

A242308


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


3



1, 1, 2, 2, 1, 3, 3, 2, 5, 3, 5, 3, 1, 3, 4, 8, 5, 4, 8, 5, 2, 5, 3, 7, 13, 7, 8, 4, 7, 13, 7, 8, 4, 1, 3, 4, 8, 5, 5, 11, 11, 21, 12, 7, 13, 7, 5, 11, 11, 21, 12, 7, 13, 7, 2, 5, 3, 7, 13, 7, 8, 4, 9, 18, 10, 19, 34, 18, 19, 9, 12, 21, 11, 11, 5, 9, 18, 10
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,3


COMMENTS

Decree that row 1 is (1) and row 2 is (1/2). For n >=3, row n consists of numbers in increasing order generated as follows: 1/(x + 1) for each x in row n1 together with x + 1 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

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


EXAMPLE

First 6 rows of the array of rationals:
1/1
1/2
2/3 ... 2/1
1/3 ... 3/5 ... 3/2
2/5 ... 5/8 ... 3/4 ... 5/3 ... 3/1
1/4 ... 3/8 ... 4/7 ... 8/13 .. 5/7 .. 4/3 .. 8/5 .. 5/2
The numerators, by rows: 1,1,2,2,1,3,3,2,5,3,5,3,1,3,4,8,5,4,8,5,...


MATHEMATICA

z = 18; g[1] = {1}; f1[x_] := 1/x; f2[x_] := 1/(x + 1); 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]]]
u = Table[g[n], {n, 1, z}]; v = Flatten[u];
Denominator[v]; (* A243574 *)
Numerator[v]; (* A242308 *)


CROSSREFS

Cf. A242574, A242360, A000045.
Sequence in context: A029266 A325035 A035387 * A011373 A321783 A327035
Adjacent sequences: A242305 A242306 A242307 * A242309 A242310 A242311


KEYWORD

nonn,easy,tabf,frac


AUTHOR

Clark Kimberling, Jun 07 2014


STATUS

approved



