login
This site is supported by donations to The OEIS Foundation.

 

Logo

Please make a donation to keep the OEIS running. We are now in our 55th year. In the past year we added 12000 new sequences and reached 8000 citations (which often say "discovered thanks to the OEIS"). We need to raise money to hire someone to manage submissions, which would reduce the load on our editors and speed up editing.
Other ways to donate

Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A210468 Collatz (3x+1) problem with rational numbers: number of steps to reach the end of the cycle starting with 1/(2n+1). 7
0, 2, 3, 6, 5, 7, 4, 17, 8, 31, 10, 76, 24, 8, 5, 35, 28, 17, 16, 11, 27, 13, 22, 44, 72, 54, 90, 15, 16, 38, 6, 35, 39, 113, 86, 99, 92, 33, 53, 63, 13, 54, 170, 79, 56, 71, 41, 107, 23, 11, 96, 67, 30, 158, 87, 9, 40, 49, 22, 116, 62, 60, 7, 54, 71, 44, 67 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

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.

If x is of the form 1/2n, each trajectory is divergent because the numerator is always odd and tends into infinity.

If x is of the form x = m/(2n+1) where m is an integer, it is conjectured that the number of steps tends into the end of a finite cycle.

In this sequence, it appears that the last number of the cycle is 1 when 2n+1 is a power of 3. For example, starting with 1/9, the trajectory is 1/9 -> 4/3 -> 2/3 -> 1/3 -> 2 -> 1 with 5 iterations.

Observations: the last number of each trajectory has three possible forms: 1, 2/(2n+1) or a form m/(2n+1) where m> 2.

LINKS

Michel Lagneau, Table of n, a(n) for n = 0..10000

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

EXAMPLE

For n = 3, a(3) = 6 because the corresponding trajectory of 1/7 requires 6 iterations to reach the last term of the cycle:

1/7 -> 10/7 -> 5/7 -> 22/7 -> 11/7 -> 40/7 -> 20/7 and 20/7 is the last term because 20/7 -> 10/7 is already in the trajectory. The rational 10/7 has two antecedents: 1/7 and 20/7 are in the same trajectory (this property is conjecturally impossible in the classical 3x + 1 problem with x integer). The periodic nontrivial loop contains 6 distinct rational numbers (20/7 ->10/7->5/7 ->22/7 -> 11/7 -> 40/7).

MAPLE

with(numtheory): z:={1}:for m from 0 to 80 do: n:=2*m+1:lst:={1/n}:x0:=1: x:=x0*3+n: lst:=lst union {x/n}:for i from 1 to 10000 do: x:=x/2: lst:=lst union {x/n}:  if irem(x, 2)=1 then x0:=x:x:=x0*3+n: lst:=lst union {x/n}:else fi:od: n0:=nops(lst):if lst intersect z = {1} then n1:=n0-2: printf(`%d, `, n1): else n1:=n0-1: printf(`%d, `, n1):fi: od:

MATHEMATICA

Collatz[n_] := NestWhileList[If[EvenQ[Numerator[#]], #/2, 3 # + 1] &, n, UnsameQ, All]; Join[{0}, Table[s = Collatz[1/(2 n + 1)]; len = Length[s] - 2; If[s[[-1]] == 2, len = len - 1]; len, {n, 100}]] (* T. D. Noe, Jan 22 2013 *)

CROSSREFS

Cf. A006577.

Sequence in context: A097723 A187831 A087786 * A080950 A023852 A048750

Adjacent sequences:  A210465 A210466 A210467 * A210469 A210470 A210471

KEYWORD

nonn,nice

AUTHOR

Michel Lagneau, Jan 22 2013

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified December 15 04:23 EST 2019. Contains 329991 sequences. (Running on oeis4.)