What are the initial conditions for the Associated Stirling numbers of the second kind?

Question

I'm trying to compute the Associated Stirling numbers of the second kind from the recurrence relation http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind#Associated_Stirling_numbers_of_the_second_kind but I'm not sure what the initial condition are for K <= 1 and N <= 1. I thought if either K <= 1 or N <= 1 the number of ways is just 1 but that doesn't give me the right answers.

My code is just:

import scipy.misc
def stirling_helper(R, N, K):

    if N <= 1 or K <= 1:
        return 1
    Z = K * stirling_helper(R, N-1, K) + scipy.misc.comb(N-1, R-1) * stirling_helper  (R, N-1-R+1, K-1)
    return Z


def stirling(R, N, K): # Added because the recurrence is for S_r(N+1, K)
    return stirling_helper(R, N-1, K)

R = 2
N = 5
K = 2
print stirling(R, N, K) # prints 19, but should be 20?

Answer 1

In general you should choose these initial conditions so that they give the right answers for $n$ and $k$ positive. That is, you force them to work.

If $k>0$ and $n=0$ then $S_r(n,k) = 0$, since there are zero partitions of the empty set into $k$ blocks or size at least $r>1$.

If $k=0$ and $n>0$ then $S_r(n,k)=0$, since there will be no way to partition a non-empty set into zero blocks.

Now define $S_r(0,0)$ to be $1$. Why? We will want $$ 1= S_r(r,1)= 1 \cdot S_r(r-1,1) + \binom{r-1}{r-1} S_r(0,0) $$ to hold.