A224166 Number of halving and tripling steps to reach the last number of the cycle in the Collatz (3x+1) problem for the negative integers (the initial term is counted).

2, 2, 5, 3, 5, 6, 5, 4, 12, 5, 7, 7, 10, 5, 10, 5, 18, 13, 8, 5, 24, 8, 19, 8, 18, 11, 8, 6, 11, 11, 22, 6, 29, 18, 14, 14, 18, 9, 14, 6, 18, 25, 9, 9, 23, 20, 17, 9, 27, 18, 12, 12, 17, 9, 18, 7, 30, 12, 12, 12, 18, 23, 20, 7, 38, 30, 21, 18, 20, 15, 20, 15

Offset: 1

Author

Michel Lagneau, Apr 01 2013

Keywords

• nonn

Examples

			a(10) = 5 because the trajectory of -10 is -10 -> -5 -> -14 -> -7 -> -20 -> -10 and -10 is the last term of the cycle, hence 5 iterations where the first term -10 is counted.
		

Crossrefs

Cf. A006577, A224183.

Programs

Mathematica

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, UnsameQ, All]; Table[s = Collatz[-n]; len = Length[s] - 2; If[s[[-1]] == 2, len = len - 1]; len+1, {n, 1, 100}]

Formula

a(n) = A224183(n) + 1.

Extensions

a(1) changed to 2 by Pontus von Brömssen, Jan 24 2021