About Collatz 3n+3? While trying to prove the Collatz conjecture I came across the following Lemma :
Lemma $1$ : 
All variables are positive integers.
Define $Collatz(n)$ as the result of the (repeated) collatz iteration $x/2$ for even , $3x+1$ for uneven halt at $1$.
Define $Collatz2(n)$ as the result of the (repeated) iteration $x/2$ for even , $3x+3$ for uneven halt at $3$.
Lemma 1 says for each positive integer $n$ $:$ $Collatz(n) = 1 <=> Collatz2(n) = 3$
How to prove that ?
Edit : $n$ is the starting value for both $Collatz$ and $Collatz2$ , Imho its obvious but it seems it was not clear enough. $x$ is just used to describe the process of the iterations.
http://en.wikipedia.org/wiki/Collatz_conjecture
Edit : Bonus question ( not related to my attempt at proving Collatz )
On average in a race between $Collatz(n)$ and $Collatz2(n)$ which reaches its halting value( $1$ or $3$) first ( less iterations required = faster ) ?
Too clarify consider the hypothetical example :
We define $f(n) = 1$ if $Collatz$ wins the race and $= 0$ if $Collatz2$ wins the race.
Let $F(n) = f(1) + f(2) + ... + f(n)$.
Then the limit as $n$ goes to infinity of $F(n)/n$ gives for example $\pi/10$.
Edit : related question : What is known about Collatz like 3n + k?
 A: 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 $\forall n C(n)$.
A: If $n$ is odd, then $3n+3$ is a multiple of $3$, and if $n$ is an even multiple of $3$, $n/2$ is still a multiple of $3$. So unless you happen to pick a number where every term in the sequence is even (which can only happen with $n=0$ but then $n$ is a multiple of $3$ too), you will eventually end up with a sequence containing only multiples of $3$.
And if you look closely, for $n$ even, $3n$ is sent to $3(n/2)$ and for $n$ odd, $3n$ is sent to $3(3n+1)$ so your sequence from $3n$ is simply the normal Collatz sequence starting from $n$ but multiplied by $3$.
You can think of the Collatz2 sequences as Collatz sequences that are allowed to start from numbers of the form $n/3$. Then those sequence always end up being sequences of integers, i.e. regular Collatz sequences (and conjecturally end up at $1$). There is no new branch in the Collatz graph if you add all the $n/3$ numbers.
