A243613 - OEIS

%I #7 Jun 11 2014 21:19:48

%S 1,1,1,2,1,2,3,1,2,3,4,3,1,2,3,4,3,5,5,5,1,2,3,4,3,5,5,5,6,7,8,7,4,1,

%T 2,3,4,3,5,5,5,6,7,8,7,4,7,9,11,11,7,9,8,7,1,2,3,4,3,5,5,5,6,7,8,7,4,

%U 7,9,11,11,7,9,8,7,8,11,14,15,10,14,13,12

%N Irregular triangular array of denominators of the positive rational numbers ordered as in Comments.

%C Let F = A000045 (the Fibonacci numbers).  Decree that (row 1) = (1) and (row 2) = (2).  Thereafter, row n consists of F(n) numbers in decreasing order, specifically, F(n-1) numbers x+1 from x in row n-1, together with F(n-2) numbers x/(x+1) from x in row n-2.  The resulting array is also obtained by deleting from the array at A243611 all except the positive numbers and then reversing the rows.

%H Clark Kimberling, <a href="/A243613/b243613.txt">Table of n, a(n) for n = 1..1500</a>

%e First 6 rows of the array of all positive rationals:

%e 1/1

%e 2/1

%e 3/1 .. 1/2

%e 4/1 .. 3/2 .. 2/3

%e 5/1 .. 5/2 .. 5/3 .. 3/4 .. 1/3

%e 6/1 .. 7/2 .. 8/3 .. 7/4 .. 4/3 .. 4/5 .. 3/5 .. 2/5

%e The denominators, by rows:  1,1,1,2,1,2,3,1,2,3,4,3,1,2,3,4,3,5,5,5,...

%t z = 12; g[1] = {0}; f1[x_] := x + 1; 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}]

%t v = Table[Reverse[Drop[g[n], Fibonacci[n - 1]]], {n, 2, z}]

%t Delete[Flatten[Denominator[u]], 6]  (* A243611 *)

%t Delete[Flatten[Numerator[u]], 6]    (* A243612 *)

%t Delete[Flatten[Denominator[v]], 2]  (* A243613 *)

%t Delete[Flatten[Numerator[v]], 2]    (* A243614 *)

%Y Cf. A243611, A243612, A243614, A000045.

%K nonn,easy,tabf,frac

%O 1,4

%A _Clark Kimberling_, Jun 08 2014