How do you show that number of permutations of $\{1,2,3,\ldots,n\}$ such that image of no two consecutive numbers is consecutive is

$$n! + \sum_{k = 1}^{n}(-1)^k\sum_{i = 1}^{k}\dbinom{k - 1}{i - 1}\dbinom{n - k}{i}2^i(n - k)!$$

In short we need to find number of permutations of $\{1,2,3,\ldots,n\}$ such that none of the following occur: $12, 23, \ldots, (n-1)n \quad $ and $ \quad21, 32, \ldots, n(n-1)$ that is no adjacent numbers should be consecutive.

I tried proving the formula but didn't get any satisfactory result, It seems to be inclusion exclusion principle would work, but there are too many cases to count. I tried to find a recurrence relation, but couldn't do it either. Afterwards I tried to get a generating function for the same but didn't succeed. I don't see any other approach to get through this but I think the most useful tool would be PIE however I'm not finding a good way to use PIE since number of cases are too much. Any help or hint would be highly appreciated. Thanks!

Note that:

I have read almost all the references related to the problem from OEIS.

I have read the whole paper https://projecteuclid.org/journals/annals-of-mathematical-statistics/volume-38/issue-4/Permutations-without-Rising-or-Falling-omega-Sequences/10.1214/aoms/1177698793.full but in the paper there aren't any rigorous proofs and most of the proofs are just excluded simply by saying that 'use basic PIE to derive this'. I am looking for a more direct poof using enumerative combinatorics or generating functions.

I highly appreciate your time and efforts. Thanks.