A210471 Collatz (3x+1) problem with rational numbers: number of steps to reach the end of the cycle starting with 1/(2n+1).

1, 3, 4, 7, 6, 8, 5, 18, 9, 32, 11, 77, 25, 9, 6, 36, 29, 18, 17, 12, 28, 14, 23, 45, 73, 55, 91, 16, 17, 39, 7, 36, 40, 114, 87, 100, 93, 34, 54, 64, 14, 55, 171, 80, 57, 72, 42, 108, 24, 12, 97, 68, 31, 159, 88, 10, 41, 50, 23, 117, 63, 61, 8, 55, 72, 45, 68

Offset: 0

Author

Michel Lagneau, Jan 22 2013

Keywords

• nonn

Comments

In this sequence, the initial value is counted and a(n) = A210468(n) + 1.

This variation of the "3x+1" problem with a class of rational numbers is as follows: start with any number 1/(2n+1). If the numerator is even, divide it by 2, otherwise multiply it by 3 and add 1. Do we always reach the end of a cycle with a rational number? It is conjectured that the answer is yes.

Links

• J. C. Lagarias, The set of rational cycles for the 3x+1 problem, Acta Arith. 56 (1990), 33-53.

Crossrefs

Cf. A210468.

Programs

Mathematica

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

Formula

a(n) = A210468(n) + 1.