Color the top row of points arbitrarily ($ 2^{n+1}$ ways). If it alternates red-blue-red-blue for its entire length (2 possibilities), proceed as follows: coloring the leftmost dot of the second row (2 ways) determines the entire second row, and it is also alternating. Then coloring the leftmost dot of the third row (2 ways) determines the entire third row, and it is alternating. And so on. Thus, we have from these possibilities a total of $ 2^{n+1}$ colorings. If the top row does not alternate red-blue-red-blue for its entire length ($ 2^{n+1}-2$ ways), there are two consecutive dots of the same color. Note that these dots determine the color of the two dots directly below them, and as a result the first row determines the second row. Moreover, the second row now also has two consecutive dots of the same color, so the third row and all future rows are determined. Thus, from these options we have in total $ 2^{n+1}-2$ ways. All together, this gives us $ 2^{n+2}-2$ total valid colorings. (In an $ m \times n$ rectangle, this answer would be $ 2^{m+1}+2^{n+1}-2$ colorings.) A follow-up question: how many ways are there to color the vertices with 4 colors so that each unit square has one vertex of each color?

I think it's 2^(n+1) - 2. Here is my solution. Let the vertices of the top row be assigned a random coloring. Then suppose that two vertices that are next to each other have the same color. Then obviously the coloring of the rest of the vertices is fixed. There are 2^n - 2 colorings of the top row so that there are at least two vertices next to each other with the same color.(total=2^n and there are two that alternate the colorings)If the vertices of the top row are colored alternately we have 2 ways so in total we have 2^n ways. So the total number is 2^n - 2 + 2^n = 2^(n+1)

JBL wrote: Color the top row of points arbitrarily ($ 2^{n+1}$ ways). If it alternates red-blue-red-blue for its entire length (2 possibilities), proceed as follows: coloring the leftmost dot of the second row (2 ways) determines the entire second row, and it is also alternating. Then coloring the leftmost dot of the third row (2 ways) determines the entire third row, and it is alternating. And so on. Thus, we have from these possibilities a total of $ 2^{n+1}$ colorings. If the top row does not alternate red-blue-red-blue for its entire length ($ 2^{n+1}-2$ ways), there are two consecutive dots of the same color. Note that these dots determine the color of the two dots directly below them, and as a result the first row determines the second row. Moreover, the second row now also has two consecutive dots of the same color, so the third row and all future rows are determined. Thus, from these options we have in total $ 2^{n+1}-2$ ways. All together, this gives us $ 2^{n+2}-2$ total valid colorings. (In an $ m \times n$ rectangle, this answer would be $ 2^{m+1}+2^{n+1}-2$ colorings.) A follow-up question: how many ways are there to color the vertices with 4 colors so that each unit square has one vertex of each color? Just to not cause confusion to the readers, here n was taken as n+1.

Nice bijective action!