User:Pinazza/sandbox


 * $$\left[{n\atop 4}\right] = \frac{1}{3!}(n-1)! \left[ (H_{n-1})^3 - 3H_{n-1}H_{n-1}^{(2)}+2H_{n-1}^{(3)} \right], $$


 * $$\left[{n\atop 5}\right] = \frac{1}{4!}(n-1)! \left[ (H_{n-1})^4 - 6(H_{n-1})^{2}H_{n-1}^{(2)}+ 8H_{n-1}H_{n-1}^{(3)} + 3(H_{n-1}^{(2)})^2 -6H_{n-1}^{(4)} \right], $$


 * $$\left[{n\atop 6}\right] = \frac{1}{5!}(n-1)! \left[ (H_{n-1})^5 -10(H_{n-1})^{3}H_{n-1}^{(2)} +20(H_{n-1})^{2}H_{n-1}^{(3)} +15H_{n-1}(H_{n-1}^{(2)})^2 -30H_{n-1}H_{n-1}^{(4)} -20H_{n-1}^{(2)}H_{n-1}^{(3)} +24H_{n-1}^{(5)} \right], $$


 * $$ \left[{n\atop 7}\right] = \frac{1}{6!}(n-1)! \left[ \begin{align} & (H_{n-1})^6 -15(H_{n-1})^{4}H_{n-1}^{(2)} +45(H_{n-1})^{2}(H_{n-1}^{(2)})^2 +40(H_{n-1})^{3}H_{n-1}^{(3)} -90(H_{n-1})^{2}H_{n-1}^{(4)} -120H_{n-1}H_{n-1}^{(2)}H_{n-1}^{(3)} \\& -15(H_{n-1}^{(2)})^3 +144H_{n-1}H_{n-1}^{(5)} +90H_{n-1}^{(2)}H_{n-1}^{(4)} +40(H_{n-1}^{(3)})^2 -120H_{n-1}^{(6)} \\ \end{align}\right], $$

Note that the sum of the coefficents into the brackets is always zero.

Stirling numbers with negative integral values
The Stirling numbers can be extended to negative integral values. It is worth noting that Stirling numbers of first and second kind are connected by the relations:


 * $$ \left[{n \atop k}\right]=\left\{\begin{matrix} -k \\ -n \end{matrix}\right\}\, \qquad\text{or}\qquad \left\{\begin{matrix} n \\ k \end{matrix}\right\}=\left[{-k \atop -n}\right]$$

The relation continues to hold if n is positive and k negative. e.g.


 * $$ \left[{n \atop -k}\right]=\left\{\begin{matrix} k \\ -n \end{matrix}\right\}\, \qquad\text{or}\qquad \left\{\begin{matrix} -n \\ k \end{matrix}\right\}=\left[{-k \atop n}\right]$$

So we have following table for $$\left[{-n \atop -k}\right]$$:

and from


 * $$ \left[{-n \atop k}\right]=\frac{(-1)^{n+1}}{n!}\sum_{i=1}^{n}(-1)^{i+1}\frac{n \choose i}{i^k} $$

which is a generalizarion of the Recurrence relation of the Stirling numbers of the second kind, e.g. $$ \left[{-5 \atop k}\right]=(5-10/2^k+10/3^k-5/4^k+1/5^k)/120 $$

we have following table for $$\left[{-n \atop k}\right]$$:

Note that: $$\sum_{n=-1}^{-\infty}\left[{-n \atop k}\right]=B_{-k} \qquad\text{and}\qquad \sum_{n=-1}^{-\infty}\left[{-n \atop -k}\right]=B_{k} $$

where $$B_{k}$$ is the Bell number of $$k$$, e.g. $$\sum_{n=-1}^{-\infty}\left[{-n \atop 2}\right]=B_{-2}=0.421773\ldots$$