%I #5 Jun 11 2014 21:19:24
%S 1,1,1,2,1,1,3,2,1,2,3,4,3,2,1,1,3,5,5,5,3,4,3,2,1,2,3,4,4,7,8,7,6,5,
%T 5,5,3,4,3,2,1,1,3,5,5,5,7,8,9,7,11,11,9,7,4,7,8,7,6,5,5,5,3,4,3,2,1,
%U 2,3,4,4,7,8,7,6,5,10,13,12,11,12,13,14
%N Irregular triangular array of denominators of all rational numbers ordered as in Comments.
%C Let F = A000045 (the Fibonacci numbers). Row n of the array to be generated consists of F(n-1) nonnegative rationals together with F(n-1) negative rationals. The nonnegatives, for n >=3, are x + 1 from the F(n-2) nonnegative numbers x in row n-1, together with x/(x + 1) from the F(n-3) nonnegative numbers x in row n-2. The negatives in row n are the negative reciprocals of the positives in row n.
%H Clark Kimberling, <a href="/A243611/b243611.txt">Table of n, a(n) for n = 1..3000</a>
%e First 6 rows of the array of all rationals:
%e 0/1
%e -1/1 .. 1/1
%e -1/2 .. 2/1
%e -2/1 .. -1/3 .. 1/2 ... 3/1
%e -3/2 .. -2/3 .. -1/4 .. 2/3 ... 3/2 ... 4/1
%e -3/1 .. -4/3 .. -3/5 .. -2/5 .. -1/5 .. 1/3 . 3/4 . 5/3 . 5/2 . 5/1
%e The denominators, by rows: 1,1,1,2,1,1,3,2,1,2,3,4,3,2,1,1,3,5,5,3,4,3,2,1,...
%t z = 12; g[1] = {0}; f1[x_] := x + 1; f2[x_] := -1/(x + 1); h[1] = g[1];
%t 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}]
%t v = Table[Reverse[Drop[g[n], Fibonacci[n - 1]]], {n, 2, z}]
%t Delete[Flatten[Denominator[u]], 6] (* A243611 *)
%t Delete[Flatten[Numerator[u]], 6] (* A243612 *)
%t Delete[Flatten[Denominator[v]], 2] (* A243613 *)
%t Delete[Flatten[Numerator[v]], 2] (* A243614 *)
%t ListPlot[g[20]]
%Y Cf. A243612, A243613, A243614, A226130, A000045.
%K nonn,easy,tabf,frac
%O 1,4
%A _Clark Kimberling_, Jun 08 2014