Question
Asked 8th Jun, 2011

3n+1 problem solved?

Hallo,
there is a new proof of the 3n+1-Problem !
The paper ist available
?
Perhaps there is a flaw in the proof
?
What is your opinon?
The Collatz-conjecture (the famous 3n+1-problem):
?we construct a sequence of integers starting with integer n = a_0
If a_j is even, the next number ist a_(j+1) = a_j/2.
If a_j is odd, the next number ist a_(j+1) = 3*a_j+1.
Example n = 6:
6, 3, 10, 5, 16, 8, 4, 2, 1
?The Collatz-conjecture: the sequence with a every positive starting-integer ends always in the sequence 4,2 1

Most recent answer

6th Jul, 2020
Alessio Lunghi
AlessioLunghi
Hello, I observed just this fact, don't know whether or not it can be interesting.
Just analysing the variations: in "even" case the iteration converges (0.5n) and in the odd case the iteration diverges (3n). So calling the f(n)=n/2 the "C" phase and f(n)=3n+1 the "D" phase we have initially the same probability that algorithm start with C or D phase (because even and odds are equally distributed).
The C1 phase will equally (probabilistically) call another C2 phase or a D2 phase (4 multipliers are equally distributed among even numers) and so does the D1 phase: numbers that are not multipliers of 3 are equally distributed among even and odds.
The answer seems to be strongly initial conditions dependant, so the algorithm for me is unstable, hence the conjecture is false.

All Answers (6)

Deleted profile
This is affirmative. ? Please consider the proof below.
Deleted profile
2 * 7 *11 ?= 2 * 77 implies largest odd factor is 77 and not 11.
17th Apr, 2016
Gunter Lassmann
Deutsche Telekom Berlin (retired)
Hallo David,
the proof is wrong. See the new comment.
Deleted profile
Hello Gunter, ?
Hmm. ?My proof is probabilistic, and it demonstrates the Collatz is true by contradiction.
It states there is no Collatz sequence which does not converge to one or the
Probability(any Collatz sequence does not converge to one) ?= 0.
Furthermore, ?1. ?3n + 1 = (2^l) *k with k = 2j+1 for some positive integers, j and l.?This must be true always for Collatz sequence to continue indefinitely. ?Thus, 3n+1 is always a multiple of 2 which I assumed. ?This is fact if the Collatz process is to continue indefinitely.
So, given 3n+1 is always a multiple of two where n?> 1?is a? largest odd factor,?
?or Prob(3n+1 is always a multiple of two) =1. ?Then
Prob(3n+1 is ?a multiple of four |?3n+1 is always a multiple of two) =?1/2.
Prob(3n+1 is a multiple of eight?| 3n+1 is a multiple of four) =?1/2.
Prob(3n+1 is a multiple of sixteen?| 3n+1 is a multiple of eight) = 1/2.
...
These conditional probabilities depend on the size of l ≥ 1?of equation one.
So, (3n + 1) / 2^l ?= ?k (the largest odd factor > 1 of 3n+1), ?and the Collatz sequence continues ...
?But, my proof shows by contradiction that the Collatz?process must end eventually with a convergence to one. ?So, the Collatz Conjecture is true!?
Can you help by adding an answer?

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