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%I #5 Mar 22 2024 17:40:18
%S 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,
%T 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,
%U 5,5,1,5,5,2,9,15,3,5,5,4,9,15,5,4,10,5,8,3
%N 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.
%C 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.
%C Conjecture: every rational number in the interval (0,1] occurs exactly once in T.
%e Successive generations:
%e g(1) = (1)
%e g(2) = (1/2)
%e g(3) = (2/3, 1/4, 1/3)
%e 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)
%e Let U(n) = ordered union of g(1), g(2), ..., g(n).
%e U(1) = (1)
%e U(2) = (1, 1/2)
%e U(3) = (1, 1/2, 2/3, 1/4, 1/3)
%e 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)
%e Numerators in U(4): 1, 1, 2, 1, 1, 3, 4, 3, 3, 2, 3, 4, 8, 1, 1, 3, 1, 3, 2, 4.
%t (* In the remarks below, U(n) = ordered union of generations g(1), g(2),...g(n) *)
%t x = {1};
%t x = DeleteDuplicates[Join[x, Map[x[[#[[1]]]]/(1 + x[[#[[2]]]]) &, Tuples[Range[Length[x]], {2}]]]] (* U(2) *)
%t x = DeleteDuplicates[Join[x, Map[x[[#[[1]]]]/(1 + x[[#[[2]]]]) &, Tuples[Range[Length[x]], {2}]]]] (* U(3) *)
%t x = DeleteDuplicates[Join[x, Map[x[[#[[1]]]]/(1 + x[[#[[2]]]]) &, Tuples[Range[Length[x]], {2}]]]] (* U(4) *)
%t Numerator[x] (* this sequence *)
%t Denominator[x] (* A371280 *)
%t (* _Peter J. C. Moses_, Mar 16 2024 *)
%Y Cf. A226080, A371280.
%K nonn,tabf,frac
%O 1,3
%A _Clark Kimberling_, Mar 18 2024
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