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What is the best way to write a Stirling number of the second kind? Isn't there any standard command in LaTeX? For example, a command \binom is for binomial coefficients.

In the wikipedia article on Stirling number of the second kind, they used \atop command. But people say \atop is not recommended.

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1  
Even putting any technical reasons aside, \atop is a bad choice as it left-aligns the "numerator" and "denominator", rather than centring them. –  David Richerby Mar 28 at 16:33

3 Answers 3

up vote 11 down vote accepted

The following is taken from amsmath and uses \genfrac - a generic fraction function:

enter image description here

\documentclass{article}
\usepackage{amsmath}% http://ctan.org/pkg/amsmath
\DeclareRobustCommand{\stirling}{\genfrac\{\}{0pt}{}}
\begin{document}
% Source: http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind
In mathematics, particularly in combinatorics, a Stirling number of the second kind 
is the number of ways to partition a set of $n$~objects into $k$~non-empty subsets and 
is denoted by~$S(n,k)$ or~$\stirling{a}{b}$. Stirling numbers of the second kind occur 
in the field of mathematics called combinatorics and the study of partitions.
\end{document}

How does this work? \genfrac takes five arguments to create a structure (from the amsmath documentation; section 4.11.3 The \genfrac command, p 14):

The last two correspond to \frac’s numerator and denominator; the first two are optional delimiters [...]; the third is a line thickness override [0 implies an invisible rule]; and the fourth argument is a mathstyle override: integer values 0-3 select respectively \displaystyle, \textstyle, \scriptstyle, and \scriptscriptstyle. If the third argument is left empty, the line thickness defaults to ‘normal’.

So \genfrac\{\}{0pt}{} creates a fraction with an invisible horizontal rule (third argument is 0pt), left and right delimiter given by \{ and \}, respectively and no specific math style (an empty {} fourth argument). \stirling doesn't include a fifth and sixth argument for \genfrac (numerator and denominator), because this is implicitly supplied by the user as the two "arguments" to \stirling.

In a similar manner (perhaps for reference), amsmath defines

\newcommand{\frac}[2]{\genfrac{}{}{}{}{#1}{#2}}
\newcommand{\tfrac}[2]{\genfrac{}{}{}{1}{#1}{#2}}
\newcommand{\binom}[2]{\genfrac{(}{)}{0pt}{}{#1}{#2}}

using \genfrac.

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Here is a plain pdfTeX solution, just to illustrate what type of typesetting Knuth used for such numbers (he wrote a few papers on this topic)

% ========= Fonts
\font\sc=cmcsc10
% ========== Heading macros
\magnification =\magstep 1
\overfullrule =0pt
%

\noindent 1. {\sc Stirling numbers} ---
Stirling cycle numbers ${ n\brack m}$ are defined  by
$$ \ln^m(1+z) = m! \sum_n (-1)^{n+m} { n\brack m} {z^n\over n!}
  \ .\leqno(1a) $$
The numbers $(-1)^{n+m}{n\brack m}$ are also called Stirling numbers of the first kind.
Stirling subset numbers ${n\brace m}$, also called Stirling numbers
of the second kind, are defined by
$$ \left( e^z-1\right)^m = m! \sum_n {n\brace m} {z^n\over n!}
  \ ,\leqno(1b) $$
and 2-associated Stirling subset numbers ${n\brace m}_{\ge 2}$ are
defined by 

$$ \left( e^z-1-z\right)^m = m!\sum_n {n\brace m}_{\!\ge 2} {z^n\over n!}
  \ .\leqno(1c) $$

\bye

enter image description here

You can have a look at more examples at Knuth Papers

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Wow. I am tempted to learn plain typesetting. –  hpesoj626 Dec 8 '12 at 6:37
    
@hpesoj626 Not really necessary. If you place the code between \begin{document}...\end{document} it will work with LaTeX as well; amsmath will complain about the \over though, if you using it. They can easily be converted to amsmath styles. –  Yiannis Lazarides Dec 8 '12 at 7:38
1  
Take a look specially at Knuth's P137 from the link above, it's called "Two notes on notation" and it's precisely about the Stirling numbers. –  Mafra Dec 8 '12 at 14:29
    
@Mafra Thanks for the reference. –  Yiannis Lazarides Dec 8 '12 at 17:44

There is a Bmatrix environment in amsmath.

\documentclass{article}

\usepackage{amsmath}

\begin{document}
\begin{equation*}
\begin{Bmatrix}
x\\
y
\end{Bmatrix}
=\frac{1}{k!}\sum\limits_{j=0}{k}(-1)^{k-j}\binom{k}{j}j^n
\end{equation*}
\end{document}

enter image description here

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