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%I #5 Jun 14 2014 21:40:56
%S 1,1,1,1,3,1,3,2,1,3,2,5,5,1,3,2,5,5,3,4,3,1,3,2,5,5,3,4,7,11,5,11,7,
%T 1,3,2,5,5,3,4,7,11,5,11,7,7,7,6,8,7,7,4,1,3,2,5,5,3,4,7,11,5,11,7,7,
%U 7,6,8,7,9,17,4,9,21,17,11,5,17,21,9,17,9
%N Irregular triangular array of denominators of the positive rational numbers ordered as in Comments.
%C Decree that (row 1) = (1), (row 2) = (2), and (row 3) = (3). For n >= 4, row n consists of numbers in decreasing order generated as follows: x+1 for each x in row n-1 together with 2/x for each x in row n-1, and duplicates are rejected as they occur. Every positive rational number occurs exactly once in the resulting array. Let c(n) be the number of numbers in (row n); it appears that (c(n)) is not linearly recurrent.
%H Clark Kimberling, <a href="/A243848/b243848.txt">Table of n, a(n) for n = 1..3000</a>
%e First 6 rows of the array of rationals:
%e 1/1
%e 2/1
%e 3/1
%e 4/1 ... 2/3
%e 5/1 ... 5/3 ... 1/2
%e 6/1 ... 8/3 ... 3/2 ... 6/5 ... 2/5
%e The denominators, by rows: 1,1,1,1,3,1,3,2,1,3,3,2,5,5.
%t z = 12; 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[Reverse[g[n]], {n, 1, z}]; v = Flatten[u];
%t Denominator[v] (* A243848 *)
%t Numerator[v] (* A243849 *)
%t Table[Length[g[n]], {n, 1, z}] (* A243850 *)
%Y Cf. A243849, A243850, A243571.
%K nonn,easy,tabf,frac
%O 1,5
%A _Clark Kimberling_, Jun 12 2014
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