The Stirling numbers of the second kind obey the well-known recurrence relation:
$$S(n,k) = k \cdot S(n-1,k) + S(n-1,k-1).$$
This recurrence relation reduces the value of the first argument in the Stirling numbers by one. I would like to apply this recurrence relation iteratively, allowing me to reduce the first argument by some arbitrary value $t \in \mathbb{N}$. This will give me a generalised recurrence of the form:
$$S(n,k) = \sum_{i=0}^t a_{t,i} \cdot S(n-t,k-i),$$
where the coefficients $a_{t,i}$ also depend implicitly on the fixed value $k$. So far I have not found an explicit expression for these coefficients, and hence for the generalised recurrence. I have managed to characterise the coefficients recursively by:
$$a_{t,0} = k^t \quad \quad \quad a_{t,t} = 1 \quad \quad \quad a_{t+1,i} = a_{t,i-1} + (k-i) \cdot a_{t,i}.$$
Can anyone identify the coefficients in this generalise recursion, and give an explicit formula for these coefficients?