Why application of Sharkovskii's Theorem to Collatz is wrong? I thought about this years ago but never had it resolved. Sharkovskii's Theorem has a corollary according to Wikipedia that if there exists a 3-cycle on a map on $\mathbb{R}$, then the map has $n$-cycles for every $n\in\mathbb{N}$. Now, the Collatz map $f(x)$ has a 3-cycle: $f(f(f(4))) = 4$... 

Now, I assume the conclusion that the Collatz map has cycles of every size is wrong (otherwise, it would have been solved) but why does it not apply here?
I don't understand how it could not be wrong?

As compu.sci. I work with base-2 and integers all the time.
Could you not map two to the power of n: $2^n$ in $\mathbb{R}$ ? and use those mappings
to map any $n$ to $f^3(n) = n$. where $n$ (the exponent) is mapped to all positive integers in $\mathbb{N^+}$ ?

With a slight modification to that latter mapping we get:

$f^3(2^{n\mod3}) = 2^{n\mod3}.$
so..

$$
\begin{array}{|c|c|} 
\text{$n$} &amp; \text{$f^3(2^{n\mod3})$} \\ \hline
 0 &amp; 1 \\ \hline
 1 &amp; 2 \\ \hline
 2 &amp; 4 \\ \hline
 3 &amp; 1 \\ \hline
 4 &amp; 2 \\ \hline
 5 &amp; 4 \\ \hline
 . &amp; . \\ \hline
\end{array}$$
Consider this as mapping-function.

Any power of two number would reach the 3-cycle at any time this way.
Assuming all Collatz-numbers will eventually reach 4 this would indeed be correct. Not sure if the interpretation is easy to follow, so bear with me, im learning math every day. And this is not considered to be a solution, only an idea. -R
I don't understand how it could not be wrong?

As compu.sci. I work with base-2 and integers all the time.
Could you not map two to the power of n: $2^n$ in $\mathbb{R}$ ? and use those mappings
to map any $n$ to $f^3(n) = n$. where $n$ (the exponent) is mapped to all positive integers in $\mathbb{N^+}$ ?

With a slight modification to that latter mapping we get:

$f^3(2^{n\mod3}) = 2^{n\mod3}.$
so..

$$
\begin{array}{|c|c|} 
\text{$n$} &amp; \text{$f^3(2^{n\mod3})$} \\ \hline
 0 &amp; 1 \\ \hline
 1 &amp; 2 \\ \hline
 2 &amp; 4 \\ \hline
 3 &amp; 1 \\ \hline
 4 &amp; 2 \\ \hline
 5 &amp; 4 \\ \hline
 . &amp; . \\ \hline
\end{array}$$
Consider this as mapping-function.

Any power of two number would reach the 3-cycle at any time this way.
Assuming all Collatz-numbers will eventually reach 4 this would indeed be correct. Not sure if the interpretation is easy to follow, so bear with me, im learning math every day. And this is not considered to be a solution, only an idea. -R