To explain mercio's answer in greater detail: suppose that $n$ is a number of the form $3k$; then obviously $n$ and $k$ are either both odd or both even. Now, imagine running one step of Collatz2 on $n$: if $n$ is even, then $\mathrm{Collatz2}(n) = n/2 = 3k/2 = 3(k/2) = 3\cdot\mathrm{Collatz}(k)$. Similarly, if $n$ is odd, then $\mathrm{Collatz2}(n) = 3n+3 = 3(3k)+3 = 3(3k+1) = 3\cdot\mathrm{Collatz}(k)$. This means that the sequence of Collatz2 iterates for a number $n=3k$ is exactly the sequence of 'original' Collatz iterates for $k$, multiplied by $3$.
Now, suppose that we have some odd $n$ (if $n$ is even, then obviously we can divide out all the factors of 2 and eventually get an odd starting point); then the first Collatz2 step takes us to $3n+3$ — but this is a number of the form $3k$ (with $k=n+1$), and so now everything in the first paragraph applies. This means that for $n$ odd running Collatz2($n$) is exactly like running Collatz($n+1)$ — this is why you see Collatz2($31$) converging so quickly, because it's identical to running Collatz($32$).
At heart, this means that there's nothing to be gained by studying Collatz2, but there's nothing to be lost either; it is exactly equivalent to the original Collatz problem.
One further point that bears raising, though, is your Lemma 1: 'for all positive integers $n$, $\mathrm{Collatz}(n) \Leftrightarrow \mathrm{Collatz2}(n)$'. As it's stated, this is precisely equivalent to the Collatz conjecture itself - but this is a different statement from one saying that 'Collatz($n$) is true for all positive integers $n$ if and only off Collatz2($n$) is true for all positive integers $n$'. To see why, consider replacing Collatz($n$) with '$n$ is even' and Collatz2($n$) with '$n$ is odd'; then 'all positive integers are even if and only if all positive integers are odd' is true (since both sides are false!), but 'for all positive integers $n$, $n$ is even if and only if $n$ is odd' is false. Universal quantification (the 'for all' statement) doesn't distribute over 'if and only if' in the way that you're using here.
ADDED: To see how Lemma 1 implies the Collatz conjecture, we start by breaking down into cases. (For convenience here, I'm going to abbreviate 'the Collatz sequence starting from $n$ converges' as $C(n)$ and 'the Collatz2 sequence starting from $n$ converges' as $C_2(n)$.) First, if $n$ is even, then our first step of Collatz takes us to $n/2$, and so (almost trivially) $C(n)\Leftrightarrow C(n/2)$, and in particular $C(n/2)\Rightarrow C(n)$. On the other hand, if $n$ is odd, then we use our lemma: we know that $C_2(n)\Leftrightarrow C(n+1)$ by the equivalence shown above, and the lemma states that $C(n)\Leftrightarrow C_2(n)$; putting the two together, we get $C(n)\Leftrightarrow C(n+1)$. But since $n+1$ is even (because we're looking at the $n$ odd case), $C(n+1)\Leftrightarrow C((n+1)/2)$, so we have the equivalency $C(n)\Leftrightarrow C((n+1)/2)$ and in particular $C((n+1)/2)\Rightarrow C(n)$. Now we can induct, using the strong induction principle: suppose we have $\forall k\lt n C(k)$. Then by specializing we have (either) $C(n/2)$ or $C((n+1)/2)$; therefore $C(n)$ by (one of) the two implications we proved. Therefore, $(\forall k\lt n C(k))\Rightarrow C(n)$, and since we know $C(1)$ we get $\forall n C(n)$.
Best Answer
I did some "independent and unpaid research" on this (much against all advice an undergrad student in mathematics receives regarding this particular problem) and arrived at the following:
The Collatz conjecture is true if and only if all positive integers of the forms $4r+3$ or $8r+3$, $r$ being odd, eventually reach lower values by iterating the Collatz function.
So indeed, you may restrict your attention to odd numbers, specifically to those of the form I just mentioned, and drop the computations when they reach lower values.
(My analysis can be pushed even further to filter away more odd numbers that will reach lower points, but I saw no clear pattern in my analysis, hence paused working on it)
A short version of my proof of the above statement: We just show that all other positive integers eventually reach lower values. For even numbers this is clear. For odd numbers, write them in the form $2^kn+1$, $n$ odd and $k>0$. Iterate the Collatz function. For even exponents $k$, 3 iterations will be enough (details left to the reader). For odd exponents other than 1, 3 iterations will again suffice. Thus the conjecture is true iff all positive integers of the form $2n+1$, $n$ odd, reach lower values.
Now, substitute $n$ by $2^xy+1$, $y$ odd and $x>1$, repeat the above analysis.
As for what happens with $4r+3$ and $8r+3$, $r$ odd…I realized that I probably wouldn't be able to finish any kind of proof with this, but that I could go on forever with these substitutions…
Also, as all others have said, please make sure that all integers less than those that you consider have already been verified, otherwise you might fool yourself (even though I feel it wouldn't matter in the long run, but it could cause some delays in finding the truth in those scenarios…).