%I #5 Jun 19 2014 11:18:17
%S 1,1,1,1,1,3,1,1,2,3,1,1,5,2,3,1,1,1,2,3,5,2,3,1,3,1,3,7,7,5,3,5,2,3,
%T 1,3,4,5,5,4,7,7,5,3,5,2,3,1,1,5,7,5,13,7,13,9,5,4,7,7,7,5,3,5,2,3,1,
%U 1,1,3,5,4,6,6,4,9,9,7,8,5,13,7,13,9,5
%N Irregular triangular array of denominators of the positive rational numbers ordered as in Comments.
%C Decree that (row 1) = (1). For n >=2, row n consists of numbers in increasing order generated as follows: x+1 for each x in row n1 together with 2/x for each nonzero x in row n1, where duplicates are deleted as they occur. The number of numbers in row n is A243927(n). Conjecture: every rational number occurs exactly once in the array.
%H Clark Kimberling, <a href="/A243925/b243925.txt">Table of n, a(n) for n = 1..2500</a>
%e First 7 rows of the array of rationals:
%e 1/1
%e 2/1 ... 2/1
%e 1/1 ... 3/1
%e 2/3 ... 0/1 ... 4/1
%e 1/2 ... 1/3 ... 5/1
%e 6/1 ... 2/5 .. 1/2 ... 4/3 ... 6/1
%e 5/1 ... 4/1 .. 3/2 .. 1/3 .. 3/5 .. 3/2 .. 7/3 .. 7/1
%e The denominators, by rows: 1,1,1,1,1,3,1,1,2,3,1,1,5,2,3,1,1,1,2,3,5,2,3,1.
%t z = 13; g[1] = {1}; f1[x_] := x + 1; f2[x_] := 2/x; 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 = Delete[Flatten[u], 12]
%t Denominator[v] (* A243925 *)
%t Numerator[v] (* A243926 *)
%Y Cf. A243926, A243927, A242359, A243712, A243928.
%K nonn,easy,tabf,frac
%O 1,6
%A _Clark Kimberling_, Jun 15 2014
