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Regarding the secondary question: If we start the process from a large $N$, it will always reach 1 sooner or later. The probability that it passes through the number 2 is $1/2$, since as long as the numbers are larger than 2, going in the next step to 2 is as likely as going to 1, and when the process reaches 1 or 2 for the first time, where it went decides whether it ever passes through 2. Similarly the probability that the process ever visits a number $n$ is $1/n$, since this happens precisely when the first visit to any of $1,\dots,n$ is to $n$.

Now it's easy in principle to see what the inverse is: For each pair of numbers $m < n$, the probability that a process starting at a larger $N$ will include a step from $n$ to $m$ is $$\frac1{n(n-1)},$$ since it will reach $n$ with probability $1/n$, and the next number distinct from $n$ is uniform on $1,\dots,n-1$. In particular, for "infinite $N$", the last number visited before reaching 1 is $n$ with this probability.

For $m>1$, the probability that $m$ was reached from $n$ given that the process reached $m$ is $m/(n(n-1))$, since we get a factor $m$ from conditioning on the process ever reaching $m$. It might be easier to sort out the details if we assume that the process never repeats the same number.

ADDED: The inverse process constructed this way has the property that at any $m$, the expectation of the next (previous in the original process) step is infinite. But it does have the nice property that the probability of going from $m$ to a number $\leq 2m$ is $1/2$, so the median growth factor over one step is 2.

NEW UPDATE: If we discard repetitions, then one way to understand the inverse is that from a number $m$, the next step is $$m\mapsto \left\lceil\frac{m}{U}\right\rceil,$$ where $U$ is uniform on the interval $[0,1]$. When $m$ is large, the growth factor is therefore asymptotically the reciprocal of a uniform $[0,1]$, which is the same thing as an "exponential of an exponential" ($e$ to an exponential variable). The median growth factor over one step is 2, but over a large number of steps approaches $e$.

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Regarding the secondary question: If we start the process from a large $N$, it will always reach 1 sooner or later. The probability that it passes through the number 2 is $1/2$, since as long as the numbers are larger than 2, going in the next step to 2 is as likely as going to 1, and when the process reaches 1 or 2 for the first time, where it went decides whether it ever passes through 2. Similarly the probability that the process ever visits a number $n$ is $1/n$, since this happens precisely when the first visit to any of $1,\dots,n$ is to $n$.

Now it's easy in principle to see what the inverse is: For each pair of numbers $m < n$, the probability that a process starting at a larger $N$ will include a step from $n$ to $m$ is $$\frac1{n(n-1)},$$ since it will reach $n$ with probability $1/n$, and the next number distinct from $n$ is uniform on $1,\dots,n-1$. In particular, for "infinite $N$", the last number visited before reaching 1 is $n$ with this probability.

For $m>1$, the probability that $m$ was reached from $n$ given that the process reached $m$ is $m/(n(n-1))$, since we get a factor $m$ from conditioning on the process ever reaching $m$. It might be easier to sort out the details if we assume that the process never repeats the same number.

ADDED: The inverse process constructed this way has the property that at any $m$, the expectation of the next (previous in the original process) step is infinite. But it does have the nice property that the probability of going from $m$ to a number $\leq 2m$ is $1/2$, so the median growth factor over one step is 2.

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Regarding the secondary question: If we start the process from a large $N$, it will always reach 1 sooner or later. The probability that it passes through the number 2 is $1/2$, since as long as the numbers are larger than 2, going in the next step to 2 is as likely as going to 1, and when the process reaches 1 or 2 for the first time, where it went decides whether it ever passes through 2. Similarly the probability that the process ever visits a number $n$ is $1/n$, since this happens precisely when the first visit to any of $1,\dots,n$ is to $n$.

Now it's easy in principle to see what the inverse is: For each pair of numbers $m < n$, the probability that a process starting at a larger $N$ will include a step from $n$ to $m$ is $$\frac1{n(n-1)},$$ since it will reach $n$ with probability $1/n$, and the next number distinct from $n$ is uniform on $1,\dots,n-1$. In particular, for "infinite $N$", the last number visited before reaching 1 is $n$ with this probability.

For $m>1$, the probability that $m$ was reached from $n$ given that the process reached $m$ is $m/(n(n-1))$, since we get a factor $m$ from conditioning on the process ever reaching $m$. It might be easier to sort out the details if we assume that the process never repeats the same number.