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INTEGERS 11 (2011)
#A18
THE (EXPONENTIAL) BIPARTITIONAL POLYNOMIALS AND
POLYNOMIAL SEQUENCES OF TRINOMIAL TYPE: PART I
Miloud Mihoubi1
Universit´e des Sciences et de la Technologie Houari Boumediene, Faculty of
Mathematics, El Alia, Bab Ezzouar, Algeria
mmihoubi@usthb.dz, miloudmihoubi@gmail.com
Hac`
ene Belbachir2
Universit´e des Sciences et de la Technologie Houari Boumediene, Faculty of
Mathematics, El Alia, Bab Ezzouar, Algeria
hbelbachir@usthb.dz, hacenebelbachir@gmail.com
Received: 3/29/10, Revised: 10/3/10, Accepted: 12/31/10, Published: 2/4/11
Abstract
The aim of this paper is to investigate and present the general properties of the
(exponential) bipartitional polynomials. After establishing relations between bipartitional polynomials and polynomial sequences of binomial and trinomial type, a
number of identities are deduced.
1. Introduction
For (m, n) ∈ N2 with N = {0, 1, 2, . . .}, the complete bipartitional polynomial
Am,n ≡ Am,n (x0,1 , x1,0 , x0,2 , x1,1 , x2,0 , . . . , xm,n )
with variables x0,1 , x1,0 , . . . , xm,n is defined by the sum
�x
� x �k0,1 � x �k1,0
�km,n
�
m!n!
1,0
m,n
0,1
···
,
Am,n :=
k0,1 !k1,0 ! · · · km,n ! 0!1!
1!0!
m!n!
(1)
where the summation is carried out over all partitions of the bipartite number
(m, n), that is, over all nonnegative integers k0,1 , k1,0 , . . . , km,n which are solution
to the equations
m
n �
n
m �
�
�
(2)
j ki,j = n.
i ki,j = m,
i=1 j=0
1 The
2 The
j=1 i=0
research is partially supported by LAID3 Laboratory of USTHB University.
research is partially supported by LAID3 Laboratory and CMEP project 09 MDU 765.
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INTEGERS: 11 (2011)
The partial bipartitional polynomial Am,n,k ≡ Am,n,k (x0,1 , . . . , xm,n ) with variables
x0,1 , x1,0 , . . . , xm,n , of degree k ∈ N, is defined by the sum
�x
� x �k0,1 � x �k1,0
�km,n
�
m!n!
m,n
1,0
0,1
Am,n,k :=
···
, (3)
k0,1 !k1,0 ! · · · km,n ! 0!1!
1!0!
m!n!
where the summation is carried out over all partitions of the bipartite number
(m, n) into k parts, that is, over all nonnegative integers k0,1 , k1,0 , . . . , km,n which
are solution to the equations
m �
n
n �
m
m �
n
�
�
�
i ki,j = m,
j ki,j = n,
ki,j = k, with the convention k0,0 = 0.
i=1 j=0
j=1 i=0
i=0 j=0
(4)
These polynomials were introduced in [1, pp. 454] with properties such as those
given in (5), (6), (7), (8) below. Indeed, for all real numbers α, β, γ, we have
�
�
Am,n,k βγx0,1 , αγx1,0 , β 2 γx0,2 , . . . , αm β n γxm,n =
β m αn γ k Am,n,k (x0,1 , x1,0 , . . . , xm,n ) , (5)
and
Am,n (αx0,1 , βx1,0 , . . . , αm β n xm,n ) = αm β n Am,n (x0,1 , x1,0 , . . . , xm,n ) .
(6)
Moreover, we can check that the exponential generating functions for Am,n and
Am,n,k are respectively provided by
i j
�
�
t
u
tm un
,
(7)
= exp
xi,j
1+
Am,n (x0,1 , . . .)
m! n!
i! j!
i+j≥1
m+n≥1
and
�
m+n≥k
k
tm un
1 �
ti uj
Am,n,k (x0,1 , . . .)
.
=
xi,j
m! n!
k!
i! j!
(8)
i+j≥1
From the above definitions and properties, we attempt to present more properties and identities for the partial and complete bipartitional polynomials. We also
consider the connection between these polynomials and polynomials (fm,n (x)) of
trinomial type, defined by
x
i j
�
�
tm un
t u
=
fm,n (x)
, with f0,0 (x) := 1.
(9)
fi,j (1)
i! j!
m! n!
i,j≥0
m,n≥0
For the investigation of polynomials of trinomial and multinomial type, we can refer
to [8].
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We use the notations Am,n,k (xi,j ) or Am,n,k (xI,J ) for Am,n,k (x0,1 , x1,0 , . . .), and
Am,n (xi,j ) or Am,n (xI,J ) for Am,n (x0,1 , . . .). Moreover, we represent respectively
by Bn,k (xj ) and Bn (xj ), the partial and complete Bell polynomials Bn,k (x1 , x2 , . . .)
and Bn (x1 , x2 , . . .). We recall that these polynomials are defined by their generating
functions
� ∞
�k
∞
�
tm
1 �
tn
xm
=
,
Bn,k (xj )
n!
k! m=1 m!
n=k
�
� ∞
∞
�
�
tn
tm
Bn (xj )
1+
xm
,
= exp
n!
m!
n=1
m=1
which admit the following explicit expressions:
k1 +k2 +···=k,
Bn (xj ) =
� x �k1 � x �k2
n!
2
1
··· ,
k1 !k2 ! · · · 1!
2!
k1 +2k2 +···=n
� x �k1 � x �k2
n!
2
1
··· .
k1 !k2 ! · · · 1!
2!
�
Bn,k (xj ) =
�
k1 +2k2 +3k3 +···=n
For m < 0 or n < 0, we set fm,n (x) = 0 and Am,n (xi,j ) = 0, and for m < 0 or
n < 0 or k < 0, we set Am,n,k (xi,j ) = 0. For x ∈ C, where C is the set of complex
numbers, we set
� �
1
x
:= x (x − 1) · · · (x − k + 1) for k = 1, 2, . . . ,
k!
k
� �
� �
x
x
= 0 otherwise,
= 1 and
k
0
� �
which gives nk = 0 for k > n, n ∈ N.
2. Properties of Bipartitional Polynomials
In this section, we give some recurrence relations for the bipartitional polynomials
from which we deduce connections with Bell polynomials. In the following, we
consider a sequence (xn,m ) of real numbers with x0,0 = 0.
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INTEGERS: 11 (2011)
Theorem 1. We start with
� �� �
�
m n
xi,j Am−i,n−j,k−1 (xI,J ) = kAm,n,k (xI,J ) ,
j
i
i+j≤m+n−k+1
� �� �
�
m n
ixi,j Am−i,n−j,k−1 (xI,J ) = mAm,n,k (xI,J ) ,
j
i
i+j≤m+n−k+1
� �� �
�
m n
jxi,j Am−i,n−j,k−1 (xI,J ) = nAm,n,k (xI,J ) ,
j
i
(10)
(11)
(12)
i+j≤m+n−k+1
and
��m��n�
ixi,j Am−i,n−j (xI,J ) = mAm,n (xI,J ) ,
i
j
i,j
��m��n�
jxi,j Am−i,n−j (xI,J ) = nAm,n (xI,J ) .
j
i
i,j
(13)
(14)
Proof. From (8), we obtain
k
k−1
i j
�
d 1 �
1
tr us
ti uj
t
u
=
xi,j
xi,j
dxr,s k!
i! j!
(k − 1)!
i! j!
r! s!
i+j≥1
i+j≥1
�
��
�
tm+r
m+r n+s
un+s
Am,n,k−1 (xI,J )
s
r
(m + r)! (n + s)!
m+n≥k−1
�
��
�
�
tm un
m n
,
Am−r,n−s,k−1 (xI,J )
=
m! n!
s
r
=
�
m+n≥r+s+k−1
we also have
k
�
d 1 �
d
tm un
ti uj
=
Am,n,k (xI,J )
.
xi,j
dxr,s k!
i! j!
dxr,s
m! n!
i+j≥1
m+n≥k
From these two expressions, we deduce that
� �� �
m n
d
Am−r,n−s,k−1 (xI,J ) .
Am,n,k (xI,J ) =
s
r
dxr,s
(15)
For α = β = 1, we take the derivatives on both sides of (5) with respect to γ and,
by means of (15), we obtain (10). The same goes for identities (11) and (12). When
we sum over all possible values of k for both sides of each of the identities (11) and
(12), we obtain identities (13) and (14) respectively.
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INTEGERS: 11 (2011)
The next theorem provides the link between the bipartitional polynomials and the
classical Bell polynomials.
Theorem 2. We have
�
� j � �
n � �
� j
�
n
xi,j−i ,
Aj,n−j,k (xI,J ) = Bn,k
i
j
i=0
j=0
(16)
A0,n,k (xI,J ) = Bn,k (x0,j ) ,
(17)
Am,0,k (xI,J ) = Bm,k (xj,0 ) ,
(18)
and
�
� j � �
n � �
� j
�
n
xi,j−i ,
Aj,n−j (xI,J ) = Bn
i
j
i=0
j=0
(19)
A0,n (xI,J ) = Bn (x0,j ) ,
(20)
Am,0 (xI,J ) = Bm (xj,0 ) .
(21)
Proof. From (8), if we set t = u, we obtain
k
i+j
�
�
1
um+n
u
=
Am,n,k (xI,J )
,
xi,j
k!
i!j!
m!n!
i,j≥0
i.e.
1
k!
m+n≥k
�k
�∞
s � �
s � �
� us �
� us �
s
s
=
xi,s−i
Am,s−m,k (xI,J ) ,
i
m
s!
s!
s=1
m=0
i=0
s≥k
which, by identification, gives the identity (16). The identities (17) and (18) result
by setting respectivelly t = 0 and u = 0 in (8).
Summing over all possible values of k on both sides of each of the identities (16),
(17) and (18), we obtain identities (19), (20) and (21) respectively.
For a sequence of real numbers (xn ), we denote by Am,n,k (xi+j ) the quantity obtained by replacing xi,j by xi+j in Am,n,k (xi,j ). The same goes for Am,n (xi+j ) .
Theorem 3. Let (xn ) be a sequence of real numbers. The following holds
Am,n,k (xi+j ) = Bm+n,k (xj ) ,
Am,n (xi+j ) = Bm+n (xj ) ,
�−1
��
�
m+n m+n+k
Bm+n+k,k (jxj ) ,
Am+k,n,k (ixi+j ) =
n
n
�−1
��
�
m+n m+n+k
Bm+n+k,k (jxj ) .
Am,n+k,k (jxi+j ) =
m
m
(22)
(23)
(24)
(25)
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INTEGERS: 11 (2011)
Proof. From (8), we get for xi,j = xi+j that
k
�
tm un
ti uj
1 �
=
Am,n,k (xi+j )
;
xi+j
k!
i! j!
m! n!
i+j≥1
we also get
m+n≥k
k
k
1 � xs
ti uj
1 �
s
=
(t + u)
xi+j
k!
i! j!
k!
s!
i+j≥1
s≥1
s
�
(t + u)
=
Bs,k (xj )
s!
s≥k
=
�
Bm+n,k (xj )
m+n≥k
tm un
.
m! n!
By identification, we obtain (22). Identity (23) follows by summing over all possible
values of k on both sides of identity (22).We have
k
k
1 �
1 � xs � ti uj
ti uj
=
i
ixi+j
k!
i! j!
k!
s! i+j=s i! j!
i+j≥1
s≥1
k
tk � xs d
s
=
(t + u)
k!
s! dt
s≥1
1
=
k!
�
t
t+u
�k
k
s
�
(t
+
u)
sxs
s!
s≥1
r−k
= tk
�
Br,k (jxj )
r≥k
=
�
m+n≥k
�m+n−k�
(t + u)
r!
m−k
� Bm+n−k,k (jxj )
�m+n
n
tm un
.
m! n!
Then, by identification with (8) and making use of the symmetry of m, n, we obtain
(24) and (25).
Note that, for Theorem 3, using the identities on partial and complete Bell
polynomials given in [3] and [4], we can deduce several identities for the partial and
complete bipartitional polynomials.
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INTEGERS: 11 (2011)
3. Polynomial Sequences of Trinomial Type
We know that the sequences of binomial type have a strong relationship with the
partial and complete Bell polynomials. If (fn (x)) is a sequence of binomial type,
x
fn (an + x) is also of binomial type.
the sequence (hn (x)) defined by hn (x) := an+x
For sequences (fn (x))
satisfying
the
property
of
convolution, the following is holds:
�n � �
fn (x + y) = k=0 nk fk (x) fn−k (y) with f0 (x) �= 1.
Theorem 4. Let (fm,n (x)) be a sequence of trinomial type and let a, b be real
numbers. Then the sequence (hm,n (x)) defined by
hm,n (x) :=
x
fm,n (am + bn + x)
am + bn + x
(26)
is of trinomial type.
Proof. The definition of (fm,n (x)), given in (9), states that
x
�∞
�
∞
∞
∞
n
i
j
�
�
�
�
u
t
tm
u
fi,j (1) =
fm,n (x)
,
j! i!
n! m!
m=0 n=0
j=0
i=0
which can also be written as follows:
�x
�∞
∞
∞
�
�
�
un
tm
ti
fm,n (x)
um (x)
with um (x) :=
.
=
ui (1)
i!
m!
n!
n=0
m=0
i=0
This implies that the sequence (um (x)) satisfies the property of convolution. Hence,
the sequence (Um (x)) given by
Um (x) :=
∞
un
x �
x
fm,n (am + x)
um (am + x) =
,
am + x
am + x n=0
n!
has the same property of convolution. We have
�∞
�
i
Ui (1) ti!
i=0
�x
=
∞
�
m
Um (x) tm! ,
m=0
or similarly
�∞
�x
∞
∞
�
�
�
ui
tm
un
x
vi (1)
, with vn (x) :=
fm,n (am + x)
.
vn (x)
=
i!
n!
am + x
m!
m=0
n=0
i=0
This implies that the sequence (vm (x)) satisfies the property of convolution. Moreover, the sequence (Vm (x)) given by
Vn (x) :=
∞
�
tm
x
x
vn (bn + x) =
fm,n (am + bn + x)
,
bn + x
am + bn + x
m!
m=0
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INTEGERS: 11 (2011)
has the same property of convolution. We have
�
∞
�
j
Vj (1) uj!
j=0
or similarly
x
i j
�
�
t u
tm un
hi,j (1)
=
hm,n (x)
i! j!
m! n!
i,j≥0
�x
=
∞
�
n
Vn (x) un! ,
n=0
with h0,0 (x) = 1.
m,n≥0
Consequently, the last identity shows that the sequence (hm,n (x)) is of trinomial
type.
Remark. From (9), we can verify that if a sequence (fm,n (x)) is of trinomial type,
then
n � �� �
m �
�
m n
fm,n (x + y) =
fi,j (x) fm−i,n−j (y) .
i
j
i=0 j=0
If in (9) we set t = 0, we deduce that the sequence (fn (x)), defined by fn (x) :=
f0,n (x), is of binomial type.
Theorem 5. Let (fn (x)) and (gn (x)) be two sequences of binomial type with
f0 (x) = g0 (x) = 1. Then the sequences (pm,n (x)) and (qm,n (x)) defined by
pm,n (x) := fm+n (x) ,
qm,n (x) := fm (x) gn (x)
are sequences of trinomial type.
Proof. We have p0,0 (x) = q0,0 (x) = 1 and
�
tm un
pm,n (x)
m! n!
=
m,n≥0
k � �
� fk (x) �
k m k−m
t u
k! m=0 m
k≥0
k
=
�
fk (x)
k≥0
=
k
�
fk (1)
k≥0
�
tm un
qm,n (x)
m! n!
m,n≥0
(t + u)
k!
x
(t + u)
,
k!
n
m
�
u
t
=
gn (x)
fm (x)
m!
n!
n≥0
m≥0
x
m
n
�
�
t
u
.
=
fm (1)
gn (1)
m!
n!
�
m≥0
n≥0
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INTEGERS: 11 (2011)
Remark. Let (fn (x)) and (gn (x)) be two sequences of binomial type with f0 (x) =
g0 (x) = 1. Then, from [3], the sequences (Fn (x)) and (Gn (x)) defined by
Fn (x) :=
x
fn ((c − a) n + x) ,
(c − a) n + x
Gn (x) :=
x
gn ((d − b) n + x) ,
(d − b) n + x
and
are of binomial type for all real numbers a, b, c, d. This implies, according to Theorem 4, that, for all a, b, c, d, the sequences
Pm,n (x) := x
and
Qm,n (x) := x (am + bn + x)
fm+n (am + bn + x)
,
am + bn + x
fm (cm + bn + x) gn (am + dn + x)
,
cm + bn + x
am + dn + x
are of trinomial type.
� � ��
Example 6. We know that sequences (xn ) , n! nx and (Bn (x)) are of binomial
type, where Bn (x) is the single variable Bell polynomial. The sequences
�
�
x
an + x
n!x
n−1
,
Bn (an + x) ,
x (an + x)
,
n
an + x
an + x
are also of binomial type.
Combining these sequences, Theorem 5 states that the following sequences are also
of trinomial type:
m−1
x2 (cm + x)
n−1
(dn + x)
,
�
m−1 �
2
x (cm + x)
dn + x
n!
,
n
dn + x
�
��
�
x2
cm + x dn + x
m!n!
,
n
m
(cm + x) (dn + x)
m−1
x2 (cm + x)
Bn (dn + x) ,
dn + x
�
�
x2
cm + x
m!
Bn (dn + x) ,
(cm + x) (dn + x)
m
�
�
x
(m + n)!
,
m+n
Bm+n (x) .
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INTEGERS: 11 (2011)
Combining these sequences and using Theorem 4, we deduce that the following
sequences are of trinomial type:
m−1
am,n (x) = x (am + bn + x) (cm + bn + x)
n−1
(am + dn + x)
,
�
m−1 �
am + dn + x
x (am + bn + x) (cm + bn + x)
,
bm,n (x) = n!
am + dn + x
n
�
��
�
x (am + bn + x)
cm + bn + x am + dn + x
cm,n (x) = m!n!
,
n
m
(cm + bn + x) (am + dn + x)
m−1
x (am + bn + x) (cm + bn + x)
Bn (am + dn + x) ,
am + dn + x
�
�
x (am + bn + x)
cm + bn + x
Bn (am + dn + x) ,
em,n (x) = m!
m
(cm + bn + x) (am + dn + x)
�
�
(m + n)!x am + bn + x
,
fm,n (x) =
m+n
am + bn + x
x
gm,n (x) =
Bm+n (am + bn + x) .
am + bn + x
dm,n (x) =
4. Bipartitional Polynomials and Polynomials of Binomial and Trinomial
Type
Roman [7, p. 82] proves that any polynomial sequence of binomial type (fn (x)),
with f0 (x) = 1, is connected with the partial Bell polynomials
fn (x) =
n
�
Bn,k (Dα=0 fj (α)) xk = Bn (xDα=0 fj (α)) , n ≥ 1.
(27)
k=1
Similarly, we establish that any polynomial sequence of trinomial type (fm,n (x)),
with f0,0 (x) = 1, admits a connection with the partial bipartitional Bell polynomials.
Theorem 7. Let (fm,n (x)) be a sequence of trinomial type. Then
fm,n (x) =
m+n
�
Am,n,k (Dα=0 fi,j (α)) xk = Am,n (xDα=0 fi,j (α)) , m+n ≥ 1, (28)
k=1
where Dα=0 ≡
�
d �
dα α=0
.
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INTEGERS: 11 (2011)
Proof. Since Dα=0 f0,0 (α) = 0, then for xi,j = Dα=0 fi,j (α), we have
�
� tm un m+n
Am,n,k (xi,j ) xk
m! n!
m,n≥0
=
k=0
=
=
=
=
=
∞
�
tm un
m! n!
k=0
m+n≥k
k
∞ k
�
x �
ti uj
Dα=0 fi,j (α)
k!
i! j!
i,j≥0
k=0
i j
�
t
u
exp xDα=0
fi,j (α)
i! j!
i,j≥0
α
i j
�
t
u
exp xDα=0
fi,j (1)
i! j!
i,j≥0
i j
�
t
u
fi,j (1)
exp x ln
i! j!
i,j≥0
x
i j
�
u
t
fi,j (1)
i! j!
�
xk
Am,n,k (xi,j )
i,j≥0
=
�
fm,n (x)
m,n≥0
tm un
.
m! n!
Therefore, the desired identity follows by identification.
Some identities related to these polynomials are given by the following theorem.
Theorem 8. Let (fm,n (x)) be a sequence of trinomial type. We have
�
�
fi−1,j−1 (a (i − 1) + b (j − 1) + x)
Am+k,n+k,k xij
a (i − 1) + b (j − 1) + x
�
��
�
m + k n + k fm,n (am + bn + kx)
. (29)
= k!kx
am + bn + kx
k
k
Proof. We have
�
m+n≥k
Am,n,k (ijfi−1,j−1 (x))
tm tn
m! n!
12
INTEGERS: 11 (2011)
k
1 �
ti uj
=
ijfi−1,j−1 (x)
k!
i! j!
i+j≥1
k k
=
t u
k!
�
i
fi,j (x)
i,j≥0
=
j
k
t u
i! j!
tm un
tk uk �
fm,n (kx)
k!
m! n!
m,n≥0
1 � (m + k)! (n + k)!
tm+k
un+k
=
fm,n (kx)
k!
m!
n!
(m + k)! (n + k)!
m,n≥0
�
��
�
� m n
tm un
,
fm−k,n−k (kx)
=k!
m! n!
k
k
m,n≥k
which gives
� �� �
m n
fm−k,n−k (kx) ,
Am,n,k (ijfi−1,j−1 (x)) = k!
k
k
m, n ≥ k.
(30)
To finish the proof, we substitute the trinomial type polynomials (hm,n (x)) given
by (26), instead of (fm,n (x)), and replace m by m + k and n by n + k.
Example 9. From relation (29), for r = i − 1 and s = j − 1, we have
� x �
,
1. fm,n (x) = (m + n)! m+n
�
�
�ar+bs+x��
�
(m+n
m ) (m+k)!(n+k)! am+bn+kx
.
= x (k−1)!
Am+k,n+k,k x (r+1)(s+1)(r+s)!
r+s
m+n
ar+bs+x
am+bn+kx
2. fm,n (x) = Bm+n (x) ,
�
�
(ar+bs+x)
Bm+n (am+bn+kx)
Am+k,n+k,k x (r + 1) (s + 1) Br+s
.
= x (m+k)!(n+k)!
ar+bs+x
(k−1)!m!n!
am+bn+kx
m−1
n−1
3. fm,n (x) = x2 ((c − a) m + x)
((d − b) n + x)
,
�
�
r−1
s−1
Am+k,n+k,k x (r + 1) (s + 1) (ar + bs + x) (cr + bs + x)
(ar + ds + x)
��n+k� (am+bn+kx)(cm+bn+kx)m−1
�
.
= k!kx m+k
k
k
(am+dn+kx)1−n
(c−a)m+x
m!n!x2
m
((c−a)m+x)((d−b)n+x)
��(d−b)n+x�
,
n
�
�
�
��
�
cr+bs+x ar+ds+x
Am+k,n+k,k x (r+1)!(s+1)!(ar+bs+x)
s
r
(cr+bs+x)(ar+ds+x)
x(m+k)!(n+k)!(am+bn+kx) �cm+bn+kx��am+dn+kx�
.
= (k−1)!(cm+bn+kx)(am+dn+kx)
n
m
4. fm,n (x) =
�
13
INTEGERS: 11 (2011)
�
2
m−1 �
(d−b)n+x
,
5. fm,n (x) = n! x ((c−a)m+x)
n
(d−b)n+x
�
��
i−2 �
ar+ds+x
Am+k,n+k,k x(r+1)(s+1)!(ar+bs+x)(cr+bs+x)
s
(ar+ds+x)
�
m−1 �
am+dn+kx
.
= x(m+k)!(n+k)!(am+bn+kx)(cm+bn+kx)
n
(k−1)!m!(am+dn+kx)
6. fm,n (x) =
x2 ((c−a)m+x)m−1
Bn
(d−b)n+x
Am+k,n+k,k
=
�
((d − b) n + x) ,
x(r+1)(s+1)(ar+bs+x)(cr+bs+x)r−1
Bs
(ar+ds+x)
�
(ar + ds + x)
�m+k��n+k�
k!kx(am+bn+kx)
k Bn
k
(am+dn+kx)(cm+bn+kx)1−m
(am + dn + kx) .
�(c−a)m+x�
x2
Bn ((d − b) n + x) ,
7. fm,n (x) = m! ((c−a)m+x)((d−b)n+x)
m
�
��
�
s (ar+ds+x) cr+bs+x
Am+k,n+k,k x(s+1)(r+1)!(ar+bs+x)B
r
(cr+bs+x)(ar+ds+x)
�
�
n (am+dn+kx) cm+bn+kx
= x(m+k)!(n+k)!(am+bn+kx)B
.
m
(k−1)!n!(cm+bn+kx)(am+dn+kx)
Theorem 10. Let (fm,n (x)) be a sequence of trinomial type. We have
�
�
fi−1,j (a (i − 1) + bj + x)
Am+k,n,k xi
a (i − 1) + bj + x
�
�
m + k fm,n (am + bn + kx)
= xk
, (31)
k
am + bn + kx
�
�
fi,j−1 (ai + b (j − 1) + x)
Am,n+k,k xj
ai + b (j − 1) + x
= xk
�
�
n + k fm,n (am + bn + kx)
. (32)
am + bn + kx
k
Proof. Similarly to the proof in Theorem 8, we obtain
� �
m
fm−k,n (kx) , m ≥ k, n ≥ 0,
Am,n,k (ifi−1,j (x)) =
k
� �
n
fm,n−k (kx) , m ≥ 0, n ≥ k.
Am,n,k (jfi,j−1 (x)) =
k
It will be sufficient to use the trinomial type polynomials (hm,n (x)) given in (26)
instead of (fm,n (x)) and replace m by m + k and n by n + k respectively.
Example 11. From relations (31) and (32), we have the following:
� x �
,
1. fm,n (x) = (m + n)! m+n
�
��
�
kx(m+n)! �m+k��am+bn+kx�
(i+j−1)!
a(i−1)+bj+x
,
= am+bn+kx
Am+k,n,k xi a(i−1)+bj+x
m+n
i+j−1
k
�
�
kx(m+n)! �n+k��am+bn+kx�
(i+j−1)! �ai+b(j−1)+x�
.
= am+bn+kx
Am,n+k,k xj ai+b(j−1)+x
m+n
i+j−1
k
14
INTEGERS: 11 (2011)
2. fm,n (x) = Bm+n (x) ,
�
�
� Bm+n(am+bn+kx)
�
B
(a(i−1)+bj+x)
,
= xk m+k
Am+k,n,k xi i+j−1
a(i−1)+bj+x
am+bn+kx
k
�
�
�
�
B
(ai+b(j−1)+x)
Bm+n (am+bn+kx)
Am,n+k,k xj i+j−1
.
= xk n+k
ai+b(j−1)+x
am+bn+kx
k
m−1
n−1
3. fm,n (x) = x2 ((c − a) m + x)
((d − b) n + x)
,
�
�
i−2
j−2
Am+k,n,k xi (a (i − 1) + bj + x) (c (i − 1) + bj + x)
(a (i − 1) + dj + x)
�
�
m−1
n−1
(am + bn + kx) (cm + bn + kx)
(am + dn + kx)
,
= xk m+k
k
�
�
i−2
j−2
Am,n+k,k xj (ai + b (j − 1) + x) (ci + b (j − 1) + x)
(ai + d (j − 1) + x)
�
�
m−1
n−1
(am + bn + kx) (cm + bn + kx)
(am + dn + kx)
.
= xk n+k
k
�
��
�
(c−a)m+x (d−b)n+x
x2
,
4. fm,n (x) = m!n! ((c−a)m+x)((d−b)n+x)
n
m
�
�c(i−1)+bj+x��a(i−1)+dj+x��
xi!(j−1)!(a(i−1)+bj+x)
Am+k,n,k (c(i−1)+bj+x)(a(i−1)+dj+x)
j
i−1
�cm+bn+kx��am+dn+kx�
(m+k)!n!
am+bn+kx
= x (k−1)! (cm+bn+kx)(am+dn+kx)
,
n
m
�
�
�ci+b(j−1)+x��ai+d(j−1)+x�
x(i−1)!j!(ai+b(j−1)+x)
Am+k,n,k (ci+b(j−1)+x)(ai+d(j−1)+x)
j−1
i
�
��
�
cm+bn+kx am+dn+kx
am+bn+kx
.
= x m!(n+k)!
n
m
(k−1)! (cm+bn+kx)(am+dn+kx)
�
2
m−1 �
(d−b)n+x
5. fm,n (x) = n! x ((c−a)m+x)
,
n
(d−b)n+x
�
�a(i−1)+dj+x��
xij!(a(i−1)+bj+x)
Am+k,n,k (a(i−1)+dj+x)(c(i−1)+bj+x)
2−i
j
x(m+k)!n!(am+bn+kx)(cm+bn+kx)m−1 �am+dn+kx�
=
,
n
(k−1)!m!(am+dn+kx)
�
�
�ai+d(j−1)+x�
xi!j(ai+b(j−1)+x)
Am,n+k,k (ai+d(j−1)+x)(ci+b(j−1)+x)
2−j
j−1
xm!(n+k)!(am+bn+kx)(cm+bn+kx)m−1 �am+dn+kx�
=
.
n
(k−1)!n!(am+dn+kx)
2
m−1
6. fm,n (x) = x ((c−a)m+x)
Bn ((d − b) n + x) ,
(d−b)n+x
�
�
i−2
B
(a
(i
−
1)
+
dj
+
x)
Am+k,n,k xi(a(i−1)+bj+x)(c(i−1)+bj+x)
j
a(i−1)+dj+x
�m+k� (am+bn+kx)(cm+bn+kx)m−1
Bn (am + dn + kx) ,
= xk k
am+dn+kx
�
�
i−1
Am,n+k,k xj(ai+b(j−1)+x)(ci+b(j−1)+x)
B
(ai
+
d
(j
−
1)
+
x)
j−1
ai+d(j−1)+x
�n+k� (am+bn+kx)(cm+bn+kx)m−1
Bn (am + dn + kx) .
= xk k
am+dn+kx
15
INTEGERS: 11 (2011)
�(c−a)m+x�
x2
Bn ((d − b) n + x) ,
7. fm,n (x) = m! ((c−a)m+x)((d−b)n+x)
m
�
�
�
xi!(a(i−1)+bj+x)Bj (a(i−1)+dj+x) c(i−1)+bj+x�
Am+k,n,k
i−1
(c(i−1)+bj+x)(a(i−1)+dj+x)
�cm+bn+kx�
x(m+k)!(am+bn+kx)
Bn (am + dn + kx) ,
= (k−1)!(cm+bn+kx)(am+dn+kx)
m
�
�
�
xj!(ai+b(j−1)+x)Bj−1 (ai+d(j−1)+x) ci+b(j−1)+x�
Am,n+k,k
i
(ci+b(j−1)+x)(ai+d(j−1)+x)
�cm+bn+kx�
x(n+k)!(am+bn+kx)
Bn (am + dn + kx) .
= (k−1)!(cm+bn+kx)(am+dn+kx)
m
Theorem 12. Let (fm,n (x)) be a sequence of trinomial type. We have
�
�
x
x
Am,n
fi,j (ai + bj) =
fm,n (am + bn + x) .
ai + bj
am + bn + x
(33)
Proof. From (28), we get
Am,n (xDα=0 fi,j (α)) = fm,n (x) ,
which gives the desired identity by replacing fm,n (x) by
x
fm,n (am + bn + x) .
hm,n (x) :=
am + bn + x
given by (26).
Example 13. From relation (33), we have the following:
� x �
,
1. fm,n (x) = (m + n)! m+n
�
��
�
�
�
(i + j)! ai + bj
(m + n)!
(am + bn + x)
Am,n x
=x
.
m+n
ai + bj i + j
am + bn + x
2. fm,n (x) = Bm+n (x) ,
�
�
x
x
Bi+j (ai + bj) =
Bm+n (am + bn + x) .
Am,n
ai + bj
am + bn + x
m−1
n−1
3. fm,n (x) = x2 ((c − a) m + x)
((d − b) n + x)
�
�
i−1
j−1
Am,n x (ai + bj) (ci + bj)
(ai + dj)
m−1
= x (am + bn + x) (cm + bn + x)
2
x
4. fm,n (x) = m!n! ((c−a)m+x)((d−b)n+x)
,
n−1
(am + dn + x)
�(c−a)m+x��(d−b)n+x�
,
n
m
.
��
��
�
ai + dj
ci + bj
(ai + bj)
j
i
(ci + bj) (ai + dj)
�
��
�
cm + bn + x am + dn + x
(am + bn + x)
.
= xm!n!
n
m
(cm + bn + x) (am + dn + x)
�
Am,n xi!j!
16
INTEGERS: 11 (2011)
2
5. fm,n (x) = n! x
Am,n
�
((c−a)m+x)m−1 �(d−b)n+x�
,
n
(d−b)n+x
i−1 �
= xn!
6. fm,n (x) =
Am,n
ai + dj
j
(ai + bj) (ci + bj)
xj!
ai + dj
�
��
m−1 �
(am + bn + x) (cm + bn + x)
am + dn + x
x2 ((c−a)m+x)m−1
Bn
(d−b)n+x
((d − b) n + x) ,
i−1
x (ai + bj) (ci + bj)
ai + dj
�
am + dn + x
.
n
�
Bj (ai + dj)
m−1
=
x (am + bn + x) (cm + bn + x)
am + dn + x
2
x
7. fm,n (x) = m! ((c−a)m+x)((d−b)n+x)
Am,n
Bn (am + dn + x) .
�(c−a)m+x�
Bn ((d − b) n + x) ,
m
�
�
�
(ai + bj) ci+bj
i
Bj (ai + dj)
xi!
(ci + bj) (ai + dj)
�
�
(am + bn + x) cm+bn+x
m
= m!
Bn (am + dn + x) .
(cm + bn + x) (am + dn + x)
�
Remark. For m = 0, identity (17) and Theorem 10 give Proposition 1 given in [3],
and identity (20) and Theorem 12 give Proposition 3 given in [3].
Acknowledgments The authors wish to warmly thank their referee and the editor
for their valuable advice and comments which helped to greatly improve the quality
of this paper.
References
[1] Charalambos A. Charalambides, Enumerative Combinatorics. Chapman & Hall/CRC, A CRC
Press Company, (2001).
[2] L. Comtet, Advanced Combinatorics. D. Reidel Publishing Company, Dordrecht-Holland /
Boston-U.S.A, (1974).
[3] M. Mihoubi, Bell polynomials and binomial type sequences. Discrete Math. 308 (2008), 2450–
2459.
INTEGERS: 11 (2011)
17
[4] M. Mihoubi, The role of binomial type sequences in determination identities for Bell polynomials. To appear, Ars Combin. Preprint available at http://arxiv.org/abs/0806.3468v1.
[5] M. Mihoubi, Some congruences for the partial Bell polynomials. J. Integer Seq. 12 (2009),
Article 09.4.1.
[6] J. Riordan, Combinatorial Identities. Huntington, New York, (1979).
[7] S. Roman, The Umbral Calculus. Academic Press (1984).
[8] J. Zeng, Multinomial Convolution Polynomials, Discrete Mathematics 160 (1996), 219–228.
#A18
THE (EXPONENTIAL) BIPARTITIONAL POLYNOMIALS AND
POLYNOMIAL SEQUENCES OF TRINOMIAL TYPE: PART I
Miloud Mihoubi1
Universit´e des Sciences et de la Technologie Houari Boumediene, Faculty of
Mathematics, El Alia, Bab Ezzouar, Algeria
mmihoubi@usthb.dz, miloudmihoubi@gmail.com
Hac`
ene Belbachir2
Universit´e des Sciences et de la Technologie Houari Boumediene, Faculty of
Mathematics, El Alia, Bab Ezzouar, Algeria
hbelbachir@usthb.dz, hacenebelbachir@gmail.com
Received: 3/29/10, Revised: 10/3/10, Accepted: 12/31/10, Published: 2/4/11
Abstract
The aim of this paper is to investigate and present the general properties of the
(exponential) bipartitional polynomials. After establishing relations between bipartitional polynomials and polynomial sequences of binomial and trinomial type, a
number of identities are deduced.
1. Introduction
For (m, n) ∈ N2 with N = {0, 1, 2, . . .}, the complete bipartitional polynomial
Am,n ≡ Am,n (x0,1 , x1,0 , x0,2 , x1,1 , x2,0 , . . . , xm,n )
with variables x0,1 , x1,0 , . . . , xm,n is defined by the sum
�x
� x �k0,1 � x �k1,0
�km,n
�
m!n!
1,0
m,n
0,1
···
,
Am,n :=
k0,1 !k1,0 ! · · · km,n ! 0!1!
1!0!
m!n!
(1)
where the summation is carried out over all partitions of the bipartite number
(m, n), that is, over all nonnegative integers k0,1 , k1,0 , . . . , km,n which are solution
to the equations
m
n �
n
m �
�
�
(2)
j ki,j = n.
i ki,j = m,
i=1 j=0
1 The
2 The
j=1 i=0
research is partially supported by LAID3 Laboratory of USTHB University.
research is partially supported by LAID3 Laboratory and CMEP project 09 MDU 765.
2
INTEGERS: 11 (2011)
The partial bipartitional polynomial Am,n,k ≡ Am,n,k (x0,1 , . . . , xm,n ) with variables
x0,1 , x1,0 , . . . , xm,n , of degree k ∈ N, is defined by the sum
�x
� x �k0,1 � x �k1,0
�km,n
�
m!n!
m,n
1,0
0,1
Am,n,k :=
···
, (3)
k0,1 !k1,0 ! · · · km,n ! 0!1!
1!0!
m!n!
where the summation is carried out over all partitions of the bipartite number
(m, n) into k parts, that is, over all nonnegative integers k0,1 , k1,0 , . . . , km,n which
are solution to the equations
m �
n
n �
m
m �
n
�
�
�
i ki,j = m,
j ki,j = n,
ki,j = k, with the convention k0,0 = 0.
i=1 j=0
j=1 i=0
i=0 j=0
(4)
These polynomials were introduced in [1, pp. 454] with properties such as those
given in (5), (6), (7), (8) below. Indeed, for all real numbers α, β, γ, we have
�
�
Am,n,k βγx0,1 , αγx1,0 , β 2 γx0,2 , . . . , αm β n γxm,n =
β m αn γ k Am,n,k (x0,1 , x1,0 , . . . , xm,n ) , (5)
and
Am,n (αx0,1 , βx1,0 , . . . , αm β n xm,n ) = αm β n Am,n (x0,1 , x1,0 , . . . , xm,n ) .
(6)
Moreover, we can check that the exponential generating functions for Am,n and
Am,n,k are respectively provided by
i j
�
�
t
u
tm un
,
(7)
= exp
xi,j
1+
Am,n (x0,1 , . . .)
m! n!
i! j!
i+j≥1
m+n≥1
and
�
m+n≥k
k
tm un
1 �
ti uj
Am,n,k (x0,1 , . . .)
.
=
xi,j
m! n!
k!
i! j!
(8)
i+j≥1
From the above definitions and properties, we attempt to present more properties and identities for the partial and complete bipartitional polynomials. We also
consider the connection between these polynomials and polynomials (fm,n (x)) of
trinomial type, defined by
x
i j
�
�
tm un
t u
=
fm,n (x)
, with f0,0 (x) := 1.
(9)
fi,j (1)
i! j!
m! n!
i,j≥0
m,n≥0
For the investigation of polynomials of trinomial and multinomial type, we can refer
to [8].
3
INTEGERS: 11 (2011)
We use the notations Am,n,k (xi,j ) or Am,n,k (xI,J ) for Am,n,k (x0,1 , x1,0 , . . .), and
Am,n (xi,j ) or Am,n (xI,J ) for Am,n (x0,1 , . . .). Moreover, we represent respectively
by Bn,k (xj ) and Bn (xj ), the partial and complete Bell polynomials Bn,k (x1 , x2 , . . .)
and Bn (x1 , x2 , . . .). We recall that these polynomials are defined by their generating
functions
� ∞
�k
∞
�
tm
1 �
tn
xm
=
,
Bn,k (xj )
n!
k! m=1 m!
n=k
�
� ∞
∞
�
�
tn
tm
Bn (xj )
1+
xm
,
= exp
n!
m!
n=1
m=1
which admit the following explicit expressions:
k1 +k2 +···=k,
Bn (xj ) =
� x �k1 � x �k2
n!
2
1
··· ,
k1 !k2 ! · · · 1!
2!
k1 +2k2 +···=n
� x �k1 � x �k2
n!
2
1
··· .
k1 !k2 ! · · · 1!
2!
�
Bn,k (xj ) =
�
k1 +2k2 +3k3 +···=n
For m < 0 or n < 0, we set fm,n (x) = 0 and Am,n (xi,j ) = 0, and for m < 0 or
n < 0 or k < 0, we set Am,n,k (xi,j ) = 0. For x ∈ C, where C is the set of complex
numbers, we set
� �
1
x
:= x (x − 1) · · · (x − k + 1) for k = 1, 2, . . . ,
k!
k
� �
� �
x
x
= 0 otherwise,
= 1 and
k
0
� �
which gives nk = 0 for k > n, n ∈ N.
2. Properties of Bipartitional Polynomials
In this section, we give some recurrence relations for the bipartitional polynomials
from which we deduce connections with Bell polynomials. In the following, we
consider a sequence (xn,m ) of real numbers with x0,0 = 0.
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INTEGERS: 11 (2011)
Theorem 1. We start with
� �� �
�
m n
xi,j Am−i,n−j,k−1 (xI,J ) = kAm,n,k (xI,J ) ,
j
i
i+j≤m+n−k+1
� �� �
�
m n
ixi,j Am−i,n−j,k−1 (xI,J ) = mAm,n,k (xI,J ) ,
j
i
i+j≤m+n−k+1
� �� �
�
m n
jxi,j Am−i,n−j,k−1 (xI,J ) = nAm,n,k (xI,J ) ,
j
i
(10)
(11)
(12)
i+j≤m+n−k+1
and
��m��n�
ixi,j Am−i,n−j (xI,J ) = mAm,n (xI,J ) ,
i
j
i,j
��m��n�
jxi,j Am−i,n−j (xI,J ) = nAm,n (xI,J ) .
j
i
i,j
(13)
(14)
Proof. From (8), we obtain
k
k−1
i j
�
d 1 �
1
tr us
ti uj
t
u
=
xi,j
xi,j
dxr,s k!
i! j!
(k − 1)!
i! j!
r! s!
i+j≥1
i+j≥1
�
��
�
tm+r
m+r n+s
un+s
Am,n,k−1 (xI,J )
s
r
(m + r)! (n + s)!
m+n≥k−1
�
��
�
�
tm un
m n
,
Am−r,n−s,k−1 (xI,J )
=
m! n!
s
r
=
�
m+n≥r+s+k−1
we also have
k
�
d 1 �
d
tm un
ti uj
=
Am,n,k (xI,J )
.
xi,j
dxr,s k!
i! j!
dxr,s
m! n!
i+j≥1
m+n≥k
From these two expressions, we deduce that
� �� �
m n
d
Am−r,n−s,k−1 (xI,J ) .
Am,n,k (xI,J ) =
s
r
dxr,s
(15)
For α = β = 1, we take the derivatives on both sides of (5) with respect to γ and,
by means of (15), we obtain (10). The same goes for identities (11) and (12). When
we sum over all possible values of k for both sides of each of the identities (11) and
(12), we obtain identities (13) and (14) respectively.
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INTEGERS: 11 (2011)
The next theorem provides the link between the bipartitional polynomials and the
classical Bell polynomials.
Theorem 2. We have
�
� j � �
n � �
� j
�
n
xi,j−i ,
Aj,n−j,k (xI,J ) = Bn,k
i
j
i=0
j=0
(16)
A0,n,k (xI,J ) = Bn,k (x0,j ) ,
(17)
Am,0,k (xI,J ) = Bm,k (xj,0 ) ,
(18)
and
�
� j � �
n � �
� j
�
n
xi,j−i ,
Aj,n−j (xI,J ) = Bn
i
j
i=0
j=0
(19)
A0,n (xI,J ) = Bn (x0,j ) ,
(20)
Am,0 (xI,J ) = Bm (xj,0 ) .
(21)
Proof. From (8), if we set t = u, we obtain
k
i+j
�
�
1
um+n
u
=
Am,n,k (xI,J )
,
xi,j
k!
i!j!
m!n!
i,j≥0
i.e.
1
k!
m+n≥k
�k
�∞
s � �
s � �
� us �
� us �
s
s
=
xi,s−i
Am,s−m,k (xI,J ) ,
i
m
s!
s!
s=1
m=0
i=0
s≥k
which, by identification, gives the identity (16). The identities (17) and (18) result
by setting respectivelly t = 0 and u = 0 in (8).
Summing over all possible values of k on both sides of each of the identities (16),
(17) and (18), we obtain identities (19), (20) and (21) respectively.
For a sequence of real numbers (xn ), we denote by Am,n,k (xi+j ) the quantity obtained by replacing xi,j by xi+j in Am,n,k (xi,j ). The same goes for Am,n (xi+j ) .
Theorem 3. Let (xn ) be a sequence of real numbers. The following holds
Am,n,k (xi+j ) = Bm+n,k (xj ) ,
Am,n (xi+j ) = Bm+n (xj ) ,
�−1
��
�
m+n m+n+k
Bm+n+k,k (jxj ) ,
Am+k,n,k (ixi+j ) =
n
n
�−1
��
�
m+n m+n+k
Bm+n+k,k (jxj ) .
Am,n+k,k (jxi+j ) =
m
m
(22)
(23)
(24)
(25)
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INTEGERS: 11 (2011)
Proof. From (8), we get for xi,j = xi+j that
k
�
tm un
ti uj
1 �
=
Am,n,k (xi+j )
;
xi+j
k!
i! j!
m! n!
i+j≥1
we also get
m+n≥k
k
k
1 � xs
ti uj
1 �
s
=
(t + u)
xi+j
k!
i! j!
k!
s!
i+j≥1
s≥1
s
�
(t + u)
=
Bs,k (xj )
s!
s≥k
=
�
Bm+n,k (xj )
m+n≥k
tm un
.
m! n!
By identification, we obtain (22). Identity (23) follows by summing over all possible
values of k on both sides of identity (22).We have
k
k
1 �
1 � xs � ti uj
ti uj
=
i
ixi+j
k!
i! j!
k!
s! i+j=s i! j!
i+j≥1
s≥1
k
tk � xs d
s
=
(t + u)
k!
s! dt
s≥1
1
=
k!
�
t
t+u
�k
k
s
�
(t
+
u)
sxs
s!
s≥1
r−k
= tk
�
Br,k (jxj )
r≥k
=
�
m+n≥k
�m+n−k�
(t + u)
r!
m−k
� Bm+n−k,k (jxj )
�m+n
n
tm un
.
m! n!
Then, by identification with (8) and making use of the symmetry of m, n, we obtain
(24) and (25).
Note that, for Theorem 3, using the identities on partial and complete Bell
polynomials given in [3] and [4], we can deduce several identities for the partial and
complete bipartitional polynomials.
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INTEGERS: 11 (2011)
3. Polynomial Sequences of Trinomial Type
We know that the sequences of binomial type have a strong relationship with the
partial and complete Bell polynomials. If (fn (x)) is a sequence of binomial type,
x
fn (an + x) is also of binomial type.
the sequence (hn (x)) defined by hn (x) := an+x
For sequences (fn (x))
satisfying
the
property
of
convolution, the following is holds:
�n � �
fn (x + y) = k=0 nk fk (x) fn−k (y) with f0 (x) �= 1.
Theorem 4. Let (fm,n (x)) be a sequence of trinomial type and let a, b be real
numbers. Then the sequence (hm,n (x)) defined by
hm,n (x) :=
x
fm,n (am + bn + x)
am + bn + x
(26)
is of trinomial type.
Proof. The definition of (fm,n (x)), given in (9), states that
x
�∞
�
∞
∞
∞
n
i
j
�
�
�
�
u
t
tm
u
fi,j (1) =
fm,n (x)
,
j! i!
n! m!
m=0 n=0
j=0
i=0
which can also be written as follows:
�x
�∞
∞
∞
�
�
�
un
tm
ti
fm,n (x)
um (x)
with um (x) :=
.
=
ui (1)
i!
m!
n!
n=0
m=0
i=0
This implies that the sequence (um (x)) satisfies the property of convolution. Hence,
the sequence (Um (x)) given by
Um (x) :=
∞
un
x �
x
fm,n (am + x)
um (am + x) =
,
am + x
am + x n=0
n!
has the same property of convolution. We have
�∞
�
i
Ui (1) ti!
i=0
�x
=
∞
�
m
Um (x) tm! ,
m=0
or similarly
�∞
�x
∞
∞
�
�
�
ui
tm
un
x
vi (1)
, with vn (x) :=
fm,n (am + x)
.
vn (x)
=
i!
n!
am + x
m!
m=0
n=0
i=0
This implies that the sequence (vm (x)) satisfies the property of convolution. Moreover, the sequence (Vm (x)) given by
Vn (x) :=
∞
�
tm
x
x
vn (bn + x) =
fm,n (am + bn + x)
,
bn + x
am + bn + x
m!
m=0
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INTEGERS: 11 (2011)
has the same property of convolution. We have
�
∞
�
j
Vj (1) uj!
j=0
or similarly
x
i j
�
�
t u
tm un
hi,j (1)
=
hm,n (x)
i! j!
m! n!
i,j≥0
�x
=
∞
�
n
Vn (x) un! ,
n=0
with h0,0 (x) = 1.
m,n≥0
Consequently, the last identity shows that the sequence (hm,n (x)) is of trinomial
type.
Remark. From (9), we can verify that if a sequence (fm,n (x)) is of trinomial type,
then
n � �� �
m �
�
m n
fm,n (x + y) =
fi,j (x) fm−i,n−j (y) .
i
j
i=0 j=0
If in (9) we set t = 0, we deduce that the sequence (fn (x)), defined by fn (x) :=
f0,n (x), is of binomial type.
Theorem 5. Let (fn (x)) and (gn (x)) be two sequences of binomial type with
f0 (x) = g0 (x) = 1. Then the sequences (pm,n (x)) and (qm,n (x)) defined by
pm,n (x) := fm+n (x) ,
qm,n (x) := fm (x) gn (x)
are sequences of trinomial type.
Proof. We have p0,0 (x) = q0,0 (x) = 1 and
�
tm un
pm,n (x)
m! n!
=
m,n≥0
k � �
� fk (x) �
k m k−m
t u
k! m=0 m
k≥0
k
=
�
fk (x)
k≥0
=
k
�
fk (1)
k≥0
�
tm un
qm,n (x)
m! n!
m,n≥0
(t + u)
k!
x
(t + u)
,
k!
n
m
�
u
t
=
gn (x)
fm (x)
m!
n!
n≥0
m≥0
x
m
n
�
�
t
u
.
=
fm (1)
gn (1)
m!
n!
�
m≥0
n≥0
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INTEGERS: 11 (2011)
Remark. Let (fn (x)) and (gn (x)) be two sequences of binomial type with f0 (x) =
g0 (x) = 1. Then, from [3], the sequences (Fn (x)) and (Gn (x)) defined by
Fn (x) :=
x
fn ((c − a) n + x) ,
(c − a) n + x
Gn (x) :=
x
gn ((d − b) n + x) ,
(d − b) n + x
and
are of binomial type for all real numbers a, b, c, d. This implies, according to Theorem 4, that, for all a, b, c, d, the sequences
Pm,n (x) := x
and
Qm,n (x) := x (am + bn + x)
fm+n (am + bn + x)
,
am + bn + x
fm (cm + bn + x) gn (am + dn + x)
,
cm + bn + x
am + dn + x
are of trinomial type.
� � ��
Example 6. We know that sequences (xn ) , n! nx and (Bn (x)) are of binomial
type, where Bn (x) is the single variable Bell polynomial. The sequences
�
�
x
an + x
n!x
n−1
,
Bn (an + x) ,
x (an + x)
,
n
an + x
an + x
are also of binomial type.
Combining these sequences, Theorem 5 states that the following sequences are also
of trinomial type:
m−1
x2 (cm + x)
n−1
(dn + x)
,
�
m−1 �
2
x (cm + x)
dn + x
n!
,
n
dn + x
�
��
�
x2
cm + x dn + x
m!n!
,
n
m
(cm + x) (dn + x)
m−1
x2 (cm + x)
Bn (dn + x) ,
dn + x
�
�
x2
cm + x
m!
Bn (dn + x) ,
(cm + x) (dn + x)
m
�
�
x
(m + n)!
,
m+n
Bm+n (x) .
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INTEGERS: 11 (2011)
Combining these sequences and using Theorem 4, we deduce that the following
sequences are of trinomial type:
m−1
am,n (x) = x (am + bn + x) (cm + bn + x)
n−1
(am + dn + x)
,
�
m−1 �
am + dn + x
x (am + bn + x) (cm + bn + x)
,
bm,n (x) = n!
am + dn + x
n
�
��
�
x (am + bn + x)
cm + bn + x am + dn + x
cm,n (x) = m!n!
,
n
m
(cm + bn + x) (am + dn + x)
m−1
x (am + bn + x) (cm + bn + x)
Bn (am + dn + x) ,
am + dn + x
�
�
x (am + bn + x)
cm + bn + x
Bn (am + dn + x) ,
em,n (x) = m!
m
(cm + bn + x) (am + dn + x)
�
�
(m + n)!x am + bn + x
,
fm,n (x) =
m+n
am + bn + x
x
gm,n (x) =
Bm+n (am + bn + x) .
am + bn + x
dm,n (x) =
4. Bipartitional Polynomials and Polynomials of Binomial and Trinomial
Type
Roman [7, p. 82] proves that any polynomial sequence of binomial type (fn (x)),
with f0 (x) = 1, is connected with the partial Bell polynomials
fn (x) =
n
�
Bn,k (Dα=0 fj (α)) xk = Bn (xDα=0 fj (α)) , n ≥ 1.
(27)
k=1
Similarly, we establish that any polynomial sequence of trinomial type (fm,n (x)),
with f0,0 (x) = 1, admits a connection with the partial bipartitional Bell polynomials.
Theorem 7. Let (fm,n (x)) be a sequence of trinomial type. Then
fm,n (x) =
m+n
�
Am,n,k (Dα=0 fi,j (α)) xk = Am,n (xDα=0 fi,j (α)) , m+n ≥ 1, (28)
k=1
where Dα=0 ≡
�
d �
dα α=0
.
11
INTEGERS: 11 (2011)
Proof. Since Dα=0 f0,0 (α) = 0, then for xi,j = Dα=0 fi,j (α), we have
�
� tm un m+n
Am,n,k (xi,j ) xk
m! n!
m,n≥0
=
k=0
=
=
=
=
=
∞
�
tm un
m! n!
k=0
m+n≥k
k
∞ k
�
x �
ti uj
Dα=0 fi,j (α)
k!
i! j!
i,j≥0
k=0
i j
�
t
u
exp xDα=0
fi,j (α)
i! j!
i,j≥0
α
i j
�
t
u
exp xDα=0
fi,j (1)
i! j!
i,j≥0
i j
�
t
u
fi,j (1)
exp x ln
i! j!
i,j≥0
x
i j
�
u
t
fi,j (1)
i! j!
�
xk
Am,n,k (xi,j )
i,j≥0
=
�
fm,n (x)
m,n≥0
tm un
.
m! n!
Therefore, the desired identity follows by identification.
Some identities related to these polynomials are given by the following theorem.
Theorem 8. Let (fm,n (x)) be a sequence of trinomial type. We have
�
�
fi−1,j−1 (a (i − 1) + b (j − 1) + x)
Am+k,n+k,k xij
a (i − 1) + b (j − 1) + x
�
��
�
m + k n + k fm,n (am + bn + kx)
. (29)
= k!kx
am + bn + kx
k
k
Proof. We have
�
m+n≥k
Am,n,k (ijfi−1,j−1 (x))
tm tn
m! n!
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INTEGERS: 11 (2011)
k
1 �
ti uj
=
ijfi−1,j−1 (x)
k!
i! j!
i+j≥1
k k
=
t u
k!
�
i
fi,j (x)
i,j≥0
=
j
k
t u
i! j!
tm un
tk uk �
fm,n (kx)
k!
m! n!
m,n≥0
1 � (m + k)! (n + k)!
tm+k
un+k
=
fm,n (kx)
k!
m!
n!
(m + k)! (n + k)!
m,n≥0
�
��
�
� m n
tm un
,
fm−k,n−k (kx)
=k!
m! n!
k
k
m,n≥k
which gives
� �� �
m n
fm−k,n−k (kx) ,
Am,n,k (ijfi−1,j−1 (x)) = k!
k
k
m, n ≥ k.
(30)
To finish the proof, we substitute the trinomial type polynomials (hm,n (x)) given
by (26), instead of (fm,n (x)), and replace m by m + k and n by n + k.
Example 9. From relation (29), for r = i − 1 and s = j − 1, we have
� x �
,
1. fm,n (x) = (m + n)! m+n
�
�
�ar+bs+x��
�
(m+n
m ) (m+k)!(n+k)! am+bn+kx
.
= x (k−1)!
Am+k,n+k,k x (r+1)(s+1)(r+s)!
r+s
m+n
ar+bs+x
am+bn+kx
2. fm,n (x) = Bm+n (x) ,
�
�
(ar+bs+x)
Bm+n (am+bn+kx)
Am+k,n+k,k x (r + 1) (s + 1) Br+s
.
= x (m+k)!(n+k)!
ar+bs+x
(k−1)!m!n!
am+bn+kx
m−1
n−1
3. fm,n (x) = x2 ((c − a) m + x)
((d − b) n + x)
,
�
�
r−1
s−1
Am+k,n+k,k x (r + 1) (s + 1) (ar + bs + x) (cr + bs + x)
(ar + ds + x)
��n+k� (am+bn+kx)(cm+bn+kx)m−1
�
.
= k!kx m+k
k
k
(am+dn+kx)1−n
(c−a)m+x
m!n!x2
m
((c−a)m+x)((d−b)n+x)
��(d−b)n+x�
,
n
�
�
�
��
�
cr+bs+x ar+ds+x
Am+k,n+k,k x (r+1)!(s+1)!(ar+bs+x)
s
r
(cr+bs+x)(ar+ds+x)
x(m+k)!(n+k)!(am+bn+kx) �cm+bn+kx��am+dn+kx�
.
= (k−1)!(cm+bn+kx)(am+dn+kx)
n
m
4. fm,n (x) =
�
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INTEGERS: 11 (2011)
�
2
m−1 �
(d−b)n+x
,
5. fm,n (x) = n! x ((c−a)m+x)
n
(d−b)n+x
�
��
i−2 �
ar+ds+x
Am+k,n+k,k x(r+1)(s+1)!(ar+bs+x)(cr+bs+x)
s
(ar+ds+x)
�
m−1 �
am+dn+kx
.
= x(m+k)!(n+k)!(am+bn+kx)(cm+bn+kx)
n
(k−1)!m!(am+dn+kx)
6. fm,n (x) =
x2 ((c−a)m+x)m−1
Bn
(d−b)n+x
Am+k,n+k,k
=
�
((d − b) n + x) ,
x(r+1)(s+1)(ar+bs+x)(cr+bs+x)r−1
Bs
(ar+ds+x)
�
(ar + ds + x)
�m+k��n+k�
k!kx(am+bn+kx)
k Bn
k
(am+dn+kx)(cm+bn+kx)1−m
(am + dn + kx) .
�(c−a)m+x�
x2
Bn ((d − b) n + x) ,
7. fm,n (x) = m! ((c−a)m+x)((d−b)n+x)
m
�
��
�
s (ar+ds+x) cr+bs+x
Am+k,n+k,k x(s+1)(r+1)!(ar+bs+x)B
r
(cr+bs+x)(ar+ds+x)
�
�
n (am+dn+kx) cm+bn+kx
= x(m+k)!(n+k)!(am+bn+kx)B
.
m
(k−1)!n!(cm+bn+kx)(am+dn+kx)
Theorem 10. Let (fm,n (x)) be a sequence of trinomial type. We have
�
�
fi−1,j (a (i − 1) + bj + x)
Am+k,n,k xi
a (i − 1) + bj + x
�
�
m + k fm,n (am + bn + kx)
= xk
, (31)
k
am + bn + kx
�
�
fi,j−1 (ai + b (j − 1) + x)
Am,n+k,k xj
ai + b (j − 1) + x
= xk
�
�
n + k fm,n (am + bn + kx)
. (32)
am + bn + kx
k
Proof. Similarly to the proof in Theorem 8, we obtain
� �
m
fm−k,n (kx) , m ≥ k, n ≥ 0,
Am,n,k (ifi−1,j (x)) =
k
� �
n
fm,n−k (kx) , m ≥ 0, n ≥ k.
Am,n,k (jfi,j−1 (x)) =
k
It will be sufficient to use the trinomial type polynomials (hm,n (x)) given in (26)
instead of (fm,n (x)) and replace m by m + k and n by n + k respectively.
Example 11. From relations (31) and (32), we have the following:
� x �
,
1. fm,n (x) = (m + n)! m+n
�
��
�
kx(m+n)! �m+k��am+bn+kx�
(i+j−1)!
a(i−1)+bj+x
,
= am+bn+kx
Am+k,n,k xi a(i−1)+bj+x
m+n
i+j−1
k
�
�
kx(m+n)! �n+k��am+bn+kx�
(i+j−1)! �ai+b(j−1)+x�
.
= am+bn+kx
Am,n+k,k xj ai+b(j−1)+x
m+n
i+j−1
k
14
INTEGERS: 11 (2011)
2. fm,n (x) = Bm+n (x) ,
�
�
� Bm+n(am+bn+kx)
�
B
(a(i−1)+bj+x)
,
= xk m+k
Am+k,n,k xi i+j−1
a(i−1)+bj+x
am+bn+kx
k
�
�
�
�
B
(ai+b(j−1)+x)
Bm+n (am+bn+kx)
Am,n+k,k xj i+j−1
.
= xk n+k
ai+b(j−1)+x
am+bn+kx
k
m−1
n−1
3. fm,n (x) = x2 ((c − a) m + x)
((d − b) n + x)
,
�
�
i−2
j−2
Am+k,n,k xi (a (i − 1) + bj + x) (c (i − 1) + bj + x)
(a (i − 1) + dj + x)
�
�
m−1
n−1
(am + bn + kx) (cm + bn + kx)
(am + dn + kx)
,
= xk m+k
k
�
�
i−2
j−2
Am,n+k,k xj (ai + b (j − 1) + x) (ci + b (j − 1) + x)
(ai + d (j − 1) + x)
�
�
m−1
n−1
(am + bn + kx) (cm + bn + kx)
(am + dn + kx)
.
= xk n+k
k
�
��
�
(c−a)m+x (d−b)n+x
x2
,
4. fm,n (x) = m!n! ((c−a)m+x)((d−b)n+x)
n
m
�
�c(i−1)+bj+x��a(i−1)+dj+x��
xi!(j−1)!(a(i−1)+bj+x)
Am+k,n,k (c(i−1)+bj+x)(a(i−1)+dj+x)
j
i−1
�cm+bn+kx��am+dn+kx�
(m+k)!n!
am+bn+kx
= x (k−1)! (cm+bn+kx)(am+dn+kx)
,
n
m
�
�
�ci+b(j−1)+x��ai+d(j−1)+x�
x(i−1)!j!(ai+b(j−1)+x)
Am+k,n,k (ci+b(j−1)+x)(ai+d(j−1)+x)
j−1
i
�
��
�
cm+bn+kx am+dn+kx
am+bn+kx
.
= x m!(n+k)!
n
m
(k−1)! (cm+bn+kx)(am+dn+kx)
�
2
m−1 �
(d−b)n+x
5. fm,n (x) = n! x ((c−a)m+x)
,
n
(d−b)n+x
�
�a(i−1)+dj+x��
xij!(a(i−1)+bj+x)
Am+k,n,k (a(i−1)+dj+x)(c(i−1)+bj+x)
2−i
j
x(m+k)!n!(am+bn+kx)(cm+bn+kx)m−1 �am+dn+kx�
=
,
n
(k−1)!m!(am+dn+kx)
�
�
�ai+d(j−1)+x�
xi!j(ai+b(j−1)+x)
Am,n+k,k (ai+d(j−1)+x)(ci+b(j−1)+x)
2−j
j−1
xm!(n+k)!(am+bn+kx)(cm+bn+kx)m−1 �am+dn+kx�
=
.
n
(k−1)!n!(am+dn+kx)
2
m−1
6. fm,n (x) = x ((c−a)m+x)
Bn ((d − b) n + x) ,
(d−b)n+x
�
�
i−2
B
(a
(i
−
1)
+
dj
+
x)
Am+k,n,k xi(a(i−1)+bj+x)(c(i−1)+bj+x)
j
a(i−1)+dj+x
�m+k� (am+bn+kx)(cm+bn+kx)m−1
Bn (am + dn + kx) ,
= xk k
am+dn+kx
�
�
i−1
Am,n+k,k xj(ai+b(j−1)+x)(ci+b(j−1)+x)
B
(ai
+
d
(j
−
1)
+
x)
j−1
ai+d(j−1)+x
�n+k� (am+bn+kx)(cm+bn+kx)m−1
Bn (am + dn + kx) .
= xk k
am+dn+kx
15
INTEGERS: 11 (2011)
�(c−a)m+x�
x2
Bn ((d − b) n + x) ,
7. fm,n (x) = m! ((c−a)m+x)((d−b)n+x)
m
�
�
�
xi!(a(i−1)+bj+x)Bj (a(i−1)+dj+x) c(i−1)+bj+x�
Am+k,n,k
i−1
(c(i−1)+bj+x)(a(i−1)+dj+x)
�cm+bn+kx�
x(m+k)!(am+bn+kx)
Bn (am + dn + kx) ,
= (k−1)!(cm+bn+kx)(am+dn+kx)
m
�
�
�
xj!(ai+b(j−1)+x)Bj−1 (ai+d(j−1)+x) ci+b(j−1)+x�
Am,n+k,k
i
(ci+b(j−1)+x)(ai+d(j−1)+x)
�cm+bn+kx�
x(n+k)!(am+bn+kx)
Bn (am + dn + kx) .
= (k−1)!(cm+bn+kx)(am+dn+kx)
m
Theorem 12. Let (fm,n (x)) be a sequence of trinomial type. We have
�
�
x
x
Am,n
fi,j (ai + bj) =
fm,n (am + bn + x) .
ai + bj
am + bn + x
(33)
Proof. From (28), we get
Am,n (xDα=0 fi,j (α)) = fm,n (x) ,
which gives the desired identity by replacing fm,n (x) by
x
fm,n (am + bn + x) .
hm,n (x) :=
am + bn + x
given by (26).
Example 13. From relation (33), we have the following:
� x �
,
1. fm,n (x) = (m + n)! m+n
�
��
�
�
�
(i + j)! ai + bj
(m + n)!
(am + bn + x)
Am,n x
=x
.
m+n
ai + bj i + j
am + bn + x
2. fm,n (x) = Bm+n (x) ,
�
�
x
x
Bi+j (ai + bj) =
Bm+n (am + bn + x) .
Am,n
ai + bj
am + bn + x
m−1
n−1
3. fm,n (x) = x2 ((c − a) m + x)
((d − b) n + x)
�
�
i−1
j−1
Am,n x (ai + bj) (ci + bj)
(ai + dj)
m−1
= x (am + bn + x) (cm + bn + x)
2
x
4. fm,n (x) = m!n! ((c−a)m+x)((d−b)n+x)
,
n−1
(am + dn + x)
�(c−a)m+x��(d−b)n+x�
,
n
m
.
��
��
�
ai + dj
ci + bj
(ai + bj)
j
i
(ci + bj) (ai + dj)
�
��
�
cm + bn + x am + dn + x
(am + bn + x)
.
= xm!n!
n
m
(cm + bn + x) (am + dn + x)
�
Am,n xi!j!
16
INTEGERS: 11 (2011)
2
5. fm,n (x) = n! x
Am,n
�
((c−a)m+x)m−1 �(d−b)n+x�
,
n
(d−b)n+x
i−1 �
= xn!
6. fm,n (x) =
Am,n
ai + dj
j
(ai + bj) (ci + bj)
xj!
ai + dj
�
��
m−1 �
(am + bn + x) (cm + bn + x)
am + dn + x
x2 ((c−a)m+x)m−1
Bn
(d−b)n+x
((d − b) n + x) ,
i−1
x (ai + bj) (ci + bj)
ai + dj
�
am + dn + x
.
n
�
Bj (ai + dj)
m−1
=
x (am + bn + x) (cm + bn + x)
am + dn + x
2
x
7. fm,n (x) = m! ((c−a)m+x)((d−b)n+x)
Am,n
Bn (am + dn + x) .
�(c−a)m+x�
Bn ((d − b) n + x) ,
m
�
�
�
(ai + bj) ci+bj
i
Bj (ai + dj)
xi!
(ci + bj) (ai + dj)
�
�
(am + bn + x) cm+bn+x
m
= m!
Bn (am + dn + x) .
(cm + bn + x) (am + dn + x)
�
Remark. For m = 0, identity (17) and Theorem 10 give Proposition 1 given in [3],
and identity (20) and Theorem 12 give Proposition 3 given in [3].
Acknowledgments The authors wish to warmly thank their referee and the editor
for their valuable advice and comments which helped to greatly improve the quality
of this paper.
References
[1] Charalambos A. Charalambides, Enumerative Combinatorics. Chapman & Hall/CRC, A CRC
Press Company, (2001).
[2] L. Comtet, Advanced Combinatorics. D. Reidel Publishing Company, Dordrecht-Holland /
Boston-U.S.A, (1974).
[3] M. Mihoubi, Bell polynomials and binomial type sequences. Discrete Math. 308 (2008), 2450–
2459.
INTEGERS: 11 (2011)
17
[4] M. Mihoubi, The role of binomial type sequences in determination identities for Bell polynomials. To appear, Ars Combin. Preprint available at http://arxiv.org/abs/0806.3468v1.
[5] M. Mihoubi, Some congruences for the partial Bell polynomials. J. Integer Seq. 12 (2009),
Article 09.4.1.
[6] J. Riordan, Combinatorial Identities. Huntington, New York, (1979).
[7] S. Roman, The Umbral Calculus. Academic Press (1984).
[8] J. Zeng, Multinomial Convolution Polynomials, Discrete Mathematics 160 (1996), 219–228.