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%I #5 Jun 11 2014 21:20:01
%S 1,2,3,1,4,3,2,5,5,5,3,1,6,7,8,7,4,4,3,2,7,9,11,11,7,9,8,7,5,5,5,3,1,
%T 8,11,14,15,10,14,13,12,11,12,13,10,5,6,7,8,7,4,4,3,2,9,13,17,19,13,
%U 19,18,17,17,19,21,17,9,13,16,19,18,11,13,11,9,7
%N Irregular triangular array of numerators of the positive rational numbers ordered as in Comments.
%C Let F = A000045 (the Fibonacci numbers). Decree that (row 1) = (1) and (row 2) = (2). Thereafter, row n consists of F(n) numbers in decreasing order, specifically, F(n-1) numbers x+1 from x in row n-1, together with F(n-2) numbers x/(x+1) from x in row n-2. The resulting array is also obtained by deleting from the array at A243611 all except the positive numbers and then reversing the rows.
%H Clark Kimberling, <a href="/A243614/b243614.txt">Table of n, a(n) for n = 1..1500</a>
%e First 6 rows of the array of all positive rationals:
%e 1/1
%e 2/1
%e 3/1 .. 1/2
%e 4/1 .. 3/2 .. 2/3
%e 5/1 .. 5/2 .. 5/3 .. 3/4 .. 1/3
%e 6/1 .. 7/2 .. 8/3 .. 7/4 .. 4/3 .. 4/5 .. 3/5 .. 2/5
%e The numerators, by rows: 1,2,3,1,4,3,2,5,5,5,3,1,6,7,8,7,4,4,3,2...
%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 *)
%Y Cf. A243611, A243612, A243613, A000045.
%K nonn,easy,tabf,frac
%O 1,2
%A _Clark Kimberling_, Jun 08 2014
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