How to find Stirling coefficient of the second order ?

Question

Can someone please explain how can I find the coefficient of the Stirling number of the second kind for S(8,k) ?

After checking here , I found that :

S(8, k) =

    1
    127
    966
    1701
    1050
    266
    28
    1

So , if we use the Stirling formula :

$$ x^8 = S(8,0) \cdot (x)_{0} + S(8,1) \cdot (x)_{1} + S(8,2) \cdot (x)_{2} + S(8,3) \cdot (x)_{3} + S(8,4) \cdot (x)_{4} + S(8,5) \cdot (x)_{5} + S(8,6) \cdot (x)_{6} + S(8,7) \cdot (x)_{7} + S(8,8) \cdot (x)_{8}$$

By properties of Stirling number of the 2nd kind : $$S(n,n) = 1$$ so $$S(8,8)=1$$ .

But how can I find the rest of the coefficients ?

Does $S(n,k)$ has an easy form for calculating the rest of the coefficients $$S(8,0) , S(8,1) , ... S(8,7) $$ ?

Thanks

Answer 1

I would use the recurrence $$S(n,k) = S(n-1,k-1) + k S(n-1,k)$$ starting with $S(0,0)=1$ and $S(n,0)=S(0,n)=0$ for $n\gt 0$. That easily gives this triangle, with each column derived from the previous one:

1   0   0   0   0   0   0   0   0
0   1   1   1   1   1   1   1   1
0   0   1   3   7   15  31  63  127
0   0   0   1   6   25  90  301 966
0   0   0   0   1   10  65  350 1701
0   0   0   0   0   1   15  140 1050
0   0   0   0   0   0   1   21  266
0   0   0   0   0   0   0   1   28
0   0   0   0   0   0   0   0   1

Answer 2

You can use the identity

$$ \left\{\begin{matrix} n \\ k \end{matrix}\right\} =\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}{k \choose j} j^n, $$

where

$$ {k \choose j}=\frac{k!}{j!(k-j)!}. $$