cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A226130 Denominators 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.)

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 4, 3, 2, 1, 1, 5, 4, 3, 2, 2, 3, 1, 6, 5, 4, 3, 3, 5, 2, 5, 3, 1, 7, 6, 5, 4, 4, 7, 3, 8, 5, 2, 7, 5, 3, 1, 1, 8, 7, 6, 5, 5, 9, 4, 11, 7, 3, 11, 8, 5, 2, 2, 9, 7, 5, 3, 3, 4, 1, 9, 8, 7, 6, 6, 11, 5, 14, 9, 4, 15, 11, 7, 3, 3
Offset: 1

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Author

Clark Kimberling, May 28 2013

Keywords

Comments

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.
A226130: Denominators of terms of S'
A226131: Numerators of terms of S'
A226136: Positions of positive integers in S'
A226137: Positions of integers in S'
The length of row n is given by A226275(n-1). - Peter Kagey, Jan 17 2022

Examples

			The denominators and numerators are read from the rationals in S':
  1/1, 2/1, -1/1, 3/1, -1/2, 0/1, 4/1, -1/3, 1/2, ...
Table begins:
  n |
  --+-----------------------------------------------
  1 | 1;
  2 | 1, 1;
  3 | 1, 2, 1;
  4 | 1, 3, 2;
  5 | 1, 4, 3, 2, 1;
  6 | 1, 5, 4, 3, 2, 2, 3;
  7 | 1, 6, 5, 4, 3, 3, 5, 2, 5, 3;
  8 | 1, 7, 6, 5, 4, 4, 7, 3, 8, 5, 2, 7, 5, 3, 1;
		

Crossrefs

Cf. A226080 (rabbit ordering of positive rationals).
Cf. A226247 (analogous with "0 is in S").

Programs

  • Mathematica
    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 *)
    Map[g, Range[z]]; Table[Length[g[i]], {i, 1, z}] (* cf. A003410 *)
    f = Flatten[Map[g, Range[z]]];
    Take[Denominator[f], 100] (* A226130 *)
    Take[Numerator[f], 100]   (* A226131 *)
    p1 = Flatten[Table[Position[f, n], {n, 1, z}]] (* A226136 *)
    p2 = Flatten[Table[Position[f, -n], {n, 0, z}]];
    Union[p1, p2]  (* A226137 *)  (* Peter J. C. Moses, May 26 2013 *)
  • Python
    from fractions import Fraction
    from itertools import count, islice
    def agen():
        rats = [Fraction(1, 1)]
        seen = {Fraction(1, 1)}
        for n in count(1):
            yield from [r.denominator for r in rats]
            newrats = []
            for r in rats:
                f = 1+r
                if f not in seen:
                    newrats.append(1+r)
                    seen.add(f)
                if r != 0:
                    g = -1/r
                    if g not in seen:
                        newrats.append(-1/r)
                        seen.add(g)
            rats = newrats
    print(list(islice(agen(), 84))) # Michael S. Branicky, Jan 17 2022

A226274 Position of 1/n in the ordering of the rationals given by application of the map t -> (1+t,-1/t), cf. A226130.

Original entry on oeis.org

1, 9, 31, 100, 317, 1000, 3150, 9918, 31223, 98289, 309406, 973981, 3065996, 9651448, 30381786, 95638797, 301061279, 947710512, 2983297009, 9391117780, 29562290606, 93059106094, 292940670339, 922147653681, 2902827709802, 9137808548505, 28764898718296, 90548996937472
Offset: 1

Views

Author

M. F. Hasler, Jun 01 2013

Keywords

Comments

An analog of the Fibonacci ordering of the positive rationals (cf. A226080) is the sequence of rationals produced from the initial vector [1] by appending iteratively the new rationals obtained by applying the map t-> (t+1, -1/t) to each term of the vector (cf. example).
It is seen that the unit fraction 1/n appears as the last term produced in the (3n-3)th iteration, therefore the indices a(n) equal every third terms in the partial sums of A226275 (= new terms produced during the respective iteration), cf. formula.

Examples

			Starting with [1], applying the map t->(1+t,-1/t) to the (most recently obtained) vector and discarding the numbers occurring earlier, one gets the sequence (grouped by "generation"): [1], [2, -1], [3, -1/2, 0], [4, -1/3, 1/2], [5, -1/4, 2/3, 3/2, -2], [6, -1/5, 3/4, 5/3, -3/2, 5/2, -2/3], [7, -1/6, 4/5, 7/4, -4/3, 8/3, -3/5, 7/2, -2/5, 1/3], [8, -1/7, 5/6, 9/5, -5/4, 11/4, -4/7, 11/3, -3/8, 2/5, 9/2, -2/7, 3/5, 4/3, -3],...
The unit fractions 1/1, 1/2, 1/3, 1/4,... occur at positions 1, 9(=1+2+3+3), 31(=9+5+7+10), 100(=31+15+22+32), ...
		

Crossrefs

Programs

  • PARI
    {print1([s=1]", ");U=Set(g=[1]); for(n=1,29,U=setunion(U,Set(g=select(f->!setsearch(U,f), concat(apply(t->[t+1,if(t,-1/t)],g))))); for(i=1,#g, numerator(g[i])==1&&print1(s+i/*",g[i],*/","));s+=#g)} /* illustrative purpose only */

Formula

a(n) = s(3n-3) where s(k) = Sum_{j=0..k} A226275(j).
O.g.f.: x(1 + 4*x - 7*x^2 + 4*x^3 - x^4)/((1 - x)(1 - 4*x + 3*x^2 - x^3)).
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