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In [[mathematics]], especially in [[combinatorics]], '''Stirling numbers of the first kind''' arise in the study of permutations.  In particular, the Stirling numbers of the first kind count [[permutation]]s according to their number of [[Cycles and fixed points|cycles]] (counting fixed points as cycles of length one).

(The Stirling numbers of the first and [[Stirling numbers of the second kind|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 first kind. Identities linking the two kinds appear in the article on [[Stirling numbers]] in general.)

==Definitions==
The original definition of Stirling numbers of the first kind was algebraic.  These numbers, usually written ''s''(''n'',&nbsp;''k''), are signed integers whose sign, positive or negative, depends on the [[parity (mathematics)|parity]] of ''n'' &minus; ''k''.  Afterwards, the absolute values of these numbers, |''s''(''n'',&nbsp;''k'')|, which are known as unsigned Stirling numbers of the first kind, were found to count permutations, so in combinatorics the (signed) Stirling numbers of the first kind, ''s''(''n'',&nbsp;''k''), are often defined as counting numbers multiplied by a sign factor.  That is the approach taken on this page.

Most identities on this page are stated for unsigned Stirling numbers.  Note that the notations on this page are not universal.

=== Unsigned Stirling numbers of the first kind ===
[[Image:Stirling number of the first kind s(4,2).svg|350px|thumb|s(4,2)=11]]
The unsigned Stirling numbers of the first kind are denoted in various ways by different authors.  Common notations are <math>c(n,k),</math> <math>|s(n,k)|</math> and <math>\left[{n\atop k}\right]</math>.  (The last is also common notation for the [[Gaussian binomial coefficient|Gaussian coefficient]]s.)  They count the number of [[permutation]]s of ''n'' elements with ''k'' disjoint [[cyclic permutation|cycles]].  For example, of the <math>3! = 6</math> permutations of three elements, there is one permutation with three cycles (the [[identity permutation]], given in [[Permutation#Notation|one-line notation]] by <math>123</math> or in [[cycle notation]] by <math>(1)(2)(3)</math>), three permutations with two cycles (<math>132 = (1)(23)</math>, <math>213 = (3)(12)</math>, and <math>321 = (2)(13)</math>) and two permutations with one cycle (<math>312 = (132)</math> and <math>231 = (123)</math>).  Thus, <math>\left[{3\atop 3}\right] = 1</math>, <math>\left[{3\atop 2}\right] = 3</math> and <math>\left[{3\atop 1}\right] = 2</math>.  

As a second example, the image at right shows that <math>\left[{4\atop 2}\right] = 11</math>: the [[symmetric group]] on 4 objects has 3 permutations of the form

:<math> (\bullet\bullet)(\bullet\bullet)</math>—2 orbits, each of size 2,

and 8 permutations of the form

:<math> (\bullet\bullet\bullet)(\bullet)</math>—1 orbit of size 3 and 1 orbit of size 1.

The unsigned Stirling numbers also arise as coefficients of the [[rising factorial]], i.e.,
:<math> (x)^{(n)} = x(x+1)\cdots(x+n-1)=\sum_{k=0}^n \left[{n\atop k}\right] x^k</math>.
Thus, for example, <math>(x)^{(3)} = x(x + 1)(x + 2) = 1 \cdot x^3 + 3 \cdot x^2 + 2 \cdot x</math>, which matches the computations in the preceding paragraph.

=== Stirling numbers of the first kind ===
Stirling numbers of the first kind (sometimes with the qualifying adjective ''signed'') are given by

:<math>s(n,k) = (-1)^{n-k}  \left[{n \atop k}\right] .</math>

They are the coefficients in the expansion

:<math>(x)_n = \sum_{k=0}^n s(n,k) x^k,</math>

where <math>(x)_n</math> is the [[falling factorial]]

:<math>(x)_n = x(x-1)(x-2)\cdots(x-n+1).</math>

Note that

:<math> \left[{n \atop k}\right] = \left|s(n,k)\right| .</math>

==Recurrence relation==
The unsigned Stirling numbers of the first kind can be calculated by the [[recurrence relation]]

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

for <math>k > 0</math>, with the 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.

It follows immediately that the (signed) Stirling numbers of the first kind satisfy the recurrence

: <math> s(n + 1, k) = -n s(n, k) + s(n, k - 1)</math>.

===Algebraic proof===
We prove the recurrence relation using the definition of Stirling numbers in terms of rising factorials.  Distributing the last term of the product, we have

:<math>(x)^{(n+1)} = x(x+1)\cdots (x+n-1)(x+n)=n(x)^{(n)}+x(x)^{(n)}.</math>

The coefficient of ''x''<sup>''k''</sup> on the left-hand side of this equation is <math>\left[{n+1\atop k}\right]</math>.  The coefficient of ''x''<sup>''k''</sup> in ''n''(''x'')<sup>(''n'')</sup> is <math>n \cdot \left[{n\atop k}\right]</math>, while the coefficient of ''x''<sup>''k''</sup> in ''x'' (''x'')<sup>(''n'')</sup> is <math>\left[{n\atop k - 1}\right]</math>.  Since the two sides are equal as polynomials, the coefficients of ''x''<sup>''k''</sup> on both sides must be equal, and the result follows.

===Combinatorial proof===
We prove the recurrence relation using the definition of Stirling numbers in terms of permutations with a given number of cycles (or equivalently, [[orbit (group theory)|orbits]]).

Consider forming a permutation of ''n''&nbsp;+&nbsp;1 objects from a permutation of ''n'' objects by adding a distinguished object. There are exactly two ways in which this can be accomplished. We could do this by forming a [[singleton (mathematics)|singleton]] cycle, i.e., leaving the extra object alone. This increases the number of cycles by 1 and so accounts for the <math>\left[{n\atop k-1}\right]</math> term in the recurrence formula. We could also insert the new object into one of the existing cycles. Consider an arbitrary permutation of ''n'' objects with ''k'' cycles, and [[vertex weighted digraph|label]] the objects ''a''<sub>1</sub>,&nbsp;...,&nbsp;''a''<sub>''n''</sub>, so that the permutation is represented by

:<math>\displaystyle\underbrace{(a_1 \ldots a_{j_1})(a_{j_1+1} \ldots a_{j_2})\ldots(a_{j_{k-1}+1} \ldots a_n)}_{ k\ \mathrm{cycles}}.</math>

To form a new permutation of ''n''&nbsp;+&nbsp;1 objects and ''k'' cycles one must insert the new object into this array. There are ''n'' ways to perform this insertion, inserting the new object immediately following any of the ''n'' already present. This explains the <math>n \left[{n\atop k}\right]</math> term of the recurrence relation.  These two cases include all possibilities, so the recurrence relation follows.

==Table of values for small ''n'' and ''k''==

Below is a [[triangular array]] of unsigned values for the Stirling numbers of the first kind, similar in form to [[Pascal's triangle]].  These values are easy to generate using the recurrence relation in the previous section.

{| cellspacing="0" cellpadding="3" style="text-align:right;"
|-
| '''n'''&nbsp;\&nbsp;''k''
| style="width:4%;"| ''0''
| style="width:10%;"| ''1''
| style="width:11%;"| ''2''
| style="width:12%;"| ''3''
| style="width:13%;"| ''4''
| style="width:12%;"| ''5''
| style="width:11%;"| ''6''
| style="width:10%;"| ''7''
| style="width:9%;"| ''8''
| style="width:4%;"| ''9''
|-
|'''0'''
| 1
|-
|'''1'''
| 0
| 1
|-
|'''2'''
| 0
| 1
| 1
|-
|'''3'''
| 0
| 2
| 3
| 1
|-
|'''4'''
| 0
| 6
| 11
| 6
| 1
|-
|'''5'''
| 0
| 24
| 50
| 35
| 10
| 1
|-
|'''6'''
| 0
| 120
| 274
| 225
| 85
| 15
| 1
|-
|'''7'''
| 0
| 720
| 1764
| 1624
| 735
| 175
| 21
| 1
|-
|'''8'''
| 0
| 5040
| 13068
| 13132
| 6769
| 1960
| 322
| 28
| 1
|-
|'''9'''
| 0
| 40320
| 109584
| 118124
| 67284
| 22449
| 4536
| 546
| 36
| 1
|-
 |}

==Identities involving Stirling numbers of the first kind==
===Simple identities===
Note that although

:<math>\left[{0 \atop 0}\right] = 1</math>, we have <math>\left[{n\atop 0}\right] = 0</math> if ''n'' > 0

and

:<math>\left[{0\atop k}\right] = 0</math> if ''k'' > 0, or more generally <math>\left[{n\atop k}\right] = 0</math> if ''k'' > ''n''.

Also

:<math>\left[{n \atop 1}\right] = (n-1)!, 
\quad 
\left[{n\atop n}\right] = 1,
\quad
\left[{n\atop n-1}\right] = {n \choose 2},</math>

and

:<math>\left[{n\atop n-2}\right] = \frac{1}{4} (3n-1) {n \choose 3}\quad\mbox{ and }\quad\left[{n\atop n-3}\right] = {n \choose 2} {n \choose 4}.</math>

Similar relationships involving the Stirling numbers hold for the [[Bernoulli polynomials]].  Many relations for the Stirling numbers shadow similar relations on the [[binomial coefficient]]s.  The study of these 'shadow relationships' is termed [[umbral calculus]] and culminates in the theory of [[Sheffer sequences]].

====Combinatorial proofs====
These identities may be derived by enumerating permutations directly.
For example, how many permutations on [''n''] are there that consist of ''n''&nbsp;&minus;&nbsp;3 cycles?
There are three possibilities:
* ''n''&nbsp;&minus;&nbsp;6 fixed points and three two-cycles
* ''n''&nbsp;&minus;&nbsp;5 fixed points, a three-cycle and a two-cycle, and
* ''n''&nbsp;&minus;&nbsp;4 fixed points and a four-cycle.

We enumerate the three types, as follows:

* choose the six elements that go into the two-cycles, decompose them into two-cycles and take into account that the order of the cycles is not important:
::<math>{n \choose 6} {6 \choose 2, 2, 2} \frac{1}{6}</math>
* choose the five elements that go into the three-cycle and the two-cycle, choose the elements of the three-cycle and take into account that three elements generate two three-cycles:
::<math>{n \choose 5} {5 \choose 3} \times 2</math>
* choose the four elements of the four-cycle and take into account that four elements generate six four-cycles:
::<math>{n \choose 4} \times 6.</math>

Sum the three contributions to obtain
:<math>
{n \choose 6} {6 \choose 2, 2, 2} \frac{1}{6} +
{n \choose 5} {5 \choose 3} \times 2 +
{n \choose 4} \times 6 =
{n \choose 2} {n \choose 4}.</math>

===Other relations===
Other relations involving Stirling numbers of the first kind include

:<math>\left[{n\atop 2}\right] = (n-1)!\; H_{n-1},</math>

where ''H''<sub>''n''</sub> is a [[harmonic number]], and

:<math>\left[{n\atop 3}\right] = \frac{1}{2} (n-1)! \left[ (H_{n-1})^2 - H_{n-1}^{(2)} \right]</math>

:<math>\left[{n\atop 4}\right] = \frac{1}{3!}(n-1)! \left[ (H_{n-1})^3 - 3H_{n-1}H_{n-1}^{(2)}+2H_{n-1}^{(3)} \right]</math>

where ''H''<sub>''n''</sub><sup>(''m'')</sup> is a [[harmonic number#Introduction|generalized harmonic number]]. For a generalization of this relation, see below.

===Generating function===
A variety of identities may be derived by manipulating the [[generating function]]:

:<math>H(z,u)= (1+z)^u = \sum_{n=0}^\infty {u \choose n} z^n = \sum_{n=0}^\infty \frac{z^n}{n!} \sum_{k=0}^n s(n,k) u^k
= \sum_{k=0}^\infty u^k \sum_{n=k}^\infty \frac {z^n}{n!} s(n,k).</math> 

Using the equality

:<math>(1+z)^u = e^{u\log(1+z)} = \sum_{k = 0}^\infty (\log(1 + z))^k \frac{u^k}{k!},</math> 

it follows that 

:<math>\sum_{n=k}^\infty (-1)^{n-k} \left[{n\atop k}\right] \frac{z^n}{n!} = \frac{\left(\log (1+z)\right)^k}{k!}.</math>

(This identity is valid for [[formal power series]], and the sum [[convergent series|converges]] in the [[complex plane]] for |''z''| < 1.) Other identities arise by exchanging the order of summation, taking derivatives, making substitutions for ''z'' or ''u'', etc.  For example, we may derive:

:<math>(1-z)^{-u}
= \sum_{k=0}^\infty u^k \sum_{n=k}^\infty \frac {z^n}{n!} \left[{n\atop k}\right] = e^{u\log(1/(1-z))}</math>

and

:<math>\sum_{n=i}^\infty \frac{\left[{n\atop i}\right]}{n (n!)}  = \zeta(i+1) </math>
where <math>\zeta(k)</math> is the [[Riemann zeta function]].

===Finite sums===
A simple sum is

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

or in a more general relationship,

:<math>\sum_{k=0}^a \left[{n\atop k}\right] = n! - \sum_{k=0}^n \left[{n\atop k+a+1}\right].</math>

The identity
:<math>\sum_{p=k}^{n} {\left[{n\atop p}\right]\binom{p}{k}} = \left[{n+1\atop k+1}\right]</math>
can be proved by the techniques on the page
[[Stirling numbers and exponential generating functions]].

===Explicit formula===
The Stirling number ''s(n,n-p)'' can be found from the formula<ref> [http://arxiv.org/abs/1103.1585/ J. Malenfant, "Finite, Closed-form Expressions for the Partition Function and for Euler, Bernoulli, and Stirling Numbers"]</ref>
:<math>
\begin{align}
        s(n,n-p)  &= \frac{1}{(n-p-1)!} \sum_{0 \leq k_1, \ldots , k_p : \sum_2^p mk_m = p}  (-1)^K 
                            \frac{(n+K-1)!}{k_2! \cdots  k_p! ~ 1!^{k_1} 2!^{k_2} 3!^{k_3} \cdots p!^{k_p}} ,
\end{align}
</math>
where <math>K =k_2 + \cdots + k_p.</math>  The sum is a sum over all [[Partition (number theory)|partitions]] of ''p''.

===Relation to harmonic numbers===
Stirling numbers of the first kind can be expressed in terms of the [[harmonic number]]s

:<math>H^{(m)}_n=\sum_{k=1}^n \frac{1}{k^m}</math>

and the [[Gamma function]], <math>\Gamma(x),</math> as follows:

:<math>\left[{n \atop k}\right]= \frac{\Gamma(n)}{\Gamma(k)}w(n,k-1)</math>

where ''w''(''n'',&nbsp;0)&nbsp;=&nbsp;1 and

:<math>w(n,k)=\sum_{m=0}^{k-1}\frac{\Gamma(1-k+m)}{\Gamma(1-k)}H_{n-1}^{(m+1)} w(n,k-1-m)</math>

for ''k'' &gt; 0.

==References==
{{reflist}}
* [[The Art of Computer Programming]]
* [[Concrete Mathematics]]
* {{cite book |editor=M. Abramowitz, I. Stegun |chapter=§24.1.3. Stirling Numbers of the First Kind |title=Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables |edition=9th |place=New York |publisher=Dover |page=824 |year=1972}}
* {{planetmath reference|id=2809|title=Stirling numbers of the first kind, s(n,k)}}.
* {{SloanesRef |sequencenumber=A008275|name=Triangle read by rows of Stirling numbers of first kind}}

{{PlanetMath attribution|id=2809|title=Stirling numbers of the first kind}}

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

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