I didn't make any mistake i guess I find a sequence which is f(n) = 3f(n-2) + 2f(n-4) + 2f(n-6) + 2f(n-8) ..... + 2f(1) + 1 For all odd numbers n this f(n) sequence denotes the number of ways to color a 3xn board with one corner square removed. (n is odd.) If you control if it is correct i will be happy, thanks.

Btw I meant If I didn't made any mistake.

By the way the answer is m=0.

Here's my idea. Let $t_{2k+1}$ represent the number of ways to cover a $3\times (2k+1)$ chessboard with one corner square removed, with no gaps or overlaps. Let $s_{2k}$ represent the number of ways to cover a $3\times 2k$ chessboard with no corner squares removed, with no gaps or overlaps. If we draw a tree of possibilites (see attached image), we can see $t_{2k+1} = s_{2k} + t_{2k-1}$ $s_{2k} = s_{2k-2} + 2t_{2k-1}$ From visual inspection, we have $t_1 = 1$ and $t_3 = 4$. We are asked to find $t_{2021}\ (\bmod\ 19)$. Manipulating the two equations above, we can derive $t_{2k+1} = 4t_{2k-1} - t_{2k-3}$ Consider the sequence $\{u_n\}$, defined by $u_n = t_{2n-1}$. So we're looking for $u_{1011}\ (\bmod\ 19)$. Restate the above as $u_{n+1} = 4u_n - u_{n-1}$. We know $u_1 = 1$ and $u_2 = 4$. What is $u_{n}\ (\bmod\ 19)$ for each n? If $n=1,2,3,4,5,6,7,...$, we can calculate $u_{n}\equiv 1,4,15,18,0,1,4,... \ (\bmod\ 19)$ in a repeating pattern. We can use this pattern to say that $u_{1011}\equiv 1\ (\bmod\ 19)$, so the answer is $1$. $\square$

Attachments: