Simplistic Odd Collatz formulas I've been experimenting with the Collatz for a couple of months. 
I just want to give you some formulas and show tables that give the same results for odd inputs. I have also several new functions, but ill only provide a few of them here. I tweaked the formulas. It seems very strange that all of them give the same outputs. My notes are a bit messy, so bear with me if I didnt provide all the details.
\begin{array}{|c|c|c|c|}\hline
n  & 3n & 3n+1 & \frac{3n+1}{2} \\ \hline
1  & 3  & 4  & 2  \\ \hline
3  & 9  & 10 & 5  \\ \hline
5  & 15 & 16 & 8  \\ \hline
7  & 21 & 22 & 11 \\ \hline
9  & 27 & 28 & 14 \\ \hline
11 & 33 & 34 & 17 \\ \hline
13 & 39 & 40 & 20 \\ \hline
\end{array} orginal collatz formula, dividing by two.
\begin{array}{|c|c|c|c|}\hline
n  & 2n & \lfloor\frac{n}{2}\rfloor & 2n-\lfloor\frac{n}{2}\rfloor \\ \hline
1  & 2  & 0 & 2  \\ \hline
3  & 6  & 1 & 5  \\ \hline
5  & 10 & 2 & 8  \\ \hline
7  & 14 & 3 & 11 \\ \hline
9  & 18 & 4 & 14 \\ \hline
11 & 22 & 5 & 17 \\ \hline
13 & 26 & 6 & 20 \\ \hline
\end{array} a new(?) formula.
\begin{array}{|c|c|c|c|}\hline
n  & \lfloor\frac{n}{2}\rfloor & n+1 & \lfloor\frac{n}{2}\rfloor + n+1 \\ \hline
1  & 0 & 2  & 2  \\ \hline
3  & 1 & 4  & 5  \\ \hline
5  & 2 & 6  & 8  \\ \hline
7  & 3 & 8  & 11 \\ \hline
9  & 4 & 10 & 14 \\ \hline
11 & 5 & 12 & 17 \\ \hline
13 & 6 & 14 & 20 \\ \hline
\end{array} yet another.
Here $n$ is a positive integer. The floor function omits the fractional part if the result is real.
The functions from the tables above:
$f(n) = \frac{3n+1}{2}$
$f(n) = 2n-\lfloor\frac{n}{2}\rfloor$
$f(n) = \lfloor\frac{n}{2}\rfloor + n+1$
The latter formula is the one i've been working on lately.
We can change it again,
$\lfloor\frac{n}{2}\rfloor+n+1 = \frac{n}{2}+\frac{2n}{2}+\frac{2}{2} = \frac{3n+2}{2}=\lfloor\frac{3n}{2}\rfloor+1$.
That gets us back to the formula in the first table. They look quite similar.
The function
$f(n) = 2n-\lfloor\frac{n}{2}\rfloor$
is also quite interesting. In a "geometric" standpoint we can look at it like doubling the $n$ and then removing the half of the total, here's a simple figure showing this:

I have never seen this formula before, and wonder if I can use this notation:
$f(n)=\lfloor\frac{n}{2}\rfloor+\begin{cases} n+1 & n\equiv 1\pmod2 \\ 0 & n\equiv 0\pmod2 \end{cases}$
See, for odd and even we can halve the input in one go. And then add the result of the cases.

There is nothing wrong with using an alternate form of the original Collatz ($3x+1$) rule, as long as you have a reason for the modification.
Without using a floor function, it is much easier to show that your method is equivalent to the $\frac{(3n+1)}{2}$ rule. The floor function truncates 0.5 after dividing an odd integer in half, therefore removing 1 from the odd integer in the first place and then dividing by 2 will provide the same effect. (ex. $\lfloor$9/2$\rfloor$ = 4.5 - 0.5 = 4, and (9-1)/2 = 4.)
With that in mind, here is a proof to validate your method:
$\frac{n-1}{2}$ + $n$ + $1$
= $\frac{n-1}{2}$ + $2n\over2$ + $2\over2$
= $\frac{n - 1 + 2n + 2}{2}$
= $\frac{n + 2n + (-1) + 2}{2}$
= $\frac{3n+1}{2}$
The expressions $\frac{3n+1}{2}$ and $\frac{n-1}{2}$ + $n$ + $1$ are equivalent.
