Lamps that never match

Treat the grid as $4$ columns of $3$ lamps. A column is either of the form "outer lamps equal" or "all different".

1. A valid column of $3$ lamps has its middle lamp different from both others, so it has the form ABA (outer lamps equal, $6$ of them) or ABC (all different, $6$ of them).
2. For a column of the form ABA, the next column can be chosen in $3$ ways of the form ABA and $2$ ways of the form ABC; for a column of the form ABC it is $2$ and $2$.
3. Let $u$ and $v$ count colourings of the first columns that end in the form ABA and ABC. Then $u=6$ and $v=6$ for the first column, and each new column gives $u^{\prime }=3u+2v$ and $v^{\prime }=2u+2v$.
4. The pairs are $(6,6)$, $(30,24)$, $(138,108)$ and $(630,492)$ for $1$ to $4$ columns, so the total is $630+492=1122$.

Answer: $1122$