There are $2n$ cards in a stack. At any stage, if the cards are $a_1,a_2,\ldots,a_{2n}$ from top to bottom, then it becomes $a_{2n},a_1,a_{2n-1},a_2,\ldots,a_{n+1},a_n$. In how many steps will the cards return to the original order (if it ever happens)?
$n=1$: $12\rightarrow 21\rightarrow 12$. ($2$ steps)
$n=2$: $1234\rightarrow 4132\rightarrow 2431\rightarrow 1234$. ($3$ steps)
$n=3$: $123456\rightarrow 615243\rightarrow 364125\rightarrow 532614\rightarrow 451362\rightarrow 246531\rightarrow 123456$ ($6$ steps)
We might be able to track each card individually, and see when it returns to the original position. For example, $a_1$ goes to position $2$, then position $4$, ..., but when it gets to the second half, the rule is more complicated.
This permutation is
$$\pmatrix{1&2&3&4&\ldots&2n-3&2n-2&2n-1&2n\\ 2&4&6&8&\ldots&7&5&3&1}\;.$$
If you number the cards from the bottom of the deck, the same permutation is
$$\pmatrix{1&2&3&4&\ldots&2n-3&2n-2&2n-1&2n\\ 2n&2n-2&2n-4&2n-6&\ldots&2n-7&2n-5&2n-3&2n-1}\;.$$
OEIS A019567 gives the order of this permutation, which is also the number of Mongean shuffles needed to return a deck to its original order. OEIS gives no recurrence or generating function but does note that $a(n)$ is the smallest $m$ such that either $2^m+1$ or $2^m-1$ is divisible by $4n+1$.
Note that if $n=2^k$, then $4n+1=2^{k+2}+1$, so $a(2^k)=k+2$.