

A224360


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


2



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
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OFFSET

1,4


COMMENTS

This sequence is an extension of A210516 with negative values.
We consider the triangle T(n,k) = (nk)/(2k+1) for n = 1,2,... and k = 0..n1.
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;
...


LINKS



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 *).


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AUTHOR



EXTENSIONS



STATUS

approved



