OFFSET
1,4
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;
...
EXAMPLE
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.
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 *).
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Michel Lagneau, Apr 04 2013
EXTENSIONS
Better definition from Michel Marcus, Sep 14 2017
STATUS
approved