We have the following nice relations for Striling numbers of the first kind
$$\left[n\atop 2\right] = \Gamma(n) H_{n-1}$$

$$\left[n\atop 3\right] = \frac{\Gamma(n)}{2} ((H_{n-1})^2-H_{n-1}^{(2)})$$


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

$$\left[n\atop k\right] = ?$$

  • $\begingroup$ What does $H_{n-1}^{(2)}$ mean? Shouldn't it be $H_{n-1}^2$? $\endgroup$ – Yuriy S Jan 13 '17 at 13:42
  • $\begingroup$ @YuriyS, see my edits. $\endgroup$ – Zaid Alyafeai Jan 13 '17 at 13:48
  • 1
    $\begingroup$ @YuriyS: $H_n^{(2)}=\sum_{k=1}^{n}\frac{1}{k^2}$. We are talking about the Taylor coefficients of $\log(1-x)^m$, so we may find the expression of ${n\brack s+1}$ by considering the expression of ${n\brack s}$, removing the $\Gamma$ part, dividing by $n$ and summing over $n=1,2,\ldots,N$ through summation by parts. $\endgroup$ – Jack D'Aurizio Jan 13 '17 at 13:50
  • $\begingroup$ @ZaidAlyafeai, I see, thanks $\endgroup$ – Yuriy S Jan 13 '17 at 13:50
  • $\begingroup$ These are discussed and proved at the following MSE link. $\endgroup$ – Marko Riedel Jan 13 '17 at 23:09

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