--description--

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.
• $n = 4$: There are 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 $1 ≤ n ≤ 1\,000\,000$, how many values of $n$ are there for which the first player can force a win?

--hints--

paperStripGame() should return 852938.

assert.strictEqual(paperStripGame(), 852938);

--seed--

--seed-contents--

function paperStripGame() {

  return true;
}

paperStripGame();

--solutions--

// solution required