Particular Values of Unsigned Stirling Numbers of the First Kind

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

• Particular Values of Signed Stirling Numbers of the First Kind
• Particular Values of Stirling Numbers of the Second Kind