TITLE: Simple/efficient representation of Stirling numbers of the first kind
QUESTION [13 upvotes]: Stirling numbers of the second kind can be expressed by means of a simple hypergeometric (considering $n$ fixed) sum
$$S_2(n,k) = \frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}{k \choose j} j^n. \qquad (1)$$
This can be used for direct calculation of $S_2(n,k)$, without the need to compute any preceding values. But for Stirling numbers of the first kind, one seems to need a nested sum or a recurrence over preceding values, the most direct known representation perhaps being
$$S_1(n,k) = \sum_{j=0}^{n-k} (-1)^j {n+j-1\choose n-k+j} {2n-k \choose n-k-j} S_2(n-k+j,j). \qquad (2)$$
Is there a reason to believe that no formula similar to (1) exists for Stirling numbers of the first kind? Does a formula better than (2)+(1) for calculations exist (assume that I have no interest in generating a table of all preceding values)?

REPLY [4 votes]: $S_1(n,0)=S_1(0,n)= \delta_n \; \;$ and, for $n > 0$, 
$$S_1(n,k)= \lim_{y \to 0} \;  \frac{y^{-k}}{k!} \; \sum_{j=1}^k (-1)^j \binom{k}{j} \; \frac{(-j \; y)!}{(-j \; y-n)!} \;$$
$$ = \lim_{y \to 0} \;  \frac{y^{-k}}{k!} \; \sum_{j=1}^k (-1)^{n-j} \binom{k}{j} \; \frac{(j \; y - 1 + n)!}{(j \; y-1)!} \; $$
$$= \sum_{j=k}^n \; S_1(n,j)\; (-y)^{j-k}\;S_2(j,k) \; |_{y=0} \; . $$
For a derivation, see A class of differential operators and the Stirling numbers. Note that with $y$ small enough taking the nearest integer generates $S_1$.