dp[i] is the contribution to the answer if we have i consecutive white cells in a row/column.

Let's take the case of row. If last cell if blue then it's equivalent to dp[i-1]. If last cell is red but 2nd last cell is blue it is equivalent to dp[i-2]. And the remaining case, if both last 2 cells are red, it is equivalent to dp[i-2] and a tile on last 2 positions. That tile will contribute for all the 1<<(i-2) states of the previous cells.

Hence, dp[i]=dp[i-1]+2*dp[i-2]+a[i-2] where a[x]=2^x.

That was the main intuition behind the solution.