OFFSET
1,3
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 trajectory of k/(2n+1) in 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 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, flattened
J. C. Lagarias, The set of rational cycles for the 3x+1 problem, Acta Arith. 56 (1990), 33-53.
EXAMPLE
The triangle of lengths begins
0;
1, 2;
7, 3, 3;
2, 0, 3, 6;
5, 4, 15, 7, 5;
...
Individual numbers have the following Collatz sequences:
[1] => [0] (0 iteration);
[2 1/3] => [1, 2] because: 2 -> 1 => 1 iteration; 1/3 -> 2 -> 1 => 2 iterations;
[3 2/3 1/5] => [7, 3, 3] because: 3->10->5->16->8->4->2->1 => 7 iterations; 2/3 -> 1/3 -> 2 -> 1 => 3 iterations; 1/5 -> 8/5 -> 4/5 -> 2/5 => 3 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] - 1, {n, 12}, {k, 0, n - 1}]; Flatten[t] (* T. D. Noe, Jan 28 2013 *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Michel Lagneau, Jan 26 2013
STATUS
approved