In how many ways are we able to permute a set $[n]$ such that for each $2\le k\le n$:
$$\sigma(k) \ne k-1$$
This is like the partial derangement,however $1$ is not fixed,clearly $1$ has $n$ places which can be mapped to ,but I could't find a formula.
The number of those is all permutations minus the ones in which this happens. Meaning $$\left | S_n\setminus \bigcup _{i=2}^nA_i\right |,$$ where $$A_i = \{ \sigma \in S_n:\sigma(i)=i-1 \},$$ Notice that $|A_i|=(n-1)!$ because we are just forcing one variable to be settle. Same happens for $|A_i\cap A_j|=(n-2)!$ by the same reasoning and so, in genera, $$\left |\bigcap _{i\in I} A_i\right |=(n-|I|)!.$$ The inclusion-exclusion principle tells you that $$|\bigcup _{i=2}^nA_i|=\sum _{k = 1}^{n-1}(-1)^{k-1}\sum _{\substack{I\subseteq \{2,3,\cdots ,n\}\\|I|=k}}\left |\bigcap _{i\in I} A_i\right |=\sum _{k = 1}^{n-1}(-1)^{k-1}\sum _{\substack{I\subseteq \{2,3,\cdots ,n\}\\|I|=k}}(n-|I|)!=\sum _{k = 1}^{n-1}(-1)^{k-1}(n-k)!\sum _{\substack{I\subseteq \{2,3,\cdots ,n\}\\|I|=k}}1=\sum _{k = 1}^{n-1}(-1)^{k-1}(n-k)!\binom{n-1}{k}$$ and so your formula looks like $$n!+\sum _{i=1}^{n-1}(-1)^i\binom{n-1}{i}(n-i)!=\sum _{i=0}^{n-1}(-1)^i\binom{n-1}{i}(n-i)!$$
Looking up the numbers, they are here
