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

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

1,2

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;

...

Links

Michel Lagneau, Rows n = 1..100 of triangle, flattened

J. C. Lagarias, The set of rational cycles for the 3x+1 problem, Acta Arith. 56 (1990), 33-53.

Formula

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

Example

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.

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 *)

Crossrefs

Cf. A210516.

Sequence in context: A349003 A079555 A100870 * A328428 A195794 A145605

Adjacent sequences: A210685 A210686 A210687 * A210689 A210690 A210691

Keyword

nonn,tabl

Author

Michel Lagneau, Jan 29 2013

Status

approved