Just for the sake of interest, I have realized that the additive step in the Collatz function can be technically avoided when computing the function iterates.
Rather than defining the Collatz function as $$ T_0(n) = \begin{cases} (3n + 1)/2 &\quad \text{ if $n \equiv 1 \pmod{2}$,} \\ n/2 &\quad \text{ if $n \equiv 0 \pmod{2}$,} \end{cases} $$ and tracking the trajectory directly on $n$, one can track the same trajectory on $n+1$ with the function $$ T_1(n) = \begin{cases} (n+1)/2 &\quad \text{ if $n \equiv 1 \pmod{2}$,} \\ 3n/2 &\quad \text{ if $n \equiv 0 \pmod{2}$.} \end{cases} $$ Thus the "multiplying by 3" just moved to the "even" branch.
The trick is that, when calculating the function iterates, we switch between $n$ and $n+1$ in such a way that we always use only the "even" branch of either $T_0$ or $T_1$. Therefore, the above functions can be expressed as $$ T_0(n) = \begin{cases} T_1(n+1)-1 &\quad \text{ if $n \equiv 1 \pmod{2}$,} \\ n/2 &\quad \text{ if $n \equiv 0 \pmod{2}$,} \end{cases} $$ and $$ T_1(n) = \begin{cases} T_0(n-1)+1 &\quad \text{ if $n \equiv 1 \pmod{2}$,} \\ 3n/2 &\quad \text{ if $n \equiv 0 \pmod{2}$.} \end{cases} $$
There are seemingly still two additions. However, considering the binary representation of the $n$, these additions can be avoided using right shifts and operations which count the number of one/zero bits following the least significant zero/non-zero bit. The powers of three can be precomputed in a look-up table.
I am just interested in knowing whether this formulation has appeared before? Is this a special case of some other known (more general) recurrence relation? Any feedback is welcome.
UPDATE: Some simple code to illustrate my idea can be found here. Currently, I am able to verify the convergence of all numbers below $2^{40}$ in approximatelly 4 minutes (single-threaded program running at 2.40GHz CPU).
Answer to @BaBler's comment. The formula for finding trailing zeros is: $n\oplus(n-1)$, i.e., in Boolean form. To get a pow$2$ number we must add $+1$ to that or else it will just give trailing $1$'s. We must then subtract $1$ from the result and take the base $2$ logarithm. Here is the complete form:
$$v_2(n)=\log_2(n\oplus(n-1))$$
$$C(n) = \frac{3n+1}{2^{v_2(3n+1)}}$$
Input to this function must (only) be odd numbers: $2n+1$.
Code
Since $n\oplus(n-1)$ gives $2^x-1$, the $\log_2$ rounds off to nearest exponent which is actually the number of trailing zeroes.
This is also known as the Reduced Collatz Function.