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 A242308 Irregular triangular array of numerators of the positive rational numbers ordered as in Comments. 3
 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, 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, 9, 18, 10, 19, 34, 18, 19, 9, 12, 21, 11, 11, 5, 9, 18, 10 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS 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. LINKS Clark Kimberling, Table of n, a(n) for n = 1..6000 EXAMPLE First 6 rows of the array of rationals: 1/1 1/2 2/3 ... 2/1 1/3 ... 3/5 ... 3/2 2/5 ... 5/8 ... 3/4 ... 5/3 ... 3/1 1/4 ... 3/8 ... 4/7 ... 8/13 .. 5/7 .. 4/3 .. 8/5 .. 5/2 The numerators, by rows:  1,1,2,2,1,3,3,2,5,3,5,3,1,3,4,8,5,4,8,5,... MATHEMATICA z = 18; g[1] = {1}; f1[x_] := 1/x; f2[x_] := 1/(x + 1); h[1] = g[1]; b[n_] := b[n] = DeleteDuplicates[Union[f1[g[n - 1]], f2[g[n - 1]]]]; 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}]; v = Flatten[u]; Denominator[v]; (* A243574 *) Numerator[v];   (* A242308 *) CROSSREFS Cf. A242574, A242360, A000045. Sequence in context: A029266 A325035 A035387 * A011373 A321783 A327035 Adjacent sequences:  A242305 A242306 A242307 * A242309 A242310 A242311 KEYWORD nonn,easy,tabf,frac AUTHOR Clark Kimberling, Jun 07 2014 STATUS approved

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Last modified September 21 09:52 EDT 2021. Contains 347597 sequences. (Running on oeis4.)