

A243730


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


3



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

1,6


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(n1), {x/(x+1) : x in h(nm)}), 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(n1). (For m = 3, see A243712.)


LINKS



EXAMPLE

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 denominators, by rows: 1,1,1,1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,1,2,3,4,5,6,3,...


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}]];


CROSSREFS



KEYWORD

nonn,easy,tabf,frac


AUTHOR



STATUS

approved



