This is problem 5.31 from Introduction to the Theory of Computation, 3rd Edition by Michael Sipser.
Let $$ f(x) = \begin{cases} 3x+1, & \text{for odd $x$} \\ x/2, & \text{for even $x$} \end{cases} $$ for any natural number $x$. If you start with an integer $x$ and iterate $f$, you obtain a sequence, $x$, $f(x)$, $f(f(x))$, $\dots$ . Stop if you ever hit 1. For example, if $x = 17$, you get the sequence 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1. Extensive computer tests have shown that every starting point between 1 and a large positive integer gives a sequence that ends in 1. But the question of whether all positive starting points end up at 1 is unsolved; it is called the $3x + 1$ problem.
Suppose that $A_{TM}$ = $\{\langle M,w\rangle$∣ $M$ is a Turing Machine and $M$ accepts $w\}$ (Acceptance Turing machine) were decidable by a Turing Machine $H$. Use $H$ to describe a Turing Machine that is guaranteed to state the answer to the $3x + 1$ problem.
I suppose the question is: if we can solve the halting problem, can we figure out the answer to the $3x+1$ (Collatz) problem?
Well, if we have a program $H$ that solves the halting problem (that is, it can decide whther some Turing machine $M$ with input $I$ halts or not), then do the following:
Create a Turing machine $F$ that takes a number $x$ and iterates through the $f(x)$ sequence, and stops once it reaches $1$. It is easy to show such a Turing machine $F$ exists .. here is its pseudocode:
F (input $i$)
$Begin$
$\quad$ While $i \not = 1$
$\quad \quad $ Case $i$ is even: $i = i/2$
$\quad \quad $ Case $i$ is odd: $i = 3*i+1$
$End$
(So this F routine will halt whenever the sequence ends with a 1 ... otherwise it will go on forever)
Now create a new Turing machine $C$ that starts with a given $i$, and that calls halting program $H$ on $F$ and $i$. If $H$ says that $F$ does not halt on $i$, then stop. If $H$ says that $F$ does halt $i$, then increase $i$ by $1$, and repeat the process. Again, it is easy to show that such a Turing machine $C$ exists, assuming $H$ exists .. here is its pseudocode:
C (input $i$)
$Begin$
$\quad$ While $H(F,i)$
$\quad \quad$ $i=i+1$
$End$
(so this routine $C$ will only stop if for some $i$, $H$ finds that $F$ does not halt on $i$. Effectively, $C$ looks for a counterexample to the Collatz conjecture. If there is one, then $C$ will eventually run into it. If there isn't, then $C$ will run forever)
Finally, call $H$ on $C$ and $1$. If $H$ says that $C$ with $1$ will not halt, then apparently the solution to the $3x+1$ is that the sequence will always end up with $1$ for any $x$ (that is, there is no counterexample to the Collatz conjecture ... so the Collatz Conjecture is true). If $H$ says that $C$ with $1$ does halt, then apparently the solution is that for some $x$ the sequence does not end up with $1$ (i.e. there is a counterexample to the Collatz conjecture). So, either way, you have solved the $3x+1$ Collatz problem.