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We have the following nice relations for Striling numbers of the first kind
$${n\brack 2} = \Gamma(n) H_{n-1}$$ $${n\brack 3} = \frac{\Gamma(n)}{2} \big((H_{n-1})^2-H_{n-1}^{(2)}\big)$$

Where

$$H^{(p)}_n = \sum^n_{k=1} \frac{1}{k^p}, \,\,\,H^{(1)}_n \equiv H_n$$

Questions

  • I want an algebraic proof (not combinatorial) for the previous relations.
  • Is there a "simple" general formula in terms of the harmonic numbers for

$${n\brack k} = {?}$$

metamorphy
  • 39,111
Zaid Alyafeai
  • 14,343

1 Answers1

1

An idea to get a general formula is given in this Wikipedia article: we write $$\frac{1}{n!}\sum_{k=0}^n{n+1\brack k+1}x^k=\prod_{k=1}^n\left(1+\frac xk\right)=\exp\sum_{k=1}^n\log\left(1+\frac xk\right)\\=\exp\sum_{k=1}^n\sum_{j=1}^\infty\frac{(-1)^{j-1}}{j}\left(\frac xk\right)^j=\exp\sum_{j=1}^\infty\frac{(-1)^{j-1}}{j}H_n^{(j)}x^j$$ and apply exponentiation of a power series. The result can be written as $${n+1\brack k+1}=n!(-1)^k\sum_{\substack{a_1,\,a_2,\,\ldots,\,a_k\,\geqslant\,0\\a_1+2a_2+\ldots+ka_k=k}}\prod_{j=1}^k\frac{1}{a_j!}\left(-\frac{H_n^{(j)}}{j}\right)^{a_j}.$$

metamorphy
  • 39,111