Stirling numbers of the first kind

Stirling numbers of the first kind

Introduction.

The

Stirling numbers

of the first kind, frequently denoted as

s

⁢

(

n

,

k

)

=

[

n

k

]

,

k

,

n

∈

ℕ

,

1

≤

k

≤

n

,

are the

integer

coefficients

of the

falling factorial

polynomials

. To be more precise,
the

defining relation

for the Stirling numbers of the first kind is:

x

n

¯

=

x

⁢

(

x

-

1

)

⁢

(

x

-

2

)

⁢

…

⁢

(

x

-

n

+

1

)

=

∑

k

=

1

n

s

⁢

(

n

,

k

)

⁢

x

k

.

Here is the table of some initial values.

n

\

k

1

2

3

4

5

1

1

2

-1

1

3

2

-3

1

4

-6

11

-6

1

5

24

-50

35

-10

1

Recurrence Relation.

The evident observation that

x

n

+

1

¯

=

x

⁢

x

n

¯

-

n

⁢

x

n

¯

.

leads to the following

equivalent

characterization

of the

s

⁢

(

n

,

k

)

, in
terms of a 2-place
recurrence formula:

s

⁢

(

n

+

1

,

k

)

=

s

⁢

(

n

,

k

-

1

)

-

n

⁢

s

⁢

(

n

,

k

)

,

1

≤

k

<

n

,

subject to the
following initial conditions:

s

⁢

(

n

,

0

)

=

0

,

s

⁢

(

1

,

1

)

=

1

.

Generating Function.

There is also a strong connection with the generalized binomial
formula, which furnishes us with the following

generating function

:

(

1

+

t

)

x

=

∑

n

=

0

∞

∑

k

=

1

n

s

⁢

(

n

,

k

)

⁢

x

k

⁢

t

n

n

!

.

This generating function implies a number of

identities

.
Taking the

derivative

of both sides with respect to

t

and equating
powers, leads to the

recurrence relation

described above.
Taking the derivative of both sides with respect to

x

gives

(

k

+

1

)

⁢

s

⁢

(

n

,

k

+

1

)

=

∑

j

=

k

n

(

-

1

)

n

-

j

⁢

(

n

-

j

)

!

⁢

(

n

+

1

j

)

⁢

s

⁢

(

j

,

k

)

This is because the derivative of the left side of the
generating funcion

equation

with respect to

x

is

(

1

+

t

)

x

⁢

ln

⁡

(

1

+

t

)

=

(

1

+

t

)

x

⁢

∑

k

=

1

∞

(

-

1

)

k

-

1

⁢

t

k

k

.

The

relation

(

1

+

t

)

x

1

⁢

(

1

+

t

)

x

2

=

(

1

+

t

)

x

1

+

x

2

yields the following family of

summation

identities. For any given

k

1

,

k

2

,

d

≥

1

we have

(

k

1

+

k

2

k

1

)

⁢

s

⁢

(

d

+

k

1

+

k

2

,

k

1

+

k

2

)

=

∑

d

1

+

d

2

=

d

(

d

+

k

1

+

k

2

k

1

+

d

1

)

⁢

s

⁢

(

d

1

+

k

1

,

k

1

)

⁢

s

⁢

(

d

2

+

k

2

,

k

2

)

.

Enumerative interpretation.

The

absolute value

of the Stirling number of the first kind,

s

⁢

(

n

,

k

)

,
counts the number of

permutations

of

n

objects with exactly

k

orbits (equivalently, with exactly

k

cycles). For example,

s

⁢

(

4

,

2

)

=

11

, corresponds to the fact that
the

symmetric group

on 4 objects has

3

permutations of the form

(

*

*

)

(

*

*

)

— 2 orbits of size 2 each

,

and

8

permutations
of the form

(

*

*

*

)

— 1 orbit of size 3, and 1 orbit of size
1

,

(see the entry on

cycle notation

for the meaning of the above
expressions.)

Let us prove this. First, we can remark that the unsigned Stirling
numbers of the first are characterized by the following recurrence
relation:

|

s

⁢

(

n

+

1

,

k

)

|

=

|

s

⁢

(

n

,

k

-

1

)

|

+

n

⁢

|

s

⁢

(

n

,

k

)

|

,

1

≤

k

<

n

.

To see why the above recurrence relation matches the count of
permutations with

k

cycles, consider forming a permutation of

n

+

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 cycle, i.e. leaving the extra
object alone. This accounts for the

s

⁢

(

n

,

k

-

1

)

term in the recurrence
formula. We could also insert the new object into one of the existing
cycles. Consider an arbitrary permutation of

n

object with

k

cycles, and label the objects

a

1

,

…

,

a

n

, so that the
permutation is represented by

(

a

1

⁢

…

⁢

a

j

1

)

⁢

(

a

j

1

+

1

⁢

…

⁢

a

j

2

)

⁢

…

⁢

(

a

j

k

-

1

+

1

⁢

…

⁢

a

n

)

⏟

k

⁢

cycles

.

To form a new permutation of

n

+

1

objects and

k

cycles one must
insert the new object into this array. There are, evidently

n

ways
to perform this insertion. This explains the

n

⁢

s

⁢

(

n

,

k

)

term of the
recurrence relation. Q.E.D.

Title

Stirling numbers of the first kind

Canonical name

StirlingNumbersOfTheFirstKind

Date of creation

2013-03-22 12:33:44

Last modified on

2013-03-22 12:33:44

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

6

Author

rmilson (146)

Entry type

Definition

Classification

msc 05A15

Synonym

Stirling numbers

Related topic

StirlingNumbersSecondKind

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