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A243574
Irregular triangular array of denominators of the positive rational numbers ordered as in Comments.
5
1, 2, 3, 1, 3, 5, 2, 5, 8, 4, 3, 1, 4, 8, 7, 13, 7, 3, 5, 2, 7, 13, 7, 12, 21, 11, 11, 5, 5, 8, 4, 3, 1, 5, 11, 11, 21, 12, 9, 19, 18, 34, 19, 10, 18, 9, 4, 8, 7, 13, 7, 3, 5, 2, 9, 18, 10, 19, 34, 18, 19, 9, 16, 31, 17, 31, 55, 29, 30, 14, 17, 29, 15, 14, 6
OFFSET
1,2
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
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 denominators, by rows: 1,2,3,1,3,5,2,5,8,4,3,1,4,8,7,13,7,3,5,2,...
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
KEYWORD
nonn,easy,tabf,frac
AUTHOR
Clark Kimberling, Jun 07 2014
STATUS
approved