The general problem is this. I try to find a positive integer $\delta_n$ such $qv_n^2 +\delta_n = p\cdot 4^n$. More precisely, I am looking for a lower bound (depending on $n$) for $\delta_n$ as $n\rightarrow\infty$. Here $v_n$ is the largest positive odd integer such that $qv_n^2 \leq p\cdot 4^n$, given $n$. In short, $\delta_n = p \cdot 4^n - qv_n^2\geq 0$ is the approximation error. For simplicity, focus on $q=2, p=1$. More generally, $0 < p/q < 1$ is not the square of a rational number.
Background
Assume $q=2, p=1$. I expect $\delta_n$ to possibly be something like $O(2^n/\log_2 n)$. So far (I just started) all I have is $\delta_n > 0$. This trivial bound actually leads to a remarkable result: the longest run of zeros in in the binary digits of $\sqrt{p/q}$, starting at position $n+1$ assuming the digit in position $n$ is equal to 1 and $n$ is large enough, can not be longer than $n + C$ where $C$ does not depend on $n$.
The general result, explaining my interest in the problem and why I have a good idea (but no proof) about the asymptotics of $\delta_n$, is the following (the result below is proved and verified empirically):
If the binary digit of $\sqrt{p/q}$ at position $n$ is equal to $1$, then it is followed by a run of exactly $R_n$ digits all equal to zero (with $R_n=0$ in case the next digit is also $1$), with
$$R_n = \Big\lfloor n+1 +\frac{1}{2}\log_2(pq) - \log_2 \delta_n \Big\rfloor.$$
It is an exact formula, not an asymptotic relationship or approximation.
