View source for Stirling numbers of the second kind

{{Use American English|date = March 2019}}
{{Short description|Numbers parameterizing ways to partition a set}}
[[File:Set partitions 4; Hasse; circles.svg|thumb|right|300px|The [[Bell number|15]] partitions of a 4-element set ordered in a [[Hasse diagram]] {{paragraph}}There are ''S''(4,1), ..., ''S''(4, 4) = 1, 7, 6, 1 partitions containing 1, 2, 3, 4 sets.]]
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 \left\{{n\atop k}\right\}</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]]. They are named after [[James Stirling (mathematician)|James Stirling]].

The Stirling numbers of the [[Stirling numbers of the first kind|first]] and second kind can be understood as inverses of one another when viewed as [[triangular matrix|triangular matrices]]. This article is devoted to specifics of Stirling numbers of the second kind. Identities linking the two kinds appear in the article on [[Stirling number|Stirling numbers]].

==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 of a set|partition]] a [[Set (mathematics)|set]] of <math>n</math> labelled objects into <math>k</math> nonempty unlabelled subsets. Equivalently, they count the number of different [[equivalence relation]]s with precisely <math>k</math> equivalence classes that can be defined on an <math>n</math> element set. In fact, there is a [[bijection]] between the set of partitions and the set of equivalence relations on a given set. Obviously,

:<math>\left\{ {n \atop n} \right\} = 1</math> for ''n'' ≥ 0,  and  <math>\left\{ {n \atop 1}\right\} = 1</math>  for ''n'' ≥ 1,

as the only way to partition an ''n''-element set into ''n'' parts is to put each element of the set into its own part, and the only way to partition a nonempty set into one part is to put all of the elements in the same part. Unlike [[Stirling numbers of the first kind]], they can be calculated using a one-sum formula:<ref>{{Cite web|url=https://discrete.openmathbooks.org/more/mdm/sec_adv-stirling.html|title=Stirling Numbers of the Second Kind, Theorem 3.4.1|last=|first=|date=|website=|publisher=|access-date=}}</ref>
:<math>\left\{ {n \atop k}\right\} = \frac{1}{k!}\sum_{i=0}^k (-1)^{k-i} \binom{k}{i} i^n = \sum_{i=0}^k \frac{(-1)^{k-i} i^n}{(k-i)!i!}.</math>

The Stirling numbers of the first kind may be characterized as the numbers that arise when one expresses powers of an indeterminate ''x'' in terms of the [[falling factorial]]s<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>

:<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]].)

Stirling numbers of the second kind satisfy the relation

:<math>\sum_{k=0}^n \left\{ {n \atop k} \right\}(x)_k=x^n.</math>

==Notation==
Various notations have been used for Stirling numbers of the second kind.  The brace notation <math display="inline">\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>{{citation |author-link=Donald Knuth|first=D.E.|last=Knuth| journal = Amer. Math. Monthly | title=Two notes on notation| year=1992 | volume=99 |issue=5| pages=403–422 | arxiv=math/9205211 | doi=10.2307/2325085 | jstor= 2325085|bibcode=1992math......5211K|s2cid=119584305 }}</ref><ref>Donald E. Knuth, ''Fundamental Algorithms'', Reading, Mass.: Addison–Wesley, 1968.</ref>  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]]'' and also, much earlier, by many other writers.<ref name=tnn />

The notations used on this page for Stirling numbers are not universal, and may conflict with notations in other sources.

==Relation to Bell numbers==
{{main|Bell number}}

Since the Stirling number <math>\left\{ {n \atop k} \right\}</math> counts set partitions of an ''n''-element set into ''k'' parts, the sum 

:<math>B_n=\sum_{k=0}^n \left\{ {n \atop k} \right\}</math>

over all values of ''k'' is the total number of partitions of a set with ''n'' members.  This number is known as the ''n''th [[Bell numbers|Bell number]]. 

Analogously, the [[ordered Bell number]]s can be computed from the Stirling numbers of the second kind via
:<math>a_n = \sum_{k=0}^n k! \left\{ {n \atop k}\right\}.</math><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| url = http://dml.cz/bitstream/handle/10338.dmlcz/143149/CommentatMathUnivCarolRetro_54-2013-1_2.pdf
 | doi-access = free
 }}</ref>

==Table of values==
Below is a [[triangular array]] of values for the Stirling numbers of the second kind {{OEIS|A008277}}:
{| style="text-align:right;" class="wikitable"
|-
! {{diagonal split header|''n''|''k''}}
! {{rh|align=right}} | 0
! {{rh|align=right}} | 1
! {{rh|align=right}} | 2
! {{rh|align=right}} | 3
! {{rh|align=right}} | 4
! {{rh|align=right}} | 5
! {{rh|align=right}} | 6
! {{rh|align=right}} | 7
! {{rh|align=right}} | 8
! {{rh|align=right}} | 9
! {{rh|align=right}} | 10
|-
! {{rh|align=right}} | 0
| 1
|-
! {{rh|align=right}} | 1
| 0
| 1
|-
! {{rh|align=right}} | 2
| 0
| 1
| 1
|-
! {{rh|align=right}} | 3
| 0
| 1
| 3
| 1
|-
! {{rh|align=right}} | 4
| 0
| 1
| 7
| 6
| 1
|-
! {{rh|align=right}} | 5
| 0
| 1
| 15
| 25
| 10
| 1
|-
! {{rh|align=right}} | 6
| 0
| 1
| 31
| 90
| 65
| 15
| 1
|-
! {{rh|align=right}} | 7
| 0
| 1
| 63
| 301
| 350
| 140
| 21
| 1
|-
! {{rh|align=right}} | 8
| 0
| 1
| 127
| 966
| 1701
| 1050
| 266
| 28
| 1
|-
! {{rh|align=right}} | 9
| 0
| 1
| 255
| 3025
| 7770
| 6951
| 2646
| 462
| 36
| 1
|-
! {{rh|align=right}} | 10
| 0
| 1
| 511
| 9330
| 34105
| 42525
| 22827
| 5880
| 750
| 45
| 1
|}

As with the [[binomial coefficients]], this table could be extended to&nbsp;{{math|''k'' > ''n''}}, but the 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\}
\quad \mbox{for} \; 0<k<n </math>
with initial conditions
:<math>\left\{{ n \atop n }\right\} = 1 \quad  \mbox{ for} \; n \geq 0 \quad \text{ and } \quad
\left\{{ n \atop 0 }\right\} = \left\{{ 0 \atop n }\right\} = 0 \quad \text{ for } n>0 \text{.} </math>

For instance, the number 25 in column ''k''&nbsp;=&nbsp;3 and row ''n''&nbsp;=&nbsp;5 is given by 25&nbsp;=&nbsp;7&nbsp;+&nbsp;(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&nbsp;6.

To prove this recurrence, observe that a partition of the {{tmath|n+1}} objects into ''k'' nonempty subsets either contains the {{tmath|(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 {{mvar|n}} objects into the available {{tmath|k-1}} subsets. In the other case the {{tmath|(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 {{tmath|(n+1)}}-th into ''k'' subsets, and then we are left with ''k'' choices for inserting object {{tmath|n+1}}. Summing these two values gives the desired result.

Another recurrence relation is given by

<math> \left\lbrace\begin{matrix} n \\ k \end{matrix}\right\rbrace =
\frac{k^n}{k!}-\sum_{r=1}^{k-1} \frac{\left\lbrace\begin{matrix} n \\ r\end{matrix}\right\rbrace}{(k-r)!}.
</math>

which follows from evaluating <math>\sum_{r=0}^n \left\{ {n \atop r} \right\}(x)_r=x^n </math> at <math>x=k</math>.

===Simple identities===
Some simple identities include

:<math>\left\{ {n \atop n-1}\right\} = \binom{n}{2}. </math>

This is because dividing ''n'' elements into {{math|''n''&nbsp;&minus;&nbsp;1}} sets necessarily means dividing it into one set of size 2 and {{math|''n''&nbsp;&minus;&nbsp;2}} sets of size 1.  Therefore we need only pick those two elements;

and

:<math>\left\{ {n \atop 2}\right\} = 2^{n-1}-1.</math>

To see this, first note that there are 2{{sup|''n''}} ''ordered'' pairs of complementary subsets ''A'' and ''B''.  In one case, ''A'' is empty, and in another ''B'' is empty, so {{math|2{{sup|''n''}}&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 expansion of the recurrence-relation gives identities in the spirit of the above example.

=== Identities ===
The table in section 6.1 of ''Concrete Mathematics'' provides a plethora of generalized forms of finite sums involving the Stirling numbers. Several particular finite sums relevant to this article include

:<math>
\begin{align}
\left\{{n+1\atop k+1}\right\} &= \sum_{j=k}^n {n \choose j} \left\{{ j \atop k }\right\}\\
\left\{{n+1\atop k+1}\right\} &= \sum_{j=k}^n (k+1)^{n-j} \left\{{j \atop k}\right\}\\
\left\{{n+k+1 \atop k}\right\} &= \sum_{j=0}^k j \left\{{ n+j \atop j }\right\} \\
\left\{{n \atop \ell+m } \right\} \binom{\ell+m}{\ell} &= \sum_k \left\{{k \atop \ell} \right\} \left\{{n-k \atop m } \right\} \binom{n}{k}
\end{align}
</math>

===Explicit formula===
The Stirling numbers of the second kind are given by the explicit formula:

:<math>\left\{ {n \atop k} \right\}
=\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}{k \choose j} j^n
=\sum_{j=0}^k \frac{(-1)^{k-j} j^n}{(k-j)!j!}
.</math>

This can be derived by using [[Inclusion–exclusion principle|inclusion-exclusion]] to count the surjections from ''n'' to ''k'' and using the fact that the number of such surjections is <math display="inline"> k! \left\{ {n \atop k} \right\} </math>.

Additionally, 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\{ {m \atop k} \right\}. </math>

The evaluation of incomplete exponential [[Bell polynomial]] ''B''<sub>''n'',''k''</sub>(''x''<sub>1</sub>,''x''<sub>2</sub>,...) on the sequence of ones equals a Stirling number of the second kind:

:<math>\left\{{n\atop k}\right\} = B_{n,k}(1,1,\dots,1).</math>

Another explicit formula given in the ''NIST Handbook of Mathematical Functions'' is

:<math>
\left\{ {n \atop k} \right\} = \sum_{
\begin{array}{c}
c_1 + \ldots + c_k = n-k\\
c_1, \ldots,\ c_k\ \geq\ 0
\end{array}
} 1^{c_1} 2^{c_2} \cdots k^{c_k} 
</math>

===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 same as the parity of a related [[binomial coefficient]]:
:<math>\left\{ {n\atop k}\right\}\equiv \binom{z}{w}\ \pmod{2},</math> where <math>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]].

More 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>

One can 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[\left( \left(n-k\right)\ \And\ \left( \left(k-1\right)\,\mathrm{div}\,2 \right)\right) = 0\right];
</math>
where <math>
\left[b\right]
</math> is the [[Iverson bracket]].

The parity of a central Stirling number of the second kind <math>\textstyle \left\{{2n\atop n}\right\}</math> is odd if and only if <math>n</math> is a [[fibbinary number]], a number whose [[binary representation]] has no two consecutive 1s.<ref>{{citation
 | last1 = Chan | first1 = O-Yeat
 | last2 = Manna | first2 = Dante
 | contribution = Congruences for Stirling numbers of the second kind
 | contribution-url = http://www.oyeat.com/papers/stirling_final.pdf
 | doi = 10.1090/conm/517/10135
 | mr = 2731094
 | pages = 97–111
 | publisher = American Mathematical Society | location = Providence, Rhode Island
 | series = Contemporary Mathematics
 | title = Gems in Experimental Mathematics
 | volume = 517
 | year = 2010}}</ref>

===Generating functions===
For a fixed integer ''n'', the [[ordinary generating function]] for Stirling numbers of the second kind <math>\left\{ {n\atop 0} \right\}, \left\{ {n\atop 1} \right\}, \ldots</math> is given by
:<math> \sum_{k=0}^n \left\{ {n\atop k} \right\} x^k = T_n(x),</math>
where <math>T_n(x)</math> are [[Touchard polynomials]]. If one sums the Stirling numbers against the falling factorial instead, one can show the following identities, among others:
:<math> \sum_{k=0}^n \left\{ {n\atop k} \right\} (x)_k = x^n</math>
and
:<math> \sum_{k=1}^{n+1} \left\{ {n+1 \atop k} \right\} (x-1)_{k-1} = x^n,</math>
which has special case
:<math> \sum_{k=0}^n \left\{ {n\atop k} \right\} (n)_k = n^n.</math>

For a fixed integer ''k'', the Stirling numbers of the second kind have rational ordinary generating function
:<math>\sum_{n=k}^\infty \left\{ {n\atop k} \right\} x^{n - k} = \prod_{r=1}^k \frac{1}{1-rx} = \frac{1}{x^{k + 1} (1/x)_{k+1}}</math>
and have an [[exponential generating function]] given by
:<math> \sum_{n=k}^\infty \left\{ {n \atop k}\right\} \frac{x^n}{n!} = \frac{(e^x-1)^k}{k!}.</math>

A mixed bivariate generating function for the Stirling numbers of the second kind is
:<math> \sum_{k=0}^{\infty} \sum_{n=k}^{\infty} \left\{ {n \atop k} \right\} \frac{x^n}{n!} y^k = e^{y(e^x-1)}.</math>

===Lower and upper bounds===
If <math>n \geq 2</math> and <math>1 \leq k \leq n-1</math>, then

:<math>\frac{1}{2}(k^2+k+2)k^{n-k-1}-1 \leq \left\{{n \atop k}\right\} \leq \frac{1}{2}{n \choose k} k^{n-k} </math> <ref name="RennieDobson1969">{{cite journal |last1=Rennie |first1=B.C. |last2=Dobson |first2=A.J. |year=1969 |title=On stirling numbers of the second kind |journal=[[Journal of Combinatorial Theory]] |volume=7 |issue=2 |pages=116–121 |doi=10.1016/S0021-9800(69)80045-1 |issn=0021-9800 |doi-access=free}}</ref>

===Asymptotic approximation===
For fixed value of <math>k,</math> the asymptotic value of the Stirling numbers of the second kind as <math> n\rightarrow \infty </math> is given by
:<math> \left\{{n \atop k}\right\} \underset{n \to \infty}{\sim} \frac{k^n}{k!}.</math>

If <math> k = o(\sqrt{n}) </math> (where ''o'' denotes the [[little o notation]]) then
:<math> \left\{{n+k \atop n}\right\} \underset{n \to \infty}{\sim} \frac{n^{2k}}{2^k k!}.</math><ref>[[Leetsch C. Hsu|L. C. Hsu]], Note on an Asymptotic Expansion of the nth Difference of Zero, AMS Vol.19 NO.2 1948, pp. 273--277</ref>

A uniformly valid approximation also exists: for all {{mvar|k}} such that {{math|1 < ''k'' < ''n''}}, one has

:<math> \left\{{n \atop k}\right\} \sim \sqrt{\frac{v-1}{v(1-G)}} \left(\frac{v-1}{v-G}\right)^{n-k} \frac{k^n}{n^k} e^{k(1-G)} \left({n \atop k}\right),</math>

where <math > v=n/k </math>, and <math> G\in (0,1) </math> is the unique solution to <math>   G = v e^{G-v} </math>.<ref>N. M. Temme, Asymptotic Estimates of Stirling Numbers, STUDIES IN APPLIED MATHEMATICS 89:233-243 (1993), Elsevier Science Publishing.</ref> Relative error is bounded by about <math> 0.066/n </math>.

===Unimodality===
For fixed <math>n</math>, <math>\left\{{n \atop k}\right\}</math> is unimodal, that is, the sequence increases and then decreases. The maximum is attained for at most two consecutive values of ''k''. That is, there is an integer <math>k_n</math> such that

:<math>\left\{{n \atop 1}\right\} < \left\{{n \atop 2}\right\} < \cdots < \left\{{n \atop k_n}\right\} \geq \left\{{n \atop k_n+1}\right\} > \cdots > \left\{{n \atop n}\right\}.</math>

Looking at the table of values above, the first few values for <math>k_n</math> are <math>0, 1, 1, 2, 2, 3, 3, 4, 4, 4, 5, \ldots</math>

When <math>n</math> is large

:<math>k_n \underset{n\to\infty}{\sim} \frac{n}{\log n},</math>

and the maximum value of the Stirling number can be approximated with
:<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="RennieDobson1969" />

==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=0}^n \left\{ {n \atop k} \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 [[Dobiński'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=0}^m \left\{ {n \atop k} \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==

===''r''-Stirling numbers of the second kind===
The ''r''-Stirling number of the second kind <math>\left\{{n\atop k}\right\}_r </math> counts the number of partitions of a set of ''n'' objects into ''k'' non-empty disjoint subsets, such that the first ''r'' elements are in distinct subsets.<ref>Broder, A. (1984). The r-Stirling numbers. Discrete Mathematics 49, 241-259</ref> These numbers satisfy the recurrence relation

:<math>\left\{{n\atop k}\right\}_r = k\left\{{n-1\atop k}\right\}_r + \left\{{n-1\atop k-1}\right\}_r</math>

Some combinatorial identities and a connection between these numbers and context-free grammars can be found in <ref>Triana, J. (2022). r-Stirling numbers of the second kind through context-free grammars. Journal of automata, languages and combinatorics 27(4), 323-333</ref>

===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==
* [[Stirling number]]
* [[Stirling numbers of the first kind]]
* [[Bell number]] &ndash; the number of partitions of a set with ''n'' members
* [[Stirling polynomials]]
* [[Twelvefold way]]
* {{Wikiversity-inline|Partition related number triangles}}

==References==

{{reflist|colwidth=30em}}

* {{cite journal|last=Boyadzhiev|first=Khristo|title=Close encounters with the Stirling numbers of the second kind|year=2012|journal=Mathematics Magazine|volume=85|number=4|pages=252–266|doi=10.4169/math.mag.85.4.252|arxiv=1806.09468|s2cid=115176876 }}.

* {{planetmath reference |urlname=StirlingNumbersOfTheSecondKind|title=Stirling numbers of the second kind}}.
* {{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]
* {{cite book |author=Jack van der Elsen |title=Black and white transformations |publisher=Maastricht |year=2005 |isbn=90-423-0263-1}}

{{Classes of natural numbers}}

[[Category:Permutations]]
[[Category:Factorial and binomial topics]]
[[Category:Triangles of numbers]]
[[Category:Operations on numbers]]

[[pl:Liczby Stirlinga#Liczby Stirlinga II rodzaju]]