A243614 Irregular triangular array of numerators of the positive rational numbers ordered as in Comments.

1, 2, 3, 1, 4, 3, 2, 5, 5, 5, 3, 1, 6, 7, 8, 7, 4, 4, 3, 2, 7, 9, 11, 11, 7, 9, 8, 7, 5, 5, 5, 3, 1, 8, 11, 14, 15, 10, 14, 13, 12, 11, 12, 13, 10, 5, 6, 7, 8, 7, 4, 4, 3, 2, 9, 13, 17, 19, 13, 19, 18, 17, 17, 19, 21, 17, 9, 13, 16, 19, 18, 11, 13, 11, 9, 7

Offset

1,2

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 numerators, by rows:  1,2,3,1,4,3,2,5,5,5,3,1,6,7,8,7,4,4,3,2...

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, A243613, A000045.

Sequence in context: A006021 A002186 A125936 * A200942 A286234 A161621

Adjacent sequences: A243611 A243612 A243613 * A243615 A243616 A243617

Keyword

nonn,easy,tabf,frac

Author

Clark Kimberling, Jun 08 2014

Status

approved