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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)?

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    $\begingroup$ Is it really simpler/faster to use (1) instead of the usual recurrence formula to compute $S_2(n,k)$? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Aug 1, 2010 at 20:48
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    $\begingroup$ What's wrong with the first formula in the Wikipedia article? One can easily extract a particular coefficient from it without a recurrence. $\endgroup$
    – Qiaochu Yuan
    Commented Aug 1, 2010 at 22:05
  • $\begingroup$ Mariano: yes, for large $n$. Qiaochu: this is a good method, but even expanding the polynomial using a balanced product (I tried it using Sage) is considerably slower for large n than evaluating (1), and of course requires much more memory. I'm interested in whether there exists a formula that does not amount to computing all $k$ numbers. $\endgroup$
    – Fredrik Johansson
    Commented Aug 2, 2010 at 1:23
  • $\begingroup$ Fredrik, so what's wrong with Eq. (17) on mathworld.wolfram.com/StirlingNumberoftheFirstKind.html ? (You don't need to compute SNs of the 2nd kind.) In view of your comments to Mariano and Qiaochu, I am trying to understand what is exactly unsatisfactory in all these classical formulae... You can't get something better, because everything is too classical. $\endgroup$
    – Wadim Zudilin
    Commented Aug 2, 2010 at 6:27
  • $\begingroup$ Wadim: I'm asking whether there is a formula that does not involve nested Stirling numbers. $\endgroup$
    – Fredrik Johansson
    Commented Aug 2, 2010 at 17:08

4 Answers 4

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Would you, or would you not, consider as "simple" integral and/or series representations that work for complex values, suitably restricted?

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    $\begingroup$ That article gives a description of Stirling numbers of the first kind as a hypergeometric function up to a Gamma factor. $\endgroup$
    – Gjergji Zaimi
    Commented Aug 30, 2010 at 8:41
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$S_1(n,0)=S_1(0,n)= \delta_n \; \;$ 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 $S_1$.

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  • $\begingroup$ This looks to me like using an asymptotical inverse of the infinite Vandermondematrix $V_{r,c}=r^c$ (which of course does not exist). Like $V= S_2 \cdot \ ^dF \cdot P$ then $V \cdot P^{-1} \cdot \ ^dF^{-1} \cdot V = S_2 $ and then the inversion: $V^{-1} \cdot \ ^dF \cdot P =S_1 $ where of course we cannot exactly use $V$ because the inversion would produce singularities. Did you get your formula by something like this? ($P$: upper triangular binomialmatrix, $S_2$ Stirling numbers 2nd kind, $ \ ^dF$ diagonalmatrix of factorials) $\endgroup$
    – Gottfried Helms
    Commented Sep 4, 2015 at 10:47
  • $\begingroup$ @Gottfried, I included a link to a simple derivation of the formula, including an equivalent matrix formula. $\endgroup$
    – Tom Copeland
    Commented Sep 4, 2015 at 17:34
  • $\begingroup$ Ahh, thanks, that looks very promising. $\endgroup$
    – Gottfried Helms
    Commented Sep 4, 2015 at 17:56
  • $\begingroup$ How would you evaluate the limit to get a single series? $\endgroup$
    – Тyma Gaidash
    Commented Aug 2, 2023 at 12:05
  • $\begingroup$ @TymaGaidash, can you give me an example of how you would obtain a number other than $S_1(n,k)$ in the limit? $\endgroup$
    – Tom Copeland
    Commented Aug 2, 2023 at 18:04
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You ask "Is there a reason to believe that no formula similar to (1) exists for Stirling numbers of the first kind?" One reason to believe that there is no such formula is that Louis Comtet (Advanced Combinatorics, p. 216) says so: "... the Stirling number of the second kind s(n, k) can be expressed as a single summation of elementary terms, that is, which are themselves products and quotients of factorials and powers. There does not exist an analogous formula for the numbers of the first kind, the ‘shortest formula’ [7a, a’] below being a double summation of elementary terms."

A proof of Comtet's nonexistence assertion would be interesting.

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http://members.lycos.co.uk/sobalian/index.html

OEIS A008275

a(n,k) = s(k,n) = (-1)^(k-n) * S1(k,n) = ( (-1)^(k-n) ) * ( k!/{(n-1)!*2^(k-n)} ) * [ { 1/(k-n)! }k^(k-n-1) - { (1/6)(1/(k-n-2)!) }k^(k-n-2) + { (1/72)(1/(k-n-4)!) }k^(k-n-3) - { (1/6480)(5/(k-n-6)! -36/(k-n-4)!) }k^(k-n-4) + { (1/155520)(5/(k-n-8)!-144/(k-n-6)!) }k^(k-n-5) - { (1/6531840)(7/(k-n-10)! -504/(k-n-8)!+2304/(k-n-6)!) }k^(k-n-6) + { (1/1175731200)(35/(k-n-12)!-5040/(k-n-10)!+87264/(k-n-8)!) }k^(k-n-7) - { (1/7054387200)(5/(k-n-14)!-1260/(k-n-12)!+52704/(k-n-10)!-186624/ (k-n-8)!) }k^(k-n-8) + { (1/338610585600)(5/(k-n-16)!-2016/(k-n-14)!+164736/ (k-n-12)!-2156544/(k-n-10)!) }*k^(k-n-9) - ..... ]

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    $\begingroup$ Can you clarify this a bit more? $\endgroup$
    – Stefan Waldmann
    Commented Jun 27, 2013 at 10:57

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