Someone on Art of Problem Solving claims to know how to calculate the $2^{2020}$th decimal place of $\sqrt{2},$ and will tell us if everyone gives up. Brute force will not work, nor will a BBP style formula (for the reason that one does not exist, and the ones known so far are for base $2^k$ expansions; $10$ is not a power of $2$).
Is there a feasible solution, or am I being trolled? Trying to search various queries relating to my question online results in nothing except spigot algorithms which spit out the digits one by one from the start. As you may know, they will not work. I am operating under the assumption that if a clever solution exists, it is not a new discovery, for such a simple question has surely been considered before.
Update: The problem poster has promised to post the solution in 2021 if no one finds it before then. This is not the first time they've done this, but every time they've done this in previous years, they have delivered on the promises.
2nd Update: 2 weeks left and I've forgotten to show the source. Soon we'll see if we have been trolled or not.
Required digits equals to $d_n=\bar d(2^n),\;$ where $$\bar d(k)=\left\lfloor 10^{k}\sqrt 2\right\rfloor\hspace{-10pt}\mod 10,\tag1$$ $$\{d_n\}=\{{1, 2, 6, 0, 9, 9, 5, 3, 6, 5, 9, 4, 7, 4, 0, 9, 4, 9, 0, 6, \dots}\}$$
Alternative way is to use the expression $$d_n=\left\lfloor 10^{2^n}a(n)\right\rfloor\hspace{-10pt}\mod 10,\quad n=1,2\dots\tag2$$ where $a(n)$ can be defined via recurrence relation $$a(0)=\dfrac32,\quad a(n+1)=\frac12a(n)+a(n)^{-1},\tag3$$ or from the closed form $$a(n)=\sqrt2\coth\left(2^{n}\ln\left(3+2\sqrt2\right)\right),\quad n=1,2,\dots\tag4$$ Formulas $(2),(3)$ allow to operate with the rational numbers. This provides exact result.
EDIT of 30.07.23
At the same time, is possible improved algorithm in the form of $$\begin{align} &d_n=b_n\hspace{-8pt} \mod 10,\quad n=0, 1, 2\dots,\\[4pt] &b_0=14,\quad b_{n+1} = \dfrac12 10^{2^n}\, b_n+\genfrac\lfloor\rfloor{}{}{1000^{2^n}}{b_n}, \end{align}\tag5$$ which stores the intermediate results in the integer format.