A233696 -id:A233696 - cp's OEIS frontend

A245264 Number of numbers in n-th generation of the tree of Gaussian rationals at A233696.

1, 2, 6, 12, 26, 53, 110, 231, 483, 1013, 2125, 4445, 9307, 19487, 40802, 85439, 178910, 374622, 784426, 1642522

Offset: 1

Clark Kimberling, Jul 15 2014

A233696 gives rules for constructing the Gaussian rationals in generations as follows: g(1) = {0}, and for n > 1, if x is in g(n-1) then x+1, i*x, and 1/x (for x not 0) are in g(n), except for those that are in g(j) for some j < n. Conjecture: 2 < lim(|g(n+1)|/|g(n)|) < 3.

			g(1) = {1}, so a(n) = 1; g(2) = {2,i}, so a(n) = 2; g(3) = {3, 1/2, 2*i, 1+i, -i, -1}, so a(3) = 6.

Cf. A233696.

x = {0}; lenX = {}; Off[Power::infy]; Do[{x = DeleteDuplicates[    Flatten[Transpose[{x, x + 1, 1/x, I*x} /. ComplexInfinity -> 0]]], AppendTo[lenX, Length[x]]}, {15}]; On[Power::infy]; Join[{1}, Differences[lenX]]  (* Peter J. C. Moses, Dec 21 2013 *)

A226080 Denominators in the Fibonacci (or rabbit) ordering of the positive rational numbers.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 2, 1, 4, 3, 2, 3, 1, 5, 4, 3, 4, 2, 5, 3, 1, 6, 5, 4, 5, 3, 7, 4, 2, 7, 5, 3, 5, 1, 7, 6, 5, 6, 4, 9, 5, 3, 10, 7, 4, 7, 2, 9, 7, 5, 7, 3, 8, 5, 1, 8, 7, 6, 7, 5, 11, 6, 4, 13, 9, 5, 9, 3, 13, 10, 7, 10, 4, 11, 7, 2, 11, 9, 7, 9, 5, 12, 7

Offset: 1

Clark Kimberling, May 25 2013

Let S be the set of numbers defined by these rules: 1 is in S, and if x is in S, then x+1 and 1/x are in S. Then S is the set of positive rational numbers, which arise in generations as follows: g(1) = (1/1), g(2) = (1+1) = (2), g(3) = (2+1, 1/2) = (3/1, 1/2), g(4) = (4/1, 1/3, 3/2), ... . Once g(n-1) = (g(1), ..., g(z)) is defined, g(n) is formed from the vector (g(1) + 1, 1/g(1), g(2) + 1, 1/g(2), ..., g(z) + 1, 1/g(z)) by deleting all elements that are in a previous generation. A226080 is the sequence of denominators formed by concatenating the generations g(1), g(2), g(3), ... . It is easy to prove the following:
(1) Every positive rational is in S.
(2) The number of terms in g(n) is the n-th Fibonacci number, F(n) = A000045(n).
(3) For n > 2, g(n) consists of F(n-2) numbers < 1 and F(n-1) numbers > 1, hence the name "rabbit ordering" since the n-th generation has F(n-2) reproducing pairs and F(n-1) non-reproducing pairs, as in the classical rabbit-reproduction introduction to Fibonacci numbers.
(4) The positions of integers in S are the Fibonacci numbers.
(5) The positions of 1/2, 3/2, 5/2, ..., are Lucas numbers (A000032).
(6) Continuing from (4) and (5), suppose that n > 0 and 0 < r < n, where gcd(n,r) = 1.  The positions in A226080 of the numbers congruent to r mod n comprise a row of the Wythoff array, W = A035513. The correspondence is sampled here:
   row 1 of W: positions of n+1 for n>=0,
   row 2 of W: positions of n+1/2,
   row 3 of W: positions of n+1/3,
   row 4 of W: positions of n+1/4,
   row 5 of W: positions of n+2/3,
   row 6 of W: positions of n+1/5,
   row 7 of W: positions of n+3/4.
(7) If the numbers <=1 in S are replaced by 1 and those >1 by 0, the resulting sequence is the infinite Fibonacci word A003849 (except for the 0-offset first term).
(8) The numbers <=1 in S occupy positions -1 + A001950, where A001950 is the upper Wythoff sequence; those > 1 occupy positions given by -1 + A000201, where A000201 is the lower Wythoff sequence.
(9) The rules (1 is in S, and if x is in S, then 1/x and 1/(x+1) are in S) also generate all the positive rationals.
A variant which extends this idea to an ordering of all rationals is described in A226130. - M. F. Hasler, Jun 03 2013
The updown and downup zigzag limits are (-1 + sqrt(5))/2 and (1 + sqrt(5))/2; see A020651. - Clark Kimberling, Nov 10 2013
From Clark Kimberling, Jun 19 2014: (Start)
Following is a guide to related trees and sequences; for example, the tree A226080 is represented by (1, x+1, 1/x), meaning that 1 is in S, and if x is in S, then x+1 and 1/x are in S (except for x = 0).
All the positive integers:
  A243571, A243572, A232559 (1, x+1, 2x)
  A232561, A242365, A243572 (1, x+1, 3x)
  A243573 (1, x+1, 4x)
All the integers:
  A243610 (1, 2x, 1-x)
  A232723, A242364
All the positive rationals:
  A226080, A226081, A242359, A242360 (1, x+1, 1/x)
  A243848, A243849, A243850 (1, x+1, 2/x)
  A243851, A243852, A243853 (1, x+1, 3/x)
  A243854, A243855, A243856 (1, x+1, 4/x)
  A243574, A242308 (1, 1/x, 1/(x+1))
  A241837, A243575 ({1,2,3}, x+4, 12/x)
  A242361, A242363 (1, 1 + 1/x, 1/x)
  A243613, A243614 (0, x+1, x/(x+1))
All the rationals:
  A243611, A243612 (0, x+1, -1/(x+1))
  A226130, A226131 (1, x+1, -1/x)
  A243712, A243713 ({1,2,3}, x+1, 1/(x+1))
  A243730, A243731 ({1,2,3,4}, x+1, 1/(x+1))
  A243732, A243733 ({1,2,3,4,5}, x+1, 1/(x+1))
  A243714, A243715
  A243925, A243926, A243927 (1, x+1, -2/x)
  A243928, A243929, A243930 (1, x+1, -3/x)
All the Gaussian integers:
  A243924 (1, x+1, i*x)
All the Gaussian rational numbers:
  A233694, A233695, A233696 (1, x+1, i*x, 1/x).
(End)

			The denominators are read from the rationals listed in "rabbit order":
1/1, 2/1, 3/1, 1/2, 4/1, 1/3, 3/2, 5/1, 1/4, 4/3, 5/2, 2/3, 6/1, ...

Clark Kimberling, Table of n, a(n) for n = 1..6000
Clark Kimberling, The infinite Fibonacci tree and other trees generated by rules, Proceedings of the 16th International Conference on Fibonacci Numbers and Their Applications, Fibonacci Quarterly 52 (2014), no. 5, pp. 136-149.
Index entries for fraction trees

Cf. A000045, A035513, A226081 (numerators), A226130, A226247, A020651.

z = 10; g[1] = {1}; g[2] = {2}; g[3] = {3, 1/2};
j[3] = Join[g[1], g[2], g[3]]; j[n_] := Join[j[n - 1], g[n]];
d[s_List, t_List] := Part[s, Sort[Flatten[Map[Position[s, #] &, Complement[s, t]]]]]; j[3] = Join[g[1], g[2], g[3]]; n = 3; While[n <= z, n++; g[n] = d[Riffle[g[n - 1] + 1, 1/g[n - 1]], g[n - 2]]];
Table[g[n], {n, 1, z}]; j[z] (* rabbit-ordered rationals *)
Denominator[j[z]]  (* A226080 *)
Numerator[j[z]]    (* A226081 *)
Flatten[NestList[(# /. x_ /; x > 1 -> Sequence[x, 1/x - 1]) + 1 &, {1}, 9]] (* rabbit-ordered rationals, Danny Marmer, Dec 07 2014 *)

A226080_vec(N=100)={my(T=[1],S=T,A=T); while(N>#A=concat(A, apply(denominator, T=select(t->!setsearch(S,t), concat(apply(t->[t+1,1/t],T))))), S=setunion(S,Set(T)));A} \\ M. F. Hasler, Nov 30 2018

(A226080(n)=denominator(RabbitOrderedRational(n))); ROR=List(1); RabbitOrderedRational(n)={if(n>#ROR, local(S=Set(ROR), i=#ROR*2\/(sqrt(5)+1), a(t)=setsearch(S,t)||S=setunion(S,[listput(ROR,t)])); until( type(ROR[i+=1])=="t_INT" && n<=#ROR, a(ROR[i]+1); a(1/ROR[i])));ROR[n]} \\ M. F. Hasler, Nov 30 2018

A233694 Position of n in the sequence (or tree) S generated in order by these rules: 0 is in S; if x is in S then x + 1 is in S; if nonzero x is in S then 1/x is in S; if x is in S, then i*x is in S; where duplicates are deleted as they occur.

Original entry on oeis.org

1, 2, 3, 5, 11, 23, 49, 102, 212, 443, 926, 1939, 4064, 8509, 17816, 37303, 78105, 163544, 342454, 717076, 1501502, 3144024, 6583334, 13784969

Offset: 0

Clark Kimberling, Dec 19 2013

It can be proved using the division algorithm for Gaussian integers that S is the set of Gaussian rational numbers: (b + c*i)/d, where b,c,d are integers and d is not 0.

The differences of this sequence give the number of elements in each level of the tree. This means that d(n) = a(n) - a(n-1) is at least 1, and is bounded by 3*d(n-1), since there are three times as many elements in each level, before we exclude repetitions. - Jack W Grahl, Aug 10 2018

			The first 16 numbers generated are as follows: 0, 1, 2, i, 3, 1/2, 2 i, 1 + i, -i, -1, 4, 1/3, 3 i, 3/2, i/2, 1 + 2 i. The positions of the nonnegative integers are 1, 2, 3, 5, 11.

Cf. A233695, A233696, A232559, A226130, A232723, A226080.

Off[Power::infy]; x = {0}; Do[x = DeleteDuplicates[Flatten[Transpose[{x, x + 1, 1/x, I*x} /. ComplexInfinity -> 0]]], {18}]; On[Power::infy]; t1 = Flatten[Position[x, _?(IntegerQ[#] && NonNegative[#] &)]]    (* A233694 *)
t2 = Flatten[Position[x, _?(IntegerQ[#] && Negative[#] &)]]  (* A233695 *)
t = Union[t1, t2]  (* A233696 *)
(* Peter J. C. Moses, Dec 21 2013 *)

More terms from Jack W Grahl, Aug 10 2018

A233695 a(n) gives the position of -n in the sequence (or tree) S generated in order by these rules: 0 is in S; if x is in S then x + 1 is in S; if nonzero x is in S then 1/x is in S; if x is in S, then i*x is in S; where duplicates are deleted as they occur.

Original entry on oeis.org

10, 18, 30, 56, 109, 219, 450, 933, 1946, 4071, 8516, 17823, 37310, 78112, 163551, 342461, 717083, 1501509, 3144031, 6583341, 13784976

Offset: 1

Clark Kimberling, Dec 19 2013

It can be proved using the division algorithm for Gaussian integers that S is the set of Gaussian rational numbers: (b + c*i)/d, where b,c,d are integers and d is not 0.

Empirically, it appears that a(n) = A233694(n+2) + 7 for n > 2. It seems clear that positive integers appear for the first time at the start of a new level of the tree. If this is always the case, then the row starting with n will be followed by a row starting n+1, 1/n, ni, followed by a row starting n+2, 1/(n+1), (n+1)i, 1+1/n, n+1, i/(n+1), 1+ni, -i/n, -n. It may be possible to show that of these 9 values, only n+1 has ever appeared before. If so, then -n will always appear exactly 7 places after n + 2 in the sequence. - Jack W Grahl, Aug 10 2018

			The first 16 numbers generated are as follows:  0, 1, 2, i, 3, 1/2, 2 i, 1 + i, -i, -1, 4, 1/3, 3 i, 3/2, i/2, 1 + 2 i. -1 appears in the 10th place, so a(1) = 10.

Jack W Grahl, Haskell code to generate this sequence

Cf. A233694, A233696, A232559, A226130, A232723, A226080.

Off[Power::infy]; x = {0}; Do[x = DeleteDuplicates[Flatten[Transpose[{x, x + 1, 1/x, I*x} /. ComplexInfinity -> 0]]], {18}]; On[Power::infy]; t1 = Flatten[Position[x, _?(IntegerQ[#] && NonNegative[#] &)]]   (*A233694*)
t2 = Flatten[Position[x, _?(IntegerQ[#] && Negative[#] &)]] (* A233695 *)
t = Union[t1, t2]  (* A233696 *)
(* Peter J. C. Moses, Dec 21 2013 *)

More terms by Jack W Grahl, Aug 10 2018

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A245264 Number of numbers in n-th generation of the tree of Gaussian rationals at A233696.

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A226080 Denominators in the Fibonacci (or rabbit) ordering of the positive rational numbers.

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A233694 Position of n in the sequence (or tree) S generated in order by these rules: 0 is in S; if x is in S then x + 1 is in S; if nonzero x is in S then 1/x is in S; if x is in S, then i*x is in S; where duplicates are deleted as they occur.

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A233695 a(n) gives the position of -n in the sequence (or tree) S generated in order by these rules: 0 is in S; if x is in S then x + 1 is in S; if nonzero x is in S then 1/x is in S; if x is in S, then i*x is in S; where duplicates are deleted as they occur.

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