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A243574 Irregular triangular array of denominators of the positive rational numbers ordered as in Comments.

%I A243574 #4 Jun 08 2014 09:18:29
%S A243574 1,2,3,1,3,5,2,5,8,4,3,1,4,8,7,13,7,3,5,2,7,13,7,12,21,11,11,5,5,8,4,
%T A243574 3,1,5,11,11,21,12,9,19,18,34,19,10,18,9,4,8,7,13,7,3,5,2,9,18,10,19,
%U A243574 34,18,19,9,16,31,17,31,55,29,30,14,17,29,15,14,6
%N A243574 Irregular triangular array of denominators of the positive rational numbers ordered as in Comments.
%C A243574 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 n-1 together with x + 1 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.
%H A243574 Clark Kimberling, <a href="/A243574/b243574.txt">Table of n, a(n) for n = 1..6000</a>
%e A243574 First 6 rows of the array of rationals:
%e A243574 1/1
%e A243574 1/2
%e A243574 2/3 ... 2/1
%e A243574 1/3 ... 3/5 ... 3/2
%e A243574 2/5 ... 5/8 ... 3/4 ... 5/3 ... 3/1
%e A243574 1/4 ... 3/8 ... 4/7 ... 8/13 .. 5/7 .. 4/3 .. 8/5 .. 5/2
%e A243574 The denominators, by rows:  1,2,3,1,3,5,2,5,8,4,3,1,4,8,7,13,7,3,5,2,...
%t A243574 z = 18; g[1] = {1}; f1[x_] := 1/x; f2[x_] := 1/(x + 1); h[1] = g[1];
%t A243574 b[n_] := b[n] = DeleteDuplicates[Union[f1[g[n - 1]], f2[g[n - 1]]]];
%t A243574 h[n_] := h[n] = Union[h[n - 1], g[n - 1]];
%t A243574 g[n_] := g[n] = Complement [b[n], Intersection[b[n], h[n]]]
%t A243574 u = Table[g[n], {n, 1, z}]; v = Flatten[u];
%t A243574 Denominator[v]; (* A243574 *)
%t A243574 Numerator[v];   (* A242308 *)
%Y A243574 Cf. A242308, A242359, A000045.
%K A243574 nonn,easy,tabf,frac
%O A243574 1,2
%A A243574 _Clark Kimberling_, Jun 07 2014