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.

Showing 1-3 of 3 results.

A210688 The length of the Collatz (3k+1) sequence for all odd fractions and integers.

Original entry on oeis.org

1, 2, 3, 8, 4, 4, 3, 1, 4, 7, 6, 5, 16, 8, 6, 9, 10, 4, 12, 7, 8, 17, 2, 1, 9, 3, 8, 5, 4, 5, 17, 6, 8, 26, 5, 18, 20, 6, 14, 13, 7, 8, 18, 19, 9, 7, 8, 4, 1, 4, 23, 5, 4, 9, 32, 15, 11, 7, 10, 12, 27, 13, 20, 22, 33, 11, 10, 11, 2, 32, 9, 8, 19, 3, 9, 17, 12
Offset: 1

Views

Author

Michel Lagneau, Jan 29 2013

Keywords

Comments

This sequence is the unification, in the limit, of the length of Collatz sequences for all fractions whose denominator is odd, and also all integers.
The sequence A210483 gives the triangle read by rows giving the trajectory of k/(2n+1) in the Collatz problem, k = 1..2n, but particular attention should be paid to numbers in the triangle T(n,k) = (n-k)/(2k+1) for n = 1,2,... and k = 0..n-1.
The example shown below gives a general idea of this regular triangle. This contains all fractions whose denominator is odd and all integers. Now, from T(n,k) we could introduce a 3D triangle in order to produce a complete Collatz sequence starting from each rational T(n,k).
The initial triangle T(n,k) begins
1;
2, 1/3;
3, 2/3, 1/5,;
4, 3/3, 2/5, 1/7;
5, 4/3, 3/5, 2/7, 1/9;
6, 5/3, 4/5, 3/7, 2/9, 1/11;
...

Examples

			The triangle of lengths begins
  1;
  2,  3;
  8,  4,  4;
  3,  1,  4,  7;
  6,  5, 16,  8,  6;
  ...
Individual numbers have the following Collatz sequences (including the first term):
[1] => [1] because: 1 -> 1 with 1 iteration;
[2 1/3] => [2, 3] because: 2 -> 2 -> 1 => 2 iterations; 1/3 -> 1/3 -> 2 -> 1 => 3 iterations;
[3 2/3 1/5] => [8, 4, 4] because: 3 -> 3->10->5->16->8->4->2->1 => 8 iterations; 2/3 -> 2/3 -> 1/3 -> 2 -> 1 => 4 iterations; 1/5 -> 1/5 -> 8/5 -> 4/5 -> 2/5 => 4 iterations.

Links

Crossrefs

Cf. A210516.

Programs

  • Mathematica
    Collatz2[n_] := Module[{lst = NestWhileList[If[EvenQ[Numerator[#]], #/2, 3 # + 1] &, n, Unequal, All]}, If[lst[[-1]] == 1, lst = Drop[lst, -3], If[lst[[-1]] == 2, lst = Drop[lst, -2], If[lst[[-1]] == 4, lst = Drop[lst, -1], If[MemberQ[Rest[lst], lst[[-1]]], lst = Drop[lst, -1]]]]]]; t = Table[s = Collatz2[(n - k)/(2*k + 1)]; Length[s] , {n, 12}, {k, 0, n - 1}]; Flatten[t] (* T. D. Noe, Jan 28 2013 *)

Formula

a(n) = A210516(n) + 1.

A224360 Triangle read by rows: T(n,k) = -1 + length of the Collatz sequence of -(n-k)/(2k+1) for n >= 1 and k >= 0.

Original entry on oeis.org

0, 1, 1, 4, 2, 4, 2, 0, 5, 9, 4, 3, 7, 10, 4, 5, 3, 6, 11, 5, 8, 4, 1, 0, 11, 1, 9, 14, 3, 6, 8, 13, 6, 8, 15, 4, 11, 4, 9, 12, 3, 10, 5, 5, 17, 4, 4, 7, 0, 2, 11, 16, 4, 18, 36, 6, 4, 14, 12, 4, 9, 16, 6, 9, 37, 13, 6, 5, 1, 16, 7, 13, 6, 1, 19, 16, 14, 7, 9
Offset: 1

Views

Author

Michel Lagneau, Apr 04 2013

Keywords

Comments

This sequence is an extension of A210516 with negative values.
We consider the triangle T(n,k) = -(n-k)/(2k+1) for n = 1,2,... and k = 0..n-1.
The example shown below gives a general idea of this regular triangle. This contains all negative fractions whose denominator is odd and all integers. Now, from T(n,k) we could introduce a 3D triangle in order to produce a complete Collatz sequence starting from each rational T(n,k).
The initial triangle T(n,k) begins
-1,
-2, -1/3;
-3, -2/3, -1/5;
-4, -3/3, -2/5, -1/7;
-5, -4/3, -3/5, -2/7, -1/9;
-6, -5/3, -4/5, -3/7, -2/9, -1/11;
...

Examples

			The triangle of lengths begins
  1;
  2, 2;
  5, 3, 5;
  3, 1, 6, 10;
  5, 4, 8, 11, 5;
  ...
Individual numbers have the following Collatz sequences (the first term is not counted):
[-1] => [1] because: -1 -> -1 with 0 iterations;
[-2 -1/3] => [1, 1] because: -2 -> -1 => 1 iteration; -1/3 -> 0 => 1 iteration;
[-3 -2/3 -1/5] => [4, 2, 4] because: -3 -> -8 -> -4 -> -2 -> -1 => 4 iterations; -2/3 -> -1/3 -> 0 => 2 iterations; -1/5 -> 2/5 -> 1/5 -> 8/5 -> 4/5 => 4 iterations.

Crossrefs

Cf. A210516, A210688, A224299, A224300, A224361.

Programs

  • Mathematica
    Collatz2[n_] := Module[{lst = NestWhileList[If[EvenQ[Numerator[#]], #/2, 3 # + 1] &, n, Unequal, All]}, If[lst[[-1]] == -1, lst = Drop[lst, -2], If[lst[[-1]] == 2, lst = Drop[lst, -2], If[lst[[-1]] == 4, lst = Drop[lst, -1], If[MemberQ[Rest[lst], lst[[-1]]], lst = Drop[lst, -1]]]]]]; t = Table[s = Collatz2[-(n - k)/(2*k + 1)]; Length[s] - 1, {n, 13}, {k, 0, n - 1}]; Flatten[t] (* program from T. D. Noe, adapted for this sequence - see A210516 *)

Extensions

Better definition from Michel Marcus, Sep 14 2017

A224361 The length of the Collatz (3k+1) sequence for all odd negative fractions and integers.

Original entry on oeis.org

1, 2, 2, 5, 3, 5, 3, 1, 6, 10, 5, 4, 8, 11, 5, 6, 4, 7, 12, 6, 9, 5, 2, 1, 12, 2, 10, 15, 4, 7, 9, 14, 7, 9, 16, 5, 12, 5, 10, 13, 4, 11, 6, 6, 18, 5, 5, 8, 1, 3, 12, 17, 5, 19, 37, 7, 5, 15, 13, 5, 10, 17, 7, 10, 38, 14, 7, 6, 2, 17, 8, 14, 7, 2, 20, 17, 15
Offset: 1

Views

Author

Michel Lagneau, Apr 04 2013

Keywords

Comments

This sequence is the extension of A210688 with negative values.
We consider the triangle T(n,k) = -(n-k)/(2k+1) for n = 1,2,... and k = 0..n-1.
The example shown below gives a general idea of this regular triangle. This contains all negative fractions whose denominator is odd and all integers. Now, from T(n,k) we could introduce a 3D triangle in order to produce a complete Collatz sequence starting from each rational T(n,k).
The initial triangle T(n,k) begins
-1;
-2, -1/3;
-3, -2/3, -1/5;
-4, -3/3, -2/5, -1/7;
-5, -4/3, -3/5, -2/7, -1/9;
-6, -5/3, -4/5, -3/7, -2/9, -1/11;
...
Needs a more precise definition. - N. J. A. Sloane, Sep 14 2017

Examples

			The triangle of lengths begins
  1;
  2, 2;
  5, 3, 5;
  3, 1, 6, 10;
  5, 4, 8, 11, 5;
  ...
Individual numbers have the following Collatz sequences (including the first term):
  [-1] => [1] because -1 -> -1 with 1 iteration;
  [-2 -1/3] => [2, 2] because: -2 -> -1 => 2 iterations; -1/3 -> 0 => 2 iterations;
  [-3 -2/3 -1/5] => [5, 3, 5] because: -3 -> -8 -> -4 -> -2 -> -1 => 5 iterations; -2/3 -> -1/3 -> 0 => 3 iterations; -1/5 -> 2/5 -> 1/5 -> 8/5 -> 4/5 => 5 iterations.

Crossrefs

Cf. A210516, A210688, A224299, A224300, A224360.

Programs

  • Mathematica
    Collatz2[n_] := Module[{lst = NestWhileList[If[EvenQ[Numerator[#]], #/2, 3 # + 1] &, n, Unequal, All]}, If[lst[[-1]] == -1, lst = Drop[lst, -2], If[lst[[-1]] == 2, lst = Drop[lst, -2], If[lst[[-1]] == 4, lst = Drop[lst, -1], If[MemberQ[Rest[lst], lst[[-1]]], lst = Drop[lst, -1]]]]]]; t = Table[s = Collatz2[-(n - k)/(2*k + 1)]; Length[s], {n, 13}, {k, 0, n - 1}]; Flatten[t] (* program from T. D. Noe, adapted for this sequence - see A210688 *)

Formula

a(n) = A224360(n) + 1.
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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