 4, 7, 13, 7, 8, 4, 1, 3, 4, 8, 5, 5, 11, 11, 21, 12, 7, 13, 7, 5, 11, 11, 21, 12, 7, 13, 7, 2, 5, 3, 7, 13, 7, 8, 4, 9, 18, 10, 19, 34, 18, 19, 9, 12, 21, 11, 11, 5, 9, 18, 10

Offset

1,3

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 numerators, by rows:  1,1,2,2,1,3,3,2,5,3,5,3,1,3,4,8,5,4,8,5,...

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. A242574, A242360, A000045.

Sequence in context: A325035 A348331 A035387 * A011373 A321783 A327035

Adjacent sequences: A242305 A242306 A242307 * A242309 A242310 A242311

Keyword

nonn,easy,tabf,frac

Author

Clark Kimberling, Jun 07 2014

Status

approved