A224360 -id:A224360 - cp's OEIS frontend

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

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

Michel Lagneau, Apr 04 2013

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

			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.

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

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 *)

a(n) = A224360(n) + 1.

A224367 Triangle read by rows giving trajectory of -k/(2n+1) in Collatz problem, k = 1..2n.

Original entry on oeis.org

0, 1, 2, 4, 5, 7, 6, 9, 10, 11, 11, 13, 12, 4, 5, 1, 6, 3, 2, 4, 7, 8, 9, 8, 10, 11, 9, 13, 11, 13, 12, 14, 15, 5, 16, 16, 6, 18, 17, 20, 17, 19, 7, 4, 5, 4, 6, 1, 5, 6, 7, 7, 2, 9, 6, 8, 7, 17, 18, 9, 19, 9, 10, 20, 20, 11, 10, 22, 11, 24, 21, 23, 21, 36, 37

Offset: 0

Michel Lagneau, Apr 05 2013

Extension of A210483 with negative values, and subset of A224360.

			The 2nd row [4, 5, 7, 6] gives the number of iterations of -k/5 (the first element is not counted):
   k=1 => -1/5 ->2/5 -> 1/5 -> 8/5 -> 4/5 with 4 iterations;
   k=2 => -2/5 -> -1/5 -> 2/5 -> 1/5 -> 8/5 -> 4/5 with 5 iterations;
   k=3 => -3/5 -> -4/5 -> -2/5 -> -1/5 -> 2/5 -> 1/5 -> 8/5 -> 4/5 with 7 iterations;
   k=4 => -4/5 -> -2/5 -> -1/5 -> 2/5 -> 1/5 -> 8/5 -> 4/5 with 6 iterations.
The array starts:
  [0];
  [1, 2];
  [4, 5, 7, 6];
  [9, 10, 11, 11, 13, 12];
  [4, 5, 1, 6, 3, 2, 4, 7];
  ...

Cf. A210483, A210468, A224299, A224360.

Collatz[n_] := NestWhileList[If[EvenQ[Numerator[-#]], #/2, 3 # + 1] &, n, UnsameQ, All]; t = Join[{{0}}, Table[s = Collatz[-k/(2*n + 1)]; len = Length[s] - 2; If[s[[-1]] == 2, len = len - 1]; len, {n, 10}, {k, 2*n}]]; Flatten[t] (* program from T. D. Noe, adapted for this sequence - see A210483 *)

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

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