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%I #4 Jun 08 2014 09:17:41
%S 1,2,3,1,4,3,1,5,5,4,2,1,6,7,7,5,5,3,2,1,7,9,10,8,9,7,7,6,4,3,3,2,1,8,
%T 11,13,11,13,11,12,11,9,8,10,9,7,5,5,4,4,3,3,2,1,9,13,16,14,17,15,17,
%U 16,14,13,17,16,13,11,12,11,13,11,13,11,8,6,7
%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 (2). For n >=3, row n consists of numbers in decreasing order generated as follows: x+1 for each x in row n-1 together with 1/(1+x) 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="/A242360/b242360.txt">Table of n, a(n) for n = 1..2500</a>
%e First 6 rows of the array of rationals:
%e 1/1
%e 2/1
%e 3/1 ... 1/2
%e 4/1 ... 3/2 ... 1/3
%e 5/1 ... 5/2 ... 4/3 ... 2/3 ... 1/4
%e 6/1 ... 7/2 ... 7/3 ... 5/3 ... 5/4 ... 3/4 ... 2/5 ... 1/5
%e The numeratorss, by rows: 1,2,3,1,4,3,1,5,5,4,2,1,6,7,7,5,5,3,2,1.
%t z = 12; 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]]]; t = Table[Reverse[g[n]], {n, 1, z}]
%t Denominator[Flatten[t]] (* A242359 *)
%t Numerator[Flatten[t]] (* A242360 *)
%Y Cf. A226080, A242359, A000045.
%K nonn,easy,tabf,frac
%O 1,2
%A _Clark Kimberling_, Jun 07 2014
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