Simplified variant of Collatz conjecture. I came across the Collatz conjecture. So apparently the idea is to see if all prime factors of a number can be 'annihilated' by successive steps of either removing a factor of two, if n is even or in case it is not to transform it by a simple '(e.g. linear) transformation' to an even number. Now the algorithm to accomplish that is:   
Take $n$ and generate the sequence $f^i(n)$, with 
$$f(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod{2}\\ 3n+1 & \text{if } n\equiv 1 \pmod{2} .\end{cases}$$
It seems a tough problem. I thought about the following simplified version:
$$g_0(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod{2}\\ n+1 & \text{if } n\equiv 1 \pmod{2} .\end{cases}$$
The first question is can one show that all the sequences $g_0^i$ go to orbits $...,1,2,1,2,...$?
I think the answer must be yes, since there will never be members with prime factors larger than $\sqrt{n}$, so there is only a limited reservoir of possible prime factors which all in all are getting reduced by the steps. Is that about correct? 
The next question is how far can one go with such variations on the original algorithm 'approching the normal one', For example what about:
$$g_1(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod{2}\\ n+3 & \text{if } n\equiv 1 \pmod{2} .\end{cases}$$
or 
$$g_2(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod{2}\\ n+(n+1)/2 & \text{if } n\equiv 1 \pmod{2} .\end{cases}$$
and so on.
Another interesting extension would be an $abc$ variant:
$$g^j_{abc}(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod{2}\\ n+m^j & \text{if } n\equiv 1 \pmod{2} \end{cases}$$
for $m^j =$ the next-but-$j$ prime of $n$.
I suppose $g_1$ can be handled with a similar argument (in case it is correct) as what I lined out for $g_0$. For $g_2$ I am not sure anymore, but somehow I suppose $f$ (the original) is something like the simplest of this class which cannot be handled with that kind of argument (otherwise it wouldn't be such a conundrum?). 
 A: The case of $g_0$ is immediate. You can combine two steps in one and 
$$g'_0(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod{2}\\ (n+1)/2 & \text{if } n\equiv 1 \pmod{2}\end{cases}$$ so that the function is strictly decreasing for $n>2$, and $g'_0(1)=g'_0(2)=1$.
The same argument holds for $g_1$ and $g_2$.
A: $g_0(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod{2}\\ n+1 & \text{if } n\equiv 1 \pmod{2} .\end{cases}$
Hi. That the function is decreasing is a demonstration? If so, how can we transfer it to the 3n + 1 problem? Somehow the Collatz algorithm encodes or reorders n, is a permutation from N, and in this distribution, the function can be shown decreasing?
$$ \begin{matrix}
... & ... & ... & ... & ... & ... & ... & ... & ... & ... \\ 
128 & 84 & 52 & 136 & 88 & 58 & 74 & 100 & 260 & ... \\
64 & 80 & 48 & 75 & 49 & 56 & 72 & 99 & 133 & ... \\
32 & 42 & 26 & 74 & 46 & 30 & 38 & 98 & 132 & ... \\
16 & 40 & 24 & 70 & 45 & 29 & 37 & 96 & 130 & ... \\
8 & 21 & 13 & 68 & 44 & 28 & 36 & 51 & 67 & ... \\
4 & 20 & 12 & 35 & 23 & 15 & 19 & 50 & 66 & ... \\
2 & 10 & 6 & 34 & 22 & 14 & 18 & 49 & 65 & ... \\
1 & 5 & 3 & 17 & 11 & 7 & 9 & 25 & 33 & ... \\ \hline\\
k=0 & k=1 & k=2&k=3&k=4&k=5&k=6&k=7&k=8&k=...
\end{matrix} $$
k=odd steps. The even numbers go down and the odd numbers go to the left.
