2-D random walk bounding probability of number of return times to origin within a given time interval Let $(S_n)_{n\geq 0}$ be a simple random walk on $\mathbb Z^2$. It is well known that the probability of the walk returning to $0$ at time $2n$ is given by
\begin{equation}\tag{1}\label{eq1}
p_0(2n)  = \left(\frac{1}{2}\right)^{4n} \:{2n\choose n}^{\!\!2}.
\end{equation}
Now consider $\{a,b\}\subset \mathbb N$, with $a<b$, and define a new process $(R_n)_{n\geq 1}$ as $R_n:=|\{S_m=0:an\leq m\leq bn\}|.$ That is: $R_n$ is the number of equalizations of the random walk between times $an$ and $bn$. 
As part of a homework assignment I was interested in obtaining a uniform bound of
$P(R_n>x)$. I believe this can be done via combining Markov's inequality with equation \eqref{eq1}, and applying some asymptotics. We know
\begin{align}\mathbb E[R_n]&=\sum_{k=na}^{nb}p_0(k)\sim\sum_{k=na}^{nb}\frac{1}{\pi n}\leq\log\frac{2b+1/n}{2a-1/n}\\&\leq\log\frac{2b+1}{2a-1},\end{align}
which should suffice. However, this got me interested in the following

What is known about the probability distribution of $R_n$?

I have trawled the internet for a good few hours, and have not been able to find anything, so would appreciate any references.
 A: The distribution of $R_n$ can be written explicitly, but this is very involved. 
Good news: the distribution does have a limit. 
Bad news: this limit is zero, i.e. $R_n\to 0$, $n\to\infty$, in probability (not the result you wanted to hear, I presume). 
Indeed, by Donsker's invariance principle, $\{S_{[tn]}/\sqrt{n}, t\ge 0\}$ converges in law to the 2d standard Wiener process $W_t = (W^1_t,W^2_t)$. Consequently, 
$$
\min_{an\le k\le bn} \frac{||S_{k}||}{\sqrt{n}} \overset{law}{\longrightarrow} \min_{a\le t\le b} ||W_t||, n\to \infty.
$$
Therefore, for any $x>0$,
$$
\limsup_{n\to\infty}\mathrm{P}(\exists k\in[an,bn]: S_k=0) \le \limsup_{n\to\infty}\mathrm{P}\Bigl(\exists k\in[an,bn]: ||S_k||\le x\sqrt{n}\Bigr) = \mathrm{P}(||W_t||<x).
$$
Letting $x\to0+$ and recalling that $W_t$ is non-recurrent, we get
$$
\mathrm{P}(\exists k\in[an,bn]: S_k=0) \to 0, n\to\infty. 
$$
This is equivalent to $\mathrm{P}(R_n = 0)\to 1$, $n\to\infty$. 
This seems to contradict to the fact that $R_n$ has a positive expectation. But there is no contradiction. It happens that whenever $S$ does hit zero between $an$ and $bn$, it suddenly makes a lot of hits (a random quantity of order $\log n$). I therefore conjecture that the conditional distribution of $R_n/\log n$ given that $R_n>0$ has some non-trivial limit. 
