A243613 Irregular triangular array of denominators of the positive rational numbers ordered as in Comments.

1, 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 3, 1, 2, 3, 4, 3, 5, 5, 5, 1, 2, 3, 4, 3, 5, 5, 5, 6, 7, 8, 7, 4, 1, 2, 3, 4, 3, 5, 5, 5, 6, 7, 8, 7, 4, 7, 9, 11, 11, 7, 9, 8, 7, 1, 2, 3, 4, 3, 5, 5, 5, 6, 7, 8, 7, 4, 7, 9, 11, 11, 7, 9, 8, 7, 8, 11, 14, 15, 10, 14, 13, 12

Offset

1,4

Comments

Let F = A000045 (the Fibonacci numbers). Decree that (row 1) = (1) and (row 2) = (2). Thereafter, row n consists of F(n) numbers in decreasing order, specifically, F(n-1) numbers x+1 from x in row n-1, together with F(n-2) numbers x/(x+1) from x in row n-2. The resulting array is also obtained by deleting from the array at A243611 all except the positive numbers and then reversing the rows.

Links

Clark Kimberling, Table of n, a(n) for n = 1..1500

Example

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

Mathematica

z = 12; g[1] = {0}; f1[x_] := x + 1; 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 = Table[Reverse[Drop[g[n], Fibonacci[n - 1]]], {n, 2, z}]
Delete[Flatten[Denominator[u]], 6]  (* A243611 *)
Delete[Flatten[Numerator[u]], 6]    (* A243612 *)
Delete[Flatten[Denominator[v]], 2]  (* A243613 *)
Delete[Flatten[Numerator[v]], 2]    (* A243614 *)

Crossrefs

Cf. A243611, A243612, A243614, A000045.

Sequence in context: A120418 A120853 A175022 * A194844 A138528 A364604

Adjacent sequences: A243610 A243611 A243612 * A243614 A243615 A243616

Keyword

nonn,easy,tabf,frac

Author

Clark Kimberling, Jun 08 2014

Status

approved