Let us prove that Alice has winning strategy for odd numbers, otherwise Bob wins. Let $n=2k$. Bob has following winning strategy: $1.$ If Alice makes first move (move one square forward), Bob will copy that move. $2.$ If Alice makes second move (move four squares forward), Bob will make third move. $3.$ If Alice makes third move (move two squares backwards), Bob will make second move. It is easy to show, that in each move, a pebble will be moved only 2 squares forward. If $n=4t$, we can assure that Alice will be on first move in $(n-8)-th $ position. Now, If Alice makes first move, Bob will make second, and if Alice makes second move, Bob will follow with first, and in both cases Bob wins. So, Alice must make $3rd$ move, which Bob will follow and hence Alice starts on $n-12$ move. After that, Bob can always make this strategic move on $n-8$ square, and if Alice goes always on $3rd$ move, because $n=4t$ he will always play on even squares. Similar, when $n=4t+2$, we can assure that Alice will be on first move in $(n-6)-th $ position. Same as previous consideration, Alice can not make first or second move, so he must take third move. Because $n=4t+2$ and number of remaining squares are $4t+2-6 = 4t-4 = 4(t-1)$, Bob with copying 3rd move will not impact on game, because remaining number of squares are divisible by 4. Hence, Bob wins. Let $n=2k+1$. Now, Alice will make first move. Now, Bob can not make third move because pebble is on second square, and here number of "remaining" squares are even. So, it is same situation like first case, but here Bob starts first, so Alice will follow strategy for $n=2k$, and hence Alice will always win.

Bergo1305 wrote: Let us prove that Alice has winning strategy for odd numbers, otherwise Bob wins. Let $n=2k$. Bob has following winning strategy: $1.$ If Alice makes first move (move one square forward), Bob will copy that move. $2.$ If Alice makes second move (move four squares forward), Bob will make third move. $3.$ If Alice makes third move (move two squares backwards), Bob will make second move. It is easy to show, that in each move, a pebble will be moved only 2 squares forward. Because number of squares are even, Bob will always win. Let $n=2k+1$. Now, Alice will make first move. Now, Bob can not make third move because pebble is on second square, and here number of "remaining" squares are even. So, it is same situation like first case, but here Bob starts first, so Alice will follow strategy for $n=2k$, and hence Alice will always win. I believe that your argument is not correct, because if we look at the $n-4$ -th position, by your construction the second player would win in four moves. However, at that position the first player can win in one move by moving the pebble $4$ squares forward.

Blastoor wrote: Bergo1305 wrote: Let us prove that Alice has winning strategy for odd numbers, otherwise Bob wins. Let $n=2k$. Bob has following winning strategy: $1.$ If Alice makes first move (move one square forward), Bob will copy that move. $2.$ If Alice makes second move (move four squares forward), Bob will make third move. $3.$ If Alice makes third move (move two squares backwards), Bob will make second move. It is easy to show, that in each move, a pebble will be moved only 2 squares forward. Because number of squares are even, Bob will always win. Let $n=2k+1$. Now, Alice will make first move. Now, Bob can not make third move because pebble is on second square, and here number of "remaining" squares are even. So, it is same situation like first case, but here Bob starts first, so Alice will follow strategy for $n=2k$, and hence Alice will always win. I believe that your argument is not correct, because if we look at the $n-4$ -th position, by your construction the second player would win in four moves. However, at that position the first player can win in one move by moving the pebble $4$ squares forward. Thank you, I edited my solution.