How to prove escalation to infinity (+ or -) for a few numbers with cases n/3 and 4n+1. How do I prove that following numbers 45, 61, 101 and -59 are going up to infinity, but -155 falls into a loop.
f(n)=
\begin{cases}
n/3, & \text{if $n≡ 0$ $(mod$ $3)$}  \\
4n+1, & \text{if $n≢ 0$ $(mod$ $3)$}
\end{cases}
Example: 45 - 15 - 5 - 21 - 7 - 29 - 117 - 39 - 13 - 53 - 213 - 71 - 285 - 95 - 381 - 127 - 509 - 2037 - 679 - 2717 - 10869 - 3623 - 14493 - 4831 - 77301 - 25767 - etc.
 A: While it's sure to be difficult to prove, note that heuristically we should expect this process to diverge to infinity, which is different from the $3n+1$ problem (where heuristically numbers converge).
The standard heuristic for the $3n+1$ problem is based around the notion that an 'unknown' number is equally likely to be odd or even. Naively, this would seem to suggest divergence, because if we flip a coin and then apply either $n\mapsto 3n+1$ or $n\mapsto n/2$, then the average size of the number grows: ignoring the additive factor, the number has probability $\frac12$ of growing by a factor or $3$ and probability $\frac12$ of shrinking by a factor of $\frac12$.  Since $3\cdot\frac12\gt 1$, this would seem to suggest growth on average.
But we know that any odd number is carried to an even number, so once we've applied the $n\mapsto 3n+1$ rule, we know that we're going to apply a step of $n\mapsto n/2$; after that, the status of the resulting number is again unknown. To correctly model this process heuristically, we want to make sure that we're going from unknown to unknown; this suggests that our 'random' model should flip a coin and then choose either the mapping $n\mapsto n/2$ or $n\mapsto (3n+1)/2$ with equal probability. Because $\frac32\cdot\frac12\lt1$, this means that numbers in the Collatz process are actually 'on average' decreasing.
Now, for your $4n+1$ problem, we can take the same approach and follow the process from unknown to unknown; here it'll be based on the status of $n$ mod $3$. if $n\equiv0\pmod 3$ then after we apply the  $n/3$ rule we have an unknown number; if $n\equiv 1$ then we know we'll be applying the $4n+1$ rule twice and then the $n/3$ rule, so the formula for getting to the next 'unknown' number in the process is $n\mapsto\frac13(16n+5)$; and if $n\equiv 2$ then we apply the $4n+1$ rule once and then the $n/3$ rule, so the formula for getting to the next 'unknown' number is $n\mapsto\frac13(4n+1)$.  Thus, starting from an unknown $n$, with probability $\frac13$ we multiply it by $\frac13$; with probability $\frac13$ we (roughly) multiply it by $\frac{16}3$; and with probability $\frac13$ we (roughly) multiply it by $\frac43$.  Since we have $\frac13\cdot\frac{16}3\cdot\frac43\gt 1$, we should expect 'most' numbers to diverge under this process.
