Give me such $p_1, q_1$ or $p_2,q_2$ for any $n\in\mathbb N$, $f^{i}(n)=1$ and $g^{j}(n)=1$ I'm sorry , If this is a similar question. Because, I could not get a satisfactory answer. But, I hope,this question is not similar.
Let $$ f(n)=
\begin{cases}
p_1\times n + q_1          & \text {if $n$ is odd} \\
\frac{n}{2}   & \text {if $n$ is even} 
\end{cases} ,$$ Here $p_1,q_1\in
\mathbb{N}$
Let $$ g(n) =
\begin{cases}
p_2 \times n - q_2        & \text {if $n$ is odd} \\
\frac{n}{2}   & \text {if $n$ is even} 
\end{cases} ,$$ Here $p_2,q_2\in
\mathbb{N}$
Example-1
We know that, if $p_1=3$ and $q_1=1$ for any $n\in\mathbb N$ (There is no counter-example,for now ) , we can find such $"i"$ number, which that it must be $f^{i}(n)=1$.
Example-2
If $p_1=3$ and $q_1=5$ for any $n\in\mathbb N$ , we can not find such $"i"$ number, $f^{i}(n)=1$. (Because there are many counter-examples; $\Rightarrow 5,19,23,187,347$ and for any $"i"$ number $f^{i}(n)≠1$.)
Example-3
If $p_2=3$ and $q_2=1$ for any $n\in\mathbb N$ , we can not find such $"j"$ number, $g^{j}(n)=1$. (Because there are many counter-examples; $\rightarrow 5,7; 17,25,37,55,41,61,91$ and for any $"j"$ number $g^{j}(n)≠1$.)

Finally, The question: 
Give me such $p_1, q_1$ or $p_2,q_2$ for any $n\in\mathbb N$,  $f^{i}(n)=1$ and $g^{j}(n)=1$,  as a $"3n+1$ problem .
  Are such numbers known, other than $"3n+1"$ ? 

($p_1≠3$ and $q_1≠1$,  because this conjecture is Collatz Conjecture, I already know)
Note: (If you find any mistakes ,please, edit or improve my question for me.Thanks, so much!)
 A: Unfortunatelly, to my knowledge, a $p_1$ $=$ 3 and $p_2$ $=$ 1 for $f(n)$ is the only combination with the behavior that matches the Collatz Conjecture. However, if you expand $n$ so $n \in$ $\mathbb{Z}$ , then the rule $p_1=$ 3 and $p_1=$ 1 for $g(n)$ will behave exactly like the Collatz Conjecture, only with negative numbers.
ex) 
$f(n)$ , $p_1=$ 3 and $p_1=$ 1  --> 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, ...
$g(n)$ , $p_1=$ 3 and $p_1=$ 1  --> -26, -13, -40, -20, -10, -5, -16, -8, -4, -2, -1, ...
However, the Collatz Conjecture's behavior is not exclusive to these rules. If the other loops are ignored for the rule $f(n)$ where $p_1 =$ 3 and $p_2 >$ 1, Collatz-like behavior can be found amoung the values where $n\in p_2* \mathbb{Z}$ . The only difference is the scale by $p_2$.
ex)
$f(n)$ , $p_1=$ 3 and $p_1=$ 1  --> 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, ...
$f(n)$ , $p_1=$ 3 and $p_1=$ 5  --> 130, 65, 200, 100, 50, 25, 80, 40, 20, 10, ...
If reaching 1 is desired, then $p_1$ $=$ 1 and $p_2$ $=$ 1 for $f(n)$ will reduce to 1 without any loops or counter examples to anyone's knowledge.
If looking for rules with one and only one loop is desired, then the rules $p_1$ $=$ 3 and $p_2$ $=$ 3 for $f(n)$ and $p_1$ $=$ 3 and $p_2$ $=$ 9 for $f(n)$ may work. There are no known counter examples as of now to these rules not reaching their trivial cycles.
