A243732 Irregular triangular array of denominators of the positive rational numbers ordered as in Comments.

1, 1, 1, 1, 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 7, 3, 1, 2, 3, 4, 5, 6, 7, 3, 8, 5, 5, 1, 2, 3, 4, 5, 6, 7, 3, 8, 5, 5, 9, 7, 8, 7, 1, 2, 3, 4, 5, 6, 7, 3, 8, 5, 5, 9, 7, 8, 7, 10, 9, 11, 11, 9, 1, 2, 3, 4, 5, 6

Offset

1,7

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 = 5 and (row n) = h(n); the number of numbers in h(n) is A003520(n-1). (For m = 3, see A243712.)

Links

Clark Kimberling, Table of n, a(n) for n = 1..1000

Example

First 9 rows of the array:
1/1
2/1
3/1
4/1
5/1
6/1 ... 1/2
7/1 ... 3/2 ... 2/3
8/1 ... 5/2 ... 5/3 ... 3/4
9/1 ... 7/2 ... 8/3 ... 7/4 ... 4/5
10/1 .. 9/2 ... 11/3 .. 11/4 .. 9/5 ... 5/6
11/1 .. 11/2 .. 14/3 .. 15/4 .. 14/5 .. 11/6 .. 6/7 .. 1/3
The denominators, by rows:  1,1,1,1,1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,6,1,2,3,4,5,6,7,3,...

Mathematica

z = 23; g[1] = {1}; g[2] = {2}; g[3] = {3}; g[4] = {4}; g[5] = {5};
g[n_] := Reverse[Union[1 + g[n - 1], g[n - 5]/(1 + g[n - 5])]]
Table[g[n], {n, 1, 9}]
v = Flatten[Table[g[n], {n, 1, z}]];
v1 = Denominator[v]; (* A243732 *)
v2 = Numerator[v];   (* A243733 *)

Crossrefs

Cf. A243731, A243712, A243733, A003269.

Sequence in context: A136261 A140756 A002260 * A194905 A243730 A133994

Adjacent sequences: A243729 A243730 A243731 * A243733 A243734 A243735

Keyword

nonn,easy,tabf,frac

Author

Clark Kimberling, Jun 09 2014

Status

approved