Paper-strip Game

The following game is a classic example of Combinatorial Game Theory:

Two players start with a strip of $n$ white squares and they take alternate turns.
On each turn, a player picks two contiguous white squares and paints them black.
The first player who cannot make a move loses.

• $n=1$: No valid moves, so the first player loses automatically.
• $n=2$: Only one valid move, after which the second player loses.
• $n=3$: Two valid moves, but both leave a situation where the second player loses.
• $n=4$: Three valid moves for the first player, who is able to win the game by painting the two middle squares.
• $n=5$: Four valid moves for the first player (shown below in red), but no matter what the player does, the second player (blue) wins.

So, for $1≤n≤5$, there are $3$ values of $n$ for which the first player can force a win.
Similarly, for $1≤n≤50$, there are $40$ values of $n$ for which the first player can force a win.

For $1≤n≤1000000$, how many values of $n$ are there for which the first player can force a win?