Shing-Tung Yau High School Mathematical Sciences Award 2014
New Probe into Combinatorial Identities
Author
Zhou Sheng
School
The Affiliated High School of SCNU Advisor
Gui Peng
New Probe into Combinatorial Identities November 30, 2014
Abstract
In the first part, we proof some new identities of Stirling numbers, Lah numbers and Bell numbers using generating functions and inversion formulas. In the second part, we propose a new inversion formula based on inverse functions and discuss some applications of this formula. In the third part, we generalize the definition of binomial coefficients, Stirling numbers and Lah numbers. Then we discuss some basic properties of the new sequences and show a practical way to obtain new combinatorial identities.
KEYWORDS: Combinatorial identities; Stirling numbers; Lah numbers;
Bell numbers; Binomial coefficients ; Generating functions ; Inversion
formulas
1 Introduction 1.1 Notations
[ ] x
n =x x (
− ⋅⋅⋅ − +1) ( x n 1)
forn
≥1
.[ ] x
0= 1
,[ ] x
n= 0
forn < 0
.[ ] x
n =x x (
+ ⋅⋅⋅ + −1) ( x n 1)
forn
≥1
.[ ] x
0= 1
,[ ] x
n= 0
forn < 0
.k
C
n,( )
nk =[ ] n k !
k , wheren ∈ C
,k ∈ N
.( , )
s n k
Stirling number of the first kind, see Definition 1.( , )
S n k
Stirling number of the second kind, see Definition 2.( , )
L n k
Lah number, see Definition 3.B
n Bell number, see Definition 4.δ
ij Kronecker symbol.δ
ij =0
, ifi ≠ j
;δ
ij =1
, ifi = j
.i, j, k, l, n are the indexes of a sequence, they are integers without specification.
1.2 Definitions
[1]Definition 1. Stirling number of the first kind
s n k ( , ) is the coefficient of x
kin the expansion of [ ] x
n, or [ ] x
n= ∑
nk=0s n k x ( , )
k. is defined as 0 when
or .
( , ) s n k 0
k > ≥ n k < 0
Since are linearly independent, they can be the basis of
polynomial ring
0 1 2 n
[ ] ,[ ] ,[ ] , x x x ⋅⋅⋅ ,[ ] , x ⋅⋅⋅
[ ] x
R , thus we have the following
Definition 2. Stirling number of the second kind
S n k ( , ) is the coefficient of [ ] x
kin the expansion of x
n, or x
n= ∑
nk=0S n k x ( , )[ ]
k. is defined as 0 when
or .
( , ) S n k 0
k > ≥ n k < 0
Definition 3. Lah number is the coefficient of
[ ] x
k in the expansion of[ ] − x
n , or[ − x ]
n= ∑
nk=0L n k x ( , )[ ]
k.( , ) n k
k
is defined as 0 when
k > ≥ n 0 or k < 0 . L
Definition 4. Bell number
B
n= ∑
nk=0S n ( , )
, wheren ∈ N
.1.3 Genarating Functions
[1] [2]Definition 5[1]. The ordinary generating function of
{ } a
n n≥0 is0
( )
n nn
f x a
∞
=
=
∑ x
(1.1) Definition 6[1]. The exponential generating function of{ } a
n n≥0 is0
( ) !
n n n
g x a x n
∞
=
= ∑
(1.2)(1.1) and (1.2) are called formal power series. In combinatorics we seldom deal with their convergence or other properties that are often involved in mathematical analysis. For rigorous theory of generating functions, please refer to [4][5]. We will list some basic properties[1 of generating functions without proof. Denote the ordinary generating functions of and
{ }
as and
]
{ } a
nb
n1
( )
A x B x
1( )
respectively, and the exponential generating functions asA x
2( )
and2
( )
B x
respectively. We have the followingProperty 1.
{ } { } a
n= b
n⇔ A x
1( ) = B x
1( ) ⇔ A x
2( ) = B x
2( )
Property 2. The ordinary generating function and exponential generating function of are and respectively, where is a complex constant.
{ α a
n}
1( )
α A x α A x
2( ) α
Property 3. The ordinary generating function and exponential generating function of
are and respectively.
n n
{ a + b }
1
( )
1( )
A x + B x A x
2( ) + B x
2( )
Property 4.
A x B x
1( )
1( )
n∞0c x
n n= ∑
= , wherec
n= ∑
nk=0a
k nb
−k,2
( )
2( )
0!
n
n n
A x B x d x n
∞=
= ∑
, whered
n= ∑
nk=0C a
nk k nb
−k.Definition 7[1]. If
A x ( ) = ∑
n∞=0a x
n n, B x ( ) = ∑
n∞=0b x
n n, C x ( ) = ∑
n∞=0c x
n n satisfythat
B x C x ( ) ( )
=A x ( )
, thenC x ( )
is called the quotient ofA x ( )
divided byB x ( )
, anddenoted as
( )
( ) ( )
C x A x
= B x
.Definition 8 [1] . The formal derivative of
A x ( ) = ∑
n∞=0a x
n n is denoted as1
( )
n 1 n nDA x = ∑
∞=na x
− .Definition 9[1]. The higher derivative of
A x ( )
is1
( ) 0 ( ) ( ( )) 1
n
n
A x n
D A x
D D
−A x n
=
= ⎧ ⎨
⎩ ≥
(1.3)Definition 10[1]. If
B x ( ) satisfy that A x ( )
=DB x ( ) , then B x ( ) is called the formal primitive function of A x ( ) .
2 Identities of Stirling numbers, Lah numbers and Bell Numbers 2.1 Basic properties
Theorem 1[1]. (1)
s n (
+1, ) k
=s n k ( ,
− −1) ns n k ( , )
(2.1))
n k
(2)
S n (
+1, ) k
=S n k ( ,
− +1) kS n k ( , )
(2.2) (3)L n (
+1, ) k
= − +( n k L n k ) ( , )
−L n k ( ,
−1
(2.3)(4) 1
0 n
k n
k
B
+C B
=
=
∑
(2.4) Proof. For a more generalized proof of (2.1)(2.2)(2.3), see Theorem 20. We only present a new proof of (2.4). First we proof0 0
( , )
!
n k
n
k x
k k
S n k x e k x k
− ∞
=
=
=∑ ∑
(2.5)by induction. Denote
0
( ) ( , )
n
k n
k
f x S n k
=
=
∑ x
.If
n = 0 ,
we havef x
n( ) = 1
, and0 0
! ! 1
n k k
x x x
k k
x x
e k e e e
k k
∞ ∞
− − −
=
=
== ⋅
∑ ∑
x=
.(2.5) holds for
n = 0 .
Suppose (2.5) holds for n, with (2.2) we have
1 1
0
( ) ( 1, )
n
k n
k
f x S n k x
+ +
=
=
∑
+ 10
( ( , 1) ( , ) )
n
k k
k
S n k x kS n k x
+
=
− +
∑
=
= 1 =
0 0
( , ) ( , )
n n
k k
k k
S n k x
+kS n k x
= =
∑
+∑ x f (
n( ) x + Df x
n( ))
By induction hypothesis,
1
1 0
( ) ( 1)!
n k n k
x x
n
k k
!
x x
Df x e k e k
k k
∞ − ∞
− −
= =
= −
∑ − ∑
=
0
( 1) (
!
n k
x
n k
e k x f
k
− ∞
=
+ −
∑ x )
Thus,
1
( ) ( ( ) ( ))
n n n
f
+x = x f x + Df x
=1
0
( 1)
!
n k
x k
e k x
k
∞ +
−
=
∑ +
= 1
1
0
( 1)
( 1)!
n k
x k
x k
+ +
− ∞
=
+
∑ +
e k
= 10
!
n k
x k
e k x k
∞ +
−
∑
=According to mathematical induction, (2.5) holds for arbitrary non-negative integer n.
Set x=1, we obtain
0
(1) 1
!
n
n n
k
B f k
e k
∞
=
= = ∑
(2.6) Hence,0 0 k=0
1
!
n n i
k i
n k n
k i
C B C k
e k
∞
= =
= ∑
∑ ∑
0 0
1 1
!
n i i n
k i
e k C k
∞
= =
=
∑ ∑
=0
1 ( 1)
!
n
k
k
e k
∞
=
∑ +
1
0
1 ( 1)
( 1)!
n
k
k
e k
∞ +
=
= +
∑ +
10
1
!
n
k
k
e k
∞ +
=
= ∑
=B
n+1Theorem 2[1]. (1)
0
( , ) 1 (ln(1 ))
! !
n
k n
s n k x x
n k
∞
=
= +
∑
(2.7)(2)
0
( , ) 1 ( 1)
! !
n
x k
n
S n k x e
n k
∞
=
= −
∑
(2.8)(3)
0
( , )
(1 )(1 2 ) (1 )
k n
n
S n k x x
x x k
∞
=
=
− − ⋅⋅⋅ −
∑ x
(2.9)(4)
0
( , ) 1 ( )
! ! 1
n
k n
x x
L n k
n k x
∞
=
= −
∑ +
(2.10)(5) 1
0
!
x
n e n
n
B x e n
∞ −
=
∑ =
(2.11) Proof is omitted, We remark that (2.11) implies (2.6). In fact,1
0 0 0 0
1 1 ( ) 1 1 ( ) 1
( ) (
! ! ! !
x x
n x k n
e e
n
n k k n n 0 0
! ! )
n n
k
x e kx x
B e e
n e e k e k n e n
∞ ∞ ∞ ∞ ∞
−
=
= = =
==
= ==
=∑ ∑ ∑ ∑ ∑
∞k k
∑
=Compare the coefficient of
! x
nn , we obtain (2.6).
In Theorem 7 there is a new proof for (2.7) . Theorem 3[1]. (1)
1 2
1 2 0
( , ) ( 1)
n k
n k
n k n
s n k
ε ε ε
ε ε ε
−
+ −
< < <⋅⋅⋅< <
= − ∑ ⋅⋅⋅ (n
≥k ) (2.12)
⋅⋅⋅
(2) 1 2
1 1,2, 0
( , ) 1 2
kk k
n k
S n k
ε εk
εε ε
ε ε+⋅⋅⋅+ = −ε
⋅⋅⋅ ≥
= ∑ (n
≥k ) (2.13)
(3)
1
( , ) 1 ( 1)
!
k
k i i n k i
S n k C i
k
−
=
=
∑
−( n
≥k ) (2.14)
(4)!
11( , ) ( 1)
!
n k
n
L n k n C
k
−−
= −
(n
≥k ) (2.15)
(5)
0 1
1 ( 1)
!
n k
k i i n n
k i
B C i
kk
−
= =
=
∑ ∑
−(n
≥k ) (2.16)
k i
Proof is omitted. In Theorem 22 there is a generalized proof for (2.12) and (2.13).
2.2 Identities of Stirling numbers
The following identities[2] are famous :
(Vandermonde’s identity) (2.17)
0
C = C C
k
k i
n m n m
i
− +
∑
=
C C
11 (2.18)m
n m
k n
k n
+
= +
∑
=The following two theorems proof similar identities of Stirling number.
Theorem 4.
0
( , ) ( ) ( , ) ( , )
k m
j i i
j i j i
s n m k n
−s n k i s m j C
= =
+ = ∑∑ − −
0
( ) ( , ) ( , )
k n
j i i
j i j i
m
−s m k i s n j C
= =
= ∑∑ − −
(2.19) Proof. Compare the two sides of[ ] x
n m+= [ ] [ x
nx − n ]
m ,
0
[ ] ( , )
n m
k n m
k
x s n m k x
+
+ =
=
∑
+
0 0
0 0 0
0 0
[ ] [ ] ( , ) ( , )(
( , ) ( , ) ( )
( , ) ( ) ( , )
)
n m
i j
n m
i j
n m j
i j
j
i j l
n m m
i l j l l
j
i l j l
x x n s n i s m j
s n i s m j n C
s n i n C s m j
x x n
l l l
x x
x x
= =
−
= = =
−
= = =
⎛ ⎞
⎛ ⎞
− =⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
⎛ ⎞
⎛ ⎞
=⎜⎝ ⎟⎠⎝⎜ − ⎟⎠
⎛ ⎞
⎛ ⎞
=⎜ ⎟⎜ − ⎟
⎝ ⎠⎝ ⎠
∑ ∑ −
∑ ∑ ∑
∑ ∑ ∑
0 0
( ) ( , ) ( , )
n m k m
k j i
j
k i j i
n s n k i s m j C x
+ −
= = =
=
∑ ∑∑
− − i
Compare the coefficient of
x
k, we obtain0
( , ) ( ) ( , ) ( , )
k m
j i i
j i j i
s n m k n
−s n k i s m j C
= =
+ = ∑∑ − −
,Exchange
m n ,
, we have0
( , ) ( ) ( , ) ( , )
k n
j i i
j i j i
s n m k m
−s m k i s n j C
= =
+ = ∑∑ − −
We don’t have similar identities for Stirling numbers of the second kind, for the expansion of
[ ] [ ] x
ix
j using basis{ } [ ] x
n contains and (see Theorem 8), which leads to a trivial identity.( , )
s n k S n k ( , )
Theorem 5.
( 1) ( 1)
( , ) ( 1, 1)
! !
k m
m
k n
s k n s m n
k m
=
− = − +
∑ +
(2.20)
( , ) ( 1, 1)
( 1) ( 1)
m
k k n
S k n S m n
n n
=
+ +
+ = +
∑
m (2.21) Proof. According to Theorem 1,m
( 1) !
k( , )
m( 1) !
k( ( 1, 1) ( , 1) )
k n k n
s k n s k n ks k n
k k
= =
− = − + + +
∑ ∑ +
( 1) ( 1)
1( 1, 1) ( , 1)
! ( 1)!
k k
m
k n
s k n s k n
k k
−
=
⎛ − − ⎞
= ⎜ + + − ⎟
⎝ − ⎠
∑ +
( 1) ( 1) 1 ( 1)
( 1, 1) ( , 1) ( 1, 1
! ( 1)! !
m n m
s m n s n n s m n
m n m
− − − −
= + + − + = +
− + )
( )
( , ) 1
( 1, 1) ( 1) ( , 1)
( 1) ( 1)
m m
k k
k n k n
S k n
S k n n S k n
n n
= =
= + + − +
+ +
∑ ∑
+1
( 1, 1) ( , 1)
( 1) ( 1)
m
k k
k n
S k n S k n
n n
−=
⎛ + + + ⎞
= ∑ ⎜ ⎝ + − + ⎟ ⎠
1
( 1, 1) ( , 1) ( 1, 1
( 1)m ( 1)n ( 1)
S m n S n n S m n
n n − n
+ + + + +
= − =
+ + +
)
m
]
The following theorem present an expansion of
[ x + m
n using basis{ } [ ] x
n .Theorem 6.
0
[ ]
n kj k j( , ) ( , )[
i j k n
]
ix m C m
−s n k S j
≤ ≤ ≤ ≤
+ =
∑ i x
k j j
(2.22)
Proof.
0 0 0
[ ] ( , )( ) ( , )
n n k
k j
n k
k k j
x m s n k x m s n k C m
−= = =
+ = ∑ + = ∑∑ x
i
0 0 0 0
( , ) ( , )[ ] ( , ) ( , )[ ]
n k j
j k j j k j
k i k
k j i i j k n
s n k C m
−S j i x C m
−s n k S j i x
= = = ≤ ≤ ≤ ≤
= ∑∑ ∑ = ∑
Set
x = n
, we obtain Corollary 6.1.0
1 ( , ) ( , ) !
!
m j i k j
n m k n
i j k n
C C C m s n k S
n
−
+ ≤ ≤ ≤ ≤
=
∑ j i i
(2.23)Theorem 7.
[1]
0
0 0
( , ) m=1 (2.24) ( ( )) !
! ( , )(ln )
m 1 (2.25)
! ( )!
n k
n
n k k i
n
n i
s n k x ln x m n
k x s n i m
m n k i
∞
=
∞ −
= =
⎧ ⎪ + = ⎪ ⎨
⎪ ≠
⎪ −
⎩
∑
∑ ∑
Proof.
0 0 0 0 0
[ ] 1 1
( ) ( , ) ( , )
! ! !
n n
t n t n n t n n k k t n
n n k k n
t
nx m m x m x s n k t t m s n k x
n n n
∞ ∞ ∞
− − −
= = = = =
+ =
∑
=∑∑
=∑ ∑
ln( )
0
( ) (ln( )
!
k
t t x m
k
t )
kx m e x m
k
+ ∞
=
+ = = ∑ +
If
m = 1
, then compare the coefficient oft
k we have[1]
0
( ( 1))
( , )
! !
k n
n
ln x x
s n k
k n
∞
=
+ = ∑
which proof (2.7) in Theorem 2.
If
m ≠ 1
, thenm
t depends ont ,
ln 0
(ln )
!
k
t t m
k
m e t m
k
∞
=
= = ∑
kHenceforth,
0 0 0
0 0 0
0 0 0
( , )
( ) (ln )
! !
( , ) (ln )
! ( )
( , ) (ln )
! ( )!
i j
t k
k
j k i
i i j
i k
k
i j k
k k i
k n
n
k i n
s k j t
!
x m x
im k i
s k j m
t x
m k i j
s n i m
t x
m n k i
∞ ∞ ∞ +
= = =
∞ ∞ −
= = =
∞ ∞ −
= = =
+ =
= −
= −
∑∑∑
∑ ∑∑
∑ ∑∑
m
Compare the coefficient of
t
k, we obtain0 0
(ln( )) ( , )(ln )
! ! (
k n k
n
n i
x m x s n i m
k m n k i
∞ −
= =
+ =
∑ ∑ − )!
k i Setm = e
in (2.25), we haveCorollary 7.1.
0 0
(ln( )) ( , )
! ! (
k n k
n
n i
)!
x e x s n
k e n k
∞
= =
+ =
∑ ∑ − i i
(2.26) Setk = 1
in (2.25), we haveCorollary 7.2.
1
1
ln( 1) ln( ) ln ( 1)
n n
n n
x x
x m m
m m
∞ −
=
+ = + − = ∑ − n
n
(2.27)
Which is the same as Taylor expansion of
ln( x
+1)
. Thus (2.25) is a generalization of (2.27).The following theorem present an expansion of
[ ] [ ] x
mx
using basis{ } [ ] x
n .Theorem 8.
0 0
[ ] [ ] [ ] ( , ) ( , ) ( , )
m n m n k
m n i
i k i j
x x x S k i s n j s m k
+ +
= = =
= ∑ ∑∑ − j
(2.28)Proof.
0 0
[ ] [ ] ( ( , ) )( ( , ) )
n m
i j
m n
i j
x x s n i x s m j
= =
= ∑ ∑ x
0 0
( , ) ( , )
m n k
k
k j
x s n j s m k j
+
= =
= ∑ ∑ −
0 0 0
( ( , )[ ] )( ( , ) ( , )
m n k k
i
k i j
S k i x s n j s m k j
+
= = =
= ∑ ∑ ∑ − )
0 0
[ ] ( , ) ( , ) ( , )
m n m n k
i
i k i j
x S k i s n j s m k j
+ +
= = =
= ∑ ∑∑ −
Set
x = + m n
, we haveCorollary 8.1. 2 (2.29)
0 0
! !( ) ! ( , ) ( , ) ( , )
m n m n k
n i
m n m n
i k i j
m n C i C S k i s n j s m k j
+ +
+ +
= = =
= ∑ ∑∑ −
2.3 Identities of Lah numbers
Theorem 9[1]
.
111
[ ] [ ]
! !
n n
k k
n k
x
x C
n k
−−
=
= ∑
(2.30)Proof.
0
[ ] ( 1) [ ] ( 1) ( , )[ ]
n
n n n
n k
k
x x L n k
=
= − − = −
∑ x
By Theorem 3, we have
1 1
( , ) ( 1) !
!
n k
n
L n k n C
k
−
= − −
11 11
0 0
! [ ]
[ ] ( 1) ( 1) [ ] !
! !
n n
n n n k k k
n k n
k k
n x
x C x n
k k
− −
− −
= =
= −
∑
− =∑ C
1 1 1
[ ] [ ]
! !
n n
k k
n k
x
x C
n k
−−
=
= ∑
(2.30) can be written as1
1 1
n
n
k
x n n
n k k
=
+ − −
⎛ ⎞ ⎛ ⎞⎛
⎜ ⎟ = ⎜ − ⎟⎜
⎝ ⎠ ∑ ⎝ ⎠⎝ x ⎞ ⎟
⎠
Which is the same as Vandermonde’s identity when
x ∈ N
+.
By (2.12), we know that
| ( , ) | ( 1) s n i = −
n i+s n i ( , )
, compare the coefficient ofx
i in (2.30), we obtainCorollary 9.1. 11
1
1 1
( , ) ( , )
! !
n k n k
s n i C s k i
n k
−
= −
=
∑
(0
≤ ≤i n )
(2.31) With a more generalized form of Vandermonde’s identity[2]0
k
i
x y x y
k
=i k
+
⎛ ⎞ ⎛ ⎞⎛ ⎞
⎜ ⎟ = ⎜ ⎟⎜ − ⎟
⎝ ⎠ ∑ ⎝ ⎠⎝ i ⎠
(2.32) wherex y ,
are arbitrary complex numbers, we can proof a new identity.Theorem 10. (2.33)
0 0
2 ( , ) ( , ) ( , )
k j j
k
i l
s k j s i j s k i j l C
= =
=
∑∑
− − iProof. Set
x
=y
in (2.32), we get0
2
ki
x x x
k
=i k
⎛ ⎞ ⎛ ⎞⎛ ⎞
⎜ ⎟ = ⎜ ⎟⎜ − ⎟
⎝ ⎠ ∑ ⎝ ⎠⎝ i ⎠
Compare the expansion of both sides,
0
2 [2 ] 1
2 ( , )
! !
k
j j
k j
x x
s k j x
k k k
=⎛ ⎞
= =
⎜ ⎟
⎝ ⎠ ∑
0 0 0 0 0
1 1
[ ] [ ] ( , ) ( , )
! !
k k k k j
i i j
k i k i k
i i i j l
x x
C x x C x s i l s k i j l
i k i k
−k
= = = = =
⎛ ⎞⎛ ⎞
= =
⎜ ⎟⎜ − ⎟
⎝ ⎠⎝ ⎠
∑ ∑ ∑ ∑ ∑ − −
0 0 0
1 ( , ) ( , )
!
j
k k
j i
k
j i l
x s i l s k i j l C k
= = == ∑ ∑∑ − −
− i
i
We obtain
0 0
2 ( , ) ( , ) ( , )
k j j
k
i l
s k j s i l s k i j l C
= =
=
∑∑
−In (2.33), set
j
=k
, the left hand side is2
k, the right hand side is0 0 0
( , ) ( , )
k k k
i i
k k
i l i
C s i l s k i k l C
= = =
− − =
∑ ∑ ∑
And we obtain
Corollary 10.1. (2.34)
0
2
k k
k i
C
=
=
∑
which is a well-known identity. (2.34) is a special case of (2.33).
Theorem 11. 1
0 0 0
( , ) ! ( 1)
!
i j n n
i n j n i
k i j
L n k n C C
j
∞ ∞ + +
= = = + −
= −
∑ ∑∑
(2.35)
Proof. By (2.10), we have
0 0 0 0
( ( , )) ( , )
! !
n n n
n k k n
x x
L n k L n k
n n
∞ ∞ ∞
= =
=
= =∑ ∑ ∑∑
0
1 ( )
! 1
k k
x
k x
∞
=
= −
∑
+0
1 1
( 1)
! 1
k
k
k x
∞
= +
∑
= −
0 0
1 1
( ( 1) ( ) )
! 1
i k i i
k
k i
k C x
∞ ∞
−
= =
= −
∑ ∑
+1
0 0 0
( 1) 1 ( ( 1) )
!
k i i j j j
k i
k i j
C x C
k
∞ ∞ ∞
− + −
= = =
= ∑∑ − ∑ −
j1
0 0 0
( 1)
!
i j k
i j j
k i j
i j k
C C x k
∞ ∞ ∞ + +
= = = + −
= ∑∑∑ −
1
0 0 0
( 1)
!
i j n
i n n
j n i
i n j
C C x j
∞ ∞ ∞ + +
= = = + −
= ∑∑∑ −
Compare the coefficient of
x
n, we obtain1
0 0 0
( , ) ! ( 1)
!
i j n n
i n
j n i
k i j
L n k n C C
j
∞ ∞ + +
= = = + −
= −
∑ ∑∑
Together with (2.15), we have Corollary 11.1.
1 1
1
0 0 0
( 1)
! !
k i j
n
i n
n
j n i
k i j
C C C
k j
− ∞ ∞ +
− + −
= = =
= −
∑ ∑∑
(2.36)By the relation between Lah numbers and Stirling numbers[1],
0
( , ) ( 1) ( , ) ( , )
n j j
L n k s n j S j k
=
= ∑ −
We obtain
Corollary 11.2. 1
0 0 0 0
( 1) ( , ) ( , ) ! ( 1)
!
i j n
n n
l i
j n i
k l i j
s n l S l k n C C
j
∞ ∞ + +
= = = = + −
− = −
∑∑ ∑∑
n(2.37)
Theorem 12. 1
0 0 0
( 1) ! ( , ) 1
!
i j k i k
j k i
i j k
k C C L n k j
∞ ∞ ∞
+ + + −
= = =
∑∑∑ − =
a
k(2.38) Proof. Apply the inversion formula of Lah number (Theorem 14.(3))
0 0
( , ) ( , )
n n
n k n
k k
a L n k b b L n k
= =
=
∑
⇔ =∑
to Theorem 11, where
b
n= 1
and 10 0
! ( 1)
!
i j n
i n
n j
i j
a n C C
j
∞ ∞ + +
= = + −
= ∑∑ −
n i,
we obtain
1
0 0 0
1 ( , ) ! ( 1)
!
i j k n
i k j k i
k i j
L n k k C C
j
∞ ∞ + +
= = = + −
= ∑ ∑∑ −
1
0 0 0
( 1) ! ( , )
!
i j k i k
j k i
i j k
k C C L n k j
∞ ∞ ∞
+ + + −
= = =
= ∑∑∑ −
Together with (2.15), we have
1 11
0 0 0
1 ( 1) !
!
i j k n i k k
j k i n
i j k
n C C C j
∞ ∞ ∞
+ + + −
+ − −
= = =
= ∑∑∑ −
11 1
0 0 0
( 1) !
!
n
k i j k n i
n j
k i j
C n C
j
∞ ∞
− + + +
− +
= = =
= ∑ ∑∑ − C
k ik −1
1 1
1
0 0 0
( 1) !
!
n
k i j k n i
n j
k i j
C n C
j
− ∞ ∞
+ + + + +
− +
= = =
= ∑ ∑∑ − C
k ikChange
n
inton + 1
,1
0 0 0
( 1)!
1 ( 1)
!
k i j k n i
n j
k i j
C n C
j
∞ ∞ ∞
+ + + +
= = = +