A243731 Irregular triangular array of numerators of the positive rational numbers ordered as in Comments.

1, 2, 3, 4, 5, 1, 6, 3, 2, 7, 5, 5, 3, 8, 7, 8, 7, 4, 9, 9, 11, 11, 9, 5, 1, 10, 11, 14, 15, 14, 11, 4, 6, 3, 2, 11, 13, 17, 19, 19, 17, 7, 13, 8, 7, 7, 5, 5, 3, 12, 15, 20, 23, 24, 23, 10, 20, 13, 12, 15, 12, 13, 10, 8, 7, 8, 7, 4, 13, 17, 23, 27, 29, 29

Offset: 1

Author

Clark Kimberling, Jun 09 2014

Keywords

• nonn
• easy
• tabf
• frac

Comments

Suppose that m >= 3, and define sets h(n) of positive rational numbers as follows: h(n) = {n} for n = 1..m, and thereafter, h(n) = Union({x+1: x in h(n-1)}, {x/(x+1) : x in h(n-m)}), with the numbers in h(n) arranged in decreasing order. Every positive rational lies in exactly one of the sets h(n). For the present array, put m = 4 and (row n) = h(n); the number of numbers in h(n) is A003269(n-1). (For m = 3, see A243712.)

Examples

			First 9 rows of the array:
  1/1
  2/1
  3/1
  4/1
  5/1 .. 1/2
  6/1 .. 3/2 .. 2/3
  7/1 .. 5/2 .. 5/3 ... 3/4
  8/1 .. 7/2 .. 8/3 ... 7/4 ... 4/5
  9/1 .. 9/2 .. 11/3 .. 11/4 .. 9/5 .. 5/6 .. 1/3
The numerators, by rows:  1,2,3,4,5,1,6,3,2,7,5,5,3,8,7,8,7,4,9,9,11,11,9,5,1...
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..6000

Crossrefs

Cf. A243730, A243712, A003269.

Programs

Mathematica

z = 27; g[1] = {1}; g[2] = {2}; g[3] = {3}; g[4] = {4};
g[n_] := Reverse[Union[1 + g[n - 1], g[n - 4]/(1 + g[n - 4])]]
Table[g[n], {n, 1, 12}]
v = Flatten[Table[g[n], {n, 1, z}]];
Denominator[v];  (* A243730 *)
Numerator[v];  (* A243731 *)