$n$-distant permutations more than not Let $\mathfrak{S}_{2n}$ be the permutation group of the letters $[2n]=\{1,2,\dots,2n\}$. Call a permutation $\pi\in\mathfrak{S}_{2n}$ has an $n$-distant pair if there is some $j\in [2n-1]$ such that $\vert\pi_j-\pi_{j+1}\vert=n$.
Define $A=$ set of $\pi\in\mathfrak{S}_{2n}$ with exactly one $n$-distant pair and $B=$ set of permutations without $n$-distant pairs.

Question. Which inequality is true in general: $\vert A\vert>\vert B\vert$ or $\vert A\vert<\vert B\vert$? And your proof?

 A: Let f(n), g(n) and h(n) be the number of permutations with zero, one or two occurrence of n-distant pair. The corresponding sets are denoted by F(n), G(n) and H(n).
Delete n and 2n from a permutation in F(n) leaves us a permutation in S_{n-1} with at most two n-distant pairs. Conversely, to construct a permutation in F(n), you may
1) insert n and 2n into some w in F(n-1) so that they are not adjacent, or
2) insert n and 2n into some w in G(n-1) so that one of these two number break up the existing n-distant pair, or
3) insert n and 2n into some w in H(n-1) to break up both existing n-distant pairs.
Therefore, one may derive that 
$$f(n)=(2n-1)(2n-2)f(n-1) + 2(2n-2)g(n-1)+2 h(n-1).$$
Similarly,
$$g(n)=2n(2n-1)f(n-1)+2n g(n-1).$$
I checked these recurrences for n<=5 and they seem to be correct. It then follows that g(n)>f(n) for n sufficiently large.
A: There is an injective mapping from $B$ to $A$, which can be constructed as follows:
for a permutation $\pi=(p_1,p_2,\dots,p_{2n})\in B$, find the element $p_k$ that pairs with $p_1$ (i.e., $|p_k-p_1|=n$); clearly we have $k>2$. Map $\pi$ to the permutation
$\pi'=(p_2,\dots,p_{k-1},p_1,p_k,p_{k+1},\dots,p_{2n})$ (i.e., remove $p_1$ from $\pi$ and insert it right before $p_k$). 
It can be easily seen that $\pi'\in A$, and $\pi'$ has only one pre-image in $B$. Hence, $|B|\leq |A|$. 
Furthermore, for $n>2$, there exist permutations in $A$ that have no pre-images; namely, these are the permutations that start with an $n$-distant pair. It follows that $|B|<|A|$.
In fact, we can infer from the above argument that $|A_n| = |B_n| + 2n\cdot |B_{n-1}|$.
UPDATE. Inclusion-exclusion gives
$$|B_n| = \sum_{j=0}^n (-1)^j \binom{n}{j} 2^j (2n-j)! ,$$
which matches sequence A007060 and can be further used to derive an explicit formula for $|A_n|$ (added as A285850).
