# Fungrim entry: 6189b9

$\left\{{n \atop k}\right\} = \frac{1}{k !} \sum_{i=0}^{k} {\left(-1\right)}^{i} {k \choose i} {\left(k - i\right)}^{n}$
Assumptions:$n \in \mathbb{Z}_{\ge 0} \,\mathbin{\operatorname{and}}\, k \in \mathbb{Z}_{\ge 0}$
TeX:
\left\{{n \atop k}\right\} = \frac{1}{k !} \sum_{i=0}^{k} {\left(-1\right)}^{i} {k \choose i} {\left(k - i\right)}^{n}

n \in \mathbb{Z}_{\ge 0} \,\mathbin{\operatorname{and}}\, k \in \mathbb{Z}_{\ge 0}
Definitions:
Fungrim symbol Notation Short description
StirlingS2$\left\{{n \atop k}\right\}$ Stirling number of the second kind
Factorial$n !$ Factorial
Sum$\sum_{n} f\!\left(n\right)$ Sum
Pow${a}^{b}$ Power
Binomial${n \choose k}$ Binomial coefficient
ZZGreaterEqual$\mathbb{Z}_{\ge n}$ Integers greater than or equal to n
Source code for this entry:
Entry(ID("6189b9"),
Formula(Equal(StirlingS2(n, k), Mul(Div(1, Factorial(k)), Sum(Mul(Mul(Pow(-1, i), Binomial(k, i)), Pow(Sub(k, i), n)), Tuple(i, 0, k))))),
Variables(n, k),
Assumptions(And(Element(n, ZZGreaterEqual(0)), Element(k, ZZGreaterEqual(0)))))

## Topics using this entry

Copyright (C) Fredrik Johansson and contributors. Fungrim is provided under the MIT license. The source code is on GitHub.

2019-08-25 15:30:03.056001 UTC