Paving a corridor

Look at how the cells of the last column of the floor are covered, and set up $2$ counts: full floors and floors missing one corner cell.

1. Let $a(n)$ count pavings of a $3$ by $n$ floor and $b(n)$ count pavings of such a floor from which the bottom cell of the last column is missing. A slab along the floor covers $2$ cells of one row, and a slab across covers $2$ cells of one column.
2. In the last column of a full floor, either the top $2$ cells are covered by a slab across, which forces the bottom cell to take a slab along and leaves the floor of $n-1$ columns missing a bottom corner cell, or the same happens with the bottom $2$ cells, or all $3$ cells take slabs along covering the last $2$ columns; so $a(n)=a(n-2)+2b(n-1)$.
3. For a floor missing the bottom cell of the last column, the top $2$ cells of that column are either covered by one slab across, leaving a full floor of $n-1$ columns, or each takes a slab along, which forces the bottom cell of column $n-1$ to take one too and leaves a floor of $n-2$ columns missing a corner cell; so $b(n)=a(n-1)+b(n-2)$.
4. Starting from $a(0)=1$ and $b(1)=1$, this gives $a(2)=3$, $b(3)=4$, $a(4)=11$, $b(5)=15$, $a(6)=41$, $b(7)=56$, $a(8)=153$, $b(9)=209$, and finally $a(10)=153+2\times 209=571$.

Answer: $571$