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[[File:Set partitions 4; Hasse; circles.svg|thumb|right|300px|The [[Bell number|15]] partitions of a 4-element set<br>ordered in a [[Hasse diagram]]<br><br>There are ''S''(4,1),...,''S''(4,4) = 1,7,6,1 partitions<br>containing 1,2,3,4 sets.]]
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In [[mathematics]], particularly in [[combinatorics]], a '''Stirling number of the second kind'''(or Stirling partition number) is the number of ways to [[Partition of a set|partition a set]] of ''n'' objects into ''k'' non-empty subsets and is denoted by <math>S(n,k)</math> or <math>\textstyle \lbrace{n\atop k}\rbrace</math>.<ref>Ronald L. Graham, Donald E. Knuth, Oren Patashnik (1988) ''[[Concrete Mathematics]]'', Addison–Wesley, Reading MA. ISBN 0-201-14236-8, p.&nbsp;244.</ref> Stirling numbers of the second kind occur in the field of  [[mathematics]] called [[combinatorics]] and the study of [[partition (number theory)|partitions]].
 
Stirling numbers of the second kind are one of two kinds of [[Stirling number]]s, the other kind being called [[Stirling numbers of the first kind]] (or Stirling cycle numbers). Mutually inverse (finite or infinite) [[Triangular matrix|triangular matrices]] can be formed from the Stirling numbers of each kind according to the parameters ''n'', ''k''.
 
==Definition==
The Stirling numbers of the second kind, written <math>S(n,k)</math> or <math>\lbrace\textstyle{n\atop k}\rbrace</math> or with other notations, count the number of ways to partition a set of ''n'' labelled objects into ''k'' nonempty unlabelled subsets. Equivalently, they count the number of different [[equivalence relation]]s with precisely ''k'' equivalence classes that can be defined on an ''n'' element set. Obviously,
:<math>\left\{\begin{matrix} n \\ 1 \end{matrix}\right\} = \left\{\begin{matrix} n \\ n \end{matrix}\right\} = 1. </math>
They can be calculated using the following explicit formula:<ref>{{Citation | last1=Sharp | first1=Henry | title=Cardinality of finite topologies | doi=10.1016/S0021-9800(68)80031-6 | id={{MathSciNet | id = 0226578}} | year=1968 | journal=J. Combinatorial Theory | volume=5 | pages=82–86}}
</ref>
:<math>\left\{\begin{matrix} n \\ k \end{matrix}\right\} = \frac{1}{k!}\sum_{j=0}^k (-1)^{k-j}{k \choose j} j^n.</math>
 
==Notation==
Various notations have been used for Stirling numbers of the second kind.  The brace notation <math>\textstyle \lbrace{n\atop k}\rbrace</math>  was used by Imanuel Marx and Antonio Salmeri in 1962 for variants of these numbers.<ref>Transformation of Series by a Variant of Stirling's Numbers, Imanuel Marx, ''The American Mathematical Monthly'' '''69''', #6 (June–July 1962), pp. 530–532, {{JSTOR|2311194}}.</ref><ref>Antonio Salmeri, Introduzione alla teoria dei coefficienti fattoriali, ''Giornale di Matematiche di Battaglini '''90''' (1962), pp. 44–54.</ref> This led [[Donald Ervin Knuth|Knuth]] to use it, as shown here, in the first volume of ''[[The Art of Computer Programming]]'' (1968).<ref name=tnn>p. 410–412, Two Notes on Notation, Donald E. Knuth, ''The American Mathematical Monthly'' '''99''', #5 (May 1992), pp. 403–422, {{JSTOR|2325085}}.</ref><ref>Donald E. Knuth, ''Fundamental Algorithms'', Reading, Mass.: Addison–Wesley, 1968.</ref>  However, according to the third edition of ''The Art of Computer Programming'', this notation was also used earlier by [[Jovan Karamata]] in 1935.<ref>p. 66, Donald E. Knuth, ''Fundamental Algorithms'', 3rd ed., Reading, Mass.: Addison–Wesley, 1997.</ref><ref>Jovan Karamata, Théorèmes sur la sommabilité exponentielle et d'autres sommabilités s'y rattachant, ''Mathematica'' (Cluj) '''9''' (1935), pp, 164–178.</ref>  The notation ''S(n,k)'' was used by [[Richard P. Stanley|Richard Stanley]] in his book ''[[Enumerative Combinatorics]]''.<ref name=tnn />
 
==Bell numbers==
{{main|Bell number}}
 
The sum over the values for k of the Stirling numbers of the second kind, gives us
 
:<math>B_n=\sum_{k=0}^n \left\{\begin{matrix} n \\ k \end{matrix}\right\}</math>
 
the ''n''th [[Bell numbers|Bell number]], that is the total number of [[partition of a set|partitions]] of a [[Set (mathematics)|set]] with ''n'' [[element (mathematics)|members]].
 
If we let
 
:<math>(x)_n=x(x-1)(x-2)\cdots(x-n+1) \,</math>
 
(in particular, (''x'')<sub>0</sub> = 1 because it is an [[empty product]]) be the [[falling factorial]],<ref>Confusingly, the notation that combinatorialists use for ''falling'' factorials coincides with the notation used in [[special function]]s for ''rising'' factorials; see [[Pochhammer symbol]].</ref> we can characterize the Stirling numbers of the second kind by
 
:<math>\sum_{k=0}^n \left\{\begin{matrix} n \\ k \end{matrix}\right\}(x)_k=x^n.</math>
 
Analogously, the [[ordered Bell number]]s can be computed from the Stirling numbers of the second kind as<ref>{{citation
| last = Sprugnoli | first = Renzo
| doi = 10.1016/0012-365X(92)00570-H
| issue = 1-3
| journal = Discrete Mathematics
| mr = 1297386
| pages = 267–290
| title = Riordan arrays and combinatorial sums
| volume = 132
| year = 1994}}.</ref>
:<math>a_n = \sum_{k=0}^n k! \left\{\begin{matrix} n \\ k \end{matrix}\right\}</math>
 
==Table of values==
Below is a [[triangular array]] of values for the Stirling numbers of the second kind {{OEIS|A008277}}:
{| cellspacing="0" cellpadding="5" style="text-align:right;"
|-
| '''n'''&nbsp;\&nbsp;''k''
| style="width:9%;"| ''0''
| style="width:9%;"| ''1''
| style="width:9%;"| ''2''
| style="width:9%;"| ''3''
| style="width:9%;"| ''4''
| style="width:9%;"| ''5''
| style="width:9%;"| ''6''
| style="width:9%;"| ''7''
| style="width:9%;"| ''8''
| style="width:9%;"| ''9''
| style="width:9%;"| ''10''
|-
|'''0'''
| 1
|-
|'''1'''
| 0
| 1
|-
|'''2'''
| 0
| 1
| 1
|-
|'''3'''
| 0
| 1
| 3
| 1
|-
|'''4'''
| 0
| 1
| 7
| 6
| 1
|-
|'''5'''
| 0
| 1
| 15
| 25
| 10
| 1
|-
|'''6'''
| 0
| 1
| 31
| 90
| 65
| 15
| 1
|-
|'''7'''
| 0
| 1
| 63
| 301
| 350
| 140
| 21
| 1
|-
|'''8'''
| 0
| 1
| 127
| 966
| 1701
| 1050
| 266
| 28
| 1
|-
|'''9'''
| 0
| 1
| 255
| 3025
| 7770
| 6951
| 2646
| 462
| 36
| 1
|-
|'''10'''
| 0
| 1
| 511
| 9330
| 34105
| 42525
| 22827
| 5880
| 750
| 45
| 1
|-
|}
 
As with the [[binomial coefficients]], this table could be extended to&nbsp;''k''&nbsp;>&nbsp;''n'', but those entries would all be&nbsp;0.
 
== Properties ==
 
===Recurrence relation===
Stirling numbers of the second kind obey the recurrence relation
 
:<math>\left\{{n+1\atop k}\right\} = k \left\{{ n \atop k }\right\} + \left\{{n\atop k-1}\right\}
</math>
for ''k'' > 0 with initial conditions
:<math>\left\{{ 0 \atop 0 }\right\} = 1
\quad \mbox{ and } \quad
\left\{{ n \atop 0 }\right\} = \left\{{ 0 \atop n }\right\} = 0</math>
for ''n'' > 0.
 
For instance, the number 25 in column ''k''=3 and row ''n''=5 is given by 25=7+(3×6), where 7 is the number above and to the left of 25, 6 is the number above 25 and 3 is the column containing the 6.
 
To understand this recurrence, observe that a partition of the ''n+1'' objects into ''k'' nonempty subsets either contains the ''n+1''-th object as a singleton  or it does not. The number of ways that the singleton is one of the subsets is given by
 
:<math>\left\{{ n \atop k-1 }\right\}</math>
 
since we must partition the remaining <math>n</math> objects into the available ''k-1'' subsets. In the other case the ''n+1''-th object belongs to a subset containing other objects. The number of ways is given by
 
:<math>k \left\{{ n \atop k }\right\}</math>
 
since we partition all objects other than the ''n+1''-th into ''k'' subsets, and then we are left with ''k'' choices for inserting object ''n+1''. Summing these two values gives the desired result.
 
Some more recurrences are as follows:
 
:<math>\left\{{n+1\atop k+1}\right\} = \sum_{j=k}^n {n \choose j} \left\{{ j \atop k }\right\}, </math>
 
:<math>\left\{{n+1\atop k+1}\right\} = \sum_{j=k}^n (k+1)^{n-j} \left\{{j \atop k}\right\} , </math>
 
:<math>\left\{{n+k+1 \atop k}\right\} = \sum_{j=0}^k j \left\{{ n+j \atop j }\right\}. </math>
 
===Lower and upper bounds===
If <math>n \geq 2</math> and <math>1 \leq k \leq n-1</math>, then
 
:<math>L(n,k) \leq \left\{{n \atop k}\right\} \leq U(n,k)</math>
 
where
:<math>L(n,k) = \frac{1}{2}(k^2+k+2)k^{n-k-1}-1</math>
and
:<math>U(n,k) = \frac{1}{2}{n \choose k} k^{n-k}.</math> <ref name="ac.els-cdn.com">[http://ac.els-cdn.com/S0021980069800451/1-s2.0-S0021980069800451-main.pdf?_tid=4e52b648-711a-11e2-a90d-00000aab0f26&acdnat=1360237038_c9d4f38f83a536ed624e29e3554016e2 B.C. Rennie, A.J. Dobson. "On Stirling Numbers of the Second Kind"]</ref>
 
===Maximum===
For fixed <math>n</math>, <math>\left\{{n \atop k}\right\}</math> has a single maximum. That is, there is a <math>K_n</math> such that
 
:<math>\left\{{n \atop 1}\right\} < \left\{{n \atop 2}\right\} < \cdots < \left\{{n \atop K_n}\right\},</math>
:<math>\left\{{n \atop K_n}\right\} \geq \left\{{n \atop K_n+1}\right\} > \cdots > \left\{{n \atop n}\right\}.</math>
 
When <math>n</math> is large
 
:<math>K_n \sim \frac{n}{\log n},</math>
 
and the maximum value of the Stirling number of second kind at <math>K_n</math> is
:<math>\log \left\{{n \atop K_n}\right\} = n\log n - n \log\log n - n + O(n \log\log n / \log n).</math> <ref name="ac.els-cdn.com"/>
 
===Parity===
[[Image:Stirling numbers of the second kind - Parity.svg|256px|right|thumb|Parity of Stirling numbers of the second kind.]]
 
The [[Parity (mathematics)|parity]] of a Stirling number of the second kind is equal to the parity of a related [[binomial coefficient]]:
 
:<math>
\left\{\begin{matrix}n\\k\end{matrix}\right\}\equiv\begin{pmatrix}z\\w\end{pmatrix}\ \pmod{2},\quad
z = n - \left\lceil\displaystyle\frac{k + 1}{2}\right\rceil,\ w = \left\lfloor\displaystyle\frac{k - 1}{2}\right\rfloor.
</math>
 
This relation is specified by mapping ''n'' and ''k'' coordinates onto the [[Sierpinski triangle|Sierpiński triangle]].
 
Or directly, let two sets contain positions of 1's in binary representations of results of respective expressions:
 
:<math>
\begin{align}
\mathbb{A}:\ \sum_{i\in\mathbb{A}} 2^i &= n-k,\\
\mathbb{B}:\ \sum_{j\in\mathbb{B}} 2^j &= \left\lfloor\dfrac{k - 1}{2}\right\rfloor,\\
\end{align}
</math>
 
then mimic a [[bitwise AND]] operation by intersecting these two sets:
 
:<math>
\begin{Bmatrix}n\\k\end{Bmatrix}\,\bmod\,2 =
\begin{cases}
0, & \mathbb{A}\cap\mathbb{B}\ne\empty\\
1, & \mathbb{A}\cap\mathbb{B}=\empty
\end{cases}
</math>
 
to obtain the parity of a Stirling number of the second kind in [[Big O notation|''O''(1)]] time. In [[pseudocode]]:
 
:<math>
\begin{Bmatrix}n\\k\end{Bmatrix}\,\bmod\,2 := \left(n-k\right)\!\!\And\!\!\left( \left(k-1\right)\,\mathrm{div}\,2 \right) == 0.
</math>
 
===Simple identities===
Some simple identities include
 
:<math>\left\{\begin{matrix} n \\ n-1 \end{matrix}\right\} =
{n \choose 2}. </math>
 
This is because dividing ''n'' elements into ''n''&nbsp;&minus;&nbsp;1 sets necessarily means dividing it into one set of size 2 and ''n''&nbsp;&minus;&nbsp;2 sets of size 1.  Therefore we need only pick those two elements;
 
and
 
:<math>\left\{\begin{matrix} n \\ 2 \end{matrix}\right\} = 2^{n-1}-1.</math>
 
To see this, first note that there are 2<sup>&nbsp;''n''</sup> ''ordered'' pairs of complementary subsets ''A'' and ''B''.  In one case, ''A'' is empty, and in another ''B'' is empty, so 2<sup>&nbsp;''n''</sup>&nbsp;&minus;&nbsp;2 ordered pairs of subsets remain.  Finally, since we want ''unordered'' pairs rather than ''ordered'' pairs we divide this last number by 2, giving the result above.
 
Another explicit expanding of the recurrence-relation gives identities in the spirit of the above example.
 
:<math>
\begin{align}
\left\{\begin{matrix} n \\ 2 \end{matrix}\right\} & = \frac{ \frac11 (2^{n-1}-1^{n-1}) }{0!} \\[8pt]
\left\{\begin{matrix} n \\ 3 \end{matrix}\right\} & = \frac{ \frac11 (3^{n-1}-2^{n-1})- \frac12 (3^{n-1}-1^{n-1}) }{1!} \\[8pt]
\left\{\begin{matrix} n \\ 4 \end{matrix}\right\} & = \frac{ \frac11 (4^{n-1}-3^{n-1})- \frac22 (4^{n-1}-2^{n-1}) +  \frac13 (4^{n-1}-1^{n-1})}{2!} \\[8pt]
\left\{\begin{matrix} n \\ 5 \end{matrix}\right\} & = \frac{ \frac11 (5^{n-1}-4^{n-1})- \frac32 (5^{n-1}-3^{n-1}) + \frac33 (5^{n-1}-2^{n-1}) -  \frac14 (5^{n-1}-1^{n-1}) }{3!} \\[8pt]
& {}\ \  \vdots
\end{align}
</math>
 
===Explicit formula===
The Stirling numbers of the second kind are given by the explicit formula:
 
:<math>\left\{\begin{matrix} n \\ k \end{matrix}\right\}
=\sum_{j=1}^k (-1)^{k-j} \frac{j^{n-1}}{(j-1)!(k-j)!}
=\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}{k \choose j} j^n
.</math>
 
This formula is a special case of the ''k'' 'th [[forward difference]] of the [[monomial]] <math>x^n</math> evaluated at ''x'' = 0:
 
:<math> \Delta^k x^n = \sum_{j=0}^{k}(-1)^{k-j}{k \choose j} (x+j)^n.</math>
 
Because the [[Bernoulli polynomial]]s may be written in terms of these forward differences, one immediately obtains a relation in the [[Bernoulli number]]s:
 
:<math>B_m(0)=\sum_{k=0}^m \frac {(-1)^k k!}{k+1}
\left\{\begin{matrix} m \\ k \end{matrix}\right\}. </math>
 
===Generating function===
A [[generating function]] for the Stirling numbers of the second kind is given by
:<math> \sum_{k=0}^n \left\{\begin{matrix} n \\ k \end{matrix}\right\}
(x)_k = x^n.</math>
 
A rational generating function is given by
:<math> \prod_{r=1}^k \frac{1}{1-rx} = \sum_{n=0}^\infty \left\{\begin{matrix} n \\ k \end{matrix}\right\} x^{n-k}, \quad \text{ or } \quad
\sum_{n=0}^\infty \left\{\begin{matrix} n\\ k \end{matrix}\right\} x^{n+1}= \frac{1}{(k+1)! {\frac{1}{x} \choose k+1}}.</math>
 
Two exponential generating functions are given by
:<math> \sum_{n=0}^\infty \left\{\begin{matrix} n \\ k \end{matrix}\right\} \frac{x^n}{n!} = \frac{(e^x-1)^k}{k!},</math>
 
:<math> \sum_{n,k=0}^\infty \left\{\begin{matrix} n \\ k \end{matrix}\right\} \frac{x^n}{n!} y^k = \textrm{exp}(y(e^x-1))</math>
 
Note that <math>\left\{\begin{matrix} n \\ k \end{matrix}\right\} = 0</math>  for  <math>n < k</math>.
 
===Asymptotic approximation===
For fixed value of <math>k,</math> the asymptotic value of the Stirling numbers of the second kind is given by
:<math> \left\{{n \atop k}\right\} \sim \frac{k^n}{k!}.</math>
 
On the other side, for <math> n\rightarrow \infty </math> and <math> k \sim o(\sqrt{n}) </math>
,<ref>L. C. Hsu, Note on an Asymptotic Expansion of the nth Difference of Zero, AMS Vol.19 NO.2 1948, pp. 273--277</ref>
:<math> \left\{{n \atop n-k}\right\} \sim \frac{(n-k)^{2k}}{2^k k!} \left( 1 + \frac{1}{3} \frac{2 k^2 + k}{n-k} + \frac{1}{18} \frac{4k^4-k^2-3k}{(n-k)^2} + \cdots \right).</math>
 
Uniformly valid approximation also exist
<ref>W. E. Bleick and Peter C. C. Wang, Asymptotics of Stirling Numbers of the Second Kink, AMS Vol.42 No.2, 1974.</ref>
:<ref>N. M. Temme, Asymptotic Estimates of Stirling Numbers, STUDIES IN APPLIED MATHEMATICS 89:233-243 (1993), Elsevier Science Publicshing.</ref>
 
:<math> \left\{{n \atop k}\right\} \sim \frac{\sqrt{n-k}}{\sqrt{n (1-G)}\ G^k\ (v-G)^{n-k}} \left(\frac{n-k}{e}\right)^{n-k} \left({n \atop k}\right) \quad\forall k, 1<k<n </math>
 
where <math> G = - W_0(-v e^{-v}), v=n/k </math>, <math>\ W_0(z)</math> is main branche of [[Lambert W function]]. Relative error is bounded by about <math> 0.06/n </math>.
 
==Applications==
 
===Moments of the Poisson distribution===
If ''X'' is a [[random variable]] with a [[Poisson distribution]] with [[expected value]] λ, then its ''n''th [[moment (mathematics)|moment]] is
 
:<math>E(X^n)=\sum_{k=1}^n \left\{\begin{matrix} n \\ k \end{matrix}\right\}\lambda^k.</math>
 
In particular, the ''n''th moment of the Poisson distribution with expected value 1 is precisely the number of [[partition of a set|partitions of a set]] of size ''n'', i.e., it is the ''n''th [[Bell number]] (this fact is [[Dobinski's formula]]).
 
===Moments of fixed points of random permutations===
Let the random variable ''X'' be the number of fixed points of a [[discrete uniform distribution|uniformly distributed]] [[random permutation]] of a finite set of size ''m''.  Then the ''n''th moment of ''X'' is
 
:<math>E(X^n) = \sum_{k=1}^m \left\{\begin{matrix} n \\ k \end{matrix}\right\}.</math>
 
'''Note:''' The upper bound of summation is ''m'', not ''n''.
 
In other words, the ''n''th moment of this [[probability distribution]] is the number of partitions of a set of size ''n'' into no more than ''m'' parts.
This is proved in the article on [[Random permutation statistics#Moments of fixed points|random permutation statistics]], although the notation is a bit different.
 
===Rhyming schemes===
The Stirling numbers of the second kind can represent the total number of [[rhyme scheme]]s for a poem of ''n'' lines. <math>S(n,k)</math> gives the number of possible rhyming schemes for ''n'' lines using ''k'' unique rhyming syllables. As an example, for a poem of 3 lines, there is 1 rhyme scheme using just one rhyme (aaa), 3 rhyme schemes using two rhymes (aab, aba, abb), and 1 rhyme scheme using three rhymes (abc).
 
==Variants==
 
===Associated Stirling numbers of the second kind===
An ''r''-associated Stirling number of the second kind is the number of ways to partition a set of ''n'' objects into ''k'' subsets, with each subset containing at least ''r'' elements.<ref>L. Comtet, ''Advanced Combinatorics'', Reidel, 1974, p. 222.</ref> It is denoted by <math>S_r(n,k)</math> and obeys the recurrence relation
 
:<math>S_r(n+1, k)=k\ S_r(n, k)+\binom{n}{r-1}S_r(n-r+1, k-1)</math>
 
The 2-associated numbers {{OEIS|A008299}} appear elsewhere as "Ward numbers" and as the magnitudes of the coefficients of [[Mahler polynomial]]s.
 
===Reduced Stirling numbers of the second kind===
Denote the ''n'' objects to partition by the integers 1, 2, ..., ''n''.  Define the reduced Stirling numbers of the second kind, denoted <math>S^d(n, k)</math>, to be the number of ways to partition the integers 1, 2, ..., ''n'' into ''k'' nonempty subsets such that all elements in each subset have pairwise distance at least ''d''.  That is, for any integers ''i'' and ''j'' in a given subset, it is required that <math>|i-j| \geq d</math>. It has been shown that these numbers satisfy
 
:<math>S^d(n, k) = S(n-d+1, k-d+1), n \geq k \geq d</math>
 
(hence the name "reduced").<ref>A. Mohr and T.D. Porter, ''[http://www.austinmohr.com/work/files/stirling.pdf Applications of Chromatic Polynomials Involving Stirling Numbers]'', Journal of Combinatorial Mathematics and Combinatorial Computing '''70''' (2009), 57–64.</ref> Observe (both by definition and by the reduction formula), that <math>S^1(n, k) = S(n, k)</math>, the familiar Stirling numbers of the second kind.
 
==See also==
* [[Bell number]] &ndash; the number of partitions of a set with ''n'' members
* [[Stirling numbers of the first kind]]
* [[Twelvefold way]]
* [[File:Wikiversity-logo-en.svg|20px]] [[v:Partition related number triangles|Partition related number triangles]]
 
==References==
 
<references/>
 
* Khristo N. Boyadzhiev, Close encounters with the Stirling numbers of the second kind (2012) Mathematics Magazine, 85 (4) pp 252–266
* {{planetmath reference |id=2805|title=Stirling numbers of the second kind, S(n,k)}}.
* {{MathWorld |urlname=StirlingNumberoftheSecondKind |title=Stirling Number of the Second Kind}}
* [http://austinmohr.com/home/?page_id=431 Calculator for Stirling Numbers of the Second Kind]
* [http://dlmf.nist.gov/26.8#vii Set Partitions: Stirling Numbers]
* Jack van der Elsen, ''Black and white transformations'', Maastricht 2005, ISBN 90-423-0263-1
 
[[Category:Permutations]]
[[Category:Factorial and binomial topics]]
[[Category:Triangles of numbers]]
 
[[pl:Liczby Stirlinga#Liczby Stirlinga II rodzaju]]

Latest revision as of 01:54, 8 November 2014

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