Let $c_2(n, k)$ be the number of permutations of [n] with k cycles in which each cycle length is at least two. Find a recursive formula for the numbers $c_2(n, k)$.
Thought process:
Case 1 - element n is in a cycle with one other element
Then we can pair n with any of the n-1 elements and permute the rest into $c_2(n-2, k-1)$
Case 2 - element n is in a cycle with more than one other element
Then we can permute the other n-1 elements in $c_2(n-1,k)$ ways and place n after any element in a cycle n-1 ways.
Result
This would result in
$c_2(n, k) = (n-1)c_2(n-2, k-1) + (n-1)c_2(n-1,k)$
Is this answer correct?
Seems correct to me. As a way to check, notice that $$\sum _{k=1}^{n}c_2(n,k)=D_n,$$ the derangement numbers (permutations without fixed points). What you just found is essentially that $D_n=(n-1)(D_{n-1}+D_{n-2}).$
Also, in the literature, you might find them with their generalization as Associated Factorial numbers.