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Stirling numbers of the second kind

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Numbers parameterizing ways to partition a set

The15 partitions of a 4-element set ordered in aHasse diagram
There areS(4,1), ...,S(4, 4) = 1, 7, 6, 1 partitions containing 1, 2, 3, 4 sets.

Inmathematics, particularly incombinatorics, aStirling number of the second kind (orStirling partition number) is the number of ways topartition a set ofn objects intok non-empty subsets and is denoted by $S(n,k)$ or $\textstyle \left\{{n \atop k}\right\}$ .^[1] Stirling numbers of the second kind occur in the field ofmathematics calledcombinatorics and the study ofpartitions. They are named afterJames Stirling.

The Stirling numbers of thefirst and second kind can be understood as inverses of one another when viewed astriangular matrices. This article is devoted to specifics of Stirling numbers of the second kind. Identities linking the two kinds appear in the article onStirling numbers.

Definition

The Stirling numbers of the second kind, written $S(n,k)$ or $\lbrace \textstyle {n \atop k}\rbrace$ or with other notations, count the number of ways topartition aset of $n {\displaystyle n}$ labelled objects into $k {\displaystyle k}$ nonempty unlabelled subsets. Equivalently, they count the number of differentequivalence relations with precisely $k {\displaystyle k}$ equivalence classes that can be defined on an $n {\displaystyle n}$ element set. In fact, there is abijection between the set of partitions and the set of equivalence relations on a given set. Obviously,

\left\{{n \atop n}\right\}=1

forn ≥ 0, and

\left\{{n \atop 1}\right\}=1

forn ≥ 1,

as the only way to partition ann-element set inton 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. UnlikeStirling numbers of the first kind, they can be calculated using a one-sum formula:^[2]

\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!}}.

(see alsoStirling numbers of the second kind for a proof of the latter formula)

The Stirling numbers of the first kind may be characterized as the numbers that arise when one expresses powers of an indeterminatex in terms of thefalling factorials^[3]

(x)_{n}=x(x-1)(x-2)\cdots (x-n+1).

(In particular, (x)₀ = 1 because it is anempty product.)

Stirling numbers of the second kind satisfy the relation

\sum _{k=0}^{n}\left\{{n \atop k}\right\}(x)_{k}=x^{n}.

Notation

Various notations have been used for Stirling numbers of the second kind. The brace notation ${\textstyle \textstyle \lbrace {n \atop k}\rbrace }$ was used by Imanuel Marx and Antonio Salmeri in 1962 for variants of these numbers.^[4]^[5] This ledKnuth to use it, as shown here, in the first volume ofThe Art of Computer Programming (1968).^[6]^[7] According to the third edition ofThe Art of Computer Programming, this notation was also used earlier byJovan Karamata in 1935.^[8]^[9] The notationS(n,k) was used byRichard Stanley in his bookEnumerative Combinatorics and also, much earlier, by many other writers.^[6]

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 article:Bell number

Since the Stirling number $\left\{{n \atop k}\right\}$ counts set partitions of ann-element set intok parts, the sum

B_{n}=\sum _{k=0}^{n}\left\{{n \atop k}\right\}

over all values ofk is the total number of partitions of a set withn members. This number is known as thenthBell number.

Analogously, theordered Bell numbers can be computed from the Stirling numbers of the second kind via

a_{n}=\sum _{k=0}^{n}k!\left\{{n \atop k}\right\}.

^[10]

Table of values

Below is atriangular array of values for the Stirling numbers of the second kind (sequenceA048993 in theOEIS):

k n	0	1	2	3	4	5	6	7	8	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 thebinomial coefficients, this table could be extended to k >n, but the entries would all be 0.

Properties

Recurrence relation

Stirling numbers of the second kind obey the recurrence relation (first discovered by Masanobu Saka in his 1782Sanpō-Gakkai):^[11]

\left\{{n+1 \atop k}\right\}=k\left\{{n \atop k}\right\}+\left\{{n \atop k-1}\right\}\quad {\mbox{for}}\;0<k<n

with initial conditions

\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{.}}

For instance, the number 25 in columnk = 3 and rown = 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 prove this recurrence, observe that a partition of the⁠ $n+1$ ⁠ objects intok 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

\left\{{n \atop k-1}\right\}

since we must partition the remainingn 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

k\left\{{n \atop k}\right\}

since we partition all objects other than the⁠ $(n+1)$ ⁠-th intok subsets, and then we are left withk choices for inserting object⁠ $n+1$ ⁠. Summing these two values gives the desired result.

Another recurrence relation is given by

$\left\lbrace {\begin{matrix}n\\k\end{matrix}}\right\rbrace ={\frac {k^{n}}{k!}}-\sum _{r=1}^{k-1}{\frac {\left\lbrace {\begin{matrix}n\\r\end{matrix}}\right\rbrace }{(k-r)!}}.$

which follows from evaluating $\sum _{r=0}^{n}\left\{{n \atop r}\right\}(x)_{r}=x^{n}$ at $x=k$ .

It is also conjectured that for a fixed $n {\displaystyle n}$ we have

{\begin{aligned}\left\{{n \atop k}\right\}&={\frac {1}{n-k}}\sum _{j=2}^{n-k+1}(j-2)!{\binom {-k}{j}}\left\{{n \atop k+j-1}\right\},\\\left\{{n \atop n}\right\}&=1.\end{aligned}}

Here we start with recursively computing of $\left\{{n \atop n-1}\right\}$ , then compute $\left\{{n \atop n-2}\right\}$ and so on up to $\left\{{n \atop 1}\right\}$ .

Another conjecture is that for a fixed $k {\displaystyle k}$ we have

{\begin{aligned}\left\{{n \atop k}\right\}&={\frac {1}{n-k}}\sum _{j=2}^{n-k+1}{\binom {n}{j}}\left\{{n-j+1 \atop k}\right\}(-1)^{j},\\\left\{{n \atop n}\right\}&=1.\end{aligned}}

If you swap $(j-2)!$ from the first sum and $(-1)^{j}$ from the second, you will get similar conjectures, but forStirling numbers of the first kind.

Simple identities

Some simple identities include

\left\{{n \atop n-1}\right\}={\binom {n}{2}}.

This is because dividingn elements inton − 1 sets necessarily means dividing it into one set of size 2 andn − 2 sets of size 1. Therefore we need only pick those two elements;

and

\left\{{n \atop 2}\right\}=2^{n-1}-1.

To see this, first note that there are 2ⁿordered pairs of complementary subsetsA andB. In one case,A is empty, and in anotherB is empty, so2ⁿ − 2 ordered pairs of subsets remain. Finally, since we wantunordered pairs rather thanordered 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 ofConcrete Mathematics provides a plethora of generalized forms of finite sums involving the Stirling numbers. Several particular finite sums relevant to this article include

{\begin{aligned}\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{aligned}}

Explicit formula

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

\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!}}.

This can be derived by usinginclusion-exclusion to count the surjections fromn tok and using the fact that the number of such surjections is ${\textstyle k!\left\{{n \atop k}\right\}}$ .

Additionally, this formula is a special case of thekthforward difference of themonomial $x^{n}$ evaluated atx = 0:

\Delta ^{k}x^{n}=\sum _{j=0}^{k}(-1)^{k-j}{k \choose j}(x+j)^{n}.

Because theBernoulli polynomials may be written in terms of these forward differences, one immediately obtains a relation in theBernoulli numbers:

B_{m}(0)=\sum _{k=0}^{m}{\frac {(-1)^{k}k!}{k+1}}\left\{{m \atop k}\right\}.

The evaluation of incomplete exponentialBell polynomialB_n,k(x₁,x₂,...) on the sequence of ones equals a Stirling number of the second kind:

\left\{{n \atop k}\right\}=B_{n,k}(1,1,\dots ,1).

Another explicit formula given in theNIST Handbook of Mathematical Functions is

\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}}

Parity

Parity of Stirling numbers of the second kind.

Theparity of a Stirling number of the second kind is same as the parity of a relatedbinomial coefficient:

\left\{{n \atop k}\right\}\equiv {\binom {z}{w}}\ {\pmod {2}},

where

z=n-\left\lceil \displaystyle {\frac {k+1}{2}}\right\rceil ,\ w=\left\lfloor \displaystyle {\frac {k-1}{2}}\right\rfloor .

This relation is specified by mappingn andk coordinates onto theSierpiński triangle.

More directly, let two sets contain positions of 1's in binary representations of results of respective expressions:

{\begin{aligned}\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{aligned}}

One can mimic abitwise AND operation by intersecting these two sets:

{\begin{Bmatrix}n\\k\end{Bmatrix}}\,{\bmod {\,}}2={\begin{cases}0,&\mathbb {A} \cap \mathbb {B} \neq \emptyset ;\\1,&\mathbb {A} \cap \mathbb {B} =\emptyset ;\end{cases}}

to obtain the parity of a Stirling number of the second kind inO(1) time. Inpseudocode:

{\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];

where $\left[b\right]$ is theIverson bracket.

The parity of a central Stirling number of the second kind $\textstyle \left\{{2n \atop n}\right\}$ is odd if and only if $n {\displaystyle n}$ is afibbinary number, a number whosebinary representation has no two consecutive 1s.^[12]

Generating functions

For a fixed integern, theordinary generating function for Stirling numbers of the second kind $\left\{{n \atop 0}\right\},\left\{{n \atop 1}\right\},\ldots$ is given by

\sum _{k=0}^{n}\left\{{n \atop k}\right\}x^{k}=T_{n}(x),

where $T_{n}(x)$ areTouchard polynomials. If one sums the Stirling numbers against the falling factorial instead, one can show the following identities, among others:

\sum _{k=0}^{n}\left\{{n \atop k}\right\}(x)_{k}=x^{n}

and

\sum _{k=1}^{n+1}\left\{{n+1 \atop k}\right\}(x-1)_{k-1}=x^{n},

which has special case

\sum _{k=0}^{n}\left\{{n \atop k}\right\}(n)_{k}=n^{n}.

For a fixed integerk, the Stirling numbers of the second kind have rational ordinary generating function

\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}}}

and have anexponential generating function given by

\sum _{n=k}^{\infty }\left\{{n \atop k}\right\}{\frac {x^{n}}{n!}}={\frac {(e^{x}-1)^{k}}{k!}}.

A mixed bivariate generating function for the Stirling numbers of the second kind is

\sum _{k=0}^{\infty }\sum _{n=k}^{\infty }\left\{{n \atop k}\right\}{\frac {x^{n}}{n!}}y^{k}=e^{y(e^{x}-1)}.

Lower and upper bounds

If $n\geq 2$ and $1\leq k\leq n-1$ , then

{\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}

^[13]

Asymptotic approximation

For fixed value of $k, {\displaystyle k,}$ the asymptotic value of the Stirling numbers of the second kind as $n\rightarrow \infty$ is given by

\left\{{n \atop k}\right\}{\underset {n\to \infty }{\sim }}{\frac {k^{n}}{k!}}.

If $k=o({\sqrt {n}})$ (whereo denotes thelittle o notation) then

\left\{{n+k \atop n}\right\}{\underset {n\to \infty }{\sim }}{\frac {n^{2k}}{2^{k}k!}}.

^[14]

A uniformly valid approximation also exists: for allk such that1 <k <n, one has

\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),

where $v=n/k$ , and $G\in (0,1)$ is the unique solution to $G=ve^{G-v}$ .^[15] Relative error is bounded by about $0.066/n$ .

Unimodality

For fixed $n {\displaystyle n}$ , $\left\{{n \atop k}\right\}$ is unimodal, that is, the sequence increases and then decreases. The maximum is attained for at most two consecutive values ofk. That is, there is an integer $k_{n}$ such that

\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\}.

Looking at the table of values above, the first few values for $k_{n}$ are $0,1,1,2,2,3,3,4,4,4,5,\ldots$

When $n {\displaystyle n}$ is large

k_{n}{\underset {n\to \infty }{\sim }}{\frac {n}{\log n}},

and the maximum value of the Stirling number can be approximated with

\log \left\{{n \atop k_{n}}\right\}=n\log n-n\log \log n-n+O(n\log \log n/\log n).

^[13]

Applications

Moments of the Poisson distribution

IfX is arandom variable with aPoisson distribution withexpected value λ, then itsn-thmoment is

E(X^{n})=\sum _{k=0}^{n}\left\{{n \atop k}\right\}\lambda ^{k}.

In particular, thenth moment of the Poisson distribution with expected value 1 is precisely the number ofpartitions of a set of sizen, i.e., it is thenthBell number (this fact isDobiński's formula).

Moments of fixed points of random permutations

Let the random variableX be the number of fixed points of auniformly distributed random permutation of a finite set of sizem. Then thenth moment ofX is

E(X^{n})=\sum _{k=0}^{m}\left\{{n \atop k}\right\}.

Note: The upper bound of summation ism, notn.

In other words, thenth moment of thisprobability distribution is the number of partitions of a set of sizen into no more thanm parts.This is proved in the article onrandom permutation statistics, although the notation is a bit different.

Rhyming schemes

The Stirling numbers of the second kind can represent the total number ofrhyme schemes for a poem ofn lines. $S(n,k)$ gives the number of possible rhyming schemes forn lines usingk 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

Ther-Stirling number of the second kind $\left\{{n \atop k}\right\}_{r}$ counts the number of partitions of a set ofn objects intok non-empty disjoint subsets, such that the firstr elements are in distinct subsets.^[16] These numbers satisfy therecurrence relation

\left\{{n \atop k}\right\}_{r}=k\left\{{n-1 \atop k}\right\}_{r}+\left\{{n-1 \atop k-1}\right\}_{r}

Some combinatorial identities and a connection between these numbers and context-free grammars can be found in^[17]

Associated Stirling numbers of the second kind

Anr-associated Stirling number of the second kind is the number of ways to partition a set ofn objects intok subsets, with each subset containing at leastr elements.^[18] It is denoted by $S_{r}(n,k)$ and obeys the recurrence relation

S_{r}(n+1,k)=k\ S_{r}(n,k)+{\binom {n}{r-1}}S_{r}(n-r+1,k-1)

The 2-associated numbers (sequenceA008299 in theOEIS) appear elsewhere as "Ward numbers" and as the magnitudes of the coefficients ofMahler polynomials.

Reduced Stirling numbers of the second kind

Denote then objects to partition by the integers 1, 2, ...,n. Define the reduced Stirling numbers of the second kind, denoted $S^{d}(n,k)$ , to be the number of ways to partition the integers 1, 2, ...,n intok nonempty subsets such that all elements in each subset have pairwise distance at leastd. That is, for any integersi andj in a given subset, it is required that $|i-j|\geq d$ . It has been shown that these numbers satisfy

S^{d}(n,k)=S(n-d+1,k-d+1),n\geq k\geq d

(hence the name "reduced").^[19] Observe (both by definition and by the reduction formula), that $S^{1}(n,k)=S(n,k)$ , the familiar Stirling numbers of the second kind.

See also

Stirling number
Stirling numbers of the first kind
Bell number – the number of partitions of a set withn members
Stirling polynomials
Twelvefold way
Learning materials related toPartition related number triangles at Wikiversity

References

^Ronald L. Graham, Donald E. Knuth, Oren Patashnik (1988)Concrete Mathematics, Addison–Wesley, Reading MA.ISBN 0-201-14236-8, p. 244.
^"Stirling Numbers of the Second Kind, Theorem 3.4.1".
^Confusingly, the notation that combinatorialists use forfalling factorials coincides with the notation used inspecial functions forrising factorials; seePochhammer symbol.
^Transformation of Series by a Variant of Stirling's Numbers, Imanuel Marx,The American Mathematical Monthly69, #6 (June–July 1962), pp. 530–532,JSTOR 2311194.
^Antonio Salmeri, Introduzione alla teoria dei coefficienti fattoriali,Giornale di Matematiche di Battaglini90 (1962), pp. 44–54.
^^a ^bKnuth, D.E. (1992), "Two notes on notation",Amer. Math. Monthly,99 (5):403–422,arXiv:math/9205211,Bibcode:1992math......5211K,doi:10.2307/2325085,JSTOR 2325085,S2CID 119584305
^Donald E. Knuth,Fundamental Algorithms, Reading, Mass.: Addison–Wesley, 1968.
^p. 66, Donald E. Knuth,Fundamental Algorithms, 3rd ed., Reading, Mass.: Addison–Wesley, 1997.
^Jovan Karamata, Théorèmes sur la sommabilité exponentielle et d'autres sommabilités s'y rattachant,Mathematica (Cluj)9 (1935), pp, 164–178.
^Sprugnoli, Renzo (1994),"Riordan arrays and combinatorial sums"(PDF),Discrete Mathematics,132 (1–3):267–290,doi:10.1016/0012-365X(92)00570-H,MR 1297386
^Wilson, R., & Watkins, J. J., ed. (2013).Combinatorics: Ancient & Modern. Oxford University Press. p. 26.ISBN 978-0-19-965659-2.{{cite book}}: CS1 maint: multiple names: editors list (link)
^Chan, O-Yeat; Manna, Dante (2010),"Congruences for Stirling numbers of the second kind"(PDF),Gems in Experimental Mathematics, Contemporary Mathematics, vol. 517, Providence, Rhode Island: American Mathematical Society, pp. 97–111,doi:10.1090/conm/517/10135,ISBN 978-0-8218-4869-2,MR 2731094
^^a ^bRennie, B.C.; Dobson, A.J. (1969)."On stirling numbers of the second kind".Journal of Combinatorial Theory.7 (2):116–121.doi:10.1016/S0021-9800(69)80045-1.ISSN 0021-9800.
^L. C. Hsu, Note on an Asymptotic Expansion of the nth Difference of Zero, AMS Vol.19 NO.2 1948, pp. 273--277
^N. M. Temme, Asymptotic Estimates of Stirling Numbers, STUDIES IN APPLIED MATHEMATICS 89:233-243 (1993), Elsevier Science Publishing.
^Broder, A. (1984). The r-Stirling numbers. Discrete Mathematics 49, 241-259
^Triana, J. (2022). r-Stirling numbers of the second kind through context-free grammars. Journal of automata, languages and combinatorics 27(4), 323-333
^L. Comtet,Advanced Combinatorics, Reidel, 1974, p. 222.
^A. Mohr and T.D. Porter,Applications of Chromatic Polynomials Involving Stirling Numbers, Journal of Combinatorial Mathematics and Combinatorial Computing70 (2009), 57–64.

Boyadzhiev, Khristo (2012). "Close encounters with the Stirling numbers of the second kind".Mathematics Magazine.85 (4):252–266.arXiv:1806.09468.doi:10.4169/math.mag.85.4.252.S2CID 115176876..

"Stirling numbers of the second kind".PlanetMath..
Weisstein, Eric W."Stirling Number of the Second Kind".MathWorld.
Calculator for Stirling Numbers of the Second Kind
Set Partitions: Stirling Numbers
Jack van der Elsen (2005).Black and white transformations. Maastricht.ISBN 90-423-0263-1.

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