Show number of permutations on $[n]$ where $i$ is not followed by $i+1$ is $D_n + D_{n-1}$ I have that $a_n$ is the number of permutations where $i$ is not immediately followed by $i+1$ and:
\begin{align*}
a_n = \sum_{k=0}^{n-1} (-1)^k (n-k)!
\end{align*}
Which I obtained using the inclusion exclusion principle.
I am trying to show that $a_n = D_{n} + D_{n-1}$ where $D_n$ is the number of derangements on $[n]$. 
Given that
\begin{align*}
D_n = \sum_{k=0}^n \frac{(-1)^kn!}{k!}
\end{align*}
I have tried to expand the summation for all terms but i am having trouble showing equality.
 A: Observe  that with  inclusion-exclusion  we can  count  the number  of
permutations having at least $k$ values where $q$ is followed by $q+1$
by choosing those $k$ from the range from $1$ to $n-1$ and fusing them
with the elements that follow immediately, forming at most $k$ blocks. 
We may  then permute these $n-k$ items any way we like.

This yields per inclusion-exclusion
$$\sum_{k=0}^{n-1} {n-1\choose k} (-1)^k (n-k)!
= (n-1)! \sum_{k=0}^{n-1} \frac{(-1)^k}{k!} (n-k)
\\ = (n-1)! \sum_{k=0}^{n} \frac{(-1)^k}{k!} (n-k)
= n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}
- (n-1)! \sum_{k=0}^{n} k \frac{(-1)^k}{k!}
\\ = D_n - (n-1)! \sum_{k=1}^{n} k \frac{(-1)^k}{k!}
= D_n - (n-1)! \sum_{k=1}^{n} \frac{(-1)^k}{(k-1)!}
\\ = D_n + (n-1)! \sum_{k=0}^{n-1} \frac{(-1)^k}{k!}
= D_n + D_{n-1}.$$
A block  of length $m$ reduces  the number of items  being permuted by
$m$ and then adds  one back in. On the other hand  $m-1$ is the number
of adjacent consecutive  items in the block and  summing this over all
blocks yields $k.$
