Particular Values of Unsigned Stirling Numbers of the First Kind

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Theorem

This page gathers together some particular values of unsigned Stirling numbers of the first kind.


Unsigned Stirling Number of the First Kind: $\displaystyle \left[{0 \atop n}\right]$

$\displaystyle \left[{0 \atop n}\right] = \delta_{0 n}$


Unsigned Stirling Number of the First Kind: $\displaystyle \left[{1 \atop n}\right]$

$\displaystyle {1 \brack n} = \delta_{1 n}$


Unsigned Stirling Number of the First Kind: $\displaystyle \left[{n \atop n}\right]$

$\displaystyle \left[{n \atop n}\right] = 1$


Unsigned Stirling Number of the First Kind: $\displaystyle \left[{n \atop n - 1}\right]$

$\displaystyle \left[{n \atop n - 1}\right] = \binom n 2$


Unsigned Stirling Number of the First Kind: $\displaystyle \left[{n \atop n - 2}\right]$

$\displaystyle \left[{n \atop n - 2}\right] = \binom n 4 + 2 \binom {n + 1} 4$


Unsigned Stirling Number of the First Kind: $\displaystyle \left[{n \atop n - 3}\right]$

$\displaystyle \left[{n \atop n - 3}\right] = \binom n 6 + 8 \binom {n + 1} 6 + 6 \binom {n + 2} 6$


Unsigned Stirling Number of the First Kind: $\displaystyle \left[{n + 1 \atop 0}\right]$

$\displaystyle {n + 1 \brack 0} = 0$


Unsigned Stirling Number of the First Kind: $\displaystyle \left[{n + 1 \atop 1}\right]$

$\displaystyle \left[{n + 1 \atop 1}\right] = n!$


Also see