%I #4 Jun 11 2014 21:18:54
%S 1,1,2,2,3,3,1,3,5,3,5,2,5,8,4,5,8,4,3,1,4,8,7,13,7,4,8,7,13,7,3,5,2,
%T 7,13,7,12,21,11,11,5,7,13,7,12,21,11,11,5,5,8,4,3,1,5,11,11,21,12,9,
%U 19,18,34,19,10,18,9,5,11,11,21,12,9,19,18,34
%N Irregular triangular array of denominators of the positive rational numbers ordered as in Comments.
%C Let F = A000045 (the Fibonacci numbers). To construct the array of positive rationals, decree that row 1 is (1) and row 2 is (2). Thereafter, row n consists of the following numbers in increasing order: the F(n-2) numbers 1/x from numbers x > 1 in row n-1, together with the F(n-3) numbers 1 + 1/x from numbers x < 1 in row n - 1, together with the F(n - 2) numbers (2*x + 1)/ (x + 1) from numbers x in row n-2. Row n consists of F(n) numbers ranging from 1/((n+1)/2) to n/2 if n is odd and from 2/(n-1) to (n+2)/2 if n is even.
%H Clark Kimberling, <a href="/A242361/b242361.txt">Table of n, a(n) for n = 1..5000</a>
%e First 6 rows of the array of rationals:
%e 1/1
%e 2/1
%e 1/2 ... 3/2
%e 2/3 ... 5/3 ... 3/1
%e 1/3 ... 3/5 ... 4/3 ... 8/5 ... 5/2
%e 2/5 ... 5/8 ... 3/4 ... 7/5 ... 13/8 .. 7/4 ... 8/3 ... 4/1
%e The denominators, by rows: 1,1,2,2,3,3,1,3,5,3,5,2,5,8,4,5,8,4,3,1,...
%t z = 18; g[1] = {1}; f1[x_] := 1 + 1/x; f2[x_] := 1/x; h[1] = g[1]; b[n_] := b[n] = DeleteDuplicates[Union[f1[g[n - 1]], f2[g[n - 1]]]];
%t h[n_] := h[n] = Union[h[n - 1], g[n - 1]];
%t g[n_] := g[n] = Complement [b[n], Intersection[b[n], h[n]]]
%t u = Table[g[n], {n, 1, z}]; v = Flatten[u]; Length[v]
%t Denominator[v]; (* A242361 *)
%t Numerator[v]; (* A242363 *)
%Y Cf. A242363, A242359, A243574, A000045.
%K nonn,easy,tabf,frac
%O 1,3
%A _Clark Kimberling_, Jun 08 2014