Let's say that we have the points $\{1,2,3,...,k\}$. I am looking for all the ways to create a derangement of these points with the following rule:
we define a derangement as a permutation $\sigma$ such that there is no fixed point i.e. $\sigma(m) \neq m$ for each $m \in \{1,...,k\}$. Now, let's define an integer $a \in \{1,...,k\}$. In the derangement, there must exist at least one point that is the total of $a$ numbers away from its fixed point. Also, the remaining points must be, at max, $a$ away from their fixed point i.e. there does not exist a number in the derangement of $\{1,...,k\}$ where a number is more than $a$ away from its fixed point.
Here's my approach: I started by enumeration for different values of $a \in \{1,...,k\}$. If $a=1$, this only works for derangements of an even number of points where each couple $\{1,2\}$, $\{3,4\}$ etc is flipped the other way around, i.e. $\{1,2\} \rightarrow \{2,1\}$. We don't need to consider this case further.
Now, if $a=2$, then there must exist at least one number that is $2$ numbers way from its fixed point. This could be, for example, if we had $k=4$ and $\{1,2,3,4\}$, we could have, for example, $\{3,1,4,2\}$ where $1$ is one number from its fixed point as well as $4$, meanwhile $2$ and $3$ are $a=2$ away from their fixed point. However, I am struggling to find a general combinatorial rule for this.
Edit: example: Let $k=4$. Now the derangements for $a=1$ are the following:
$\{2,1,4,3\}$
And similarly, for $a=2$:
$\{3,1,4,2\}$, $\{3,4,1,2\}$ (I could only find these, so I could only find the total of $2$ derangements that fulfill this rule for $a=2$.)
For $a=3$:
$\{4,3,2,1\}$ (I could only find this, so I only found one derangement for $a=3$).