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%I #4 Jun 08 2014 09:17:23
%S 1,1,2,2,1,3,3,2,5,3,5,3,1,3,4,8,5,4,8,5,2,5,3,7,13,7,8,4,7,13,7,8,4,
%T 1,3,4,8,5,5,11,11,21,12,7,13,7,5,11,11,21,12,7,13,7,2,5,3,7,13,7,8,4,
%U 9,18,10,19,34,18,19,9,12,21,11,11,5,9,18,10
%N Irregular triangular array of numerators of the positive rational numbers ordered as in Comments.
%C Decree that row 1 is (1) and row 2 is (1/2). For n >=3, row n consists of numbers in increasing order generated as follows: 1/(x + 1) for each x in row n-1 together with x + 1 for each x in row n-2. It is easy to prove that row n consists of F(n) numbers, where F = A000045 (the Fibonacci numbers), and that every positive rational number occurs exactly once.
%H Clark Kimberling, <a href="/A242308/b242308.txt">Table of n, a(n) for n = 1..6000</a>
%e First 6 rows of the array of rationals:
%e 1/1
%e 1/2
%e 2/3 ... 2/1
%e 1/3 ... 3/5 ... 3/2
%e 2/5 ... 5/8 ... 3/4 ... 5/3 ... 3/1
%e 1/4 ... 3/8 ... 4/7 ... 8/13 .. 5/7 .. 4/3 .. 8/5 .. 5/2
%e The numerators, by rows: 1,1,2,2,1,3,3,2,5,3,5,3,1,3,4,8,5,4,8,5,...
%t z = 18; g[1] = {1}; f1[x_] := 1/x; 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}]; v = Flatten[u];
%t Denominator[v]; (* A243574 *)
%t Numerator[v]; (* A242308 *)
%Y Cf. A242574, A242360, A000045.
%K nonn,easy,tabf,frac
%O 1,3
%A _Clark Kimberling_, Jun 07 2014
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