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
Examples
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.
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.
Crossrefs
Cf. A210516.
Programs
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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 *)
Formula
a(n) = A210516(n) + 1.
Comments