Collatz Related? $x$ is odd: If $x \bmod 3=0$ execute $x/3$, or elseIf $x \bmod 3=1$ execute $(x⋅2)+1$, or elseif $x \bmod 3=2$ execute $(x⋅4)+1$. Let $x$ be an odd number:
If $x \bmod 3=0$ : execute $x/3$, or elseIf $x \bmod 3=1$ : execute $(x⋅2)+1$, or elseif $x \bmod 3=2$ : execute $(x⋅4)+1$.
I have been playing around with my calculator for hours and it keeps coming down to either $1$ or $5$
($\begin{aligned} 1\mapsto3\mapsto1\mapsto3\mapsto\ldots \end{aligned}$)
($\begin{aligned}5\mapsto21\mapsto7\mapsto15\mapsto5\mapsto21\mapsto7\mapsto15\mapsto\ldots \end{aligned}$)
Examples:
$$
\begin{aligned}
& 127\mapsto255\mapsto85\mapsto171\mapsto57\mapsto19\mapsto39\mapsto13\mapsto27\mapsto9\mapsto3\mapsto1\mapsto\ldots
\end{aligned}
$$
$$
\begin{aligned}
& 239\mapsto957\mapsto319\mapsto639\mapsto213\mapsto71\mapsto285\mapsto95\mapsto381\mapsto127\mapsto255\mapsto85\mapsto171\mapsto57\mapsto19\mapsto39\mapsto13\mapsto27\mapsto9\mapsto3\mapsto1\mapsto\ldots
\end{aligned}
$$
$$
\begin{aligned}
& 49\mapsto99\mapsto33\mapsto11\mapsto45\mapsto15\mapsto5\mapsto\ldots
\end{aligned}
$$
$$
\begin{aligned}
& 919\mapsto1839\mapsto613\mapsto1227\mapsto409\mapsto819\mapsto273\mapsto91\mapsto183\mapsto61\mapsto123\mapsto41\mapsto165\mapsto55\mapsto111\mapsto37\mapsto75\mapsto25\mapsto51\mapsto17\mapsto69\mapsto23\mapsto93\mapsto31\mapsto63\mapsto21\mapsto7\mapsto15\mapsto5\mapsto\ldots
\end{aligned}
$$
I have found many questions in regards to the Collatz variants, and I am aware that proofs are currently out of the limits. Most of the questions seemed to involve even numbers being divided by $2$, where as in the above the results are only odd numbers, and so I am just curious to know:
Am I correct about the above as a conjecture, and is the difficulty in trying to solve it similar to the original Collatz conjecture (that involve even numbers being divided by 2)?
Any references are also appreciated.
 A: this sequence T is equivalent at:
\begin{array}{lllll}
\text{0) }&x/3&\text{with}& x=3n :&T(x)=n&\text{with probability}\quad1/3\\
\text{1) }&(2x+1)/3&\text{with}&x=3n+1 :&T(x)=2n+1&\text{with probability}\quad1/3\\
\text{2) }&(4x+1)/3&\text{with}&x=3n+2 :&T(x)=4n+3&\text{with probability}\quad1/3\\
\end{array}
The growth is:
$T(x)/x=( (1/3) (2/3) (4/3) )^{1/3}=2/3<1$ so it decreases.
and cycles occur when:
$x_1 x_2 \ldots x_{k-1} x_1 = T(x_1) T(x_2) \ldots T(x_{k-1}) T(x_k)$
Hence:
$3^k > 2^{k1+2 k2}\qquad(1)$
$k_i$ is the number of values $i\bmod(3)$ in cycle: $k=k_0+k_1+k_2$ and if $k_0=k_1=k_2=k/3$, equation (1) is ok, thus equibrobability is possible, e.g.:
$x=T(x):$
\begin{array}{llll}
\text{0) }&x=x/3 :&x=0\\
\text{1) }&x=(2x+1)/3 :&x=1:&\text{cycle of lenght 1 (fixe point)}\\
\text{2) }&x=(4x+1)/3 :&x=-1:&\text{no solution in }\mathbb{Z}+\\
\end{array}
$x=T(T(x)):$
\begin{array}{lll}
\text{00) }x=(x/3)/3 :&x=0&\\
\text{01) }x=(2x/3+1)/3 :&x=3/7:&\text{no solution in }\mathbb{Z}+\\
\text{02) }x=(4x/3+1)/3 :&x=3/5:&\text{no solution in }\mathbb{Z}+\\
\text{10) }x=((2x+1)/3)/3 :&x=1/7:&\text{no solution in }\mathbb{Z}+\\
\text{11) }x=(2(2x+1)/3+1)/3 :&x=1:&\text{previous solution}\\
\text{12) }x=(4(2x+1)/3+1)/3 :&x=7:&\text{cycle of lenght }2\quad\{7;5\}\\
\text{20) }x=((4x+1)/3)/3 :&x=1/5:&\text{no solution in }\mathbb{Z}+\\
\text{21) }x=(2(4x+1)/3+1)/3 :&x=5:&\text{cycle of lenght }2\quad\{5;7\}\\
\text{22) }x=(4(4x+1)/3+1)/3 :&x=-1:&\text{no solution in }\mathbb{Z}+\\
\end{array}
and go on to maybe find other cycle of lenght $>2$.
