TITLE: How long iterations of $x \to (p \!\!\! \mod \!\! x)$ can be?
QUESTION [9 upvotes]: Suppose that $p$ is a prime number and $x_1$ is an integer in range $[1, p - 1]$. If $x_k \not = 1$,
define $x_{k + 1} := p \!\!\! \mod \!\! x_k$. Clearly, $x_{k + 1} < x_k$ and $x_{k + 1} > 0$ 
(because $p$ is prime), so there exists some $n$ (depending on $x_1$) such that $x_n = 1$. 
Is it true that $n = O(\log p)$?
For reference, it is quite easy to prove that $n = O(\sqrt p)$. Let $q_i := [\frac{p}{x_i}]$, i. e.
$p = q_i x_i + x_{i + 1}$ for every $i \in [1, n - 1]$. Because 
$q_i x_i + x_{i + 1} = p = q_{i + 1} x_{i + 1} + x_{i + 2}$ for $i \in [1, n - 2]$ and $x_i > x_{i + 1},
x_{i + 1} > x_{i + 2}$, $q_i$ is increasing. On the other hand, $x_i$ is decreasing. Also
$q_i x_i < q_i x_i + x_{i + 1} < p$ for every $i \in [1, n - 1]$. Therefore $q_i < \sqrt{p}$ or 
$x_i < \sqrt{p}$ for each $i \in [1, n - 1]$. Because $q_i$ increases, there can be at most $[\sqrt{p}]$
$i$-s with $q_i < \sqrt{p}$. The same way there can be at most $[\sqrt{p}]$ $i$-s with $x_i < \sqrt{p}$.
Therefore $n \leqslant 2[\sqrt{p}]$.
Some ideas:

Maybe it makes sense to think backwards: i. e. how long can a sequence $y$ be if 
$y_1 = 1$ and $y_{i + 1}$ is some divisor of $p - y_i$ that is bigger than $y_i$ (in this notation $x_i = y_{n + 1 - i}$)?
One more way to look at this is to notice that the question asks to estimate how deep can a rooted tree on $p - 1$ vertices be, with root in $1$ and edges between $x$ and $p \!\!\! \mod \!\! x$ for $x = 2, 3, \ldots, p - 1$. Maybe it is possible to prove that this tree is somewhat balanced?
Two heuristics suggest that the number of steps is indeed $O(\log p)$: first one assumes that
$p \!\!\! \mod \!\! x$ is a random integer number in range $[1, x - 1]$ and second one uses 
"backwards" point of view and assumes that $p - y_i$ is divisible by $t$ with probability $t^{-1}$. 
Of course, both heuristics don't make too much sense, but they show that $n = O(\log p)$ is
at least a reasonable enough statement to consider.

In fact, any ideas on proving any bound better than $O(\sqrt{p})$ are appreciated.

REPLY [2 votes]: Here is an idea to pursue.
Fix $p$. I will call $p \bmod x$  (your dynamic) $d(x)$, and I look at $H$, the set of $x$ less than $p$ with $2*d(x) \gt x$.  The dynamic can stay out of $H$ only a logarithmic number of times, so now we ask how attractive $H$ can be as a dynamic.
$H$ looks like a collection of intervals. Dividing everything by $p$, the set is like $(1/2,2/3)$ union $(1/3,2/5)$ union intervals of the form $(1/k, 2/(2k-1))$, for enough $k$ until you get bored. This is some constant fraction of $p$, so pretty large in logarithmic terms.
However, we don't have to be afraid of $H$.  If $d$ visits $H$ and not $H$ alternately, $d$ still reaches $1$ in a logarithmic number of steps. (Indeed, we can set a number in $(1,\log p)$ as a goal to finish in logarithmic time, so I am not going to worry about the very end of the dynamic.) So let us consider when $x$ is in $H$ and $d(x)$ is also in $H$.
Suppose $x$ and $x+1$ are in $H$. Then $d(x+1)-d(x) \gt 1$ when $x \lt p/2$. So the set of $x$ less than $p/2$ which visit $H$ at least twice in a row is less than half the number of $x$ less than $p/2$ which visit $H$ at least once under $d$.  If we start from a point less than $p/2$, then the chance it lands in $H$ is small, and the chance it lands in $H$ twice in a row by iterating $d$ gets very small. The idea is to analyze this region below $p/2$, and show that landing in $H$ at least $k$ times in a row is less than $(1/2)^k$ as likely as landing once.  I leave the detail to others.
Now we turn our attention to $(2x \gt p)$. The question now is how long can one iterate $d$ and stay above $p/2$.  But this happens for no $x$ less than $p$.  So the ticket is to show the subsets of $H$ which are visited by $d$ at least $k$ times in a row grows exponentially small.
Gerhard "Seems Better Than Square Root" Paseman, 2018.07.18.