Prove the integer sequence from taking the cube root of $x$ if it's a perfect cube, else $2x+1$, grows unbounded 
Alice has an integer $N > 1$ on the blackboard. Each minute, she deletes the current number $x$ on the blackboard and writes $2x+1$ if $x$ is not the cube of an integer, or the cube root of $x$ otherwise. Prove that at some point of time, she writes a number larger than $10^{100}$.


Let, $a_{1}=x,a_{2}=2x+1,a_{3}=4x+3,\ldots,a_{n+1}= 2^{n}(x+1)-1$. Now I tried taking mod $8$ and mod $9$, but there's always a case where it's possible for the sequence to have a cube. We need to show the sequence grows unbounded. How do I proceed further?
 A: Starting with $N \gt 1$, the procedure takes the cube root as many times as required until an integer $x_0 \gt 1$ results which is no longer a perfect cube. Then, for any $i \ge 0$, starting from $x_{i}$, after applying the $2x+1$ step $0$ or more times, if $10^{100}$ is passed, we're done. Otherwise, have $x_{i+1}^3$ be the next perfect cube obtained after some $n_i \ge 0$ steps. Thus, as you've basically shown, we get
$$\begin{equation}\begin{aligned}
x_{i+1}^3 & = 2^{n_i}(x_{i}+1) - 1 \\
x_{i+1}^3 + 1 & = 2^{n_i}(x_{i}+1) \\
(x_{i+1}+1)(x_{i+1}^2-x_{i+1}+1) & = 2^{n_i}(x_{i}+1)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
with $x_{i+1}$ then being the next resulting value. Starting from $x_0$, since it's not a perfect cube, the next value is $2x_0+1$, which is odd. Also, since all odd perfect cubes are the cubes of an odd integer, plus applying $2x+1$ also always results in an odd value, this means that $x_i$ are odd for all $i \ge 1$.
Using the $p$-adic valuation function (e.g., where $\nu_2(m)$ is the exponent of $2$ in the prime factorization of $m$), since $x_{i+1}^2-x_{i+1}+1$ is odd, then from \eqref{eq1A} we have
$$\nu_2(x_{i+1} + 1) = n_i + \nu_{2}(x_{i} + 1) \tag{2}\label{eq2A}$$
Thus, $\nu_2(x_{i} + 1)$ is a non-decreasing sequence for $i \ge 1$. Note with $n_i = 0$ for any given $i$, from \eqref{eq1A} we have $x_{i+1}^3=x_{i}$, and since $x_{i}\gt 1$ (due to the procedure either applying $2x+1$ or taking the cube root of a value $\gt 1$), so $x_{i+1} \lt x_{i}$. However, this can occur for only a finite number of times (with $n_i = 0$ and $i$ incrementing each time) before the value is no longer a perfect cube. Then the $2x + 1$ process will be used until one of two things happen:

*

*We get a value $\gt 10^{100}$ before the next perfect cube. Note this can also happen because no more perfect cubes can occur. This is because, as shown in \eqref{eq1A}, the odd part of $x_{i+1}+1$ (i.e., with $m=\nu_{2}(x_{i+1}+1)$, the value $\frac{x_{i+1}+1}{2^{m}}$) is the odd part of $x_{i}+1$ divided by $x_{i+1}^2-x_{i+1}+1$. Thus, since $x_{i+1}^2-x_{i+1}+1=x_{i+1}(x_{i+1}-1)+1\gt 1$, this odd part will be strictly decreasing until it's no longer large enough for another perfect cube to occur.


*A perfect cube is obtained, so for that $i = j - 1$, then $n_{j-1} \gt 0$ means $\nu_2(x_{j} + 1)$ would increase.
If #$1$ above doesn't occur, then #$2$ happening instead shows that these $\nu_2(x_{j} + 1)$ can keep increasing. Since $k = \nu_2(x_{j} + 1) \; \to \; x_{j} + 1 \ge 2^k$, then the minimum possible value of $x_{j}$ also increases, so at some point $x_{j}^3$ (or a value before this is reached) will become larger than $10^{100}$.
Note Alice will always eventually write a value $\gt 10^{100}$ due just to the odd part of $x_{i}+1$ becoming too small, as mentioned in #$1$ above, with this actually being more likely to happen before the effect of #$2$. Nonetheless, I included #$2$ as well for completeness, since that may cause a result to go past that bound first.
