Stirling numbers of the first kind
Explicit formula

It is well-known that we don't know any one-sum formula for Stirling numbers of the first kind. A two-sum formula can be obtained using one of the symmetric formulae for Stirling numbers in conjunction with the explicit formula for Stirling numbers of the second kind.

As discussed earlier, by Vieta's formulas, one get = ∑ 0 ≤ i 1 < … < i n − k < n i 1 i 2 ⋯ i n − k. {\displaystyle \left=\sum _{0\leq i_{1}<\ldots <i_{n-k}<n}i_{1}i_{2}\cdots i_{n-k}.} The Stirling number s(n,n-p) can be found from the formula

where K = k 1 + ⋯ + k p. {\displaystyle K=k_{1}+\cdots +k_{p}.} The sum is a sum over all partitions of p.

Another exact nested sum expansion for these Stirling numbers is computed by elementary symmetric polynomials corresponding to the coefficients in x {\displaystyle x} of a product of the form (1 + c 1 x) ⋯ (1 + c n − 1 x) {\displaystyle (1+c_{1}x)\cdots (1+c_{n-1}x)}. In particular, we see that

Newton's identities combined with the above expansions may be used to give an alternate proof of the weighted expansions involving the generalized harmonic numbers already noted above.
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