Example 3 A stack of cards consists of $n$ cards, numbered from top to bottom as $1,2, \cdots, n$. Now perform the following operation: First, discard the top card, then place the next card at the bottom, then discard the top card of the current hand, and place the next card at the bottom, and so on, until only one card is left in the hand. What is the original number of this card?
(1) First, consider a simple case.
When $n$ is even and each card is moved once (either discarded or placed at the bottom) in sequence, obviously, after such a round of operations, $\frac{n}{2}$ cards are discarded, and the bottom card among the remaining $\frac{n}{2}$ cards is the last one placed at the bottom, which is still the last card before the movement.
(2) Consider the case when the number of cards $n=2^{m} (m \in \mathbf{N})$. If $m=0$, i.e., $n=1$, at this point, there is only one card in hand, so the remaining card is the first one, which is also the $n$-th card.

If $m \geqslant 1$, at this point, $n=2^{m}$ is even. According to (1), after one round of operations, $\frac{n}{2}=2^{m-1}$ cards are discarded, and the bottom card among the remaining $\frac{n}{2}=2^{m-1}$ cards is still the bottom card before the movement, i.e., the $n$-th card.

Since the number of remaining cards $2^{m-1}$ is still a power of 2, continuing this process, each round of movement reduces the exponent of 2 by 1. Therefore, after $m$ rounds of such movements, the exponent of 2 will decrease to 0, meaning there is only one card left in hand.

Since the bottom card among the remaining cards after each round of operations is always the original bottom card, the last remaining card is the one that was originally at the bottom, i.e., the $n$-th card.
(3) In the general case, when $n$ is not a power of 2, let $n=2^{m}+k\left(1 \leqslant k<2^{m}\right)$.

Perform operations on the cards. After discarding $k$ cards (and also placing $k$ cards at the bottom in sequence), the bottom card in hand is the original $2k$-th card, and at this point, there are exactly $2^{m}$ cards in hand, returning to case (2).

Continue moving the cards in hand. When only one card remains, this card should be the bottom card among these $2^{m}$ cards, i.e., the original $2k$-th card.

From the above discussion, if the number of the last remaining card when starting with $n$ cards is denoted as $a_{n}$, then
$$
a_{n}=\left\{\begin{array}{ll}
n, & n=2^{m} ; \\
2 k, & n=2^{m}+k\left(1 \leqslant k<2^{m}\right) .
\end{array}\right.
$$

Combining this, we get $a_{n}=2 n-2^{-\left[-\log _{2} n\right]}$ (or set $n=2^{m}+k$ $\left(1 \leqslant k \leqslant 2^{m}\right)$, at this point, we have $\left.a_{n}=2 k\right)$.
