**later** a mistyped digit made me miss the appropriate entry in the OEIS but, as long as this answer is here, I've revised it a bit.

The desired numbers $1,0,8,240,13824,1263360,\cdots$ satisfy the recurrence $$a_n=2n\left((2n-1)a_{n-1}+(2n-2)a_{n-2}\right).$$ Also, $a_n$ is a multiple of $2^nn!$, (explanation below)

- if we break $2n!$ into the number of permutations having $0,1,\cdots,n$ such pairs we get
- $1$
- $0,2$
- $8,8,8$
- $240, 288, 144, 48$
- $13824, 15744, 8064, 2304, 384$
- $1263360,1401600,710400,211200,38400,3840$

The final number in each list is $2^nn!$ and in fact every entry in the nth row is a multiple of $n!2^n$ because swapping the positions of $i,i+n$ does not affect the number of pairs nor does any permutation of $1\cdots n$ extended to send $i+n$ to $j+n$ when $i$ goes to $j$.

The first entries $1,0,8,240,13824,1263360$ are the desired ones.

It might be worth examining the number which do not have any pair $i,i+n$ and/or the number of circular permutations with no pair differing by $n$ (exactly or in absolute value).

Since the terms $1,0,8,240,13824,1263360$ grow like $\frac{2n!}{e}$. I wonder if this sequence (also) lurks in the OEIS as the even position terms of a sequence which grows like $\frac{n!}{e}$ (as do derangement, scrambles and other pattern avoiding permutation counts). Perhaps something like permutations of $1,2,\cdots,m$ where no adjacent pair has difference $\lfloor\frac{m}{2}\rfloor$ (or maybe $\lceil\frac{m}{2}\rceil$.) Perhaps in the odd case $m=2n-1$ one should also forbid $n$ from being at the start ( or start and end).