<s>[INST] 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.
 [/INST] Sure, there's nothing strange going on here: these formulas are all equivalent. (Can you prove it?)
Hint: if $n$ is odd then $\lfloor \frac{n}{2} \rfloor = \frac{n}{2} - \frac{1}{2}$. 
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