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 $\forall n C(n)$.
Here are some pictures for your/our intuition. I graphed the trajectories for initial values $x=5,15,25,...$ for the first $256$ steps of $x_{k+1}=(5x_k+1)/2^A$.
To get the curves to a meaningfully visual interval I show logarithmic scales. The pictures show how most trajectories begin to diverge (not really a safe indication of what characteristic the infinite curves really have) but some show cycling already at early iteration indexes $k$ .
I find $2$ cycles besides the "trivial" one.
$x=5,15,25,35,...,95$ detail of the first few iterations . At the bottom we see the "trivial" cyle (brown curve):
$x=5,15,25,35,...,95$ first $2^8 = 256$ iterations. At later iteration-indexes $k$ a first "non-trivial" cycle occurs (red line):
$x=105,115,125,135,...,195$ first $2^8 = 256$ iterations .
$x=205,215,225,235,...,295$ first $2^8 = 256$ iterations . Here a second "non-trivial" cycle becomes visible:
$x=205,215,225,235,...,295$ first $2^{11} = 2048$ iterations
It seems really that all trajectories which are divergent up to iteration $k=256$ are also divergent up to iteration $k=2048$ . In general: I doubt that there are "later" cycles:
Best Answer
The answer to your question is yes; one can work the Collatz relation backwards to build numbers that last arbitrarily long (for a simple example, just take powers of 2). However the purpose of the wikipedia records is to see if we can find SMALL numbers that last a long time.