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A371279
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Irregular triangular array of numerators of the set T of fractions generated by these rules: g(1) = (1), and if x and y are in T, then x/(y+1) is in T; see Comments.
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1
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1, 1, 2, 1, 1, 3, 4, 3, 3, 2, 3, 4, 8, 1, 1, 3, 1, 3, 2, 4, 5, 5, 4, 10, 5, 8, 9, 15, 8, 6, 20, 5, 16, 9, 15, 5, 5, 2, 5, 5, 4, 9, 15, 3, 10, 5, 8, 9, 15, 10, 8, 20, 10, 16, 6, 16, 40, 32, 6, 10, 5, 5, 1, 5, 5, 2, 9, 15, 3, 5, 5, 4, 9, 15, 5, 4, 10, 5, 8, 3
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OFFSET
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1,3
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COMMENTS
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Starting with g(1) = (1), write the numbers in the ordered union of g(1), g(2),…, g(n) as (x(1),x(2),…,x(m)). Then for i=1..m, write x(i)/(1 + x(j)) for j = 1..m, and expel all the numbers that have previously occurred. The result is ordered union of g(1), g(2),..., g(n+1). The cardinalities of the first 7 unions are 1, 2, 5, 20, 245, 38179, 1032578826.
Conjecture: every rational number in the interval (0,1] occurs exactly once in T.
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LINKS
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EXAMPLE
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Successive generations:
g(1) = (1)
g(2) = (1/2)
g(3) = (2/3, 1/4, 1/3)
g(4) = (3/5, 4/5, 3/4, 3/10, 2/5, 3/8, 4/9, 8/15, 1/8, 1/6, 3/20, 1/5, 3/16, 2/9, 4/15)
Let U(n) = ordered union of g(1), g(2), ..., g(n).
U(1) = (1)
U(2) = (1, 1/2)
U(3) = (1, 1/2, 2/3, 1/4, 1/3)
U(4) = (1, 1/2, 2/3, 1/4, 1/3, 3/5, 4/5, 3/4, 3/10, 2/5, 3/8, 4/9, 8/15, 1/8, 1/6, 3/20, 1/5, 3/16, 2/9, 4/15)
Numerators in U(4): 1, 1, 2, 1, 1, 3, 4, 3, 3, 2, 3, 4, 8, 1, 1, 3, 1, 3, 2, 4.
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MATHEMATICA
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(* In the remarks below, U(n) = ordered union of generations g(1), g(2), ...g(n) *)
x = {1};
x = DeleteDuplicates[Join[x, Map[x[[#[[1]]]]/(1 + x[[#[[2]]]]) &, Tuples[Range[Length[x]], {2}]]]] (* U(2) *)
x = DeleteDuplicates[Join[x, Map[x[[#[[1]]]]/(1 + x[[#[[2]]]]) &, Tuples[Range[Length[x]], {2}]]]] (* U(3) *)
x = DeleteDuplicates[Join[x, Map[x[[#[[1]]]]/(1 + x[[#[[2]]]]) &, Tuples[Range[Length[x]], {2}]]]] (* U(4) *)
Numerator[x] (* this sequence *)
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CROSSREFS
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KEYWORD
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nonn,tabf,frac
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AUTHOR
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STATUS
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approved
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