TITLE: A game of stones
QUESTION [42 upvotes]: How I arrived at this question is a rather long story having to do with the honors calculus class I am teaching. At this point it's sheer curiosity on my part. Here is the game.
$\newcommand{\bZ}{\mathbb{Z}}$
We start with a finite collection of stones placed  at random  somewhere on the set of nodes $\newcommand{\eN}{\mathscr{N}}$ $$\eN=\{2,3,4,\dotsc\}. $$      We can view a distribution  of stones as a  function $s:\eN\to\bZ_{\geq 0} $ with finite support, $s(n)=$ the number of stones at $n$. Its weight  is the nonnegative integer
$$|s|=\sum_{n\in\eN} s(n). $$
We say that a distribution $s$ is overcrowded  if $s(n)> 1$ for some $n\in\eN$. A node $n$ is called  occupied (with respect to $s$) if there is at least one stone at $n$, $s(n)>0$. 
We are allowed the following moves:  choose an occupied node $n$. Then   you move one stone from location $n$ to location $n+1$ and add a new stone at location $n^2$.    Note that such a move increases the weight by $1$.
Now comes the question.

Is it true that for any  initial distribution of stones $s:\eN\to\bZ_{\geq 0}$ and any positive integer $N$   there exists a finite sequence of  allowable moves   such that after these moves we obtain a  new distribution of stones  which (i) is not overcrowded, and (ii) no node $n<N$ is occupied.   

Empirical evidence leads  me to believe that the answer to this question is positive. However, I have failed to find a conclusive argument.I'm hoping  someone  in the MO community will have more luck.
Remark 1. (Inspired by David Eppstein's answer.) I want to show that if the function $n\mapsto n^2$ in the definition of an allowable move is replaced by something else the answer to the question can be  negative.   In other words, any proof for  the positive answer, would have to take into account some features of the map $n\mapsto n^2$.
Here is the example. Fix an integer $k>1$ and  define $f:\eN\to\eN$, $f(n)=n+k$.   Now change the definition of an allowable move as  follows.

Pick an occupied node $m$. Move a stone from location $m$ to $m+1$ and
  add a stone  at the node $f(m)$.  We will  denote by $T_m$ this move.

Let $s_0:\eN\to\bZ_{\geq 0}$ be the configuration consisting of a single stone located at $n=2$. I claim that there exists  $N>0$ such that $s_0$ cannot be moved past $N$ without overcrowding. 
The proof is based on a conservation law suggested by David Eppstein's answer.  Consider the polynomial $P(x)=x^k+x-1$. Note that $P(0)<0$ and $P(1)>0$ so  $P$ has at least one root  in the interval $(0,1)$. Pick one such root $\rho$.  We use $\rho$ to define the energy of a configuration $s:\eN\to\bZ_{\geq 0}$ to be
$$ E(s):=\sum_{n\in \eN} s(n)\rho^n. $$
If $m$ is an occupied location of a configuration $s:\eN\to\bZ_{\geq 0}$, then
$$E(T_m s)= E(s)-\rho^m+\rho^{m+1}+\rho^{m+k}=E(s)+\rho^mP(\rho)=E(s). $$
Thus allowable moves do not change the energy of a configuration.
Let $N$ be a positive integer such that
$$\rho^{N-2}<1-\rho. \tag{1} $$
Suppose now that using allowable move we can transform  $s_0$ to a configuration $s$ such that
$$ s(n)=0,\;\;\forall n<N,\;\;s(n)\in \{0,1\},\;\;\forall n\geq N. $$
Then
$$\rho^2= E(s_0)=E(s)\leq \sum_{n\geq N}\rho^n=\frac{\rho^N}{1-\rho}. $$
This last inequality violates the assumption (1), thus confirming my claim.
Remark 2. The example in the previous remark has the following  obvious  generalization. Suppose that $f:\eN\to\eN$ is a function  such that $f(n)>n+1$, $\forall n \in \eN$ and there exists a probability measure $\pi$ on $\eN$ such that
$$ \pi\bigl(\; f(n)\;\bigr)+\pi(n+1)-\pi(n)\geq 0,\;\;\forall n\in\eN. \tag{2}$$
Using  $f$ to define the allowable  moves, one can show that  there exists $N>0$ such that $s_0$ cannot be moved past $N$ without overcrowding.   The proof uses the entropy
$$E_\pi(s)=\sum_{n\in\eN}s(n)\pi(n). $$
Note that this entropy  is precisely the expectation of $s$ with respect to the probability measure $\pi$, and it does not decrease as we apply allowable moves
$$E(s)\leq E(T_m s), \;\;\forall s. $$
For $f(n)=n+k$ we can define
$$ \pi(n)=(1-\rho)\rho^{n-2}. $$
This remark raises the following natural  question.

Find the functions $f:\eN\to \eN$ such that $f(n)>n+1$, $\forall n\in \eN$,
  and there exists a probability measure $\pi$ on $\eN$ satisfying
  (2). How fast can such a function grow as $n\to \infty$?

Remark 3. (a) For $f(n)=n^2$ the condition (2) reads
$$ \pi(n^2)+\pi(n+1)\geq \pi(n). \tag{3} $$
One can show that a  series   $$\sum_{n\geq 1}p(n) $$ with  nonnegative terms satisfying (3) is  divergent if not all the terms are trivial. (This was the rather tricky honors calculus problem that prompted the present question.)  Hence,  for the function $f(n)=n^2$, there do not exist  probability measures  satisfying (2), suggesting indirectly that the original question could have  a positive answer.
(b) If $f(n)=n +1+ \lfloor \sqrt{n}\rfloor$,   $\alpha>1$ is sufficiently close to $1$ and 
$$\pi(n)=\frac{C}{n^\alpha}, \;\;C\sum_{n\geq 2}\frac{1}{n^\alpha}=1,$$
then the condition (2) is satisfied so  moving without overcrowding is not possible if the allowable moves use the function $f(n)$.
Remark 4. Have a look at Michael Stoll's superb answer.

REPLY [8 votes]: This does not answer the original question, but explores a more general version.
Let $\mathcal N = \{n_0,n_0+1,n_0+2.\ldots\}$ for some fixed $n_0 \in \mathbb Z$. Let $f \colon {\mathcal N} \to {\mathcal N}$ be a map such that $f(n) > n$ for all $n \in \mathcal N$. We consider the game with stones on $\mathcal N$, where a legal move replaces a stone at position $n$ with two stones at $n+1$ and $f(n)$. Let us say that $f$ is good if for any finite configuration of stones and any $N \in \mathcal N$ there is a sequence of legal moves that transforms it into one that is not overcrowded and does not occupy a position $n < N$. We say that $f$ is bad otherwise. The original question was wether $f(n) = n^2$ is good and was answered positively by Aaron Meyerowitz. The question I want to discuss here is

Question.
  Where (in terms of growth behavior of $f$) is the "phase transition" between bad and good?

I am going to argue that this is at a growth behaviour like $f(n) \sim n^2/2$.
For this, I will combine the entropy approach mentioned in the OP's Remarks and the idea of the proof in Aaron's answer.

Lemma 1.
  Assume that there exists a discrete measure $\mu$ on $\mathcal N$ such that
  (i) $0 < \mu(\mathcal N) < \infty$ and
  (ii) for all sufficiently large $n \in \mathcal N$,
  $\mu(\{n+1,f(n)\}) \ge \mu(\{n\})$.
  Then $f$ is bad.

Proof.
For a configuration $s$ we define $\mu(s) = \sum_{n \in \mathcal N} s(n)\mu(\{n\})$.
Let $n' \in \mathcal N$ satisfy $\mu(\{n'\}) > 0$ and be large enough so that (ii) holds for all $n \ge n'$. Consider a configuration $s$ of $m$ stones at $n'$ where $\mu(s) = m \mu(\{n'\}) > \mu(\mathcal N)$. Then by (ii) any sequence of legal moves leads to a configuration $s'$ with $\mu(s') \ge \mu(s) > \mu(\mathcal N)$, whereas any non-overcrowded position $s''$ has
$\mu(s'') \le \mu(\mathcal N)$. So in fact no position we can reach by legal moves from $s$ avoids overcrowding. $\Box$

Lemma 2.
  Assume that there is a constant $C$ such that $f(n+1) + C \ge f(n) + n$ for all sufficiently large $n \in  \mathcal N$.
  Then $f$ is good.

Proof.
Fix $q \in  \mathcal N$ such that $q > C+2$ and the condition on $f$ holds for $n \ge q$. Let $s_0$ be the position consisting of one stone at $q$. We let $s_{m+1}$ be the position obtained from $s_m$ by applying the legal move to every stone in $s_m$. I claim that no $s_m$ is overcrowded.
This is certainly true for $m = 0$. Assume the claim is false and let $m \ge 0$
be the smallest $m$ such that $s_{m+1}$ is overcrowded. Then there are $n_1, n_2 \in s_m$ such that $n_1+1 = f(n_2)$. There are two possibilities. Either $n_1 = q+m$; then $n_1$ is the smallest occupied place in $s_m$ and we have the contradiction $n_1+1 = f(n_2) > n_2 + 1 \ge n_1 + 1$ (we use $n_2 \ge q > C+1$ here). Or we obtain $n_1$ from $q$ by a sequence of steps $n \mapsto n+1$ and $n \mapsto f(n)$ containing at least one application of $f$. In this case, at least the last $f(n_2) - f(n_2-1) - 1 \ge n_2 - (2+C)$ steps must be $n \mapsto n+1$ (since there is no value of $f$ between $f(n_2-1)$ and $f(n_2)$). So $m \ge n_2 - (2+C)$. On the other hand, $n_2 \ge q + m$, so we get a contradiction again, since $q > C+2$.
This shows the defining property for good $f$ for single-stone positions at places $> C+2$. It has already been observed in this thread that this is sufficient to establish that $f$ is good. $\Box$

Theorem.
  (1) If $\limsup_{n \to \infty} \dfrac{f(n)}{n^2} < \frac{1}{2}$, then $f$ is bad.
  (2) If $\liminf_{n \to \infty} \bigl(f(n+1)-f(n)-n\bigr) > -\infty$, then $f$ is good. In particular, if $f(n) = n^2/2 + k n + O(1)$ for some $k \in \mathbb R$, then $f$ is good.

Proof.
Part (2) follows immediately from Lemma 2.
For part (1), fix $\alpha > 1$ such that $2^{-\alpha} > \beta = \limsup f(n)/n^2$ and
consider a measure $\mu$ given by $\mu(\{n\}) = (n (\log n)^\alpha)^{-1}$
for $n \gg 0$.
Then $\mu(\mathcal N)$ is finite. Also,
$$ \mu(\{n\}) - \mu(\{n+1\}) \sim \frac{1}{n^2 (\log n)^\alpha} ,$$
which grows more slowly than
$$ \mu(\{f(n)\}) \ge \frac{1}{\beta 2^\alpha n^2 (\log n)^\alpha}(1 + o(1)), $$
so the assumptions of Lemma 1 are satisfied. $\Box$

Corollary.
  If $f$ is a polynomial, then $f + O(1)$ is good if and only if $\deg f \ge 3$ or $\deg f = 2$ and the leading coefficient of $f$ is $\ge 1/2$.