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

Author

Clark Kimberling, Jun 08 2014

Keywords

• nonn
• easy
• tabf
• frac

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.

Examples

			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,...
		

Links

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

Crossrefs

Cf. A243611, A243612, A243614, A000045.

Programs

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 *)