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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 (list; graph; refs; listen; history; text; internal format)
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 n-1 together with 1/(1+x) for each x in row n-2.  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

Clark Kimberling, Table of n, a(n) for n = 1..2500

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

Cf. A226080, A242360, A000045.

Sequence in context: A194546 A115452 A039676 * A113126 A138060 A023121

Adjacent sequences:  A242356 A242357 A242358 * A242360 A242361 A242362

KEYWORD

nonn,easy,tabf,frac

AUTHOR

Clark Kimberling, Jun 07 2014

STATUS

approved

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Last modified May 6 00:41 EDT 2021. Contains 343579 sequences. (Running on oeis4.)