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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.

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

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.

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

fontsc=cmcsc10

% ========== Heading macros

magnification =magstep 1

overfullrule =0pt

%



noindent 1. {sc Stirling numbers} ---

Stirling cycle numbers ${ nbrack m}$ are defined  by

$$ ln^m(1+z) = m! sum_n (-1)^{n+m} { nbrack m} {z^nover n!}

   .leqno(1a) $$

The numbers $(-1)^{n+m}{nbrack m}$ are also called Stirling numbers of the first kind.

Stirling subset numbers ${nbrace m}$, also called Stirling numbers

of the second kind, are defined by

$$ left( e^z-1right)^m = m! sum_n {nbrace m} {z^nover n!}

   ,leqno(1b) $$

and 2-associated Stirling subset numbers ${nbrace m}_{ge 2}$ are

defined by 



$$ left( e^z-1-zright)^m = m!sum_n {nbrace m}_{!ge 2} {z^nover n!}

   .leqno(1c) $$



bye

You can have a look at more examples at Knuth Papers

There is a Bmatrix environment in amsmath.

documentclass{article}



usepackage{amsmath}



begin{document}

begin{equation*}

begin{Bmatrix}

x\

y

end{Bmatrix}

=frac{1}{k!}sumlimits_{j=0}{k}(-1)^{k-j}binom{k}{j}j^n

end{equation*}

end{document}

It makes sense to define all at once:

documentclass{article}

usepackage{amsmath}



newcommand{genstirlingI}[3]{%

  genfrac{[}{]}{0pt}{#1}{#2}{#3}%

}

newcommand{genstirlingII}[3]{%

  genfrac{{}{}}{0pt}{#1}{#2}{#3}%

}

newcommand{stirlingI}[2]{genstirlingI{}{#1}{#2}}

newcommand{dstirlingI}[2]{genstirlingI{0}{#1}{#2}}

newcommand{tstirlingI}[2]{genstirlingI{1}{#1}{#2}}

newcommand{stirlingII}[2]{genstirlingII{}{#1}{#2}}

newcommand{dstirlingII}[2]{genstirlingII{0}{#1}{#2}}

newcommand{tstirlingII}[2]{genstirlingII{1}{#1}{#2}}



begin{document}



The Stirling symbol of the first kind $stirlingI{n}{k}$

or of the second kind $stirlingII{n}{k}$

[

stirlingI{n}{k} ne stirlingII{n}{k}

]

We can also choose the size

[

frac{stirlingI{n}{k}+stirlingII{n}{k}}{3}quad

frac{dstirlingI{n}{k}+dstirlingII{n}{k}}{3}

]



end{document}