Is this conjecture about a stick game correct? My math teacher described the following game and for a few days I have been thinking about it.

Let $A$ and $B$ be two perfect players. Let $n$ be a positive integer and let's consider a stick of length $n$.
A move in this game is to break a single section of the stick (at the beginning there is only one section, the stick itself) into two sections of positive and distinct integer lengths. $A$ is the one who makes the first move.
For example if $n=5$, a game might be $5\to2+3\to2+1+2$.
We note $x$ the number of sections of length $1$ at the end of the game and $y$ the number of sections of length $2$.
If $x=y$, then the game is tied, if $x>y$ the player who played last wins, and if $x<y$ its opponent wins.

I came up with the following conjecture and it seems to work with small numbers but I don't know how to prove it.

*

*$A$ wins if $n\equiv-1\mod 3$

*$B$ wins if $n\equiv1\mod 3$

*no one wins if $n\equiv0\mod 3$
For example, when $n=4$ the only first move possible is $1+3$ and the next move is also forced $1+1+2$. Here $x=2>y=1$ so $B$ wins and indeed $n\equiv 1\mod3$.
Again players are supposed to play perfectly.
 A: This is not a definitive answer but a hint on how to prove the result :
Games 3n : First player and Second player have both a strategy to reach a tie in an odd number of turns (First player plays last)
Games 3n+1 : Second player has a winning strategy and First player has a mitigating strategy : first player can choose if second player wins by playing last with x=(n+1) one and y=(n) two (available for all n) or if second player wins by not playing last with x=(n-1) one and y=(n+1) two (only available for n>3, hence game 13,16,19 and so on...)
Games 3n+2 : First player has a winning strategy and can choose if he wins by playing last (except for n=0, hence game 2) or not. Second player has a mitigating strategy : if first player wins by playing last, second player can force him to wins with only x=(n+2) one and y=(n) two. if first player choose to win by not playing last, second player has a strategy to loose with only x=(n) one and y=(n+1) two.
Using those hypotheses with a strong induction should lead to the desired result.
