smc

We first prove that $x_n>4$ for all $n\ge 0$, by induction. Observe that $$x_{n+1}-4=\frac{x_n^2+5x_n+4-4(x_n+6)}{x_n+6}=\frac{(x_n-4)(x_n+5)}{x_n+6}.$$ so (since $x_n$ is clearly positive for all $n$, from the initial definition), $x_{n+1}>4$ if and only if $x_n>4$. We similarly prove that $x_n$ is decreasing: $$x_{n+1}-x_n=\frac{x_n^2+5x_n+4-x_n(x_n+6)}{x_n+6}=\frac{4-x_n}{x_n+6}<0.$$ Now we need to estimate the value of $x_{n+1}-4$, which we can do using the rearranged equation: $$x_{n+1}-4=(x_n-4)\cdot \frac{x_n+5}{x_n+6}.$$ Since $x_n$ is decreasing, $\frac{x_n+5}{x_n+6}$ is also decreasing, so we have $$\frac{9}{10}<\frac{x_n+5}{x_n+6}\le \frac{10}{11}$$ and $$\frac{9}{10}(x_n-4)<x_{n+1}-4\le \frac{10}{11}(x_n-4).$$ This becomes $${\left(\frac{9}{10}\right)}^n={\left(\frac{9}{10}\right)}^n\left(x_0-4\right)<x_n-4\le {\left(\frac{10}{11}\right)}^n\left(x_0-4\right)={\left(\frac{10}{11}\right)}^n.$$ The problem thus reduces to finding the least value of $n$ such that $${\left(\frac{9}{10}\right)}^n<x_n-4\le \frac{1}{2^{20}}\text{ and }{\left(\frac{10}{11}\right)}^{n-1}>x_{n-1}-4>\frac{1}{2^{20}}.$$ Taking logarithms, we get $n\ln \frac{9}{10}<-20\ln 2$ and $(n-1)\ln \frac{10}{11}>-20\ln 2$, i.e. $$n>\frac{20\ln 2}{\ln \frac{10}{9}}\text{ and }n-1<\frac{20\ln 2}{\ln \frac{11}{10}}.$$ As approximations, we can use $\ln \frac{10}{9}\approx \frac{1}{9}$, $\ln \frac{11}{10}\approx \frac{1}{10}$, and $\ln 2\approx 0.7$. These approximations allow us to estimate $$126<n<141,$$ which gives $\boxed{[81,242]}$.