Counting permutations whose elements are not exactly their index ± 1
Checkout Derangements

"Suppose that there are n people who are numbered 1, 2, ..., n. Let there be n hats also numbered 1, 2, ..., n. We have to find the number of ways in which no one gets the hat having same number as their number."

In applying this to your problem, there would be two hats: the element's index +1, the elements​ index -1.

I'm not certain this solves the problem, but it seems like a step in the right direction.