Stirling Number of the Second Kind of n+1 with 0

Theorem

Let $n \in \Z_{\ge 0}$.

Then:

$\ds {n + 1 \brace 0} = 0$

where $\ds {n + 1 \brace 0}$ denotes a Stirling number of the second kind.

Proof

We are given that $k = 0$.

So by definition of unsigned Stirling number of the first kind:

$\ds {n \brace k} = \delta_{n k}$

where $\delta_{n k}$ is the Kronecker delta.

Thus

\(\ds n\)

\(\ge\)

\(\ds 0\)

by hypothesis

\(\ds \leadsto \ \ \)

\(\ds n + 1\)

\(>\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds n + 1\)

\(\ne\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \delta_{\paren {n + 1} 0}\)

\(=\)

\(\ds 0\)

Hence the result.

$\blacksquare$

Also see

• Unsigned Stirling Number of the First Kind of n+1 with 0
• Signed Stirling Number of the First Kind of n+1 with 0

• Particular Values of Stirling Numbers of the Second Kind

Sources

• 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: $(50)$