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A210471 Collatz (3x+1) problem with rational numbers: number of steps to reach the end of the cycle starting with 1/(2n+1). 2
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

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

Table of n, a(n) for n=0..66.

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

FORMULA

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

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}]]

CROSSREFS

Cf. A210468.

Sequence in context: A325968 A325818 A120224 * A183107 A132009 A295565

Adjacent sequences:  A210468 A210469 A210470 * A210472 A210473 A210474

KEYWORD

nonn

AUTHOR

Michel Lagneau, Jan 22 2013

STATUS

approved

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Last modified December 6 21:48 EST 2019. Contains 329809 sequences. (Running on oeis4.)