As it states in Wikipedia: Stirling numbers of the first kind count permutations according to their number of cycles. And the recurrence relation is as follows:
$\left[{n+1\atop k}\right] = n \left[{n\atop k}\right] + \left[{n\atop k-1}\right]$
I was wondering how the recurrence relation will look like if there was a lower bound on the size of the cycle ( at least x nodes in a cycle ).
This are called Associated Stirling numbers of the first kind, they are denoted by ${n\brack k}_{\geq x}$. The problem is then that you can not just put the ${n\brack k-1}$ because that corresponds to a fix point (cycle length = 1) so you have $${n+1\brack k}_{\geq x}=n{n\brack k}_{\geq x}+{n+1-x\brack k-1}_{\geq x}\cdot \binom{n}{x-1}(x-1)!,$$ where the last multiplication corresponds to take $x-1$ elements less than $n+1$ and put them with $n+1$ in a cycle of length $x,$ the $(x-1)!$ takes care of the cyclic order.
Notice that if you take $x=1,$ you recover the normal Stirling numbers and if you set $x=2$ and add, you get the well known recurrence for derangements.