Particular Values of Unsigned Stirling Numbers of the First Kind
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Contents
- 1 Theorem
- 1.1 Unsigned Stirling Number of the First Kind: $\displaystyle \left[{0 \atop n}\right]$
- 1.2 Unsigned Stirling Number of the First Kind: $\displaystyle \left[{1 \atop n}\right]$
- 1.3 Unsigned Stirling Number of the First Kind: $\displaystyle \left[{n \atop n}\right]$
- 1.4 Unsigned Stirling Number of the First Kind: $\displaystyle \left[{n \atop n - 1}\right]$
- 1.5 Unsigned Stirling Number of the First Kind: $\displaystyle \left[{n \atop n - 2}\right]$
- 1.6 Unsigned Stirling Number of the First Kind: $\displaystyle \left[{n \atop n - 3}\right]$
- 1.7 Unsigned Stirling Number of the First Kind: $\displaystyle \left[{n + 1 \atop 0}\right]$
- 1.8 Unsigned Stirling Number of the First Kind: $\displaystyle \left[{n + 1 \atop 1}\right]$
- 2 Also see
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: $\displaystyle \left[{n + 1 \atop 1}\right]$
- $\displaystyle \left[{n + 1 \atop 1}\right] = n!$