A232559 -id:A232559 - cp's OEIS frontend

A232723 Sequence (or tree) generated by these rules: 0 is in S, and if x is in S, then 2*x and 1 - x are in S, and duplicates are deleted as they occur.

0, 1, 2, 4, -1, 8, -3, -2, 16, -7, -6, -4, 3, 32, -15, -14, -12, 7, -8, 5, 6, 64, -31, -30, -28, 15, -24, 13, 14, -16, 9, 10, 12, -5, 128, -63, -62, -60, 31, -56, 29, 30, -48, 25, 26, 28, -13, -32, 17, 18, 20, -9, 24, -11, -10, 256, -127, -126, -124, 63

Offset: 1

Clark Kimberling, Nov 28 2013

Let S be the set of numbers defined by these rules: 0 is in S, and if x is in S, then 2*x and 1 - x are in S. Then S is the set of integers, which arise in generations. Deleting duplicates as they occur, the generations are given by g(1) = (0), g(2) = (1), g(3) = (2), g(4) = (4,-1), g(5) = (8,-3,-2), etc. Concatenating these gives A232723. Every integer occurs exactly once in S. The even integers occupy the positions given by the lower Wythoff sequence, A000201; the odds, by the upper Wythoff sequence, A001950. The positive integers occupy the positions given by A189035, and the positions of the nonpositives, by A189034.

Inverse beginning with 0: 1, 2, 3, 13, 4, 20, 21, 18, 6, 31, 32, 89, 33, 28, 29, 26, 9, 49, 50, 136, 51, 143, 144, 141, 53, 44, ..., . - Robert G. Wilson v, Jun 17 2014

			Each x begets 2*x and 1 - x, and if either has already occurred it is deleted. Thus, 0 begets 1, which begets 2, which begets (4,-1), etc.

Clark Kimberling, Table of n, a(n) for n = 1..10000

Cf. A232559, A226130, A226131, A000201, A189035.

x = {0}; Do[x = DeleteDuplicates[Flatten[Transpose[{x, 2*x, 1 - x}]]], {10}]; x  (* Peter J. C. Moses, Nov 28 2013 *)
Nest[ DeleteDuplicates[ Flatten[ # /. a_Integer -> {2a, 1-a}]]&, {0}, 9] (* Robert G. Wilson v, Jun 17 2014 *)

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

A233696 Positions of integers 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, 10, 11, 18, 23, 30, 49, 56, 102, 109, 212, 219, 443, 450, 926, 933, 1939, 1946, 4064, 4071, 8509, 8516, 17816, 17823, 37303, 37310, 78105, 78112, 163544, 163551

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.

			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.  Positions of integers 0, 1, 2, 3, -1, 4,... are 1,2,3,5,10,11,....

Cf. A233694, A233695, 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 *)

Definition and example corrected. - R. J. Mathar, May 06 2017

A232563 Sequence (or tree) generated by these rules: 1 is in S, and if x is in S, then x + 1 and 4*x are in S, and duplicates are deleted as they occur.

Original entry on oeis.org

1, 2, 4, 3, 8, 5, 16, 12, 9, 32, 6, 20, 17, 64, 13, 48, 10, 36, 33, 128, 7, 24, 21, 80, 18, 68, 65, 256, 14, 52, 49, 192, 11, 40, 37, 144, 34, 132, 129, 512, 28, 25, 96, 22, 84, 81, 320, 19, 72, 69, 272, 66, 260, 257, 1024, 15, 56, 53, 208, 50, 196, 193, 768

Offset: 1

Clark Kimberling, Nov 26 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 4*x are in S. Then S is the set of all positive integers, which arise in generations. Deleting duplicates as they occur, the generations are given by g(1) = (1), g(2) = (2,4), g(3) = (3,8,5,16), g(4) = (12,9,32,6,20,17,64), etc. Concatenating these gives A232563, a permutation of the positive integers. The number of numbers in g(n) is A001631(n), the n-th tetranacci number. It is helpful to show the results as a tree with the terms of S as nodes and edges from x to x + 1 if x + 1 has not already occurred, and an edge from x to 4*x if 4*x has not already occurred.

			Each x begets x + 1 and 4*x, but if either has already occurred it is deleted.  Thus, 1 begets 2 and 4; in the next generation, 2 begets 3 and 8, and 4 begets 5 and 16.

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A232559, A232564, A001631.

z = 8; g[1] = {1}; g[2] = {2, 4}; g[n_] := Riffle[g[n - 1] + 1, 4 g[n - 1]]; j[2] = Join[g[1], g[2]]; j[n_] := Join[j[n - 1], g[n]]; g1[n_] := DeleteDuplicates[DeleteCases[g[n], Alternatives @@ j[n - 1]]]; g1[1] = g[1]; g1[2] = g[2]; t = Flatten[Table[g1[n], {n, 1, z}]]  (* A232563 *)
Table[Length[g1[n]], {n, 1, z}]  (* A001631 *)
t1 = Flatten[Table[Position[t, n], {n, 1, 200}]]  (* A232564 *)

A232868 Positions of the integers in the sequence (or tree) of complex numbers generated by these rules: 0 is in S, and if x is in S, then x + 1 and i*x are in S, where duplicates are deleted as they occur.

Original entry on oeis.org

1, 2, 3, 5, 8, 9, 12, 16, 19, 27, 30, 42, 45, 61, 64, 84, 87, 111, 114, 142, 145, 177, 180, 216, 219, 259, 262, 306, 309, 357, 360, 412, 415, 471, 474, 534, 537, 601, 604, 672, 675, 747, 750, 826, 829, 909, 912, 996, 999, 1087, 1090, 1182, 1185, 1281, 1284

Offset: 1

Clark Kimberling, Dec 01 2013

Let S be the sequence (or tree) of complex numbers defined by these rules: 0 is in S, and if x is in S, then x + 1, and i*x are in S. Deleting duplicates as they occur, the generations of S are given by g(1) = (0), g(2) = (1), g(3) = (2,i), g(4) = (3, 2i, 1+i, -1), ... Concatenating these gives 0, 1, 2, i, 3, 2*i, 1 + i, -1, 4, 3*i, 1 + 2*i, -2, 2 + i, -1 + i, -i, 5, ... A232868 is the (ordered) union of two linearly recurrent sequences: A232866 and A232867.

			Each x begets x + 1, and i*x, but if either these has already occurred it is deleted.  Thus, 0 begets (1); then 1 begets (2,i,); then 2 begets 3 and 2*i, and i begets 1 + i and -1, so that g(4) = (3, 2*i, 1 + i, -1), etc.

Cf. A232559, A232866, A232867.

x = {0}; Do[x = DeleteDuplicates[Flatten[Transpose[{x, x + 1, I*x}]]], {40}]; x;
t1 = Flatten[Table[Position[x, n], {n, 0, 30}]]   (* A232866 *)
t2 = Flatten[Table[Position[x, -n], {n, 1, 30}]]  (* A232867 *)
Union[t1, t2]  (* A232868 *)

From Chai Wah Wu, Feb 20 2018: (Start)
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n > 10 (conjectured).
G.f.: x*(-x^9 - 4*x^7 + 2*x^6 + 2*x^5 - 2*x^4 + x^2 - x - 1)/((x - 1)^3*(x + 1)^2) (conjectured). (End)

A233135 Shortest (x+1,2x)-code of n.

Original entry on oeis.org

1, 2, 21, 22, 221, 212, 2121, 222, 2221, 2212, 22121, 2122, 21221, 21212, 212121, 2222, 22221, 22212, 222121, 22122, 221221, 221212, 2212121, 21222, 212221, 212212, 2122121, 212122, 2121221, 2121212, 21212121, 22222, 222221, 222212, 2222121, 222122, 2221221

Offset: 1

Clark Kimberling, Dec 05 2013

Every positive integer is a composite of f(x) = x + 1 and g(x) = 2*x starting with x = 1. For example, 5 = f(g(g(1))), which abbreviates as fgg, or 122, which we call a (x+1,2x)-code of 5. It appears that the number of (x+1,2x)-codes of n is A040039(n), that these numbers form Guy Steele's sequence GS(4,5) at A135529, and that for k >= 1, then number of such codes is F(n-1), where F = A000045, the Fibonacci numbers. See A232559 for the uncoded positive integers in the order generated by the rules x -> x+1 and x -> 2*x.

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A040039, A135529, A232559, A000045, A233136, A233137, A233138.

b[x_] := b[x] = If[OddQ[x], x - 1, x/2]; u[n_] := 2 - Mod[Drop[FixedPointList[b, n], -3], 2]; u[1] = {1}; t = Table[u[n], {n, 1, 30}]; Table[FromDigits[u[n]], {n, 1, 50}]  (* A233137 *)
Flatten[t]  (* A233138 *)
Table[FromDigits[Reverse[u[n]]], {n, 1, 30}]  (* A233135 *)
Flatten[Table[Reverse[u[n]], {n, 1, 30}]]  (* A233136 *)

Define h(x) = x - 1 if x is odd and h(x) = x/2 if x is even, and define H(x,1) = h(x) and H(x,k) = H(H(x,k-1)). For each n > 1, the sequence (H(n,k)) decreases to 1 through two kinds of steps; write 1 when the step is x - 1 and write 2 when the step is x/2. Let c(n) be the concatenation of 1s and 2s; then A233135(n) is the reversal of c(n), as in the Mathematica program.

A233136 Concatenated shortest (x+1,2x)-codes for the positive integers.

Original entry on oeis.org

1, 2, 2, 1, 2, 2, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 1, 2, 2, 1

Offset: 1

Clark Kimberling, Dec 05 2013

Concatenate the representations of the positive integers in A233135, and then separate the digits by commas, in the manner analogous to A030302.

			A233135 = (1,2,21,22,221,212,...), so that A233136 = (1,2,2,1,2,2,2,2,1,2,1,2,...).

Clark Kimberling, Table of n, a(n) for n = 1..2500

Cf. A040039, A135529, A232559, A000045, A233135, A233137, A233138.

b[x_] := b[x] = If[OddQ[x], x - 1, x/2]; u[n_] := 2 - Mod[Drop[FixedPointList[b, n], -3], 2]; u[1] = {1}; t = Table[u[n], {n, 1, 30}]; Table[FromDigits[u[n]], {n, 1, 50}]  (* A233137 *)
Flatten[t]  (* A233138 *)
Table[FromDigits[Reverse[u[n]]], {n, 1, 30}]  (* A233135 *)
Flatten[Table[Reverse[u[n]], {n, 1, 30}]]  (* A233136 *)

A233137 Reversed shortest (x+1,2x)-code of n.

Original entry on oeis.org

1, 2, 12, 22, 122, 212, 1212, 222, 1222, 2122, 12122, 2212, 12212, 21212, 121212, 2222, 12222, 21222, 121222, 22122, 122122, 212122, 1212122, 22212, 122212, 212212, 1212212, 221212, 1221212, 2121212, 12121212, 22222, 122222, 212222, 1212222, 221222, 1221222

Offset: 1

Clark Kimberling, Dec 05 2013

(See A233135.)

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A040039, A135529, A232559, A000045, A233135, A233136, A233138.

b[x_] := b[x] = If[OddQ[x], x - 1, x/2]; u[n_] := 2 - Mod[Drop[FixedPointList[b, n], -3], 2]; u[1] = {1}; t = Table[u[n], {n, 1, 30}]; Table[FromDigits[u[n]], {n, 1, 50}]  (* A233137 *)
Flatten[t]  (* A233138 *)
Table[FromDigits[Reverse[u[n]]], {n, 1, 30}]  (* A233135 *)
Flatten[Table[Reverse[u[n]], {n, 1, 30}]]  (* A233136 *)

Define h(x) = x - 1 if x is odd and h(x) = x/2 if x is even, and define H(x,1) = h(x) and H(x,k) = H(H(x,k-1)). For each n > 1, the sequence (H(n,k)) decreases to 1 through two kinds of steps; write 1 when the step is x - 1 and write 2 when the step is x/2. A233137(n) is the concatenation of 1s and 2s, as in the Mathematica program.

A233138 Concatenated reversed shortest (x+1,2x)-codes for the positive integers.

Original entry on oeis.org

1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2

Offset: 1

Clark Kimberling, Dec 05 2013

Concatenate the representations of the positive integers in A233137, and then separate the digits by commas.

			A233137 = (1,2,12,22,122,212,...), so that A233138 = (1,2,1,2,2,2,1,2,2,2,1,2,...)

Clark Kimberling, Table of n, a(n) for n = 1..10000

Cf. A040039, A135529, A232559, A000045, A233135, A233137, A233138.

b[x_] := b[x] = If[OddQ[x], x - 1, x/2]; u[n_] := 2 - Mod[Drop[FixedPointList[b, n], -3], 2]; u[1] = {1}; t = Table[u[n], {n, 1, 30}]; Table[FromDigits[u[n]], {n, 1, 50}]  (* A233137 *)
Flatten[t]  (* A233138 *)
Table[FromDigits[Reverse[u[n]]], {n, 1, 30}]  (* A233135 *)
Flatten[Table[Reverse[u[n]], {n, 1, 30}]]     (* A233136 *)

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

cp's OEIS Frontend

A232723 Sequence (or tree) generated by these rules: 0 is in S, and if x is in S, then 2*x and 1 - x are in S, and duplicates are deleted as they occur.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

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.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Extensions

A233696 Positions of integers 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.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Extensions

A232563 Sequence (or tree) generated by these rules: 1 is in S, and if x is in S, then x + 1 and 4*x are in S, and duplicates are deleted as they occur.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A232868 Positions of the integers in the sequence (or tree) of complex numbers generated by these rules: 0 is in S, and if x is in S, then x + 1 and i*x are in S, where duplicates are deleted as they occur.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

A233135 Shortest (x+1,2x)-code of n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A233136 Concatenated shortest (x+1,2x)-codes for the positive integers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A233137 Reversed shortest (x+1,2x)-code of n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica