dang, now i get why i didnt make it to cmos

The answer is $$D_{2020} - D_{2019} + D_{2018} - \ldots + D_2 - D_1$$ For this, we use inclusion-exclusion. Replace $2021$ by $n$. For any subset $S \subseteq \{1,2, \ldots, n\}$, we want to count the number of ways of arranging $1,2, \ldots, n$ in a circle (rotations don't count as distinct) such that for each $k \in S$, its left neighbour is NOT $k+1 \mod n$. On doing so, we get certain "chains" of elements that are already pre-decided to be in that order in the circle. We get exactly $n-|S|$ such chains, so they can be arranged in the circle in $(n-|S|-1)!$ ways (we define $(-1)! = 1$ here). By inclusion-exclusion, our answer is $$a_n = \sum_{S \subseteq [n]} (-1)^{|S|} (n-|S|-1)! = \sum_{i=0}^n (-1)^i \cdot \binom{n}{i} \cdot (n-i-1)!$$ Now, \begin{align*} a_n + a_{n+1} &= \sum_{i=0}^n (-1)^i \cdot \binom{n}{i} \cdot (n-i-1)! + \sum_{i=0}^{n+1} (-1)^i \cdot \binom{n+1}{i} \cdot (n-i)! \\ &= \sum_{i=0}^n (-1)^i \cdot \binom{n}{i} \cdot (n-i-1)! + \sum_{i=-1}^{n} (-1)^{i+1} \cdot \binom{n+1}{i+1} \cdot (n-i-1)! \\ &= n! + \sum_{i=0}^n (-1)^{i+1} \left ( \binom{n+1}{i+1} - \binom{n}{i} \right ) \cdot (n-i-1)! \\ &= n! + \sum_{i=0}^n (-1)^{i+1} \cdot \binom{n}{i+1} \cdot (n-i-1)! \\ &= n! + \left ( \sum_{i=0}^{n-1} (-1)^{i+1} \cdot \frac{n!}{(i+1)!} \right ) + (-1)^{n+1} \cdot \binom{n}{n+1} \cdot 1\\ &= n! \cdot \left ((-1)^0 \frac{1}{0!} + (-1)^1 \frac{1}{1!} + (-1)^2 \frac{1}{2!} + \ldots + (-1)^n \frac{1}{n!} \right ) = D_n \end{align*} Using this, we get $$a_n = D_{n-1} - D_{n-2} + D_{n-3} + \ldots + (-1)^{n} D_1$$for all naturals $n$.

Since the King doesn't move at all, this is essentially a problem about linear permutations. In particular, find the number of permutations of $\{1,2,\dots n\}$ such that $1$ is not in the first position, $n$ is not in the last position, and $k,k+1$ are not consecutive (in that order) for all $1 \leq k \leq n-1$. (Here we have replaced $2020$ by $n$). Let $f(n)$ denote the number of such permutation. Consider $n \geq 2$ first. Now, for any of the $(n-2) \cdot (n-1)!+(n-2)!$ permutations of $[n]$ not starting with $1$ or ending with $n$, we can split it into blocks of consecutive numbers; e.g., the permutation $2 \ 1 \ 4 \ 5 \ 3$ splits into four blocks: $2$, $1$, $4 \ 5$, and $3$. To construct any such permutation, first split $[n]$ into blocks of consecutive numbers, and then rearrange those blocks such that the first block is not in the first position, the last block is not in the last position, and no two consecutive blocks remain consecutive. If there are $k$ blocks, then the splitting can be done in $\binom{n-1}{k-1}$ ways, and the rearranging can be done in $f(k)$ ways. Therefore $$(n-2) \cdot (n-1)!+(n-2)! = \sum_{k=1}^{n} \binom{n-1}{k-1}f(k)$$Write $g(k)=f(k+1)$ for $k \geq 0$ for convenience, and replace $n-1$ by $n$; then, $$(n-1) \cdot n!+(n-1)!= \sum_{k=0}^{n} \binom{n}{k} g(k)$$for all $n \geq 1$. For $n=0$, the RHS is $0$ because $g(0)=f(1)=0$ trivially. Let $G(x)$ be the egf of $g(k)$. Therefore we get $$\frac{x^2}{(1-x)^2}-\ln (1-x)=e^x G(x)$$by comparing egfs. We want to get rid of the pesky $\ln$ in the LHS. To do this, we differentiate the whole thing to get $$e^x(G(x)+G'(x))=\frac{2x}{(1-x)^3}+\frac{1}{1-x}$$$$\implies G(x)+G'(x)=\frac{e^{-x}}{1-x} \left ( \frac{2x}{(1-x)^2}+1 \right)$$Now, let $F(x)$ be the egf of $D_n$. It is well known that $$F(x)=\frac{e^{-x}}{1-x}$$Differentiating the above twice, we get $$F''(x)= \frac{e^{-x}}{1-x} \left ( \frac{2x}{(1-x)^2}+1 \right) = G(x)+G'(x)$$Comparing coefficients, we get $$f(n)+f(n+1)=g(n-1)+g(n)=D_{n+1}$$Also, $f(1)=D_1=0$, so we get $$\boxed{f(n)=\sum_{k=0}^{n-1}(-1)^k D_{n-k}}$$for all positive integers $n$.