|
%I #13 Jan 18 2022 03:37:08
%S 1,2,4,7,10,15,22,32,47,69,101,148,217,318,466,683,1001,1467,2150,
%T 3151,4618,6768,9919,14537,21305,31224,45761,67066,98290,144051,
%U 211117,309407,453458,664575,973982
%N Positions of the positive integers in the ordering of rational numbers as generated by the rules: 1 is in S, and if nonzero x is in S, then x+1 and -1/x are in S. (See Comments.)
%C Let S be the set of numbers defined by these rules: 1 is in S, and if nonzero x is in S, then x + 1 and -1/x are in S. Then S is the set of all rational numbers, produced in generations as follows: g(1) = (1), g(2) = (2, -1), g(3) = (3, -1/2, 0), g(4) = (4, -1/3, 1/2), ... For n > 4, once g(n-1) = (c(1), ..., c(z)) is defined, g(n) is formed from the vector (c(1)+1, -1/c(1), c(2)+1, -1/c(2), ..., c(z)+1, -1/c(z)) by deleting previously generated elements. Let S' denote the sequence formed by concatenating the generations.
%C A226130: Denominators of terms of S'
%C A226131: Numerators of terms of S'
%C A226136: Positions of positive integers in S'
%C A226137: Positions of integers in S'
%H Clark Kimberling, <a href="/A226136/b226136.txt">Table of n, a(n) for n = 1..35</a>
%F Conjecture: a(n) = a(n-1)+a(n-3) for n>6. G.f.: -x*(x+1) * (x^2+1)^2 / (x^3+x-1). - _Colin Barker_, Jul 03 2013
%e S' = (1/1, 2/1, -1/1, 3/1, -1/2, 0/1, 4/1, -1/3, 1/2, ...), with positive integers appearing in positions 1,2,4,7,...
%t g[1] := {1}; z = 20; g[n_] := g[n] = DeleteCases[Flatten[Transpose[{# + 1, -1/#}]]&[DeleteCases[g[n - 1], 0]], Apply[Alternatives, Flatten[Map[g, Range[n - 1]]]]]; Flatten[Map[g, Range[7]]] (* ordered rationals *)
%t Map[g, Range[z]]; Table[Length[g[i]], {i, 1, z}] (* cf A003410 *)
%t f = Flatten[Map[g, Range[z]]];
%t Take[Denominator[f], 100] (* A226130 *)
%t Take[Numerator[f], 100] (* A226131 *)
%t p1 = Flatten[Table[Position[f, n], {n, 1, z}]] (* A226136 *)
%t p2 = Flatten[Table[Position[f, -n], {n, 0, z}]];
%t Union[p1, p2] (* A226137 *) (* _Peter J. C. Moses_, May 26 2013 *)
%Y Cf. A226080 (rabbit ordering of positive rationals).
%K nonn
%O 1,2
%A _Clark Kimberling_, May 28 2013
|
