Guest wrote:

Guest wrote:You have not disproved cycles that don’t reach 1.

Your cycles that don’t reach one are fictional. Otherwise, please state one or more (cycles...) or prove they can exist. Thank you!

No, a proof of the collatz conjecture requires you to prove that there cannot exist such cycles, not the other way around. If there exists a proof that such cycles do exist, the conjecture as a whole is disproven, however a proof that does not contradict such cycles is not a proof.

You have not disproved cycles that don’t reach 1.

Rewritten

The Collatz conjecture as stated in Wikipedia says that if we define the following function over the natural numbers:


\begin{equation}
f(n)= \left \{
\begin{split}
n/2 &\text{ if } n \equiv 0 \pmod{2}\\
3n+1 &\text{ if } n\equiv 1 \pmod{2}
\end{split}
\right .
\end{equation}

And build a sequence, starting in any arbitrary natural number as follows:

\begin{equation}
a_i= \left \{
\begin{split}
n &\text{ for } i=0\\
f(a_{i-1}) &\text{ for } i>0
\end{split}
\right .
\end{equation}

that is: $a_i$ is the value of $f$ applied to $n$ recursively $i$ times; $a_i = f^{(i}(n)$.

The Collatz conjecture is: This process will eventually reach the number 1, regardless of which positive integer is chosen initially.

Let be $a_j$ an arbitrary number in a sequence built with the specified method. If $a_j$ is an odd number, then $a_{j+1}$ will be even. On the other hand, if $a_j$ is even, then $a_{j+1}$ can be even with a probability $1/2$, $a_{j+2}$ with a probability $1/4$ and so on.

Now let's build the sequence $b_i$ of odd numbers that results of applying function f. Assuming we start with an odd number, and after i iterations we will obtain:

\begin{equation}
b_i= f_{even}^{k(i)}(f_{odd}(f_{even}^{k(i-1)}(f_{odd}(...f_{even}^{k(1)}(f_{odd}(n))))))
\end{equation}

Where to obtain each odd element of the sequence from the previous one we had to compose the odd part of f which is $3n+1$ once and the even part of f or $n/2$ at least once. The resulting term is:

\begin{equation}
b_i = \frac{3^in}{2^{\sum_{j=1}^ik(j)}}+ \sum_{j=1}^i {\frac{3^{j-1}}{2^{\sum_{j=1}^ik(j)}}}
\end{equation}

Thus, we obtain, for high values of i:



\begin{equation}
\lim_{i \to \infty} b_i = \lim_{i \to \infty} \left (\frac{3^in}{2^{i<k>}}+ \sum_{j=1}^i {\frac{3^{j-1}}{2^{i<k>}}} \right )
\end{equation}

And $<k>=2$ is the average value of $k(i)$, therefore after some algebra:

\begin{equation}
\lim_{i \to \infty} a_i =\lim_{i \to \infty} b_i =
\lim_{i \to \infty} \left (\frac{3^in}{4^i}+ \sum_{j=1}^i {\frac{3^{j-1}}{4^j}} \right ) =1
\end{equation}

I am sorry, the last equation is incorrect, the correct equation is the following:

\begin{equation}
\lim_{i \to \infty} a_i =\lim_{i \to \infty} b_i =
\lim_{i \to \infty} \left (\frac{3^in}{4^i}+ \sum_{j=1}^i {\frac{3^{j-1}}{4^j}} \right ) =1
\end{equation}

Maybe a good one, sorry for bothering

The Collatz conjecture as stated in Wikipedia says that if we define the following function over the natural numbers;


\begin{equation}
f(n)= \left \{
\begin{split}
n/2 &\text{ if } n \equiv 0 \pmod{2}\\
3n+1 &\text{ if } n\equiv 1 \pmod{2}
\end{split}
\right .
\end{equation}

And build a sequence, starting in any arbitrary natural number as follows:

\begin{equation}
a_i= \left \{
\begin{split}
n &\text{ for } i=0\\
f(a_{i-1}) &\text{ for } i>0
\end{split}
\right .
\end{equation}

that is: $a_i$ is the value of $f$ applied to $n$ recursively $i$ times; $a_i = f^{(i}(n)$.

The Collatz conjecture is: This process will eventually reach the number 1, regardless of which positive integer is chosen initially.

Now, let be $a_j$ an arbitrary number in a sequence built with the specified method. If $a_j$ is an odd number, then $a_{j+1}$ will be even. On the other hand, if $a_j$ is even, then $a_{j+1}$ can be even with a probability $1/2$, $a_{j+2}$ with a probability $1/4$ and so on.

Now let's build the sequence $b_i$ of odd numbers that results of applying function f. Assuming we start with an odd number, and after i iterations we will obtain:

\begin{equation}
b_i= (f_{odd}^{(i}f_{even}^{(\sum_{j=1}^ik(j)})(n)
\end{equation}

Where to obtain each odd element of the sequence from the previous one we had to compose the odd part of f which is $3n+1$ once and the even part of f or $n/2$ at least once.


Now, to prove the Collatz conjecture it is enough proving that the limit when $i\longrightarrow\infty$ is one or lower.

This can be done if we see that:

\begin{equation}
\lim_{i \to \infty} b_i = \lim_{i \to \infty} (f_{odd}^{(i}f_{even}^{(\sum_{j=1}^ik(j)})(n) =
\lim_{i \to \infty} (f_{odd}^{(i}f_{even}^{(i<k>})(n)
\end{equation}

And $<k>=2$ is the average value of $k(i)$, therefore after some algebra:

\begin{equation}
\lim_{i \to \infty} a_i =\lim_{i \to \infty} b_i =
\lim_{i \to \infty} \left (\frac{3^in}{4^i}+ \sum_{j=1}^i {\frac{1}{4^j}} \right ) =\frac{1}{3}
\end{equation}

But the lowest number we can reach with function f is one, and therefore that is the limit of the sequence.