

A242359


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


7



1, 1, 1, 2, 1, 2, 3, 1, 2, 3, 3, 4, 1, 2, 3, 3, 4, 4, 5, 5, 1, 2, 3, 3, 4, 4, 5, 5, 5, 5, 7, 7, 6, 1, 2, 3, 3, 4, 4, 5, 5, 5, 5, 7, 7, 6, 6, 7, 7, 9, 8, 10, 9, 7, 1, 2, 3, 3, 4, 4, 5, 5, 5, 5, 7, 7, 6, 6, 7, 7, 9, 8, 10, 9, 7, 7, 9, 10, 8, 9, 11, 12, 11, 13
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OFFSET

1,4


COMMENTS

Decree that row 1 is (1) and row 2 is (2). For n >=3, row n consists of numbers in decreasing order generated as follows: x+1 for each x in row n1 together with 1/(1+x) for each x in row n2. It is easy to prove that row n consists of F(n) numbers, where F = A000045 (the Fibonacci numbers), and that every positive rational number occurs exactly once.


LINKS



EXAMPLE

First 6 rows of the array of rationals:
1/1
2/1
3/1 ... 1/2
4/1 ... 3/2 ... 1/3
5/1 ... 5/2 ... 4/3 ... 2/3 ... 1/4
6/1 ... 7/2 ... 7/3 ... 5/3 ... 5/4 ... 3/4 ... 2/5 ... 1/5
The denominators, by rows: 1,1,1,2,1,2,3,1,2,3,3,4,1,2,3,3,4,4,5,5.


MATHEMATICA

z = 12; g[1] = {1}; f1[x_] := x + 1; f2[x_] := 1/x; 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]]]; t = Table[Reverse[g[n]], {n, 1, z}]
Denominator[Flatten[t]] (* A242359 *)
Numerator[Flatten[t]] (* A242360 *)


CROSSREFS



KEYWORD

nonn,easy,tabf,frac


AUTHOR



STATUS

approved



