Subtract 1 or divide by 3

Question

Two players play a game in which at the beginning a number $n\geq 2$ is written. Each turn, a player can either subtract $1$ from the number, or divide the number by $3$ if the number is divisible by $3$. The player who can turn the number into $1$ wins. For which values of $n$ does the first player have a winning strategy?

Among the numbers from $2$ to $20$, the ones for which the first player can win are $2,3,5,7,9,11,12,14,16,18,20$. At first the winning numbers are odd, but that changes because for the number $12$, the first player can win by dividing by $3$. After that the winning numbers are even.

Answer 1

The answer is $$\small n=2,\frac{3^{2m}-3}{2},\frac{3^{2m}+1}{2},\frac{3^{2m}+5}{2},\cdots,\frac{3^{2m+1}-5}{2},\frac{3^{2m+1}-3}{2},\frac{3^{2m+1}+1}{2},\frac{3^{2m+1}+5}{2},\cdots ,\frac{3^{2m+2}-5}{2}$$ where $m\ge 1\in\mathbb Z$.

Let us prove this by induction on $m$.

The base case where $m=1$ is easy to check.

Suppose that the first player has a winning strategy for $$\small n=\frac{3^{2m}-3}{2},\frac{3^{2m}+1}{2},\frac{3^{2m}+5}{2},\cdots,\frac{3^{2m+1}-5}{2},\frac{3^{2m+1}-3}{2},\frac{3^{2m+1}+1}{2},\frac{3^{2m+1}+5}{2},\cdots ,\frac{3^{2m+2}-5}{2}$$ and that the first player does not have a winning strategy for $$\small n=\frac{3^{2m}-1}{2},\frac{3^{2m}+3}{2},\frac{3^{2m}+7}{2},\cdots,\frac{3^{2m+1}-7}{2},\frac{3^{2m+1}-1}{2},\frac{3^{2m+1}+3}{2},\frac{3^{2m+1}+7}{2},\cdots ,\frac{3^{2m+2}-7}{2}$$

Now, the first player has a winning strategy for $n=\frac{3^{2m+2}-3}{2}$ choosing to divide by $3$ to get $\frac{3^{2m+1}-1}{2}$.

So, the first player does not have a winning strategy for $n=\frac{3^{2m+2}-1}{2}$, which is not divisible by $3$, since subtracting $1$ gives $\frac{3^{2m+2}-3}{2}$.

So, the first player has a winning strategy for $n=\frac{3^{2m+2}+1}{2}$ by choosing to subtract $1$.

The first player does not have a winning strategy for $n=\frac{3^{2m+2}+3}{2}$ since subtracting $1$ gives $\frac{3^{2m+2}+1}{2}$ and dividing by $3$ gives $\frac{3^{2m+1}+1}{2}$.

Doing similarly, we have that the first player has a winning strategy for $$\small n=\frac{3^{2m+2}-3}{2},\frac{3^{2m+2}+1}{2},\frac{3^{2m+2}+5}{2},\cdots,\frac{3^{2m+3}-5}{2},\frac{3^{2m+3}-3}{2},\frac{3^{2m+3}+1}{2},\frac{3^{2m+3}+5}{2},\cdots ,\frac{3^{2m+4}-5}{2}$$ and that the first player does not have a winning strategy for $$\small n=\frac{3^{2m+2}-1}{2},\frac{3^{2m+2}+3}{2},\frac{3^{2m+2}+7}{2},\cdots,\frac{3^{2m+3}-7}{2},\frac{3^{2m+3}-1}{2},\frac{3^{2m+3}+3}{2},\frac{3^{2m+3}+7}{2},\cdots ,\frac{3^{2m+4}-7}{2}$$