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How many different ways are there to tile dominos of size 2x1 on a 3x2n grid?

I'm regards to this question Dominos ($2 \times 1$ on $2 \times n$ and on $3 \times 2n$)

I understand the logic behind the first part, but I don't understand how the function $g_{n}$ helps you, and why you would add it twice to the previous number of ways

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user403033
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1 Answers1

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Because there are two ways to cover the left-most 3 squares so that the remaining uncovered squares are in the configuration of a $3\times(2n-1)$ board with one corner removed. There is only one way to cover the left-most three squares so that what remains uncovered is a $3\times(2n-2)$ board. This gives $f_n = 2g_{n-1}+f_{n-1}$.

On the other hand, given a $3\times(2n+1)$ board with one corner removed, the other two on that edge can be either covered by one domino, and you are left with a $3\times 2n$ board, or you they are covered by two dominos, which leaves a straggler in the next row, which has to be covered to remove a corner from the remaining. So you have $g_{n}=f_n+g_{n-1}$

Thomas Andrews
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  • Why is one corner removed? – dreamin May 06 '19 at 19:26
  • Also why cant the number of ways be to cover the left most three square 3? By placing the 3 dominoes horizontally, by placing one vertically in the top 2 rows and one horizontally in the top row and by placing one horizontally in the top rows and one vertically in the bottom 2 rows? – dreamin May 06 '19 at 20:07
  • Combining the recurrence relations yields $$f_n=2g_{n-1}+f_{n-1}=2(f_{n-1}+g_{n-2})+f_{n-1}=2(f_{n-1}+(f_{n-1}-f_{n-2})/2)+f_{n-1}=4f_{n-1}-f_{n-2},$$ as in OEIS A001835. – RobPratt Oct 16 '19 at 17:40
  • Sure, or you can get $g_n=f_n+g_{n-1}=(f_{n-1}+2g_{n-1})+g_{n-1}=f_{n-1}+3g_{n-1}$ and then get that $$\begin{pmatrix}f_n\g_n\end{pmatrix}=\begin{pmatrix}1&2\1&3\end{pmatrix}^n\begin{pmatrix}1\1\end{pmatrix}$$ @Pratt Then you can diagonalize the matrix and get the explicit formula for $f_n.$ – Thomas Andrews Oct 16 '19 at 17:47
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    Wolfram alpha then gives $$f_n=\frac{3-\sqrt{3}}{6}(2-\sqrt{3})^n+\frac{3+\sqrt{3}}{6}(2+\sqrt{3})^n$$ – Thomas Andrews Oct 16 '19 at 17:55