A derangement of a list of $n$ distinct entries is a permutation of that list such that no corresponding entries match. It is well-known that the number of such derangements is the nearest integer to $n!/e$ where $e$ is the base of natural logarithms.
Let's say that a permutation of a list is a $\mu$-derangement if both it and its reverse ordering are derangements. Equivalently, iff the permutation is a derangement both of the original list and of the reverse of the original list.
How many $\mu$-derangements of the list $[1,2,..,n]$ are there? Is there an exact formula? A good approximation or bound?
There are no $\mu$-derangements for $n \lt 4$, except in the trivial case of an empty list. I got counts for $n \le 10$ below by enumerating possibilities with a bit of Prolog code:
n | # of µ-derangements ---+---------------------- 4 | 4 5 | 16 6 | 80 7 | 672 8 | 4752 9 | 48768 10 | 440192
The OEIS has this sequence as A003471, with a recurrence relation that suggests some separation into even and odd terms might simplify things.