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%I #6 Jun 11 2014 21:20:26
%S 1,1,1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,3,1,2,3,4,5,3,6,5,5,1,2,3,4,5,3,6,
%T 5,5,7,7,8,7,1,2,3,4,5,3,6,5,5,7,7,8,7,8,9,11,11,9,4,1,2,3,4,5,3,6,5,
%U 5,7,7,8,7,8,9,11,11,9,4,9,11,14,15,14,7
%N Irregular triangular array of denominators of all positive rational numbers ordered as in Comments.
%C Decree that (row 1) = (1), (row 2) = (2), and (row 3) = (3). Thereafter, row n consists of the following numbers arranged in decreasing order: 1 + x for each x in (row n-1), together with x/(x + 1) for each x in row (n-3). Every positive rational number occurs exactly once in the array. The number of numbers in (row n) is A000930(n-1), for n >= 1.
%H Clark Kimberling, <a href="/A243712/b243712.txt">Table of n, a(n) for n = 1..1000</a>
%e First 8 rows of the array of all positive rationals:
%e 1/1
%e 2/1
%e 3/1
%e 4/1 .. 1/2
%e 5/1 .. 3/2 .. 2/3
%e 6/1 .. 5/2 .. 5/3 ... 3/4
%e 7/1 .. 7/2 .. 8/3 ... 7/4 ... 4/5 .. 1/3
%e 8/1 .. 9/2 .. 11/3 .. 11/4 .. 9/5 .. 4/3 .. 5/6 .. 3/5 .. 2/5
%e The denominators, by rows: 1,1,1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,3,1,2,3,4,5,3,6,5,5,...
%t z = 13; g[1] = {1}; f1[x_] := x + 1; 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]]; g[n_] := g[n] = Complement [b[n], Intersection[b[n], h[n]]]; u = Table[g[n], {n, 1, z}]; u1 = Delete[Flatten[u], 10]
%t w[1] = 0; w[2] = 1; w[3] = 1; w[n_] := w[n - 1] + w[n - 3];
%t u2 = Table[Drop[g[n], w[n]], {n, 1, z}];
%t u3 = Delete[Delete[Flatten[Map[Reverse, u2]], 4], 4]
%t Denominator[u3] (* A243712 *)
%t Numerator[u3] (* A243713 *)
%t Denominator[u1] (* A243714 *)
%t Numerator[u1] (* A243715 *)
%Y Cf. A243713, A243714, A243715, A000930, A243613.
%K nonn,easy,tabf,frac
%O 1,5
%A _Clark Kimberling_, Jun 09 2014
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