What does proving the Collatz Conjecture entail? From the get go: i'm not trying to prove the Collatz Conjecture where hundreds of smarter people have failed. I'm just curious. 
I'm wondering where one would have to start in proving the Collatz Conjecture. That is, based on the nature of the problem, what's the starting point for attempting to prove it? I know that it can be represented in many forms as an equation(that you'd have to recurse over):
$$\begin{align*}
f(x) &= 
  \left\{
    \begin{array}{ll}
      n/2 &\text{if }n \bmod2=0 \\
      3n+1 &\text{if }n \bmod2=1
    \end{array}
  \right.\\
\strut\\
a_i&=
\left\{
\begin{array}{ll}
n &\text{if }n =0\\
f(a_i-1)&\text{if }n>0
\end{array}
\right.\\
\strut\\
a_i&=\frac{1}{2}a_{i-1} - \frac{1}{4}(5a_{i-1} + 2)((-1)^{a_i-1} - 1)
\end{align*}$$
Can you just take the equation and go from there?
Other ways I thought of would be attempting to prove for only odd or even numbers, or trying to find an equation that matches the graph of a number vs. its "Collatz length"
I'm sure there's other ways; but I'm just trying to understand what, essentially, proving this conjecture would entail and where it would begin. 

The problem can be just those two points: 
1) is there a loop?
2) is there a sequence that increases without bound? 
However, another way to solve it would be to show there cannot be two distinct < families >. Up to quite high values of n, we know empirically that starting with any given n less than that value, repeatedly choosing n/2 for even n and 3n+1 for odd n gives a sequence ending in 1. 
Call this set of sequences ending in 1 the < terminate-in-1 family >. 
The task then amounts to testing if there can be a < deviant family > where either a loop or a sequence increasing without bound would amount to a deviant family. Then the challenge (in order to prove Collatz/Ulam/Thwaites correct) is to show that any other family of sequences must somewhere produce a number that is within the terminate-in-1 family. 
The terminate-in-1 family contains, and if it existed any deviant family would contain, infinitely many natural numbers each. 
