The Stirling1(n,m) command computes the Stirling numbers of the first kind using the (implicit) generating function

Instead of Stirling1 you can also use the synonym combinat[stirling1].

Regarding combinatorial functions, ${\left(-1\right)}^{\left(n-m\right)}\mathrm{Stirling1}\left(n,m\right)$ is the number of permutations of n symbols that have exactly m cycles. The Stirling numbers also enter binomial series, Mathieu function formulas, and are relevant in physical applications.

The Stirling numbers of the first kind can be expressed as an explicit Sum with the Stirling numbers of second kind in the coefficients:

Since the Stirling numbers of the second kind also admit an explicit Sum representation,

then, an explicit double Sum representation for Stirling1 is possible by combining the two formulas above. (See the Examples section.)

$\mathrm{Stirling1}\left(m\,n\right)$

$\mathrm{Stirling1}\left(m\,n\right)$

$=\mathrm{convert}\left(\,\mathrm{Sum}\right)$

$\mathrm{Stirling1}\left(m\,n\right)=\sum _{\mathrm{\_k1}=0}^{m-n}\sum _{\mathrm{\_k2}=0}^{\mathrm{\_k1}}\frac{{\left(\mathrm{-1}\right)}^{2\mathrm{\_k1}-\mathrm{\_k2}}\left(\genfrac{}{}{0ex}{}{m-1+\mathrm{\_k1}}{m-n+\mathrm{\_k1}}\right)\left(\genfrac{}{}{0ex}{}{2m-n}{m-n-\mathrm{\_k1}}\right)\left(\genfrac{}{}{0ex}{}{\mathrm{\_k1}}{\mathrm{\_k2}}\right){\mathrm{\_k2}}^{m-n+\mathrm{\_k1}}}{\mathrm{\_k1}!}$

$\mathrm{eval}\left(\,\left[m=10\,n=5\right]\right)$

$\mathrm{-269325}=\sum _{\mathrm{\_k1}=0}^5\sum _{\mathrm{\_k2}=0}^{\mathrm{\_k1}}\frac{{\left(\mathrm{-1}\right)}^{2\mathrm{\_k1}-\mathrm{\_k2}}\left(\genfrac{}{}{0ex}{}{9+\mathrm{\_k1}}{5+\mathrm{\_k1}}\right)\left(\genfrac{}{}{0ex}{}{15}{5-\mathrm{\_k1}}\right)\left(\genfrac{}{}{0ex}{}{\mathrm{\_k1}}{\mathrm{\_k2}}\right){\mathrm{\_k2}}^{5+\mathrm{\_k1}}}{\mathrm{\_k1}!}$

$\mathrm{value}\left(\right)$

$\mathrm{-269325}=\mathrm{-269325}$